Optimal. Leaf size=66 \[ -\frac {1}{a \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}+\frac {5 \text {Shi}\left (\tanh ^{-1}(a x)\right )}{8 a}+\frac {15 \text {Shi}\left (3 \tanh ^{-1}(a x)\right )}{16 a}+\frac {5 \text {Shi}\left (5 \tanh ^{-1}(a x)\right )}{16 a} \]
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Rubi [A]
time = 0.13, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {6113, 6181,
5556, 3379} \begin {gather*} -\frac {1}{a \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}+\frac {5 \text {Shi}\left (\tanh ^{-1}(a x)\right )}{8 a}+\frac {15 \text {Shi}\left (3 \tanh ^{-1}(a x)\right )}{16 a}+\frac {5 \text {Shi}\left (5 \tanh ^{-1}(a x)\right )}{16 a} \end {gather*}
Antiderivative was successfully verified.
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Rule 3379
Rule 5556
Rule 6113
Rule 6181
Rubi steps
\begin {align*} \int \frac {1}{\left (1-a^2 x^2\right )^{7/2} \tanh ^{-1}(a x)^2} \, dx &=-\frac {1}{a \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}+(5 a) \int \frac {x}{\left (1-a^2 x^2\right )^{7/2} \tanh ^{-1}(a x)} \, dx\\ &=-\frac {1}{a \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}+\frac {5 \text {Subst}\left (\int \frac {\cosh ^4(x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=-\frac {1}{a \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}+\frac {5 \text {Subst}\left (\int \left (\frac {\sinh (x)}{8 x}+\frac {3 \sinh (3 x)}{16 x}+\frac {\sinh (5 x)}{16 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=-\frac {1}{a \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}+\frac {5 \text {Subst}\left (\int \frac {\sinh (5 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{16 a}+\frac {5 \text {Subst}\left (\int \frac {\sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{8 a}+\frac {15 \text {Subst}\left (\int \frac {\sinh (3 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{16 a}\\ &=-\frac {1}{a \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}+\frac {5 \text {Shi}\left (\tanh ^{-1}(a x)\right )}{8 a}+\frac {15 \text {Shi}\left (3 \tanh ^{-1}(a x)\right )}{16 a}+\frac {5 \text {Shi}\left (5 \tanh ^{-1}(a x)\right )}{16 a}\\ \end {align*}
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Mathematica [A]
time = 0.12, size = 56, normalized size = 0.85 \begin {gather*} \frac {-\frac {16}{\left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}+5 \left (2 \text {Shi}\left (\tanh ^{-1}(a x)\right )+3 \text {Shi}\left (3 \tanh ^{-1}(a x)\right )+\text {Shi}\left (5 \tanh ^{-1}(a x)\right )\right )}{16 a} \end {gather*}
Antiderivative was successfully verified.
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(175\) vs.
\(2(58)=116\).
time = 2.97, size = 176, normalized size = 2.67
method | result | size |
default | \(\frac {5 \arctanh \left (a x \right ) \hyperbolicSineIntegral \left (5 \arctanh \left (a x \right )\right ) a^{2} x^{2}+10 \arctanh \left (a x \right ) \hyperbolicSineIntegral \left (\arctanh \left (a x \right )\right ) a^{2} x^{2}+15 \arctanh \left (a x \right ) \hyperbolicSineIntegral \left (3 \arctanh \left (a x \right )\right ) a^{2} x^{2}-\cosh \left (5 \arctanh \left (a x \right )\right ) a^{2} x^{2}-5 \cosh \left (3 \arctanh \left (a x \right )\right ) a^{2} x^{2}-5 \hyperbolicSineIntegral \left (5 \arctanh \left (a x \right )\right ) \arctanh \left (a x \right )-10 \hyperbolicSineIntegral \left (\arctanh \left (a x \right )\right ) \arctanh \left (a x \right )-15 \hyperbolicSineIntegral \left (3 \arctanh \left (a x \right )\right ) \arctanh \left (a x \right )+\cosh \left (5 \arctanh \left (a x \right )\right )+10 \sqrt {-a^{2} x^{2}+1}+5 \cosh \left (3 \arctanh \left (a x \right )\right )}{16 a \arctanh \left (a x \right ) \left (a^{2} x^{2}-1\right )}\) | \(176\) |
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int \frac {1}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {7}{2}} \operatorname {atanh}^{2}{\left (a x \right )}}\, dx \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{{\mathrm {atanh}\left (a\,x\right )}^2\,{\left (1-a^2\,x^2\right )}^{7/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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