Optimal. Leaf size=262 \[ \frac {9 a b^2 x}{10 c^4}+\frac {b^3 x^2}{20 c^3}+\frac {9 b^3 x \tanh ^{-1}(c x)}{10 c^4}+\frac {b^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )}{10 c^2}-\frac {9 b \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c^5}+\frac {3 b x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{10 c^3}+\frac {3 b x^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{5 c^5}+\frac {1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {3 b \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-c x}\right )}{5 c^5}+\frac {b^3 \log \left (1-c^2 x^2\right )}{2 c^5}-\frac {3 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{5 c^5}+\frac {3 b^3 \text {PolyLog}\left (3,1-\frac {2}{1-c x}\right )}{10 c^5} \]
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Rubi [A]
time = 0.55, 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 = {6037, 6127,
272, 45, 6021, 266, 6095, 6131, 6055, 6205, 6745} \begin {gather*} -\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} \end {gather*}
Antiderivative was successfully verified.
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Rule 45
Rule 266
Rule 272
Rule 6021
Rule 6037
Rule 6055
Rule 6095
Rule 6127
Rule 6131
Rule 6205
Rule 6745
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 \text {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 \text {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 = 0.53, size = 383, normalized size = 1.46 \begin {gather*} \frac {-b^3+18 a b^2 c x+6 a^2 b c^2 x^2+b^3 c^2 x^2+2 a b^2 c^3 x^3+3 a^2 b c^4 x^4+4 a^3 c^5 x^5-18 a b^2 \tanh ^{-1}(c x)+18 b^3 c x \tanh ^{-1}(c x)+12 a b^2 c^2 x^2 \tanh ^{-1}(c x)+2 b^3 c^3 x^3 \tanh ^{-1}(c x)+6 a b^2 c^4 x^4 \tanh ^{-1}(c x)+12 a^2 b c^5 x^5 \tanh ^{-1}(c x)-12 a b^2 \tanh ^{-1}(c x)^2-9 b^3 \tanh ^{-1}(c x)^2+6 b^3 c^2 x^2 \tanh ^{-1}(c x)^2+3 b^3 c^4 x^4 \tanh ^{-1}(c x)^2+12 a b^2 c^5 x^5 \tanh ^{-1}(c x)^2-4 b^3 \tanh ^{-1}(c x)^3+4 b^3 c^5 x^5 \tanh ^{-1}(c x)^3-24 a b^2 \tanh ^{-1}(c x) \log \left (1+e^{-2 \tanh ^{-1}(c x)}\right )-12 b^3 \tanh ^{-1}(c x)^2 \log \left (1+e^{-2 \tanh ^{-1}(c x)}\right )+6 a^2 b \log \left (1-c^2 x^2\right )+10 b^3 \log \left (1-c^2 x^2\right )+12 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )+6 b^3 \text {PolyLog}\left (3,-e^{-2 \tanh ^{-1}(c x)}\right )}{20 c^5} \end {gather*}
Antiderivative was successfully verified.
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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order
4.
time = 2.39, size = 1188, normalized size = 4.53
method | result | size |
derivativedivides | \(\text {Expression too large to display}\) | \(1188\) |
default | \(\text {Expression too large to display}\) | \(1188\) |
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int x^{4} \left (a + b \operatorname {atanh}{\left (c x \right )}\right )^{3}\, dx \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int x^4\,{\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}^3 \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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