Optimal. Leaf size=239 \[ -\frac {3 a \text {Li}_4\left (\frac {2}{a x+1}-1\right )}{c}-\frac {3 a \text {Li}_5\left (\frac {2}{1-a x}-1\right )}{2 c}+\frac {2 a \text {Li}_2\left (\frac {2}{1-a x}-1\right ) \tanh ^{-1}(a x)^3}{c}-\frac {6 a \text {Li}_2\left (\frac {2}{a x+1}-1\right ) \tanh ^{-1}(a x)^2}{c}-\frac {3 a \text {Li}_3\left (\frac {2}{1-a x}-1\right ) \tanh ^{-1}(a x)^2}{c}-\frac {6 a \text {Li}_3\left (\frac {2}{a x+1}-1\right ) \tanh ^{-1}(a x)}{c}+\frac {3 a \text {Li}_4\left (\frac {2}{1-a x}-1\right ) \tanh ^{-1}(a x)}{c}+\frac {a \tanh ^{-1}(a x)^4}{c}-\frac {\tanh ^{-1}(a x)^4}{c x}+\frac {a \log \left (2-\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)^4}{c}+\frac {4 a \log \left (2-\frac {2}{a x+1}\right ) \tanh ^{-1}(a x)^3}{c} \]
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Rubi [A] time = 0.55, antiderivative size = 239, normalized size of antiderivative = 1.00, 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 5916
Rule 5932
Rule 5934
Rule 5948
Rule 5988
Rule 6056
Rule 6058
Rule 6060
Rule 6062
Rule 6610
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*}
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Mathematica [C] time = 0.49, size = 172, normalized size = 0.72 \[ -\frac {a \left (-2 \left (\tanh ^{-1}(a x)+3\right ) \tanh ^{-1}(a x)^2 \text {Li}_2\left (e^{2 \tanh ^{-1}(a x)}\right )+3 \left (\tanh ^{-1}(a x)+2\right ) \tanh ^{-1}(a x) \text {Li}_3\left (e^{2 \tanh ^{-1}(a x)}\right )-3 \tanh ^{-1}(a x) \text {Li}_4\left (e^{2 \tanh ^{-1}(a x)}\right )-3 \text {Li}_4\left (e^{2 \tanh ^{-1}(a x)}\right )+\frac {3}{2} \text {Li}_5\left (e^{2 \tanh ^{-1}(a x)}\right )+\frac {\tanh ^{-1}(a x)^4}{a x}+\tanh ^{-1}(a x)^4+\tanh ^{-1}(a x)^4 \left (-\log \left (1-e^{2 \tanh ^{-1}(a x)}\right )\right )-4 \tanh ^{-1}(a x)^3 \log \left (1-e^{2 \tanh ^{-1}(a x)}\right )+\frac {i \pi ^5}{160}-\frac {\pi ^4}{16}\right )}{c} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\operatorname {artanh}\left (a x\right )^{4}}{a c x^{3} - c x^{2}}, x\right ) \]
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 \[ \int -\frac {\operatorname {artanh}\left (a x\right )^{4}}{{\left (a c x - c\right )} x^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 1.03, size = 583, normalized size = 2.44 \[ -\frac {a \arctanh \left (a x \right )^{4}}{c}-\frac {\arctanh \left (a x \right )^{4}}{c x}+\frac {a \arctanh \left (a x \right )^{4} \ln \left (1-\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c}+\frac {4 a \arctanh \left (a x \right )^{3} \polylog \left (2, \frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c}-\frac {12 a \arctanh \left (a x \right )^{2} \polylog \left (3, \frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c}+\frac {24 a \arctanh \left (a x \right ) \polylog \left (4, \frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c}-\frac {24 a \polylog \left (5, \frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c}+\frac {a \arctanh \left (a x \right )^{4} \ln \left (1+\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c}+\frac {4 a \arctanh \left (a x \right )^{3} \polylog \left (2, -\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c}-\frac {12 a \arctanh \left (a x \right )^{2} \polylog \left (3, -\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c}+\frac {24 a \arctanh \left (a x \right ) \polylog \left (4, -\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c}-\frac {24 a \polylog \left (5, -\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c}+\frac {4 a \arctanh \left (a x \right )^{3} \ln \left (1-\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c}+\frac {12 a \arctanh \left (a x \right )^{2} \polylog \left (2, \frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c}-\frac {24 a \arctanh \left (a x \right ) \polylog \left (3, \frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c}+\frac {24 a \polylog \left (4, \frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c}+\frac {4 a \arctanh \left (a x \right )^{3} \ln \left (1+\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c}+\frac {12 a \arctanh \left (a x \right )^{2} \polylog \left (2, -\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c}-\frac {24 a \arctanh \left (a x \right ) \polylog \left (3, -\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c}+\frac {24 a \polylog \left (4, -\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c} \]
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 \[ -\frac {a x \log \left (-a x + 1\right )^{5} + 5 \, \log \left (-a x + 1\right )^{4}}{80 \, c x} + \frac {1}{16} \, \int -\frac {\log \left (a x + 1\right )^{4} - 4 \, \log \left (a x + 1\right )^{3} \log \left (-a x + 1\right ) + 6 \, \log \left (a x + 1\right )^{2} \log \left (-a x + 1\right )^{2} - 4 \, {\left (a x + \log \left (a x + 1\right )\right )} \log \left (-a x + 1\right )^{3}}{a c x^{3} - c x^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {atanh}\left (a\,x\right )}^4}{x^2\,\left (c-a\,c\,x\right )} \,d x \]
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 \[ - \frac {\int \frac {\operatorname {atanh}^{4}{\left (a x \right )}}{a x^{3} - x^{2}}\, dx}{c} \]
Verification of antiderivative is not currently implemented for this CAS.
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