Optimal. Leaf size=162 \[ \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{5 c^5}-\frac {2 b \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{5 c^5}+\frac {b x^2 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^3}+\frac {1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {b x^4 \left (a+b \tanh ^{-1}(c x)\right )}{10 c}-\frac {b^2 \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{5 c^5}-\frac {3 b^2 \tanh ^{-1}(c x)}{10 c^5}+\frac {3 b^2 x}{10 c^4}+\frac {b^2 x^3}{30 c^2} \]
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Rubi [A] time = 0.30, antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 9, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {5916, 5980, 302, 206, 321, 5984, 5918, 2402, 2315} \[ -\frac {b^2 \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{5 c^5}+\frac {b x^2 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^3}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{5 c^5}-\frac {2 b \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{5 c^5}+\frac {1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {b x^4 \left (a+b \tanh ^{-1}(c x)\right )}{10 c}+\frac {b^2 x^3}{30 c^2}+\frac {3 b^2 x}{10 c^4}-\frac {3 b^2 \tanh ^{-1}(c x)}{10 c^5} \]
Antiderivative was successfully verified.
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Rule 206
Rule 302
Rule 321
Rule 2315
Rule 2402
Rule 5916
Rule 5918
Rule 5980
Rule 5984
Rubi steps
\begin {align*} \int x^4 \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx &=\frac {1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {1}{5} (2 b c) \int \frac {x^5 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx\\ &=\frac {1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {(2 b) \int x^3 \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{5 c}-\frac {(2 b) \int \frac {x^3 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{5 c}\\ &=\frac {b x^4 \left (a+b \tanh ^{-1}(c x)\right )}{10 c}+\frac {1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {1}{10} b^2 \int \frac {x^4}{1-c^2 x^2} \, dx+\frac {(2 b) \int x \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{5 c^3}-\frac {(2 b) \int \frac {x \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{5 c^3}\\ &=\frac {b x^2 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^3}+\frac {b x^4 \left (a+b \tanh ^{-1}(c x)\right )}{10 c}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{5 c^5}+\frac {1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {1}{10} b^2 \int \left (-\frac {1}{c^4}-\frac {x^2}{c^2}+\frac {1}{c^4 \left (1-c^2 x^2\right )}\right ) \, dx-\frac {(2 b) \int \frac {a+b \tanh ^{-1}(c x)}{1-c x} \, dx}{5 c^4}-\frac {b^2 \int \frac {x^2}{1-c^2 x^2} \, dx}{5 c^2}\\ &=\frac {3 b^2 x}{10 c^4}+\frac {b^2 x^3}{30 c^2}+\frac {b x^2 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^3}+\frac {b x^4 \left (a+b \tanh ^{-1}(c x)\right )}{10 c}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{5 c^5}+\frac {1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {2 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{5 c^5}-\frac {b^2 \int \frac {1}{1-c^2 x^2} \, dx}{10 c^4}-\frac {b^2 \int \frac {1}{1-c^2 x^2} \, dx}{5 c^4}+\frac {\left (2 b^2\right ) \int \frac {\log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx}{5 c^4}\\ &=\frac {3 b^2 x}{10 c^4}+\frac {b^2 x^3}{30 c^2}-\frac {3 b^2 \tanh ^{-1}(c x)}{10 c^5}+\frac {b x^2 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^3}+\frac {b x^4 \left (a+b \tanh ^{-1}(c x)\right )}{10 c}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{5 c^5}+\frac {1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {2 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{5 c^5}-\frac {\left (2 b^2\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-c x}\right )}{5 c^5}\\ &=\frac {3 b^2 x}{10 c^4}+\frac {b^2 x^3}{30 c^2}-\frac {3 b^2 \tanh ^{-1}(c x)}{10 c^5}+\frac {b x^2 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^3}+\frac {b x^4 \left (a+b \tanh ^{-1}(c x)\right )}{10 c}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{5 c^5}+\frac {1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {2 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{5 c^5}-\frac {b^2 \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{5 c^5}\\ \end {align*}
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Mathematica [A] time = 0.50, size = 161, normalized size = 0.99 \[ \frac {6 a^2 c^5 x^5+3 a b c^4 x^4+6 a b c^2 x^2+6 a b \log \left (c^2 x^2-1\right )+3 b \tanh ^{-1}(c x) \left (4 a c^5 x^5+b \left (c^4 x^4+2 c^2 x^2-3\right )-4 b \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )\right )-9 a b+6 b^2 \left (c^5 x^5-1\right ) \tanh ^{-1}(c x)^2+b^2 c^3 x^3+6 b^2 \text {Li}_2\left (-e^{-2 \tanh ^{-1}(c x)}\right )+9 b^2 c x}{30 c^5} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b^{2} x^{4} \operatorname {artanh}\left (c x\right )^{2} + 2 \, a b x^{4} \operatorname {artanh}\left (c x\right ) + a^{2} x^{4}, x\right ) \]
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 \[ \int {\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{2} x^{4}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.02, size = 306, normalized size = 1.89 \[ \frac {x^{5} a^{2}}{5}+\frac {x^{5} b^{2} \arctanh \left (c x \right )^{2}}{5}+\frac {b^{2} \arctanh \left (c x \right ) x^{4}}{10 c}+\frac {b^{2} \arctanh \left (c x \right ) x^{2}}{5 c^{3}}+\frac {b^{2} \arctanh \left (c x \right ) \ln \left (c x -1\right )}{5 c^{5}}+\frac {b^{2} \arctanh \left (c x \right ) \ln \left (c x +1\right )}{5 c^{5}}+\frac {b^{2} x^{3}}{30 c^{2}}+\frac {3 b^{2} x}{10 c^{4}}+\frac {3 b^{2} \ln \left (c x -1\right )}{20 c^{5}}-\frac {3 b^{2} \ln \left (c x +1\right )}{20 c^{5}}+\frac {b^{2} \ln \left (c x -1\right )^{2}}{20 c^{5}}-\frac {b^{2} \dilog \left (\frac {1}{2}+\frac {c x}{2}\right )}{5 c^{5}}-\frac {b^{2} \ln \left (c x -1\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{10 c^{5}}-\frac {b^{2} \ln \left (c x +1\right )^{2}}{20 c^{5}}+\frac {b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{10 c^{5}}-\frac {b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{10 c^{5}}+\frac {2 x^{5} a b \arctanh \left (c x \right )}{5}+\frac {x^{4} a b}{10 c}+\frac {a b \,x^{2}}{5 c^{3}}+\frac {a b \ln \left (c x -1\right )}{5 c^{5}}+\frac {a b \ln \left (c x +1\right )}{5 c^{5}} \]
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 {1}{5} \, a^{2} x^{5} + \frac {1}{10} \, {\left (4 \, x^{5} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {c^{2} x^{4} + 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} - 1\right )}{c^{6}}\right )}\right )} a b - \frac {1}{36000} \, {\left (24 \, c^{6} {\left (\frac {2 \, {\left (3 \, c^{4} x^{5} + 5 \, c^{2} x^{3} + 15 \, x\right )}}{c^{10}} - \frac {15 \, \log \left (c x + 1\right )}{c^{11}} + \frac {15 \, \log \left (c x - 1\right )}{c^{11}}\right )} - 45 \, c^{5} {\left (\frac {c^{2} x^{4} + 2 \, x^{2}}{c^{8}} + \frac {2 \, \log \left (c^{2} x^{2} - 1\right )}{c^{10}}\right )} - 1080000 \, c^{5} \int \frac {x^{5} \log \left (c x + 1\right )}{150 \, {\left (c^{6} x^{2} - c^{4}\right )}}\,{d x} + 50 \, c^{4} {\left (\frac {2 \, {\left (c^{2} x^{3} + 3 \, x\right )}}{c^{8}} - \frac {3 \, \log \left (c x + 1\right )}{c^{9}} + \frac {3 \, \log \left (c x - 1\right )}{c^{9}}\right )} - 300 \, c^{3} {\left (\frac {x^{2}}{c^{6}} + \frac {\log \left (c^{2} x^{2} - 1\right )}{c^{8}}\right )} + 900 \, c^{2} {\left (\frac {2 \, x}{c^{6}} - \frac {\log \left (c x + 1\right )}{c^{7}} + \frac {\log \left (c x - 1\right )}{c^{7}}\right )} - 540000 \, c \int \frac {x \log \left (c x + 1\right )}{150 \, {\left (c^{6} x^{2} - c^{4}\right )}}\,{d x} - \frac {60 \, {\left (30 \, c^{5} x^{5} \log \left (c x + 1\right )^{2} + {\left (12 \, c^{5} x^{5} - 15 \, c^{4} x^{4} + 20 \, c^{3} x^{3} - 30 \, c^{2} x^{2} + 60 \, c x - 60 \, {\left (c^{5} x^{5} + 1\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )\right )}}{c^{5}} - \frac {72 \, {\left (c x - 1\right )}^{5} {\left (25 \, \log \left (-c x + 1\right )^{2} - 10 \, \log \left (-c x + 1\right ) + 2\right )} + 1125 \, {\left (c x - 1\right )}^{4} {\left (8 \, \log \left (-c x + 1\right )^{2} - 4 \, \log \left (-c x + 1\right ) + 1\right )} + 2000 \, {\left (c x - 1\right )}^{3} {\left (9 \, \log \left (-c x + 1\right )^{2} - 6 \, \log \left (-c x + 1\right ) + 2\right )} + 9000 \, {\left (c x - 1\right )}^{2} {\left (2 \, \log \left (-c x + 1\right )^{2} - 2 \, \log \left (-c x + 1\right ) + 1\right )} + 9000 \, {\left (c x - 1\right )} {\left (\log \left (-c x + 1\right )^{2} - 2 \, \log \left (-c x + 1\right ) + 2\right )}}{c^{5}} + \frac {1800 \, \log \left (150 \, c^{6} x^{2} - 150 \, c^{4}\right )}{c^{5}} - 540000 \, \int \frac {\log \left (c x + 1\right )}{150 \, {\left (c^{6} x^{2} - c^{4}\right )}}\,{d x}\right )} b^{2} \]
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 \[ \int x^4\,{\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}^2 \,d x \]
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 \[ \int x^{4} \left (a + b \operatorname {atanh}{\left (c x \right )}\right )^{2}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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