Optimal. Leaf size=154 \[ -\frac {1}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^5}-\frac {x}{10 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac {1+a^2 x^2}{30 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {x}{15 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {1+a^2 x^2}{15 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {2 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{15 a} \]
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Rubi [A]
time = 0.14, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {6113, 6143,
6181, 5556, 12, 3379} \begin {gather*} -\frac {x}{15 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {x}{10 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac {a^2 x^2+1}{15 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac {a^2 x^2+1}{30 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {1}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^5}+\frac {2 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{15 a} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 3379
Rule 5556
Rule 6113
Rule 6143
Rule 6181
Rubi steps
\begin {align*} \int \frac {1}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^6} \, dx &=-\frac {1}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^5}+\frac {1}{5} (2 a) \int \frac {x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^5} \, dx\\ &=-\frac {1}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^5}-\frac {x}{10 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac {1+a^2 x^2}{30 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}+\frac {1}{15} (2 a) \int \frac {x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3} \, dx\\ &=-\frac {1}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^5}-\frac {x}{10 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac {1+a^2 x^2}{30 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {x}{15 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {1+a^2 x^2}{15 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {1}{15} (4 a) \int \frac {x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)} \, dx\\ &=-\frac {1}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^5}-\frac {x}{10 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac {1+a^2 x^2}{30 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {x}{15 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {1+a^2 x^2}{15 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {4 \text {Subst}\left (\int \frac {\cosh (x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{15 a}\\ &=-\frac {1}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^5}-\frac {x}{10 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac {1+a^2 x^2}{30 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {x}{15 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {1+a^2 x^2}{15 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {4 \text {Subst}\left (\int \frac {\sinh (2 x)}{2 x} \, dx,x,\tanh ^{-1}(a x)\right )}{15 a}\\ &=-\frac {1}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^5}-\frac {x}{10 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac {1+a^2 x^2}{30 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {x}{15 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {1+a^2 x^2}{15 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {2 \text {Subst}\left (\int \frac {\sinh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{15 a}\\ &=-\frac {1}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^5}-\frac {x}{10 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac {1+a^2 x^2}{30 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {x}{15 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {1+a^2 x^2}{15 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {2 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{15 a}\\ \end {align*}
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Mathematica [A]
time = 0.05, size = 101, normalized size = 0.66 \begin {gather*} \frac {6+3 a x \tanh ^{-1}(a x)+\left (1+a^2 x^2\right ) \tanh ^{-1}(a x)^2+2 a x \tanh ^{-1}(a x)^3+2 \left (1+a^2 x^2\right ) \tanh ^{-1}(a x)^4+4 \left (-1+a^2 x^2\right ) \tanh ^{-1}(a x)^5 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{30 a \left (-1+a^2 x^2\right ) \tanh ^{-1}(a x)^5} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 6.88, size = 98, normalized size = 0.64
method | result | size |
derivativedivides | \(\frac {-\frac {1}{10 \arctanh \left (a x \right )^{5}}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{10 \arctanh \left (a x \right )^{5}}-\frac {\sinh \left (2 \arctanh \left (a x \right )\right )}{20 \arctanh \left (a x \right )^{4}}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{30 \arctanh \left (a x \right )^{3}}-\frac {\sinh \left (2 \arctanh \left (a x \right )\right )}{30 \arctanh \left (a x \right )^{2}}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{15 \arctanh \left (a x \right )}+\frac {2 \hyperbolicSineIntegral \left (2 \arctanh \left (a x \right )\right )}{15}}{a}\) | \(98\) |
default | \(\frac {-\frac {1}{10 \arctanh \left (a x \right )^{5}}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{10 \arctanh \left (a x \right )^{5}}-\frac {\sinh \left (2 \arctanh \left (a x \right )\right )}{20 \arctanh \left (a x \right )^{4}}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{30 \arctanh \left (a x \right )^{3}}-\frac {\sinh \left (2 \arctanh \left (a x \right )\right )}{30 \arctanh \left (a x \right )^{2}}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{15 \arctanh \left (a x \right )}+\frac {2 \hyperbolicSineIntegral \left (2 \arctanh \left (a x \right )\right )}{15}}{a}\) | \(98\) |
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 [A]
time = 0.34, size = 200, normalized size = 1.30 \begin {gather*} \frac {{\left ({\left (a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (-\frac {a x + 1}{a x - 1}\right ) - {\left (a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (-\frac {a x - 1}{a x + 1}\right )\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{5} + 4 \, a x \log \left (-\frac {a x + 1}{a x - 1}\right )^{3} + 2 \, {\left (a^{2} x^{2} + 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{4} + 24 \, a x \log \left (-\frac {a x + 1}{a x - 1}\right ) + 4 \, {\left (a^{2} x^{2} + 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} + 96}{15 \, {\left (a^{3} x^{2} - a\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{5}} \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 (a x - 1\right )^{2} \left (a x + 1\right )^{2} \operatorname {atanh}^{6}{\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 = -0.01 \begin {gather*} \int \frac {1}{{\mathrm {atanh}\left (a\,x\right )}^6\,{\left (a^2\,x^2-1\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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