$$
\begin{eqnarray}
\int x\sqrt{a x^2+b x +c} \;\mathrm{d}x
&=&\int x\sqrt{\left(\sqrt{a} x+\frac{b}{2\sqrt{a}} \right)^2-\frac{b^2}{4a}+c} \;\mathrm{d}x
\;\cdots\;平方完成
\\&&\;\cdots\;ax^2+bx+c=(\sqrt{a}x+\alpha)^2-\alpha^2+c
\\&&\;\cdots\;(\sqrt{a}x+\alpha)^2=ax^2+2\sqrt{a}\alpha x+\alpha^2,\;2\sqrt{a}\alpha=bより\alpha=\frac{b}{2\sqrt{a}}
\\&=&\int \frac{2\sqrt{a}u-b}{2a}\; \sqrt{u^2-\frac{b^2}{4a}+c} \;\frac{1}{\sqrt{a}}\mathrm{d}u
\\&&\;\cdots\;u=\sqrt{a} x+\frac{b}{2\sqrt{a}}
\\&&\;\cdots\;x=\frac{u-\frac{b}{2\sqrt{a}}}{\sqrt{a}}
=\frac{\frac{2\sqrt{a}u-b}{2\sqrt{a}}}{\sqrt{a}}
=\frac{2\sqrt{a}u-b}{2a}
\\&&\;\cdots\;\frac{\mathrm{d}x}{\mathrm{d}u}=\frac{1}{\sqrt{a}},\mathrm{d}x=\frac{1}{\sqrt{a}}\mathrm{d}u
\\&=&\frac{1}{2a^{\frac{3}{2}}} \int \left(2\sqrt{a}u-b\right)\;\sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u
\\&=& \frac{1}{2a^{\frac{3}{2}}}\int 2\sqrt{a}u\;\sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u
- \frac{1}{2a^{\frac{3}{2}}} \int b\;\sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u
\\&=& \frac{2\sqrt{a}}{2a^{\frac{3}{2}}}\int u\;\sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u
- \frac{b}{2a^{\frac{3}{2}}} \int \sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u
\\&=& \frac{1}{a}\int u\;\sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u
- \frac{b}{2a^{\frac{3}{2}}} \int \sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u
\end{eqnarray}
$$
$$
\begin{eqnarray}
\frac{1}{a}\int u\;\sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u
&=& \frac{1}{a}\int u\;\sqrt{s} \;\frac{1}{2u}\;\mathrm{d}s
\;\cdots\;s=u^2-\frac{b^2}{4a}+c,\frac{\mathrm{d}s}{\mathrm{d}u}=2u,\mathrm{d}u=\frac{1}{2u}\;\mathrm{d}s
\\&=& \frac{1}{2a}\int\sqrt{s}\;\mathrm{d}s
\\&=& \frac{1}{2a}\left[\frac{2}{3}s^{\frac{3}{2}}+C_0\right]\;\cdots\;\int x^{\frac{1}{2}}\;\mathrm{d}x=\frac{2}{3}x^{\frac{3}{2}}+C_0\;(C_0:積分定数)
\\&=& \frac{1}{3a}s^{\frac{3}{2}}+\frac{1}{2a}C_0
\\&=& \frac{1}{3a}\left(u^2-\frac{b^2}{4a}+c\right)^{\frac{3}{2}}+\frac{1}{2a}C_0
\\&=& \frac{1}{3a}\left\{\left(\sqrt{a} x+\frac{b}{2\sqrt{a}}\right)^2-\frac{b^2}{4a}+c\right\}^{\frac{3}{2}}+\frac{1}{2a}C_0
\\&=& \frac{1}{3a}\left\{\left(ax^2+bx+\frac{b^2}{4a}\right)-\frac{b^2}{4a}+c\right\}^{\frac{3}{2}}+\frac{1}{2a}C_0
\\&=& \frac{1}{3a}\left(ax^2+bx+c\right)^{\frac{3}{2}}+\frac{1}{2a}C_0
\end{eqnarray}
$$
$$
\begin{eqnarray}
-\frac{b}{2a^{\frac{3}{2}}} \int \sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u
&=&-\frac{b}{2a^{\frac{3}{2}}} \int \sqrt{\left(c-\frac{b^2}{4a}\right)\left(t^2+1\right)} \;\sqrt{c-\frac{b^2}{4a}}\;\mathrm{d}t
\\&&\;\cdots\;t=\frac{u}{\sqrt{c-\frac{b^2}{4a}}}
,\frac{\mathrm{d}t}{\mathrm{d}u}=\frac{1}{\sqrt{c-\frac{b^2}{4a}}}
,\mathrm{d}u=\sqrt{c-\frac{b^2}{4a}}\mathrm{d}t
\\&&\;\cdots\;u^2+c-\frac{b^2}{4a}
=\left(u^2+c-\frac{b^2}{4a}\right)\frac{c-\frac{b^2}{4a}}{c-\frac{b^2}{4a}}
=\left(c-\frac{b^2}{4a}\right)\frac{u^2+c-\frac{b^2}{4a}}{c-\frac{b^2}{4a}}
=\left(c-\frac{b^2}{4a}\right)\left(\frac{u^2}{c-\frac{b^2}{4a}}+1\right)
=\left(c-\frac{b^2}{4a}\right)\left\{\left(\frac{u}{\sqrt{c-\frac{b^2}{4a}}}\right)^2+1\right\}
=\left(c-\frac{b^2}{4a}\right)\left(t^2+1\right)
\\&=& - \frac{b}{2a^{\frac{3}{2}}} \left(c-\frac{b^2}{4a}\right)\int \sqrt{t^2+1} \;\mathrm{d}t
\\&=& - \frac{b}{2a^{\frac{3}{2}}} \left(c-\frac{b^2}{4a}\right)
\left\{\frac{1}{2}\left(t\sqrt{t^2+1}+\ln{ \left|t+\sqrt{t^2+1}\right| }+C_1\right)\right\}
\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/07/x2a2.html}{\int \sqrt{x^2+a^2} \;\mathrm{d}x = \frac{1}{2}\left(x\sqrt{x^2+a^2}+ a^2 \ln{ \left|x+\sqrt{x^2+a^2}\right| }+C_1\right)\;(C_1:積分定数)}
\\&=& \left(\frac{b^3}{16a^{\frac{5}{2}}}-\frac{bc}{4a^{\frac{3}{2}}}\right)
\left(t\sqrt{t^2+1}+\ln{ \left|t+\sqrt{t^2+1}\right| }+C_1\right)
\\&=& \left(\frac{b^3}{16a^{\frac{5}{2}}}-\frac{4abc}{16a^{\frac{5}{2}}}\right)
\left(t\sqrt{t^2+1}+\ln{ \left|t+\sqrt{t^2+1}\right| }+C_1\right)
\\&=& \frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}
\left(t\sqrt{t^2+1}+\ln{ \left|t+\sqrt{t^2+1}\right| }+C_1\right)
\\&=& \frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}
\left\{
\left(\frac{u}{\sqrt{c-\frac{b^2}{4a}}}\right)\sqrt{\left(\frac{u}{\sqrt{c-\frac{b^2}{4a}}}\right)^2+1}
+\ln{ \left|\left(\frac{u}{\sqrt{c-\frac{b^2}{4a}}}\right)+\sqrt{\left(\frac{u}{\sqrt{c-\frac{b^2}{4a}}}\right)^2+1}\right| }+C_1
\right\}
\\&=& \frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}
\left\{
\left(\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)\sqrt{\left(\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)^2+1}
+\ln{ \left|\left(\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)+\sqrt{\left(\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)^2+1}\right| }+C_1
\right\}
\\&=& \frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}
\left\{
\frac{2ax+b}{\sqrt{4ac-b^2}}\sqrt{\frac{4a\left(ax^2+bx+c\right)}{4ac-b^2}}
+\ln{ \left|\frac{2ax+b}{\sqrt{4ac-b^2}}+\sqrt{\frac{4a\left(ax^2+bx+c\right)}{4ac-b^2}}\right| }+C_1
\right\}
\\&&\;\cdots\;\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}=\frac{2ax+b}{\sqrt{4ac-b^2}}
\\&&\;\cdots\;\left(\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)^2=\left(\frac{2ax+b}{\sqrt{4ac-b^2}}\right)^2=\frac{4a^2x^2+4abx+b^2}{4ac-b^2}
\\&&\;\cdots\;\left(\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)^2+1=\frac{4a^2x^2+4abx+b^2}{4ac-b^2}+1=\frac{4a^2x^2+4abx+b^2+4ac-b^2}{4ac-b^2}=\frac{4a^2x^2+4abx+4ac}{4ac-b^2}=\frac{4a\left(ax^2+bx+c\right)}{4ac-b^2}
\\&=& \frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}
\left\{
\frac{2ax+b}{4ac-b^2}2\sqrt{a}\sqrt{\left(ax^2+bx+c\right)}
+\ln{ \left|\frac{1}{\sqrt{4ac-b^2}}\left\{\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right\}\right| }
+C_1
\right\}
\\&=& \frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}
\left\{
\frac{2ax+b}{4ac-b^2}2\sqrt{a}\sqrt{a\left(ax^2+bx+c\right)}
+\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| }
-\ln{\sqrt{4ac-b^2}}
+C_1
\right\}
\\&=&\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\frac{2ax+b}{4ac-b^2}2\sqrt{a}\sqrt{\left(ax^2+bx+c\right)}
+\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| }
-\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{\sqrt{4ac-b^2}}
+\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}C_1
\\&=&-\frac{b\left(2ax+b\right)}{8a^{\frac{4}{2}}}\sqrt{\left(ax^2+bx+c\right)}
+\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| }
-\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{\sqrt{4ac-b^2}}
+\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}C_1
\\&=&-\frac{b\left(2ax+b\right)}{8a^{2}}\sqrt{\left(ax^2+bx+c\right)}
+\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| }
-\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{\sqrt{4ac-b^2}}
+\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}C_1
\end{eqnarray}
$$
$$
\begin{eqnarray}
\int x\sqrt{a x^2+b x +c} \;\mathrm{d}x
&=& \frac{1}{a}\int u\;\sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u
- \frac{b}{2a^{\frac{3}{2}}} \int \sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u
\\&=& \frac{1}{3a}\left(ax^2+bx+c\right)^{\frac{3}{2}}+\frac{1}{2a}C_0
-\frac{b\left(2ax+b\right)}{8a^{2}}\sqrt{\left(ax^2+bx+c\right)}
+\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| }
-\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{\sqrt{4ac-b^2}}
+\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}C_1
\\&&\;\cdots\;\frac{1}{a}\int u\;\sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u=\frac{1}{3a}\left(ax^2+bx+c\right)^{\frac{3}{2}}+\frac{1}{2a}C_0
\\&&\;\cdots\;-\frac{b}{2a^{\frac{3}{2}}} \int \sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u=-\frac{b\left(2ax+b\right)}{8a^{2}}\sqrt{\left(ax^2+bx+c\right)}
+\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| }
-\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{\sqrt{4ac-b^2}}
+\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}C_1
\\&=& \frac{1}{3a}\left(ax^2+bx+c\right)^{\frac{3}{2}}
-\frac{b\left(2ax+b\right)}{8a^{2}}\sqrt{\left(ax^2+bx+c\right)}
+\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| }
+\frac{1}{2a}C_0
-\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{\sqrt{4ac-b^2}}
+\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}C_1
\\&=& \frac{1}{3a}\left(ax^2+bx+c\right)^{\frac{3}{2}}
-\frac{b\left(2ax+b\right)}{8a^{2}}\sqrt{\left(ax^2+bx+c\right)}
+\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| }
+C
\;\cdots\;C=\frac{1}{2a}C_0
-\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{\sqrt{4ac-b^2}}
+\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}C_1\;(C:積分定数)
\end{eqnarray}
$$
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