Optimal. Leaf size=80 \[ -\frac{1}{a \left (1-a^2 x^2\right )^{7/2} \tanh ^{-1}(a x)}+\frac{35 \text{Shi}\left (\tanh ^{-1}(a x)\right )}{64 a}+\frac{63 \text{Shi}\left (3 \tanh ^{-1}(a x)\right )}{64 a}+\frac{35 \text{Shi}\left (5 \tanh ^{-1}(a x)\right )}{64 a}+\frac{7 \text{Shi}\left (7 \tanh ^{-1}(a x)\right )}{64 a} \]
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Rubi [A] time = 0.190509, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {5966, 6034, 5448, 3298} \[ -\frac{1}{a \left (1-a^2 x^2\right )^{7/2} \tanh ^{-1}(a x)}+\frac{35 \text{Shi}\left (\tanh ^{-1}(a x)\right )}{64 a}+\frac{63 \text{Shi}\left (3 \tanh ^{-1}(a x)\right )}{64 a}+\frac{35 \text{Shi}\left (5 \tanh ^{-1}(a x)\right )}{64 a}+\frac{7 \text{Shi}\left (7 \tanh ^{-1}(a x)\right )}{64 a} \]
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 )^{9/2} \tanh ^{-1}(a x)^2} \, dx &=-\frac{1}{a \left (1-a^2 x^2\right )^{7/2} \tanh ^{-1}(a x)}+(7 a) \int \frac{x}{\left (1-a^2 x^2\right )^{9/2} \tanh ^{-1}(a x)} \, dx\\ &=-\frac{1}{a \left (1-a^2 x^2\right )^{7/2} \tanh ^{-1}(a x)}+\frac{7 \operatorname{Subst}\left (\int \frac{\cosh ^6(x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=-\frac{1}{a \left (1-a^2 x^2\right )^{7/2} \tanh ^{-1}(a x)}+\frac{7 \operatorname{Subst}\left (\int \left (\frac{5 \sinh (x)}{64 x}+\frac{9 \sinh (3 x)}{64 x}+\frac{5 \sinh (5 x)}{64 x}+\frac{\sinh (7 x)}{64 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=-\frac{1}{a \left (1-a^2 x^2\right )^{7/2} \tanh ^{-1}(a x)}+\frac{7 \operatorname{Subst}\left (\int \frac{\sinh (7 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{64 a}+\frac{35 \operatorname{Subst}\left (\int \frac{\sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{64 a}+\frac{35 \operatorname{Subst}\left (\int \frac{\sinh (5 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{64 a}+\frac{63 \operatorname{Subst}\left (\int \frac{\sinh (3 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{64 a}\\ &=-\frac{1}{a \left (1-a^2 x^2\right )^{7/2} \tanh ^{-1}(a x)}+\frac{35 \text{Shi}\left (\tanh ^{-1}(a x)\right )}{64 a}+\frac{63 \text{Shi}\left (3 \tanh ^{-1}(a x)\right )}{64 a}+\frac{35 \text{Shi}\left (5 \tanh ^{-1}(a x)\right )}{64 a}+\frac{7 \text{Shi}\left (7 \tanh ^{-1}(a x)\right )}{64 a}\\ \end{align*}
Mathematica [A] time = 0.209039, size = 65, normalized size = 0.81 \[ \frac{7 \left (5 \text{Shi}\left (\tanh ^{-1}(a x)\right )+9 \text{Shi}\left (3 \tanh ^{-1}(a x)\right )+5 \text{Shi}\left (5 \tanh ^{-1}(a x)\right )+\text{Shi}\left (7 \tanh ^{-1}(a x)\right )\right )-\frac{64}{\left (1-a^2 x^2\right )^{7/2} \tanh ^{-1}(a x)}}{64 a} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.159, size = 232, normalized size = 2.9 \begin{align*}{\frac{1}{64\,a{\it Artanh} \left ( ax \right ) \left ({a}^{2}{x}^{2}-1 \right ) } \left ( 35\,{\it Artanh} \left ( ax \right ){\it Shi} \left ({\it Artanh} \left ( ax \right ) \right ){x}^{2}{a}^{2}+63\,{\it Artanh} \left ( ax \right ){\it Shi} \left ( 3\,{\it Artanh} \left ( ax \right ) \right ){x}^{2}{a}^{2}+35\,{\it Artanh} \left ( ax \right ){\it Shi} \left ( 5\,{\it Artanh} \left ( ax \right ) \right ){x}^{2}{a}^{2}+7\,{\it Artanh} \left ( ax \right ){\it Shi} \left ( 7\,{\it Artanh} \left ( ax \right ) \right ){x}^{2}{a}^{2}-21\,\cosh \left ( 3\,{\it Artanh} \left ( ax \right ) \right ){x}^{2}{a}^{2}-7\,\cosh \left ( 5\,{\it Artanh} \left ( ax \right ) \right ){x}^{2}{a}^{2}-\cosh \left ( 7\,{\it Artanh} \left ( ax \right ) \right ){x}^{2}{a}^{2}-35\,{\it Shi} \left ({\it Artanh} \left ( ax \right ) \right ){\it Artanh} \left ( ax \right ) -63\,{\it Shi} \left ( 3\,{\it Artanh} \left ( ax \right ) \right ){\it Artanh} \left ( ax \right ) -35\,{\it Shi} \left ( 5\,{\it Artanh} \left ( ax \right ) \right ){\it Artanh} \left ( ax \right ) -7\,{\it Shi} \left ( 7\,{\it Artanh} \left ( ax \right ) \right ){\it Artanh} \left ( ax \right ) +35\,\sqrt{-{a}^{2}{x}^{2}+1}+21\,\cosh \left ( 3\,{\it Artanh} \left ( ax \right ) \right ) +7\,\cosh \left ( 5\,{\it Artanh} \left ( ax \right ) \right ) +\cosh \left ( 7\,{\it Artanh} \left ( ax \right ) \right ) \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{1}{{\left (-a^{2} x^{2} + 1\right )}^{\frac{9}{2}} \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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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{-a^{2} x^{2} + 1}}{{\left (a^{10} x^{10} - 5 \, a^{8} x^{8} + 10 \, a^{6} x^{6} - 10 \, a^{4} x^{4} + 5 \, a^{2} x^{2} - 1\right )} \operatorname{artanh}\left (a x\right )^{2}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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 )}^{\frac{9}{2}} \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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