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 = 0.19, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, 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 3298
Rule 5448
Rule 5966
Rule 6034
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*}
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Mathematica [A] time = 0.22, 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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fricas [F] time = 0.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a^{2} x^{2} + 1}}{{\left (a^{10} x^{10} - 5 \, a^{8} x^{8} + 10 \, a^{6} x^{6} - 10 \, a^{4} x^{4} + 5 \, a^{2} x^{2} - 1\right )} \operatorname {artanh}\left (a x\right )^{2}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {9}{2}} \operatorname {artanh}\left (a x\right )^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.62, size = 232, normalized size = 2.90 \[ \frac {35 \arctanh \left (a x \right ) \Shi \left (\arctanh \left (a x \right )\right ) x^{2} a^{2}+63 \arctanh \left (a x \right ) \Shi \left (3 \arctanh \left (a x \right )\right ) x^{2} a^{2}+35 \arctanh \left (a x \right ) \Shi \left (5 \arctanh \left (a x \right )\right ) x^{2} a^{2}+7 \arctanh \left (a x \right ) \Shi \left (7 \arctanh \left (a x \right )\right ) x^{2} a^{2}-21 \cosh \left (3 \arctanh \left (a x \right )\right ) x^{2} a^{2}-7 \cosh \left (5 \arctanh \left (a x \right )\right ) x^{2} a^{2}-\cosh \left (7 \arctanh \left (a x \right )\right ) x^{2} a^{2}-35 \Shi \left (\arctanh \left (a x \right )\right ) \arctanh \left (a x \right )-63 \Shi \left (3 \arctanh \left (a x \right )\right ) \arctanh \left (a x \right )-35 \Shi \left (5 \arctanh \left (a x \right )\right ) \arctanh \left (a x \right )-7 \Shi \left (7 \arctanh \left (a x \right )\right ) \arctanh \left (a x \right )+35 \sqrt {-a^{2} x^{2}+1}+21 \cosh \left (3 \arctanh \left (a x \right )\right )+7 \cosh \left (5 \arctanh \left (a x \right )\right )+\cosh \left (7 \arctanh \left (a x \right )\right )}{64 a \arctanh \left (a x \right ) \left (a^{2} x^{2}-1\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {9}{2}} \operatorname {artanh}\left (a x\right )^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\mathrm {atanh}\left (a\,x\right )}^2\,{\left (1-a^2\,x^2\right )}^{9/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {9}{2}} \operatorname {atanh}^{2}{\left (a x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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