Optimal. Leaf size=114 \[ \frac{5 \text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{16 a^2}+\frac{\text{Shi}\left (4 \tanh ^{-1}(a x)\right )}{a^2}+\frac{9 \text{Shi}\left (6 \tanh ^{-1}(a x)\right )}{16 a^2}-\frac{x}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}+\frac{5}{2 a^2 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}-\frac{3}{a^2 \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)} \]
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Rubi [A] time = 0.585044, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 22, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {6032, 6028, 5966, 6034, 5448, 3298} \[ \frac{5 \text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{16 a^2}+\frac{\text{Shi}\left (4 \tanh ^{-1}(a x)\right )}{a^2}+\frac{9 \text{Shi}\left (6 \tanh ^{-1}(a x)\right )}{16 a^2}-\frac{x}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}+\frac{5}{2 a^2 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}-\frac{3}{a^2 \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)} \]
Antiderivative was successfully verified.
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Rule 6032
Rule 6028
Rule 5966
Rule 6034
Rule 5448
Rule 3298
Rubi steps
\begin{align*} \int \frac{x}{\left (1-a^2 x^2\right )^4 \tanh ^{-1}(a x)^3} \, dx &=-\frac{x}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}+\frac{\int \frac{1}{\left (1-a^2 x^2\right )^4 \tanh ^{-1}(a x)^2} \, dx}{2 a}+\frac{1}{2} (5 a) \int \frac{x^2}{\left (1-a^2 x^2\right )^4 \tanh ^{-1}(a x)^2} \, dx\\ &=-\frac{x}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}-\frac{1}{2 a^2 \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+3 \int \frac{x}{\left (1-a^2 x^2\right )^4 \tanh ^{-1}(a x)} \, dx+\frac{5 \int \frac{1}{\left (1-a^2 x^2\right )^4 \tanh ^{-1}(a x)^2} \, dx}{2 a}-\frac{5 \int \frac{1}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2} \, dx}{2 a}\\ &=-\frac{x}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}-\frac{3}{a^2 \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+\frac{5}{2 a^2 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}-10 \int \frac{x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)} \, dx+15 \int \frac{x}{\left (1-a^2 x^2\right )^4 \tanh ^{-1}(a x)} \, dx+\frac{3 \operatorname{Subst}\left (\int \frac{\cosh ^5(x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^2}\\ &=-\frac{x}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}-\frac{3}{a^2 \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+\frac{5}{2 a^2 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac{3 \operatorname{Subst}\left (\int \left (\frac{5 \sinh (2 x)}{32 x}+\frac{\sinh (4 x)}{8 x}+\frac{\sinh (6 x)}{32 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^2}-\frac{10 \operatorname{Subst}\left (\int \frac{\cosh ^3(x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^2}+\frac{15 \operatorname{Subst}\left (\int \frac{\cosh ^5(x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^2}\\ &=-\frac{x}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}-\frac{3}{a^2 \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+\frac{5}{2 a^2 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac{3 \operatorname{Subst}\left (\int \frac{\sinh (6 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{32 a^2}+\frac{3 \operatorname{Subst}\left (\int \frac{\sinh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{8 a^2}+\frac{15 \operatorname{Subst}\left (\int \frac{\sinh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{32 a^2}-\frac{10 \operatorname{Subst}\left (\int \left (\frac{\sinh (2 x)}{4 x}+\frac{\sinh (4 x)}{8 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^2}+\frac{15 \operatorname{Subst}\left (\int \left (\frac{5 \sinh (2 x)}{32 x}+\frac{\sinh (4 x)}{8 x}+\frac{\sinh (6 x)}{32 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^2}\\ &=-\frac{x}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}-\frac{3}{a^2 \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+\frac{5}{2 a^2 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac{15 \text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{32 a^2}+\frac{3 \text{Shi}\left (4 \tanh ^{-1}(a x)\right )}{8 a^2}+\frac{3 \text{Shi}\left (6 \tanh ^{-1}(a x)\right )}{32 a^2}+\frac{15 \operatorname{Subst}\left (\int \frac{\sinh (6 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{32 a^2}-\frac{5 \operatorname{Subst}\left (\int \frac{\sinh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{4 a^2}+\frac{15 \operatorname{Subst}\left (\int \frac{\sinh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{8 a^2}+\frac{75 \operatorname{Subst}\left (\int \frac{\sinh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{32 a^2}-\frac{5 \operatorname{Subst}\left (\int \frac{\sinh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{2 a^2}\\ &=-\frac{x}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}-\frac{3}{a^2 \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+\frac{5}{2 a^2 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac{5 \text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{16 a^2}+\frac{\text{Shi}\left (4 \tanh ^{-1}(a x)\right )}{a^2}+\frac{9 \text{Shi}\left (6 \tanh ^{-1}(a x)\right )}{16 a^2}\\ \end{align*}
Mathematica [A] time = 0.358251, size = 73, normalized size = 0.64 \[ \frac{\frac{8 \left (\left (5 a^2 x^2+1\right ) \tanh ^{-1}(a x)+a x\right )}{\left (a^2 x^2-1\right )^3 \tanh ^{-1}(a x)^2}+5 \text{Shi}\left (2 \tanh ^{-1}(a x)\right )+16 \text{Shi}\left (4 \tanh ^{-1}(a x)\right )+9 \text{Shi}\left (6 \tanh ^{-1}(a x)\right )}{16 a^2} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.076, size = 121, normalized size = 1.1 \begin{align*}{\frac{1}{{a}^{2}} \left ( -{\frac{5\,\sinh \left ( 2\,{\it Artanh} \left ( ax \right ) \right ) }{64\, \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}}}-{\frac{5\,\cosh \left ( 2\,{\it Artanh} \left ( ax \right ) \right ) }{32\,{\it Artanh} \left ( ax \right ) }}+{\frac{5\,{\it Shi} \left ( 2\,{\it Artanh} \left ( ax \right ) \right ) }{16}}-{\frac{\sinh \left ( 4\,{\it Artanh} \left ( ax \right ) \right ) }{16\, \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}}}-{\frac{\cosh \left ( 4\,{\it Artanh} \left ( ax \right ) \right ) }{4\,{\it Artanh} \left ( ax \right ) }}+{\it Shi} \left ( 4\,{\it Artanh} \left ( ax \right ) \right ) -{\frac{\sinh \left ( 6\,{\it Artanh} \left ( ax \right ) \right ) }{64\, \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}}}-{\frac{3\,\cosh \left ( 6\,{\it Artanh} \left ( ax \right ) \right ) }{32\,{\it Artanh} \left ( ax \right ) }}+{\frac{9\,{\it Shi} \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 \, a x +{\left (5 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right ) -{\left (5 \, a^{2} x^{2} + 1\right )} \log \left (-a x + 1\right )}{{\left (a^{8} x^{6} - 3 \, a^{6} x^{4} + 3 \, a^{4} x^{2} - a^{2}\right )} \log \left (a x + 1\right )^{2} - 2 \,{\left (a^{8} x^{6} - 3 \, a^{6} x^{4} + 3 \, a^{4} x^{2} - a^{2}\right )} \log \left (a x + 1\right ) \log \left (-a x + 1\right ) +{\left (a^{8} x^{6} - 3 \, a^{6} x^{4} + 3 \, a^{4} x^{2} - a^{2}\right )} \log \left (-a x + 1\right )^{2}} - \int -\frac{4 \,{\left (5 \, a^{2} x^{3} + 4 \, x\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 = 1.84023, size = 1037, normalized size = 9.1 \begin{align*} \frac{{\left (9 \,{\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 ) - 9 \,{\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 ) + 16 \,{\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 ) - 16 \,{\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 \, a x + 32 \,{\left (5 \, a^{2} x^{2} + 1\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )}{32 \,{\left (a^{8} x^{6} - 3 \, a^{6} x^{4} + 3 \, a^{4} x^{2} - a^{2}\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{x}{\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{x}{{\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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