Optimal. Leaf size=121 \[ -\frac {6 \coth ^{-1}(\tanh (a+b x))^{n+4}}{b^4 (n+1) (n+2) (n+3) (n+4)}+\frac {6 x \coth ^{-1}(\tanh (a+b x))^{n+3}}{b^3 (n+1) (n+2) (n+3)}-\frac {3 x^2 \coth ^{-1}(\tanh (a+b x))^{n+2}}{b^2 (n+1) (n+2)}+\frac {x^3 \coth ^{-1}(\tanh (a+b x))^{n+1}}{b (n+1)} \]
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Rubi [A] time = 0.08, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2168, 2157, 30} \[ -\frac {3 x^2 \coth ^{-1}(\tanh (a+b x))^{n+2}}{b^2 (n+1) (n+2)}+\frac {6 x \coth ^{-1}(\tanh (a+b x))^{n+3}}{b^3 (n+1) (n+2) (n+3)}-\frac {6 \coth ^{-1}(\tanh (a+b x))^{n+4}}{b^4 (n+1) (n+2) (n+3) (n+4)}+\frac {x^3 \coth ^{-1}(\tanh (a+b x))^{n+1}}{b (n+1)} \]
Antiderivative was successfully verified.
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Rule 30
Rule 2157
Rule 2168
Rubi steps
\begin {align*} \int x^3 \coth ^{-1}(\tanh (a+b x))^n \, dx &=\frac {x^3 \coth ^{-1}(\tanh (a+b x))^{1+n}}{b (1+n)}-\frac {3 \int x^2 \coth ^{-1}(\tanh (a+b x))^{1+n} \, dx}{b (1+n)}\\ &=\frac {x^3 \coth ^{-1}(\tanh (a+b x))^{1+n}}{b (1+n)}-\frac {3 x^2 \coth ^{-1}(\tanh (a+b x))^{2+n}}{b^2 (1+n) (2+n)}+\frac {6 \int x \coth ^{-1}(\tanh (a+b x))^{2+n} \, dx}{b^2 (1+n) (2+n)}\\ &=\frac {x^3 \coth ^{-1}(\tanh (a+b x))^{1+n}}{b (1+n)}-\frac {3 x^2 \coth ^{-1}(\tanh (a+b x))^{2+n}}{b^2 (1+n) (2+n)}+\frac {6 x \coth ^{-1}(\tanh (a+b x))^{3+n}}{b^3 (1+n) (2+n) (3+n)}-\frac {6 \int \coth ^{-1}(\tanh (a+b x))^{3+n} \, dx}{b^3 (1+n) (2+n) (3+n)}\\ &=\frac {x^3 \coth ^{-1}(\tanh (a+b x))^{1+n}}{b (1+n)}-\frac {3 x^2 \coth ^{-1}(\tanh (a+b x))^{2+n}}{b^2 (1+n) (2+n)}+\frac {6 x \coth ^{-1}(\tanh (a+b x))^{3+n}}{b^3 (1+n) (2+n) (3+n)}-\frac {6 \operatorname {Subst}\left (\int x^{3+n} \, dx,x,\coth ^{-1}(\tanh (a+b x))\right )}{b^4 (1+n) (2+n) (3+n)}\\ &=\frac {x^3 \coth ^{-1}(\tanh (a+b x))^{1+n}}{b (1+n)}-\frac {3 x^2 \coth ^{-1}(\tanh (a+b x))^{2+n}}{b^2 (1+n) (2+n)}+\frac {6 x \coth ^{-1}(\tanh (a+b x))^{3+n}}{b^3 (1+n) (2+n) (3+n)}-\frac {6 \coth ^{-1}(\tanh (a+b x))^{4+n}}{b^4 (1+n) (2+n) (3+n) (4+n)}\\ \end {align*}
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Mathematica [A] time = 0.08, size = 106, normalized size = 0.88 \[ \frac {\coth ^{-1}(\tanh (a+b x))^{n+1} \left (-3 b^2 \left (n^2+7 n+12\right ) x^2 \coth ^{-1}(\tanh (a+b x))+6 b (n+4) x \coth ^{-1}(\tanh (a+b x))^2-6 \coth ^{-1}(\tanh (a+b x))^3+b^3 \left (n^3+9 n^2+26 n+24\right ) x^3\right )}{b^4 (n+1) (n+2) (n+3) (n+4)} \]
Antiderivative was successfully verified.
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fricas [B] time = 1.48, size = 411, normalized size = 3.40 \[ \frac {{\left (8 \, {\left (b^{4} n^{3} + 6 \, b^{4} n^{2} + 11 \, b^{4} n + 6 \, b^{4}\right )} x^{4} - 3 \, \pi ^{4} + 72 \, \pi ^{2} a^{2} - 48 \, a^{4} + 8 \, {\left (a b^{3} n^{3} + 3 \, a b^{3} n^{2} + 2 \, a b^{3} n\right )} x^{3} - 6 \, {\left (4 \, a^{2} b^{2} n^{2} + 4 \, a^{2} b^{2} n - \pi ^{2} {\left (b^{2} n^{2} + b^{2} n\right )}\right )} x^{2} - 12 \, {\left (3 \, \pi ^{2} a b n - 4 \, a^{3} b n\right )} x\right )} {\left (b^{2} x^{2} + 2 \, a b x + \frac {1}{4} \, \pi ^{2} + a^{2}\right )}^{\frac {1}{2} \, n} \cos \left (2 \, n \arctan \left (-\frac {2 \, b x}{\pi } - \frac {2 \, a}{\pi } + \frac {\sqrt {4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}}}{\pi }\right )\right ) - 2 \, {\left (2 \, \pi {\left (b^{3} n^{3} + 3 \, b^{3} n^{2} + 2 \, b^{3} n\right )} x^{3} + 12 \, \pi ^{3} a - 48 \, \pi a^{3} - 12 \, \pi {\left (a b^{2} n^{2} + a b^{2} n\right )} x^{2} - 3 \, {\left (\pi ^{3} b n - 12 \, \pi a^{2} b n\right )} x\right )} {\left (b^{2} x^{2} + 2 \, a b x + \frac {1}{4} \, \pi ^{2} + a^{2}\right )}^{\frac {1}{2} \, n} \sin \left (2 \, n \arctan \left (-\frac {2 \, b x}{\pi } - \frac {2 \, a}{\pi } + \frac {\sqrt {4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}}}{\pi }\right )\right )}{8 \, {\left (b^{4} n^{4} + 10 \, b^{4} n^{3} + 35 \, b^{4} n^{2} + 50 \, b^{4} n + 24 \, b^{4}\right )}} \]
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 \[ \int x^{3} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{n}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 18.41, size = 129477, normalized size = 1070.06 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [C] time = 0.53, size = 255, normalized size = 2.11 \[ \frac {{\left (8 \, {\left (n^{3} + 6 \, n^{2} + 11 \, n + 6\right )} b^{4} x^{4} - 3 \, \pi ^{4} - 24 i \, \pi ^{3} a + 72 \, \pi ^{2} a^{2} + 96 i \, \pi a^{3} - 48 \, a^{4} + {\left (-4 i \, \pi {\left (n^{3} + 3 \, n^{2} + 2 \, n\right )} b^{3} + 8 \, {\left (n^{3} + 3 \, n^{2} + 2 \, n\right )} a b^{3}\right )} x^{3} + {\left (6 \, \pi ^{2} {\left (n^{2} + n\right )} b^{2} + 24 i \, \pi {\left (n^{2} + n\right )} a b^{2} - 24 \, {\left (n^{2} + n\right )} a^{2} b^{2}\right )} x^{2} + {\left (6 i \, \pi ^{3} b n - 36 \, \pi ^{2} a b n - 72 i \, \pi a^{2} b n + 48 \, a^{3} b n\right )} x\right )} {\left (\cosh \left (-n \log \left (-i \, \pi + 2 \, b x + 2 \, a\right )\right ) - \sinh \left (-n \log \left (-i \, \pi + 2 \, b x + 2 \, a\right )\right )\right )}}{{\left (2^{n + 3} n^{4} + 5 \cdot 2^{n + 4} n^{3} + 35 \cdot 2^{n + 3} n^{2} + 25 \cdot 2^{n + 4} n + 3 \cdot 2^{n + 6}\right )} b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.50, size = 418, normalized size = 3.45 \[ -{\left (\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}{2}-\frac {\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}{2}\right )}^n\,\left (\frac {3\,{\left (\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^4}{8\,b^4\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )}-\frac {x^4\,\left (n^3+6\,n^2+11\,n+6\right )}{n^4+10\,n^3+35\,n^2+50\,n+24}+\frac {3\,n\,x\,{\left (\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^3}{4\,b^3\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )}+\frac {n\,x^3\,\left (\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )\,\left (n^2+3\,n+2\right )}{2\,b\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )}+\frac {3\,n\,x^2\,\left (n+1\right )\,{\left (\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^2}{4\,b^2\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )}\right ) \]
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 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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