Optimal. Leaf size=262 \[ -\frac {3 b^2 \text {Li}_2\left (1-\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{5 c^5}+\frac {9 a b^2 x}{10 c^4}+\frac {b^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )}{10 c^2}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{5 c^5}-\frac {9 b \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c^5}-\frac {3 b \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{5 c^5}+\frac {3 b x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{10 c^3}+\frac {1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^3+\frac {3 b x^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c}+\frac {3 b^3 \text {Li}_3\left (1-\frac {2}{1-c x}\right )}{10 c^5}+\frac {9 b^3 x \tanh ^{-1}(c x)}{10 c^4}+\frac {b^3 x^2}{20 c^3}+\frac {b^3 \log \left (1-c^2 x^2\right )}{2 c^5} \]
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Rubi [A] time = 0.77, antiderivative size = 262, normalized size of antiderivative = 1.00, number of steps used = 24, number of rules used = 11, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.786, Rules used = {5916, 5980, 266, 43, 5910, 260, 5948, 5984, 5918, 6058, 6610} \[ -\frac {3 b^2 \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{5 c^5}+\frac {3 b^3 \text {PolyLog}\left (3,1-\frac {2}{1-c x}\right )}{10 c^5}+\frac {b^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )}{10 c^2}+\frac {9 a b^2 x}{10 c^4}+\frac {3 b x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{10 c^3}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{5 c^5}-\frac {9 b \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c^5}-\frac {3 b \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{5 c^5}+\frac {1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^3+\frac {3 b x^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c}+\frac {b^3 x^2}{20 c^3}+\frac {b^3 \log \left (1-c^2 x^2\right )}{2 c^5}+\frac {9 b^3 x \tanh ^{-1}(c x)}{10 c^4} \]
Antiderivative was successfully verified.
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Rule 43
Rule 260
Rule 266
Rule 5910
Rule 5916
Rule 5918
Rule 5948
Rule 5980
Rule 5984
Rule 6058
Rule 6610
Rubi steps
\begin {align*} \int x^4 \left (a+b \tanh ^{-1}(c x)\right )^3 \, dx &=\frac {1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {1}{5} (3 b c) \int \frac {x^5 \left (a+b \tanh ^{-1}(c x)\right )^2}{1-c^2 x^2} \, dx\\ &=\frac {1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^3+\frac {(3 b) \int x^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx}{5 c}-\frac {(3 b) \int \frac {x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{1-c^2 x^2} \, dx}{5 c}\\ &=\frac {3 b x^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c}+\frac {1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {1}{10} \left (3 b^2\right ) \int \frac {x^4 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx+\frac {(3 b) \int x \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx}{5 c^3}-\frac {(3 b) \int \frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{1-c^2 x^2} \, dx}{5 c^3}\\ &=\frac {3 b x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{10 c^3}+\frac {3 b x^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{5 c^5}+\frac {1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {(3 b) \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{1-c x} \, dx}{5 c^4}+\frac {\left (3 b^2\right ) \int x^2 \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{10 c^2}-\frac {\left (3 b^2\right ) \int \frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{10 c^2}-\frac {\left (3 b^2\right ) \int \frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{5 c^2}\\ &=\frac {b^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )}{10 c^2}+\frac {3 b x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{10 c^3}+\frac {3 b x^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{5 c^5}+\frac {1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {3 b \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-c x}\right )}{5 c^5}+\frac {\left (3 b^2\right ) \int \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{10 c^4}-\frac {\left (3 b^2\right ) \int \frac {a+b \tanh ^{-1}(c x)}{1-c^2 x^2} \, dx}{10 c^4}+\frac {\left (3 b^2\right ) \int \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{5 c^4}-\frac {\left (3 b^2\right ) \int \frac {a+b \tanh ^{-1}(c x)}{1-c^2 x^2} \, dx}{5 c^4}+\frac {\left (6 b^2\right ) \int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx}{5 c^4}-\frac {b^3 \int \frac {x^3}{1-c^2 x^2} \, dx}{10 c}\\ &=\frac {9 a b^2 x}{10 c^4}+\frac {b^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )}{10 c^2}-\frac {9 b \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c^5}+\frac {3 b x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{10 c^3}+\frac {3 b x^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{5 c^5}+\frac {1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {3 b \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-c x}\right )}{5 c^5}-\frac {3 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{5 c^5}+\frac {\left (3 b^3\right ) \int \tanh ^{-1}(c x) \, dx}{10 c^4}+\frac {\left (3 b^3\right ) \int \tanh ^{-1}(c x) \, dx}{5 c^4}+\frac {\left (3 b^3\right ) \int \frac {\text {Li}_2\left (1-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx}{5 c^4}-\frac {b^3 \operatorname {Subst}\left (\int \frac {x}{1-c^2 x} \, dx,x,x^2\right )}{20 c}\\ &=\frac {9 a b^2 x}{10 c^4}+\frac {9 b^3 x \tanh ^{-1}(c x)}{10 c^4}+\frac {b^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )}{10 c^2}-\frac {9 b \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c^5}+\frac {3 b x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{10 c^3}+\frac {3 b x^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{5 c^5}+\frac {1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {3 b \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-c x}\right )}{5 c^5}-\frac {3 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{5 c^5}+\frac {3 b^3 \text {Li}_3\left (1-\frac {2}{1-c x}\right )}{10 c^5}-\frac {\left (3 b^3\right ) \int \frac {x}{1-c^2 x^2} \, dx}{10 c^3}-\frac {\left (3 b^3\right ) \int \frac {x}{1-c^2 x^2} \, dx}{5 c^3}-\frac {b^3 \operatorname {Subst}\left (\int \left (-\frac {1}{c^2}-\frac {1}{c^2 \left (-1+c^2 x\right )}\right ) \, dx,x,x^2\right )}{20 c}\\ &=\frac {9 a b^2 x}{10 c^4}+\frac {b^3 x^2}{20 c^3}+\frac {9 b^3 x \tanh ^{-1}(c x)}{10 c^4}+\frac {b^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )}{10 c^2}-\frac {9 b \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c^5}+\frac {3 b x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{10 c^3}+\frac {3 b x^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{5 c^5}+\frac {1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {3 b \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-c x}\right )}{5 c^5}+\frac {b^3 \log \left (1-c^2 x^2\right )}{2 c^5}-\frac {3 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{5 c^5}+\frac {3 b^3 \text {Li}_3\left (1-\frac {2}{1-c x}\right )}{10 c^5}\\ \end {align*}
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Mathematica [A] time = 0.81, size = 383, normalized size = 1.46 \[ \frac {4 a^3 c^5 x^5+12 a^2 b c^5 x^5 \tanh ^{-1}(c x)+3 a^2 b c^4 x^4+6 a^2 b c^2 x^2+6 a^2 b \log \left (1-c^2 x^2\right )+12 a b^2 c^5 x^5 \tanh ^{-1}(c x)^2+6 a b^2 c^4 x^4 \tanh ^{-1}(c x)+2 a b^2 c^3 x^3+12 a b^2 c^2 x^2 \tanh ^{-1}(c x)+12 b^2 \text {Li}_2\left (-e^{-2 \tanh ^{-1}(c x)}\right ) \left (a+b \tanh ^{-1}(c x)\right )+18 a b^2 c x-12 a b^2 \tanh ^{-1}(c x)^2-18 a b^2 \tanh ^{-1}(c x)-24 a b^2 \tanh ^{-1}(c x) \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )+4 b^3 c^5 x^5 \tanh ^{-1}(c x)^3+3 b^3 c^4 x^4 \tanh ^{-1}(c x)^2+2 b^3 c^3 x^3 \tanh ^{-1}(c x)+b^3 c^2 x^2+10 b^3 \log \left (1-c^2 x^2\right )+6 b^3 c^2 x^2 \tanh ^{-1}(c x)^2+6 b^3 \text {Li}_3\left (-e^{-2 \tanh ^{-1}(c x)}\right )+18 b^3 c x \tanh ^{-1}(c x)-4 b^3 \tanh ^{-1}(c x)^3-9 b^3 \tanh ^{-1}(c x)^2-12 b^3 \tanh ^{-1}(c x)^2 \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )-b^3}{20 c^5} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.69, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b^{3} x^{4} \operatorname {artanh}\left (c x\right )^{3} + 3 \, a b^{2} x^{4} \operatorname {artanh}\left (c x\right )^{2} + 3 \, a^{2} b x^{4} \operatorname {artanh}\left (c x\right ) + a^{3} x^{4}, x\right ) \]
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 \[ \int {\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{3} x^{4}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 2.22, size = 1275, normalized size = 4.87 \[ \text {result too large to display} \]
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 \[ \frac {1}{5} \, a^{3} x^{5} + \frac {3}{20} \, {\left (4 \, x^{5} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {c^{2} x^{4} + 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} - 1\right )}{c^{6}}\right )}\right )} a^{2} b - \frac {2 \, {\left (b^{3} c^{5} x^{5} - b^{3}\right )} \log \left (-c x + 1\right )^{3} - 3 \, {\left (4 \, a b^{2} c^{5} x^{5} + b^{3} c^{4} x^{4} + 2 \, b^{3} c^{2} x^{2} + 2 \, {\left (b^{3} c^{5} x^{5} + b^{3}\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )^{2}}{80 \, c^{5}} - \int -\frac {5 \, {\left (b^{3} c^{5} x^{5} - b^{3} c^{4} x^{4}\right )} \log \left (c x + 1\right )^{3} + 30 \, {\left (a b^{2} c^{5} x^{5} - a b^{2} c^{4} x^{4}\right )} \log \left (c x + 1\right )^{2} - 3 \, {\left (4 \, a b^{2} c^{5} x^{5} + b^{3} c^{4} x^{4} + 2 \, b^{3} c^{2} x^{2} + 5 \, {\left (b^{3} c^{5} x^{5} - b^{3} c^{4} x^{4}\right )} \log \left (c x + 1\right )^{2} - 2 \, {\left (10 \, a b^{2} c^{4} x^{4} - {\left (10 \, a b^{2} c^{5} + b^{3} c^{5}\right )} x^{5} - b^{3}\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{40 \, {\left (c^{5} x - c^{4}\right )}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int x^4\,{\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}^3 \,d x \]
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 \[ \int x^{4} \left (a + b \operatorname {atanh}{\left (c x \right )}\right )^{3}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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