Optimal. Leaf size=89 \[ -\frac{3 x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}-\frac{1}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}+\frac{15 \text{Chi}\left (2 \tanh ^{-1}(a x)\right )}{16 a}+\frac{3 \text{Chi}\left (4 \tanh ^{-1}(a x)\right )}{2 a}+\frac{9 \text{Chi}\left (6 \tanh ^{-1}(a x)\right )}{16 a} \]
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Rubi [A] time = 0.386036, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 7, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.368, Rules used = {5966, 6032, 6034, 5448, 3301, 5968, 3312} \[ -\frac{3 x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}-\frac{1}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}+\frac{15 \text{Chi}\left (2 \tanh ^{-1}(a x)\right )}{16 a}+\frac{3 \text{Chi}\left (4 \tanh ^{-1}(a x)\right )}{2 a}+\frac{9 \text{Chi}\left (6 \tanh ^{-1}(a x)\right )}{16 a} \]
Antiderivative was successfully verified.
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Rule 5966
Rule 6032
Rule 6034
Rule 5448
Rule 3301
Rule 5968
Rule 3312
Rubi steps
\begin{align*} \int \frac{1}{\left (1-a^2 x^2\right )^4 \tanh ^{-1}(a x)^3} \, dx &=-\frac{1}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}+(3 a) \int \frac{x}{\left (1-a^2 x^2\right )^4 \tanh ^{-1}(a x)^2} \, dx\\ &=-\frac{1}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}-\frac{3 x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+3 \int \frac{1}{\left (1-a^2 x^2\right )^4 \tanh ^{-1}(a x)} \, dx+\left (15 a^2\right ) \int \frac{x^2}{\left (1-a^2 x^2\right )^4 \tanh ^{-1}(a x)} \, dx\\ &=-\frac{1}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}-\frac{3 x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+\frac{3 \operatorname{Subst}\left (\int \frac{\cosh ^6(x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a}+\frac{15 \operatorname{Subst}\left (\int \frac{\cosh ^4(x) \sinh ^2(x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=-\frac{1}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}-\frac{3 x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+\frac{3 \operatorname{Subst}\left (\int \left (\frac{5}{16 x}+\frac{15 \cosh (2 x)}{32 x}+\frac{3 \cosh (4 x)}{16 x}+\frac{\cosh (6 x)}{32 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a}+\frac{15 \operatorname{Subst}\left (\int \left (-\frac{1}{16 x}-\frac{\cosh (2 x)}{32 x}+\frac{\cosh (4 x)}{16 x}+\frac{\cosh (6 x)}{32 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=-\frac{1}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}-\frac{3 x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+\frac{3 \operatorname{Subst}\left (\int \frac{\cosh (6 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{32 a}-\frac{15 \operatorname{Subst}\left (\int \frac{\cosh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{32 a}+\frac{15 \operatorname{Subst}\left (\int \frac{\cosh (6 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{32 a}+\frac{9 \operatorname{Subst}\left (\int \frac{\cosh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{16 a}+\frac{15 \operatorname{Subst}\left (\int \frac{\cosh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{16 a}+\frac{45 \operatorname{Subst}\left (\int \frac{\cosh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{32 a}\\ &=-\frac{1}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}-\frac{3 x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+\frac{15 \text{Chi}\left (2 \tanh ^{-1}(a x)\right )}{16 a}+\frac{3 \text{Chi}\left (4 \tanh ^{-1}(a x)\right )}{2 a}+\frac{9 \text{Chi}\left (6 \tanh ^{-1}(a x)\right )}{16 a}\\ \end{align*}
Mathematica [A] time = 0.187294, size = 83, normalized size = 0.93 \[ \frac{1}{16} \left (\frac{48 x}{\left (a^2 x^2-1\right )^3 \tanh ^{-1}(a x)}+\frac{8}{a \left (a^2 x^2-1\right )^3 \tanh ^{-1}(a x)^2}+\frac{15 \text{Chi}\left (2 \tanh ^{-1}(a x)\right )}{a}+\frac{24 \text{Chi}\left (4 \tanh ^{-1}(a x)\right )}{a}+\frac{9 \text{Chi}\left (6 \tanh ^{-1}(a x)\right )}{a}\right ) \]
Antiderivative was successfully verified.
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Maple [A] time = 0.075, size = 131, normalized size = 1.5 \begin{align*}{\frac{1}{a} \left ( -{\frac{5}{32\, \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}}}-{\frac{15\,\cosh \left ( 2\,{\it Artanh} \left ( ax \right ) \right ) }{64\, \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}}}-{\frac{15\,\sinh \left ( 2\,{\it Artanh} \left ( ax \right ) \right ) }{32\,{\it Artanh} \left ( ax \right ) }}+{\frac{15\,{\it Chi} \left ( 2\,{\it Artanh} \left ( ax \right ) \right ) }{16}}-{\frac{3\,\cosh \left ( 4\,{\it Artanh} \left ( ax \right ) \right ) }{32\, \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}}}-{\frac{3\,\sinh \left ( 4\,{\it Artanh} \left ( ax \right ) \right ) }{8\,{\it Artanh} \left ( ax \right ) }}+{\frac{3\,{\it Chi} \left ( 4\,{\it Artanh} \left ( ax \right ) \right ) }{2}}-{\frac{\cosh \left ( 6\,{\it Artanh} \left ( ax \right ) \right ) }{64\, \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}}}-{\frac{3\,\sinh \left ( 6\,{\it Artanh} \left ( ax \right ) \right ) }{32\,{\it Artanh} \left ( ax \right ) }}+{\frac{9\,{\it Chi} \left ( 6\,{\it Artanh} \left ( ax \right ) \right ) }{16}} \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*} \frac{2 \,{\left (3 \, a x \log \left (a x + 1\right ) - 3 \, a x \log \left (-a x + 1\right ) + 1\right )}}{{\left (a^{7} x^{6} - 3 \, a^{5} x^{4} + 3 \, a^{3} x^{2} - a\right )} \log \left (a x + 1\right )^{2} - 2 \,{\left (a^{7} x^{6} - 3 \, a^{5} x^{4} + 3 \, a^{3} x^{2} - a\right )} \log \left (a x + 1\right ) \log \left (-a x + 1\right ) +{\left (a^{7} x^{6} - 3 \, a^{5} x^{4} + 3 \, a^{3} x^{2} - a\right )} \log \left (-a x + 1\right )^{2}} - \int -\frac{6 \,{\left (5 \, a^{2} x^{2} + 1\right )}}{{\left (a^{8} x^{8} - 4 \, a^{6} x^{6} + 6 \, a^{4} x^{4} - 4 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right ) -{\left (a^{8} x^{8} - 4 \, a^{6} x^{6} + 6 \, a^{4} x^{4} - 4 \, a^{2} x^{2} + 1\right )} \log \left (-a x + 1\right )}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.93944, size = 1014, normalized size = 11.39 \begin{align*} \frac{192 \, a x \log \left (-\frac{a x + 1}{a x - 1}\right ) + 3 \,{\left (3 \,{\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \logintegral \left (-\frac{a^{3} x^{3} + 3 \, a^{2} x^{2} + 3 \, a x + 1}{a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1}\right ) + 3 \,{\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \logintegral \left (-\frac{a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1}{a^{3} x^{3} + 3 \, a^{2} x^{2} + 3 \, a x + 1}\right ) + 8 \,{\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \logintegral \left (\frac{a^{2} x^{2} + 2 \, a x + 1}{a^{2} x^{2} - 2 \, a x + 1}\right ) + 8 \,{\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \logintegral \left (\frac{a^{2} x^{2} - 2 \, a x + 1}{a^{2} x^{2} + 2 \, a x + 1}\right ) + 5 \,{\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \logintegral \left (-\frac{a x + 1}{a x - 1}\right ) + 5 \,{\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \logintegral \left (-\frac{a x - 1}{a x + 1}\right )\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )^{2} + 64}{32 \,{\left (a^{7} x^{6} - 3 \, a^{5} x^{4} + 3 \, a^{3} x^{2} - a\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )^{2}} \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{1}{\left (a x - 1\right )^{4} \left (a x + 1\right )^{4} \operatorname{atanh}^{3}{\left (a x \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{1}{{\left (a^{2} x^{2} - 1\right )}^{4} \operatorname{artanh}\left (a x\right )^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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