Optimal. Leaf size=197 \[ -\frac{3 i \text{PolyLog}\left (2,-\frac{i \sqrt{1-a x}}{\sqrt{a x+1}}\right )}{8 a^5}+\frac{3 i \text{PolyLog}\left (2,\frac{i \sqrt{1-a x}}{\sqrt{a x+1}}\right )}{8 a^5}+\frac{\left (1-a^2 x^2\right )^{3/2}}{12 a^5}-\frac{5 \sqrt{1-a^2 x^2}}{8 a^5}-\frac{x^3 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{4 a^2}-\frac{3 x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{8 a^4}-\frac{3 \tan ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right ) \tanh ^{-1}(a x)}{4 a^5} \]
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Rubi [A] time = 0.216797, antiderivative size = 197, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {6016, 266, 43, 261, 5950} \[ -\frac{3 i \text{PolyLog}\left (2,-\frac{i \sqrt{1-a x}}{\sqrt{a x+1}}\right )}{8 a^5}+\frac{3 i \text{PolyLog}\left (2,\frac{i \sqrt{1-a x}}{\sqrt{a x+1}}\right )}{8 a^5}+\frac{\left (1-a^2 x^2\right )^{3/2}}{12 a^5}-\frac{5 \sqrt{1-a^2 x^2}}{8 a^5}-\frac{x^3 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{4 a^2}-\frac{3 x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{8 a^4}-\frac{3 \tan ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right ) \tanh ^{-1}(a x)}{4 a^5} \]
Antiderivative was successfully verified.
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Rule 6016
Rule 266
Rule 43
Rule 261
Rule 5950
Rubi steps
\begin{align*} \int \frac{x^4 \tanh ^{-1}(a x)}{\sqrt{1-a^2 x^2}} \, dx &=-\frac{x^3 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{4 a^2}+\frac{3 \int \frac{x^2 \tanh ^{-1}(a x)}{\sqrt{1-a^2 x^2}} \, dx}{4 a^2}+\frac{\int \frac{x^3}{\sqrt{1-a^2 x^2}} \, dx}{4 a}\\ &=-\frac{3 x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{8 a^4}-\frac{x^3 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{4 a^2}+\frac{3 \int \frac{\tanh ^{-1}(a x)}{\sqrt{1-a^2 x^2}} \, dx}{8 a^4}+\frac{3 \int \frac{x}{\sqrt{1-a^2 x^2}} \, dx}{8 a^3}+\frac{\operatorname{Subst}\left (\int \frac{x}{\sqrt{1-a^2 x}} \, dx,x,x^2\right )}{8 a}\\ &=-\frac{3 \sqrt{1-a^2 x^2}}{8 a^5}-\frac{3 x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{8 a^4}-\frac{x^3 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{4 a^2}-\frac{3 \tan ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{1+a x}}\right ) \tanh ^{-1}(a x)}{4 a^5}-\frac{3 i \text{Li}_2\left (-\frac{i \sqrt{1-a x}}{\sqrt{1+a x}}\right )}{8 a^5}+\frac{3 i \text{Li}_2\left (\frac{i \sqrt{1-a x}}{\sqrt{1+a x}}\right )}{8 a^5}+\frac{\operatorname{Subst}\left (\int \left (\frac{1}{a^2 \sqrt{1-a^2 x}}-\frac{\sqrt{1-a^2 x}}{a^2}\right ) \, dx,x,x^2\right )}{8 a}\\ &=-\frac{5 \sqrt{1-a^2 x^2}}{8 a^5}+\frac{\left (1-a^2 x^2\right )^{3/2}}{12 a^5}-\frac{3 x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{8 a^4}-\frac{x^3 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{4 a^2}-\frac{3 \tan ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{1+a x}}\right ) \tanh ^{-1}(a x)}{4 a^5}-\frac{3 i \text{Li}_2\left (-\frac{i \sqrt{1-a x}}{\sqrt{1+a x}}\right )}{8 a^5}+\frac{3 i \text{Li}_2\left (\frac{i \sqrt{1-a x}}{\sqrt{1+a x}}\right )}{8 a^5}\\ \end{align*}
Mathematica [A] time = 0.504469, size = 160, normalized size = 0.81 \[ \frac{\sqrt{1-a^2 x^2} \left (-\frac{9 i \left (\text{PolyLog}\left (2,-i e^{-\tanh ^{-1}(a x)}\right )-\text{PolyLog}\left (2,i e^{-\tanh ^{-1}(a x)}\right )\right )}{\sqrt{1-a^2 x^2}}-2 a^2 x^2-6 a x \left (a^2 x^2-1\right ) \tanh ^{-1}(a x)-\frac{9 i \tanh ^{-1}(a x) \left (\log \left (1-i e^{-\tanh ^{-1}(a x)}\right )-\log \left (1+i e^{-\tanh ^{-1}(a x)}\right )\right )}{\sqrt{1-a^2 x^2}}-15 a x \tanh ^{-1}(a x)-13\right )}{24 a^5} \]
Warning: Unable to verify antiderivative.
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Maple [A] time = 0.262, size = 175, normalized size = 0.9 \begin{align*} -{\frac{6\,{a}^{3}{x}^{3}{\it Artanh} \left ( ax \right ) +2\,{a}^{2}{x}^{2}+9\,ax{\it Artanh} \left ( ax \right ) +13}{24\,{a}^{5}}\sqrt{- \left ( ax-1 \right ) \left ( ax+1 \right ) }}-{\frac{{\frac{3\,i}{8}}{\it Artanh} \left ( ax \right ) }{{a}^{5}}\ln \left ( 1+{i \left ( ax+1 \right ){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ) }+{\frac{{\frac{3\,i}{8}}{\it Artanh} \left ( ax \right ) }{{a}^{5}}\ln \left ( 1-{i \left ( ax+1 \right ){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ) }-{\frac{{\frac{3\,i}{8}}}{{a}^{5}}{\it dilog} \left ( 1+{i \left ( ax+1 \right ){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ) }+{\frac{{\frac{3\,i}{8}}}{{a}^{5}}{\it dilog} \left ( 1-{i \left ( ax+1 \right ){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ) } \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*} \int \frac{x^{4} \operatorname{artanh}\left (a x\right )}{\sqrt{-a^{2} x^{2} + 1}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{-a^{2} x^{2} + 1} x^{4} \operatorname{artanh}\left (a x\right )}{a^{2} x^{2} - 1}, x\right ) \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*} \int \frac{x^{4} \operatorname{atanh}{\left (a x \right )}}{\sqrt{- \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4} \operatorname{artanh}\left (a x\right )}{\sqrt{-a^{2} x^{2} + 1}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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