Optimal. Leaf size=66 \[ -\frac{1}{a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+\frac{15 \text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{16 a}+\frac{3 \text{Shi}\left (4 \tanh ^{-1}(a x)\right )}{4 a}+\frac{3 \text{Shi}\left (6 \tanh ^{-1}(a x)\right )}{16 a} \]
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Rubi [A] time = 0.142841, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {5966, 6034, 5448, 3298} \[ -\frac{1}{a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+\frac{15 \text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{16 a}+\frac{3 \text{Shi}\left (4 \tanh ^{-1}(a x)\right )}{4 a}+\frac{3 \text{Shi}\left (6 \tanh ^{-1}(a x)\right )}{16 a} \]
Antiderivative was successfully verified.
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Rule 5966
Rule 6034
Rule 5448
Rule 3298
Rubi steps
\begin{align*} \int \frac{1}{\left (1-a^2 x^2\right )^4 \tanh ^{-1}(a x)^2} \, dx &=-\frac{1}{a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+(6 a) \int \frac{x}{\left (1-a^2 x^2\right )^4 \tanh ^{-1}(a x)} \, dx\\ &=-\frac{1}{a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+\frac{6 \operatorname{Subst}\left (\int \frac{\cosh ^5(x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=-\frac{1}{a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+\frac{6 \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}\\ &=-\frac{1}{a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+\frac{3 \operatorname{Subst}\left (\int \frac{\sinh (6 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{16 a}+\frac{3 \operatorname{Subst}\left (\int \frac{\sinh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{4 a}+\frac{15 \operatorname{Subst}\left (\int \frac{\sinh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{16 a}\\ &=-\frac{1}{a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+\frac{15 \text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{16 a}+\frac{3 \text{Shi}\left (4 \tanh ^{-1}(a x)\right )}{4 a}+\frac{3 \text{Shi}\left (6 \tanh ^{-1}(a x)\right )}{16 a}\\ \end{align*}
Mathematica [A] time = 0.281275, size = 56, normalized size = 0.85 \[ \frac{\frac{1}{\left (a^2 x^2-1\right )^3 \tanh ^{-1}(a x)}+\frac{15}{16} \text{Shi}\left (2 \tanh ^{-1}(a x)\right )+\frac{3}{4} \text{Shi}\left (4 \tanh ^{-1}(a x)\right )+\frac{3}{16} \text{Shi}\left (6 \tanh ^{-1}(a x)\right )}{a} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.068, size = 86, normalized size = 1.3 \begin{align*}{\frac{1}{a} \left ( -{\frac{5}{16\,{\it Artanh} \left ( ax \right ) }}-{\frac{15\,\cosh \left ( 2\,{\it Artanh} \left ( ax \right ) \right ) }{32\,{\it Artanh} \left ( ax \right ) }}+{\frac{15\,{\it Shi} \left ( 2\,{\it Artanh} \left ( ax \right ) \right ) }{16}}-{\frac{3\,\cosh \left ( 4\,{\it Artanh} \left ( ax \right ) \right ) }{16\,{\it Artanh} \left ( ax \right ) }}+{\frac{3\,{\it Shi} \left ( 4\,{\it Artanh} \left ( ax \right ) \right ) }{4}}-{\frac{\cosh \left ( 6\,{\it Artanh} \left ( ax \right ) \right ) }{32\,{\it Artanh} \left ( ax \right ) }}+{\frac{3\,{\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*} -12 \, a \int -\frac{x}{{\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} + \frac{2}{{\left (a^{7} x^{6} - 3 \, a^{5} x^{4} + 3 \, a^{3} x^{2} - a\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 )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.99132, size = 954, normalized size = 14.45 \begin{align*} \frac{3 \,{\left ({\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 ) -{\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 ) + 4 \,{\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 ) - 4 \,{\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 ) + 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 )} \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}^{2}{\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 )^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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