Optimal. Leaf size=89 \[ -\frac {1}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}-\frac {3 x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+\frac {15 \text {Chi}\left (2 \tanh ^{-1}(a x)\right )}{16 a}+\frac {3 \text {Chi}\left (4 \tanh ^{-1}(a x)\right )}{2 a}+\frac {9 \text {Chi}\left (6 \tanh ^{-1}(a x)\right )}{16 a} \]
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Rubi [A]
time = 0.22, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps
used = 14, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {6113, 6179,
6181, 5556, 3382, 6115, 3393} \begin {gather*} -\frac {3 x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}-\frac {1}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}+\frac {15 \text {Chi}\left (2 \tanh ^{-1}(a x)\right )}{16 a}+\frac {3 \text {Chi}\left (4 \tanh ^{-1}(a x)\right )}{2 a}+\frac {9 \text {Chi}\left (6 \tanh ^{-1}(a x)\right )}{16 a} \end {gather*}
Antiderivative was successfully verified.
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Rule 3382
Rule 3393
Rule 5556
Rule 6113
Rule 6115
Rule 6179
Rule 6181
Rubi steps
\begin {align*} \int \frac {1}{\left (1-a^2 x^2\right )^4 \tanh ^{-1}(a x)^3} \, dx &=-\frac {1}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}+(3 a) \int \frac {x}{\left (1-a^2 x^2\right )^4 \tanh ^{-1}(a x)^2} \, dx\\ &=-\frac {1}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}-\frac {3 x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+3 \int \frac {1}{\left (1-a^2 x^2\right )^4 \tanh ^{-1}(a x)} \, dx+\left (15 a^2\right ) \int \frac {x^2}{\left (1-a^2 x^2\right )^4 \tanh ^{-1}(a x)} \, dx\\ &=-\frac {1}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}-\frac {3 x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+\frac {3 \text {Subst}\left (\int \frac {\cosh ^6(x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a}+\frac {15 \text {Subst}\left (\int \frac {\cosh ^4(x) \sinh ^2(x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=-\frac {1}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}-\frac {3 x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+\frac {3 \text {Subst}\left (\int \left (\frac {5}{16 x}+\frac {15 \cosh (2 x)}{32 x}+\frac {3 \cosh (4 x)}{16 x}+\frac {\cosh (6 x)}{32 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a}+\frac {15 \text {Subst}\left (\int \left (-\frac {1}{16 x}-\frac {\cosh (2 x)}{32 x}+\frac {\cosh (4 x)}{16 x}+\frac {\cosh (6 x)}{32 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=-\frac {1}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}-\frac {3 x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+\frac {3 \text {Subst}\left (\int \frac {\cosh (6 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{32 a}-\frac {15 \text {Subst}\left (\int \frac {\cosh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{32 a}+\frac {15 \text {Subst}\left (\int \frac {\cosh (6 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{32 a}+\frac {9 \text {Subst}\left (\int \frac {\cosh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{16 a}+\frac {15 \text {Subst}\left (\int \frac {\cosh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{16 a}+\frac {45 \text {Subst}\left (\int \frac {\cosh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{32 a}\\ &=-\frac {1}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}-\frac {3 x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+\frac {15 \text {Chi}\left (2 \tanh ^{-1}(a x)\right )}{16 a}+\frac {3 \text {Chi}\left (4 \tanh ^{-1}(a x)\right )}{2 a}+\frac {9 \text {Chi}\left (6 \tanh ^{-1}(a x)\right )}{16 a}\\ \end {align*}
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Mathematica [A]
time = 0.10, size = 83, normalized size = 0.93 \begin {gather*} \frac {1}{16} \left (\frac {8}{a \left (-1+a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}+\frac {48 x}{\left (-1+a^2 x^2\right )^3 \tanh ^{-1}(a x)}+\frac {15 \text {Chi}\left (2 \tanh ^{-1}(a x)\right )}{a}+\frac {24 \text {Chi}\left (4 \tanh ^{-1}(a x)\right )}{a}+\frac {9 \text {Chi}\left (6 \tanh ^{-1}(a x)\right )}{a}\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 2.72, size = 131, normalized size = 1.47
method | result | size |
derivativedivides | \(\frac {-\frac {5}{32 \arctanh \left (a x \right )^{2}}-\frac {15 \cosh \left (2 \arctanh \left (a x \right )\right )}{64 \arctanh \left (a x \right )^{2}}-\frac {15 \sinh \left (2 \arctanh \left (a x \right )\right )}{32 \arctanh \left (a x \right )}+\frac {15 \hyperbolicCosineIntegral \left (2 \arctanh \left (a x \right )\right )}{16}-\frac {3 \cosh \left (4 \arctanh \left (a x \right )\right )}{32 \arctanh \left (a x \right )^{2}}-\frac {3 \sinh \left (4 \arctanh \left (a x \right )\right )}{8 \arctanh \left (a x \right )}+\frac {3 \hyperbolicCosineIntegral \left (4 \arctanh \left (a x \right )\right )}{2}-\frac {\cosh \left (6 \arctanh \left (a x \right )\right )}{64 \arctanh \left (a x \right )^{2}}-\frac {3 \sinh \left (6 \arctanh \left (a x \right )\right )}{32 \arctanh \left (a x \right )}+\frac {9 \hyperbolicCosineIntegral \left (6 \arctanh \left (a x \right )\right )}{16}}{a}\) | \(131\) |
default | \(\frac {-\frac {5}{32 \arctanh \left (a x \right )^{2}}-\frac {15 \cosh \left (2 \arctanh \left (a x \right )\right )}{64 \arctanh \left (a x \right )^{2}}-\frac {15 \sinh \left (2 \arctanh \left (a x \right )\right )}{32 \arctanh \left (a x \right )}+\frac {15 \hyperbolicCosineIntegral \left (2 \arctanh \left (a x \right )\right )}{16}-\frac {3 \cosh \left (4 \arctanh \left (a x \right )\right )}{32 \arctanh \left (a x \right )^{2}}-\frac {3 \sinh \left (4 \arctanh \left (a x \right )\right )}{8 \arctanh \left (a x \right )}+\frac {3 \hyperbolicCosineIntegral \left (4 \arctanh \left (a x \right )\right )}{2}-\frac {\cosh \left (6 \arctanh \left (a x \right )\right )}{64 \arctanh \left (a x \right )^{2}}-\frac {3 \sinh \left (6 \arctanh \left (a x \right )\right )}{32 \arctanh \left (a x \right )}+\frac {9 \hyperbolicCosineIntegral \left (6 \arctanh \left (a x \right )\right )}{16}}{a}\) | \(131\) |
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 [B] Leaf count of result is larger than twice the leaf count of optimal. 435 vs.
\(2 (79) = 158\).
time = 0.36, size = 435, normalized size = 4.89 \begin {gather*} \frac {192 \, a x \log \left (-\frac {a x + 1}{a x - 1}\right ) + 3 \, {\left (3 \, {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (-\frac {a^{3} x^{3} + 3 \, a^{2} x^{2} + 3 \, a x + 1}{a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1}\right ) + 3 \, {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (-\frac {a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1}{a^{3} x^{3} + 3 \, a^{2} x^{2} + 3 \, a x + 1}\right ) + 8 \, {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (\frac {a^{2} x^{2} + 2 \, a x + 1}{a^{2} x^{2} - 2 \, a x + 1}\right ) + 8 \, {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (\frac {a^{2} x^{2} - 2 \, a x + 1}{a^{2} x^{2} + 2 \, a x + 1}\right ) + 5 \, {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (-\frac {a x + 1}{a x - 1}\right ) + 5 \, {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, 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 )^{2} + 64}{32 \, {\left (a^{7} x^{6} - 3 \, a^{5} x^{4} + 3 \, a^{3} x^{2} - a\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2}} \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 )^{4} \left (a x + 1\right )^{4} \operatorname {atanh}^{3}{\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 )}^3\,{\left (a^2\,x^2-1\right )}^4} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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