Optimal. Leaf size=68 \[ -\frac{\sqrt{\frac{1}{a x^2+1}} \sqrt{a x^2+1} \tanh ^{-1}\left (\sqrt{1-a^2 x^4}\right )}{2 a}+\frac{1}{2} x^2 e^{\text{sech}^{-1}\left (a x^2\right )}+\frac{\log (x)}{a} \]
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Rubi [A] time = 0.0425703, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {6335, 29, 259, 266, 63, 208} \[ -\frac{\sqrt{\frac{1}{a x^2+1}} \sqrt{a x^2+1} \tanh ^{-1}\left (\sqrt{1-a^2 x^4}\right )}{2 a}+\frac{1}{2} x^2 e^{\text{sech}^{-1}\left (a x^2\right )}+\frac{\log (x)}{a} \]
Antiderivative was successfully verified.
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Rule 6335
Rule 29
Rule 259
Rule 266
Rule 63
Rule 208
Rubi steps
\begin{align*} \int e^{\text{sech}^{-1}\left (a x^2\right )} x \, dx &=\frac{1}{2} e^{\text{sech}^{-1}\left (a x^2\right )} x^2+\frac{\int \frac{1}{x} \, dx}{a}+\frac{\left (\sqrt{\frac{1}{1+a x^2}} \sqrt{1+a x^2}\right ) \int \frac{1}{x \sqrt{1-a x^2} \sqrt{1+a x^2}} \, dx}{a}\\ &=\frac{1}{2} e^{\text{sech}^{-1}\left (a x^2\right )} x^2+\frac{\log (x)}{a}+\frac{\left (\sqrt{\frac{1}{1+a x^2}} \sqrt{1+a x^2}\right ) \int \frac{1}{x \sqrt{1-a^2 x^4}} \, dx}{a}\\ &=\frac{1}{2} e^{\text{sech}^{-1}\left (a x^2\right )} x^2+\frac{\log (x)}{a}+\frac{\left (\sqrt{\frac{1}{1+a x^2}} \sqrt{1+a x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-a^2 x}} \, dx,x,x^4\right )}{4 a}\\ &=\frac{1}{2} e^{\text{sech}^{-1}\left (a x^2\right )} x^2+\frac{\log (x)}{a}-\frac{\left (\sqrt{\frac{1}{1+a x^2}} \sqrt{1+a x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{1}{a^2}-\frac{x^2}{a^2}} \, dx,x,\sqrt{1-a^2 x^4}\right )}{2 a^3}\\ &=\frac{1}{2} e^{\text{sech}^{-1}\left (a x^2\right )} x^2-\frac{\sqrt{\frac{1}{1+a x^2}} \sqrt{1+a x^2} \tanh ^{-1}\left (\sqrt{1-a^2 x^4}\right )}{2 a}+\frac{\log (x)}{a}\\ \end{align*}
Mathematica [A] time = 0.0633233, size = 100, normalized size = 1.47 \[ \frac{\sqrt{\frac{1-a x^2}{a x^2+1}} \left (a x^2+1\right )+2 \log \left (a x^2\right )-\log \left (a x^2 \sqrt{\frac{1-a x^2}{a x^2+1}}+\sqrt{\frac{1-a x^2}{a x^2+1}}+1\right )}{2 a} \]
Warning: Unable to verify antiderivative.
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Maple [C] time = 0.312, size = 127, normalized size = 1.9 \begin{align*}{\frac{{x}^{2}{\it csgn} \left ({a}^{-1} \right ) }{2\,a}\sqrt{-{\frac{a{x}^{2}-1}{a{x}^{2}}}}\sqrt{{\frac{a{x}^{2}+1}{a{x}^{2}}}} \left ({\it csgn} \left ({a}^{-1} \right ) a\sqrt{-{\frac{{a}^{2}{x}^{4}-1}{{a}^{2}}}}-\ln \left ( 2\,{\frac{1}{{a}^{2}{x}^{2}} \left ({\it csgn} \left ({a}^{-1} \right ) a\sqrt{-{\frac{{a}^{2}{x}^{4}-1}{{a}^{2}}}}+1 \right ) } \right ) \right ){\frac{1}{\sqrt{-{\frac{{a}^{2}{x}^{4}-1}{{a}^{2}}}}}}}+{\frac{\ln \left ( x \right ) }{a}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\frac{1}{2} \, \sqrt{-a^{2} x^{4} + 1} - \frac{1}{2} \, \log \left (\frac{2 \, \sqrt{-a^{2} x^{4} + 1}}{x^{2}} + \frac{2}{x^{2}}\right )}{a} + \frac{\log \left (x\right )}{a} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.09933, size = 298, normalized size = 4.38 \begin{align*} \frac{2 \, a x^{2} \sqrt{\frac{a x^{2} + 1}{a x^{2}}} \sqrt{-\frac{a x^{2} - 1}{a x^{2}}} - \log \left (a x^{2} \sqrt{\frac{a x^{2} + 1}{a x^{2}}} \sqrt{-\frac{a x^{2} - 1}{a x^{2}}} + 1\right ) + \log \left (a x^{2} \sqrt{\frac{a x^{2} + 1}{a x^{2}}} \sqrt{-\frac{a x^{2} - 1}{a x^{2}}} - 1\right ) + 4 \, \log \left (x\right )}{4 \, a} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{1}{x}\, dx + \int a x \sqrt{-1 + \frac{1}{a x^{2}}} \sqrt{1 + \frac{1}{a x^{2}}}\, dx}{a} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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