Optimal. Leaf size=125 \[ -\frac {2 x}{3 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}+\frac {2}{a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac {8}{3 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}-\frac {1}{3 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3}+\frac {2 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{3 a}+\frac {4 \text {Shi}\left (4 \tanh ^{-1}(a x)\right )}{3 a} \]
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Rubi [A] time = 0.50, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {5966, 6032, 6028, 6034, 5448, 12, 3298} \[ -\frac {2 x}{3 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}+\frac {2}{a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac {8}{3 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}-\frac {1}{3 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3}+\frac {2 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{3 a}+\frac {4 \text {Shi}\left (4 \tanh ^{-1}(a x)\right )}{3 a} \]
Antiderivative was successfully verified.
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Rule 12
Rule 3298
Rule 5448
Rule 5966
Rule 6028
Rule 6032
Rule 6034
Rubi steps
\begin {align*} \int \frac {1}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^4} \, dx &=-\frac {1}{3 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3}+\frac {1}{3} (4 a) \int \frac {x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^3} \, dx\\ &=-\frac {1}{3 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3}-\frac {2 x}{3 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}+\frac {2}{3} \int \frac {1}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2} \, dx+\left (2 a^2\right ) \int \frac {x^2}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2} \, dx\\ &=-\frac {1}{3 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3}-\frac {2 x}{3 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}-\frac {2}{3 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+2 \int \frac {1}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2} \, dx-2 \int \frac {1}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2} \, dx+\frac {1}{3} (8 a) \int \frac {x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)} \, dx\\ &=-\frac {1}{3 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3}-\frac {2 x}{3 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}-\frac {8}{3 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {2}{a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {8 \operatorname {Subst}\left (\int \frac {\cosh ^3(x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{3 a}-(4 a) \int \frac {x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)} \, dx+(8 a) \int \frac {x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)} \, dx\\ &=-\frac {1}{3 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3}-\frac {2 x}{3 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}-\frac {8}{3 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {2}{a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {8 \operatorname {Subst}\left (\int \left (\frac {\sinh (2 x)}{4 x}+\frac {\sinh (4 x)}{8 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{3 a}-\frac {4 \operatorname {Subst}\left (\int \frac {\cosh (x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a}+\frac {8 \operatorname {Subst}\left (\int \frac {\cosh ^3(x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=-\frac {1}{3 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3}-\frac {2 x}{3 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}-\frac {8}{3 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {2}{a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {\operatorname {Subst}\left (\int \frac {\sinh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{3 a}+\frac {2 \operatorname {Subst}\left (\int \frac {\sinh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{3 a}-\frac {4 \operatorname {Subst}\left (\int \frac {\sinh (2 x)}{2 x} \, dx,x,\tanh ^{-1}(a x)\right )}{a}+\frac {8 \operatorname {Subst}\left (\int \left (\frac {\sinh (2 x)}{4 x}+\frac {\sinh (4 x)}{8 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=-\frac {1}{3 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3}-\frac {2 x}{3 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}-\frac {8}{3 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {2}{a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {2 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{3 a}+\frac {\text {Shi}\left (4 \tanh ^{-1}(a x)\right )}{3 a}+\frac {\operatorname {Subst}\left (\int \frac {\sinh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=-\frac {1}{3 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3}-\frac {2 x}{3 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}-\frac {8}{3 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {2}{a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {2 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{3 a}+\frac {4 \text {Shi}\left (4 \tanh ^{-1}(a x)\right )}{3 a}\\ \end {align*}
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Mathematica [A] time = 0.26, size = 108, normalized size = 0.86 \[ -\frac {-2 \left (a^2 x^2-1\right )^2 \tanh ^{-1}(a x)^3 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )-4 \left (a^2 x^2-1\right )^2 \tanh ^{-1}(a x)^3 \text {Shi}\left (4 \tanh ^{-1}(a x)\right )+6 a^2 x^2 \tanh ^{-1}(a x)^2+2 \tanh ^{-1}(a x)^2+2 a x \tanh ^{-1}(a x)+1}{3 a \left (a^2 x^2-1\right )^2 \tanh ^{-1}(a x)^3} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.51, size = 272, normalized size = 2.18 \[ \frac {{\left (2 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {log\_integral}\left (\frac {a^{2} x^{2} + 2 \, a x + 1}{a^{2} x^{2} - 2 \, a x + 1}\right ) - 2 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {log\_integral}\left (\frac {a^{2} x^{2} - 2 \, a x + 1}{a^{2} x^{2} + 2 \, a x + 1}\right ) + {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {log\_integral}\left (-\frac {a x + 1}{a x - 1}\right ) - {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {log\_integral}\left (-\frac {a x - 1}{a x + 1}\right )\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{3} - 8 \, a x \log \left (-\frac {a x + 1}{a x - 1}\right ) - 4 \, {\left (3 \, a^{2} x^{2} + 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} - 8}{3 \, {\left (a^{5} x^{4} - 2 \, a^{3} x^{2} + a\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{3}} \]
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 \[ \int -\frac {1}{{\left (a^{2} x^{2} - 1\right )}^{3} \operatorname {artanh}\left (a x\right )^{4}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.23, size = 122, normalized size = 0.98 \[ \frac {-\frac {1}{8 \arctanh \left (a x \right )^{3}}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{6 \arctanh \left (a x \right )^{3}}-\frac {\sinh \left (2 \arctanh \left (a x \right )\right )}{6 \arctanh \left (a x \right )^{2}}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{3 \arctanh \left (a x \right )}+\frac {2 \Shi \left (2 \arctanh \left (a x \right )\right )}{3}-\frac {\cosh \left (4 \arctanh \left (a x \right )\right )}{24 \arctanh \left (a x \right )^{3}}-\frac {\sinh \left (4 \arctanh \left (a x \right )\right )}{12 \arctanh \left (a x \right )^{2}}-\frac {\cosh \left (4 \arctanh \left (a x \right )\right )}{3 \arctanh \left (a x \right )}+\frac {4 \Shi \left (4 \arctanh \left (a x \right )\right )}{3}}{a} \]
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 \[ -\frac {4 \, {\left (2 \, a x \log \left (a x + 1\right ) + {\left (3 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right )^{2} + {\left (3 \, a^{2} x^{2} + 1\right )} \log \left (-a x + 1\right )^{2} - 2 \, {\left (a x + {\left (3 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right )\right )} \log \left (-a x + 1\right ) + 2\right )}}{3 \, {\left ({\left (a^{5} x^{4} - 2 \, a^{3} x^{2} + a\right )} \log \left (a x + 1\right )^{3} - 3 \, {\left (a^{5} x^{4} - 2 \, a^{3} x^{2} + a\right )} \log \left (a x + 1\right )^{2} \log \left (-a x + 1\right ) + 3 \, {\left (a^{5} x^{4} - 2 \, a^{3} x^{2} + a\right )} \log \left (a x + 1\right ) \log \left (-a x + 1\right )^{2} - {\left (a^{5} x^{4} - 2 \, a^{3} x^{2} + a\right )} \log \left (-a x + 1\right )^{3}\right )}} + \int -\frac {8 \, {\left (3 \, a^{3} x^{3} + 5 \, a x\right )}}{3 \, {\left ({\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \log \left (a x + 1\right ) - {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \log \left (-a x + 1\right )\right )}}\,{d x} \]
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 \[ -\int \frac {1}{{\mathrm {atanh}\left (a\,x\right )}^4\,{\left (a^2\,x^2-1\right )}^3} \,d x \]
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 \[ - \int \frac {1}{a^{6} x^{6} \operatorname {atanh}^{4}{\left (a x \right )} - 3 a^{4} x^{4} \operatorname {atanh}^{4}{\left (a x \right )} + 3 a^{2} x^{2} \operatorname {atanh}^{4}{\left (a x \right )} - \operatorname {atanh}^{4}{\left (a x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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