Optimal. Leaf size=52 \[ -\frac{1}{a \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}+\frac{3 \text{Shi}\left (\tanh ^{-1}(a x)\right )}{4 a}+\frac{3 \text{Shi}\left (3 \tanh ^{-1}(a x)\right )}{4 a} \]
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Rubi [A] time = 0.159893, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {5966, 6034, 5448, 3298} \[ -\frac{1}{a \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}+\frac{3 \text{Shi}\left (\tanh ^{-1}(a x)\right )}{4 a}+\frac{3 \text{Shi}\left (3 \tanh ^{-1}(a x)\right )}{4 a} \]
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 )^{5/2} \tanh ^{-1}(a x)^2} \, dx &=-\frac{1}{a \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}+(3 a) \int \frac{x}{\left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)} \, dx\\ &=-\frac{1}{a \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}+\frac{3 \operatorname{Subst}\left (\int \frac{\cosh ^2(x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=-\frac{1}{a \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}+\frac{3 \operatorname{Subst}\left (\int \left (\frac{\sinh (x)}{4 x}+\frac{\sinh (3 x)}{4 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=-\frac{1}{a \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}+\frac{3 \operatorname{Subst}\left (\int \frac{\sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{4 a}+\frac{3 \operatorname{Subst}\left (\int \frac{\sinh (3 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{4 a}\\ &=-\frac{1}{a \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}+\frac{3 \text{Shi}\left (\tanh ^{-1}(a x)\right )}{4 a}+\frac{3 \text{Shi}\left (3 \tanh ^{-1}(a x)\right )}{4 a}\\ \end{align*}
Mathematica [A] time = 0.13313, size = 45, normalized size = 0.87 \[ \frac{3 \left (\text{Shi}\left (\tanh ^{-1}(a x)\right )+\text{Shi}\left (3 \tanh ^{-1}(a x)\right )\right )-\frac{4}{\left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}}{4 a} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.157, size = 120, normalized size = 2.3 \begin{align*}{\frac{1}{4\,a{\it Artanh} \left ( ax \right ) \left ({a}^{2}{x}^{2}-1 \right ) } \left ( 3\,{\it Artanh} \left ( ax \right ){\it Shi} \left ({\it Artanh} \left ( ax \right ) \right ){x}^{2}{a}^{2}+3\,{\it Artanh} \left ( ax \right ){\it Shi} \left ( 3\,{\it Artanh} \left ( ax \right ) \right ){x}^{2}{a}^{2}-\cosh \left ( 3\,{\it Artanh} \left ( ax \right ) \right ){x}^{2}{a}^{2}-3\,{\it Shi} \left ({\it Artanh} \left ( ax \right ) \right ){\it Artanh} \left ( ax \right ) -3\,{\it Shi} \left ( 3\,{\it Artanh} \left ( ax \right ) \right ){\it Artanh} \left ( ax \right ) +3\,\sqrt{-{a}^{2}{x}^{2}+1}+\cosh \left ( 3\,{\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{5}{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^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, 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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac{5}{2}} \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 )}^{\frac{5}{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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