Optimal. Leaf size=239 \[ -\frac{3 a \text{PolyLog}\left (4,\frac{2}{a x+1}-1\right )}{c}-\frac{3 a \text{PolyLog}\left (5,\frac{2}{1-a x}-1\right )}{2 c}+\frac{2 a \tanh ^{-1}(a x)^3 \text{PolyLog}\left (2,\frac{2}{1-a x}-1\right )}{c}-\frac{6 a \tanh ^{-1}(a x)^2 \text{PolyLog}\left (2,\frac{2}{a x+1}-1\right )}{c}-\frac{3 a \tanh ^{-1}(a x)^2 \text{PolyLog}\left (3,\frac{2}{1-a x}-1\right )}{c}-\frac{6 a \tanh ^{-1}(a x) \text{PolyLog}\left (3,\frac{2}{a x+1}-1\right )}{c}+\frac{3 a \tanh ^{-1}(a x) \text{PolyLog}\left (4,\frac{2}{1-a x}-1\right )}{c}+\frac{a \tanh ^{-1}(a x)^4}{c}-\frac{\tanh ^{-1}(a x)^4}{c x}+\frac{a \log \left (2-\frac{2}{1-a x}\right ) \tanh ^{-1}(a x)^4}{c}+\frac{4 a \log \left (2-\frac{2}{a x+1}\right ) \tanh ^{-1}(a x)^3}{c} \]
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Rubi [A] time = 0.548763, antiderivative size = 239, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 10, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.526, Rules used = {5934, 5916, 5988, 5932, 5948, 6056, 6060, 6610, 6058, 6062} \[ -\frac{3 a \text{PolyLog}\left (4,\frac{2}{a x+1}-1\right )}{c}-\frac{3 a \text{PolyLog}\left (5,\frac{2}{1-a x}-1\right )}{2 c}+\frac{2 a \tanh ^{-1}(a x)^3 \text{PolyLog}\left (2,\frac{2}{1-a x}-1\right )}{c}-\frac{6 a \tanh ^{-1}(a x)^2 \text{PolyLog}\left (2,\frac{2}{a x+1}-1\right )}{c}-\frac{3 a \tanh ^{-1}(a x)^2 \text{PolyLog}\left (3,\frac{2}{1-a x}-1\right )}{c}-\frac{6 a \tanh ^{-1}(a x) \text{PolyLog}\left (3,\frac{2}{a x+1}-1\right )}{c}+\frac{3 a \tanh ^{-1}(a x) \text{PolyLog}\left (4,\frac{2}{1-a x}-1\right )}{c}+\frac{a \tanh ^{-1}(a x)^4}{c}-\frac{\tanh ^{-1}(a x)^4}{c x}+\frac{a \log \left (2-\frac{2}{1-a x}\right ) \tanh ^{-1}(a x)^4}{c}+\frac{4 a \log \left (2-\frac{2}{a x+1}\right ) \tanh ^{-1}(a x)^3}{c} \]
Antiderivative was successfully verified.
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Rule 5934
Rule 5916
Rule 5988
Rule 5932
Rule 5948
Rule 6056
Rule 6060
Rule 6610
Rule 6058
Rule 6062
Rubi steps
\begin{align*} \int \frac{\tanh ^{-1}(a x)^4}{x^2 (c-a c x)} \, dx &=a \int \frac{\tanh ^{-1}(a x)^4}{x (c-a c x)} \, dx+\frac{\int \frac{\tanh ^{-1}(a x)^4}{x^2} \, dx}{c}\\ &=-\frac{\tanh ^{-1}(a x)^4}{c x}+\frac{a \tanh ^{-1}(a x)^4 \log \left (2-\frac{2}{1-a x}\right )}{c}+\frac{(4 a) \int \frac{\tanh ^{-1}(a x)^3}{x \left (1-a^2 x^2\right )} \, dx}{c}-\frac{\left (4 a^2\right ) \int \frac{\tanh ^{-1}(a x)^3 \log \left (2-\frac{2}{1-a x}\right )}{1-a^2 x^2} \, dx}{c}\\ &=\frac{a \tanh ^{-1}(a x)^4}{c}-\frac{\tanh ^{-1}(a x)^4}{c x}+\frac{a \tanh ^{-1}(a x)^4 \log \left (2-\frac{2}{1-a x}\right )}{c}+\frac{2 a \tanh ^{-1}(a x)^3 \text{Li}_2\left (-1+\frac{2}{1-a x}\right )}{c}+\frac{(4 a) \int \frac{\tanh ^{-1}(a x)^3}{x (1+a x)} \, dx}{c}-\frac{\left (6 a^2\right ) \int \frac{\tanh ^{-1}(a x)^2 \text{Li}_2\left (-1+\frac{2}{1-a x}\right )}{1-a^2 x^2} \, dx}{c}\\ &=\frac{a \tanh ^{-1}(a x)^4}{c}-\frac{\tanh ^{-1}(a x)^4}{c x}+\frac{a \tanh ^{-1}(a x)^4 \log \left (2-\frac{2}{1-a x}\right )}{c}+\frac{4 a \tanh ^{-1}(a x)^3 \log \left (2-\frac{2}{1+a x}\right )}{c}+\frac{2 a \tanh ^{-1}(a x)^3 \text{Li}_2\left (-1+\frac{2}{1-a x}\right )}{c}-\frac{3 a \tanh ^{-1}(a x)^2 \text{Li}_3\left (-1+\frac{2}{1-a x}\right )}{c}+\frac{\left (6 a^2\right ) \int \frac{\tanh ^{-1}(a x) \text{Li}_3\left (-1+\frac{2}{1-a x}\right )}{1-a^2 x^2} \, dx}{c}-\frac{\left (12 a^2\right ) \int \frac{\tanh ^{-1}(a x)^2 \log \left (2-\frac{2}{1+a x}\right )}{1-a^2 x^2} \, dx}{c}\\ &=\frac{a \tanh ^{-1}(a x)^4}{c}-\frac{\tanh ^{-1}(a x)^4}{c x}+\frac{a \tanh ^{-1}(a x)^4 \log \left (2-\frac{2}{1-a x}\right )}{c}+\frac{4 a \tanh ^{-1}(a x)^3 \log \left (2-\frac{2}{1+a x}\right )}{c}+\frac{2 a \tanh ^{-1}(a x)^3 \text{Li}_2\left (-1+\frac{2}{1-a x}\right )}{c}-\frac{6 a \tanh ^{-1}(a x)^2 \text{Li}_2\left (-1+\frac{2}{1+a x}\right )}{c}-\frac{3 a \tanh ^{-1}(a x)^2 \text{Li}_3\left (-1+\frac{2}{1-a x}\right )}{c}+\frac{3 a \tanh ^{-1}(a x) \text{Li}_4\left (-1+\frac{2}{1-a x}\right )}{c}-\frac{\left (3 a^2\right ) \int \frac{\text{Li}_4\left (-1+\frac{2}{1-a x}\right )}{1-a^2 x^2} \, dx}{c}+\frac{\left (12 a^2\right ) \int \frac{\tanh ^{-1}(a x) \text{Li}_2\left (-1+\frac{2}{1+a x}\right )}{1-a^2 x^2} \, dx}{c}\\ &=\frac{a \tanh ^{-1}(a x)^4}{c}-\frac{\tanh ^{-1}(a x)^4}{c x}+\frac{a \tanh ^{-1}(a x)^4 \log \left (2-\frac{2}{1-a x}\right )}{c}+\frac{4 a \tanh ^{-1}(a x)^3 \log \left (2-\frac{2}{1+a x}\right )}{c}+\frac{2 a \tanh ^{-1}(a x)^3 \text{Li}_2\left (-1+\frac{2}{1-a x}\right )}{c}-\frac{6 a \tanh ^{-1}(a x)^2 \text{Li}_2\left (-1+\frac{2}{1+a x}\right )}{c}-\frac{3 a \tanh ^{-1}(a x)^2 \text{Li}_3\left (-1+\frac{2}{1-a x}\right )}{c}-\frac{6 a \tanh ^{-1}(a x) \text{Li}_3\left (-1+\frac{2}{1+a x}\right )}{c}+\frac{3 a \tanh ^{-1}(a x) \text{Li}_4\left (-1+\frac{2}{1-a x}\right )}{c}-\frac{3 a \text{Li}_5\left (-1+\frac{2}{1-a x}\right )}{2 c}+\frac{\left (6 a^2\right ) \int \frac{\text{Li}_3\left (-1+\frac{2}{1+a x}\right )}{1-a^2 x^2} \, dx}{c}\\ &=\frac{a \tanh ^{-1}(a x)^4}{c}-\frac{\tanh ^{-1}(a x)^4}{c x}+\frac{a \tanh ^{-1}(a x)^4 \log \left (2-\frac{2}{1-a x}\right )}{c}+\frac{4 a \tanh ^{-1}(a x)^3 \log \left (2-\frac{2}{1+a x}\right )}{c}+\frac{2 a \tanh ^{-1}(a x)^3 \text{Li}_2\left (-1+\frac{2}{1-a x}\right )}{c}-\frac{6 a \tanh ^{-1}(a x)^2 \text{Li}_2\left (-1+\frac{2}{1+a x}\right )}{c}-\frac{3 a \tanh ^{-1}(a x)^2 \text{Li}_3\left (-1+\frac{2}{1-a x}\right )}{c}-\frac{6 a \tanh ^{-1}(a x) \text{Li}_3\left (-1+\frac{2}{1+a x}\right )}{c}+\frac{3 a \tanh ^{-1}(a x) \text{Li}_4\left (-1+\frac{2}{1-a x}\right )}{c}-\frac{3 a \text{Li}_4\left (-1+\frac{2}{1+a x}\right )}{c}-\frac{3 a \text{Li}_5\left (-1+\frac{2}{1-a x}\right )}{2 c}\\ \end{align*}
Mathematica [C] time = 0.501071, size = 172, normalized size = 0.72 \[ -\frac{a \left (-2 \left (\tanh ^{-1}(a x)+3\right ) \tanh ^{-1}(a x)^2 \text{PolyLog}\left (2,e^{2 \tanh ^{-1}(a x)}\right )+3 \left (\tanh ^{-1}(a x)+2\right ) \tanh ^{-1}(a x) \text{PolyLog}\left (3,e^{2 \tanh ^{-1}(a x)}\right )-3 \tanh ^{-1}(a x) \text{PolyLog}\left (4,e^{2 \tanh ^{-1}(a x)}\right )-3 \text{PolyLog}\left (4,e^{2 \tanh ^{-1}(a x)}\right )+\frac{3}{2} \text{PolyLog}\left (5,e^{2 \tanh ^{-1}(a x)}\right )+\frac{\tanh ^{-1}(a x)^4}{a x}+\tanh ^{-1}(a x)^4+\tanh ^{-1}(a x)^4 \left (-\log \left (1-e^{2 \tanh ^{-1}(a x)}\right )\right )-4 \tanh ^{-1}(a x)^3 \log \left (1-e^{2 \tanh ^{-1}(a x)}\right )+\frac{i \pi ^5}{160}-\frac{\pi ^4}{16}\right )}{c} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 0.309, size = 583, normalized size = 2.4 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{a x \log \left (-a x + 1\right )^{5} + 5 \, \log \left (-a x + 1\right )^{4}}{80 \, c x} + \frac{1}{16} \, \int -\frac{\log \left (a x + 1\right )^{4} - 4 \, \log \left (a x + 1\right )^{3} \log \left (-a x + 1\right ) + 6 \, \log \left (a x + 1\right )^{2} \log \left (-a x + 1\right )^{2} - 4 \,{\left (a x + \log \left (a x + 1\right )\right )} \log \left (-a x + 1\right )^{3}}{a c x^{3} - c x^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\operatorname{artanh}\left (a x\right )^{4}}{a c x^{3} - c x^{2}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{\int \frac{\operatorname{atanh}^{4}{\left (a x \right )}}{a x^{3} - x^{2}}\, dx}{c} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{\operatorname{artanh}\left (a x\right )^{4}}{{\left (a c x - c\right )} x^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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