√(a x^2+b x +c)の積分

$$ \begin{eqnarray} \int \sqrt{a x^2+b x +c} \;\mathrm{d}x &=&\int \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 \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\;\frac{\mathrm{d}u}{\mathrm{d}x}=\sqrt{a},\mathrm{d}x=\frac{1}{\sqrt{a}}\mathrm{d}u \\&=& \frac{1}{\sqrt{a}} \int \sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u \\&=& \frac{1}{\sqrt{a}} \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{4ac-b^2}{4a^{\frac{3}{2}}} \int \sqrt{t^2+1} \;\mathrm{d}t \\&=& \frac{4ac-b^2}{4a^{\frac{3}{2}}} \left\{\frac{1}{2}\left(t\sqrt{t^2+1}+\ln{ \left|t+\sqrt{t^2+1}\right| }+C_0\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_0\right)\;(C_0:積分定数)} \\&=& \frac{4ac-b^2}{8a^{\frac{3}{2}}} \left(t\sqrt{t^2+1}+\ln{ \left|t+\sqrt{t^2+1}\right| }+C_0\right) \\&=& \frac{4ac-b^2}{8a^{\frac{3}{2}}} \left(t\sqrt{t^2+1}+\ln{ \left|t+\sqrt{t^2+1}\right| }+C_0\right) \\&=& \frac{4ac-b^2}{8a^{\frac{3}{2}}} \left(t\sqrt{t^2+1}+\ln{ \left|t+\sqrt{t^2+1}\right| }+C_0\right) \\&=& \frac{4ac-b^2}{8a^{\frac{3}{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_0 \right\} \\&=& \frac{4ac-b^2}{8a^{\frac{3}{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_0 \right\} \\&=& \frac{4ac-b^2}{8a^{\frac{3}{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_0 \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{4ac-b^2}{8a^{\frac{3}{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_0 \right\} \\&=& \frac{4ac-b^2}{8a^{\frac{3}{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_0 \right\} \\&=&\frac{4ac-b^2}{8a^{\frac{3}{2}}}\frac{2ax+b}{4ac-b^2}2\sqrt{a}\sqrt{\left(ax^2+bx+c\right)} +\frac{4ac-b^2}{8a^{\frac{3}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| } -\frac{4ac-b^2}{8a^{\frac{3}{2}}}\ln{\sqrt{4ac-b^2}} +\frac{4ac-b^2}{8a^{\frac{3}{2}}}C_0 \\&=&\frac{2ax+b}{4a}\sqrt{\left(ax^2+bx+c\right)} +\frac{4ac-b^2}{8a^{\frac{3}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| } -\frac{4ac-b^2}{8a^{\frac{3}{2}}}\ln{\sqrt{4ac-b^2}} +\frac{4ac-b^2}{8a^{\frac{3}{2}}}C_0 \\&=&\frac{2ax+b}{4a}\sqrt{\left(ax^2+bx+c\right)} +\frac{4ac-b^2}{8a^{\frac{3}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| } +C \;\cdots\;C=-\frac{4ac-b^2}{8a^{\frac{3}{2}}}\ln{\sqrt{4ac-b^2}}+\frac{4ac-b^2}{8a^{\frac{3}{2}}}C_0 \;(C:積分定数) \end{eqnarray} $$