Optimal. Leaf size=165 \[ \frac {24 \coth ^{-1}(\tanh (a+b x))^{n+5}}{b^5 (n+1) (n+2) (n+3) (n+4) (n+5)}-\frac {24 x \coth ^{-1}(\tanh (a+b x))^{n+4}}{b^4 (n+1) (n+2) (n+3) (n+4)}+\frac {12 x^2 \coth ^{-1}(\tanh (a+b x))^{n+3}}{b^3 (n+1) (n+2) (n+3)}-\frac {4 x^3 \coth ^{-1}(\tanh (a+b x))^{n+2}}{b^2 (n+1) (n+2)}+\frac {x^4 \coth ^{-1}(\tanh (a+b x))^{n+1}}{b (n+1)} \]
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Rubi [A] time = 0.13, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2168, 2157, 30} \[ -\frac {4 x^3 \coth ^{-1}(\tanh (a+b x))^{n+2}}{b^2 (n+1) (n+2)}+\frac {12 x^2 \coth ^{-1}(\tanh (a+b x))^{n+3}}{b^3 (n+1) (n+2) (n+3)}-\frac {24 x \coth ^{-1}(\tanh (a+b x))^{n+4}}{b^4 (n+1) (n+2) (n+3) (n+4)}+\frac {24 \coth ^{-1}(\tanh (a+b x))^{n+5}}{b^5 (n+1) (n+2) (n+3) (n+4) (n+5)}+\frac {x^4 \coth ^{-1}(\tanh (a+b x))^{n+1}}{b (n+1)} \]
Antiderivative was successfully verified.
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Rule 30
Rule 2157
Rule 2168
Rubi steps
\begin {align*} \int x^4 \coth ^{-1}(\tanh (a+b x))^n \, dx &=\frac {x^4 \coth ^{-1}(\tanh (a+b x))^{1+n}}{b (1+n)}-\frac {4 \int x^3 \coth ^{-1}(\tanh (a+b x))^{1+n} \, dx}{b (1+n)}\\ &=\frac {x^4 \coth ^{-1}(\tanh (a+b x))^{1+n}}{b (1+n)}-\frac {4 x^3 \coth ^{-1}(\tanh (a+b x))^{2+n}}{b^2 (1+n) (2+n)}+\frac {12 \int x^2 \coth ^{-1}(\tanh (a+b x))^{2+n} \, dx}{b^2 (1+n) (2+n)}\\ &=\frac {x^4 \coth ^{-1}(\tanh (a+b x))^{1+n}}{b (1+n)}-\frac {4 x^3 \coth ^{-1}(\tanh (a+b x))^{2+n}}{b^2 (1+n) (2+n)}+\frac {12 x^2 \coth ^{-1}(\tanh (a+b x))^{3+n}}{b^3 (1+n) (2+n) (3+n)}-\frac {24 \int x \coth ^{-1}(\tanh (a+b x))^{3+n} \, dx}{b^3 (1+n) (2+n) (3+n)}\\ &=\frac {x^4 \coth ^{-1}(\tanh (a+b x))^{1+n}}{b (1+n)}-\frac {4 x^3 \coth ^{-1}(\tanh (a+b x))^{2+n}}{b^2 (1+n) (2+n)}+\frac {12 x^2 \coth ^{-1}(\tanh (a+b x))^{3+n}}{b^3 (1+n) (2+n) (3+n)}-\frac {24 x \coth ^{-1}(\tanh (a+b x))^{4+n}}{b^4 (1+n) (2+n) (3+n) (4+n)}+\frac {24 \int \coth ^{-1}(\tanh (a+b x))^{4+n} \, dx}{b^4 (1+n) (2+n) (3+n) (4+n)}\\ &=\frac {x^4 \coth ^{-1}(\tanh (a+b x))^{1+n}}{b (1+n)}-\frac {4 x^3 \coth ^{-1}(\tanh (a+b x))^{2+n}}{b^2 (1+n) (2+n)}+\frac {12 x^2 \coth ^{-1}(\tanh (a+b x))^{3+n}}{b^3 (1+n) (2+n) (3+n)}-\frac {24 x \coth ^{-1}(\tanh (a+b x))^{4+n}}{b^4 (1+n) (2+n) (3+n) (4+n)}+\frac {24 \operatorname {Subst}\left (\int x^{4+n} \, dx,x,\coth ^{-1}(\tanh (a+b x))\right )}{b^5 (1+n) (2+n) (3+n) (4+n)}\\ &=\frac {x^4 \coth ^{-1}(\tanh (a+b x))^{1+n}}{b (1+n)}-\frac {4 x^3 \coth ^{-1}(\tanh (a+b x))^{2+n}}{b^2 (1+n) (2+n)}+\frac {12 x^2 \coth ^{-1}(\tanh (a+b x))^{3+n}}{b^3 (1+n) (2+n) (3+n)}-\frac {24 x \coth ^{-1}(\tanh (a+b x))^{4+n}}{b^4 (1+n) (2+n) (3+n) (4+n)}+\frac {24 \coth ^{-1}(\tanh (a+b x))^{5+n}}{b^5 (1+n) (2+n) (3+n) (4+n) (5+n)}\\ \end {align*}
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Mathematica [A] time = 0.11, size = 146, normalized size = 0.88 \[ \frac {\coth ^{-1}(\tanh (a+b x))^{n+1} \left (-4 b^3 \left (n^3+12 n^2+47 n+60\right ) x^3 \coth ^{-1}(\tanh (a+b x))+12 b^2 \left (n^2+9 n+20\right ) x^2 \coth ^{-1}(\tanh (a+b x))^2-24 b (n+5) x \coth ^{-1}(\tanh (a+b x))^3+24 \coth ^{-1}(\tanh (a+b x))^4+b^4 \left (n^4+14 n^3+71 n^2+154 n+120\right ) x^4\right )}{b^5 (n+1) (n+2) (n+3) (n+4) (n+5)} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.72, size = 583, normalized size = 3.53 \[ \frac {2 \, {\left (2 \, {\left (b^{5} n^{4} + 10 \, b^{5} n^{3} + 35 \, b^{5} n^{2} + 50 \, b^{5} n + 24 \, b^{5}\right )} x^{5} + 15 \, \pi ^{4} a - 120 \, \pi ^{2} a^{3} + 48 \, a^{5} + 2 \, {\left (a b^{4} n^{4} + 6 \, a b^{4} n^{3} + 11 \, a b^{4} n^{2} + 6 \, a b^{4} n\right )} x^{4} - 2 \, {\left (4 \, a^{2} b^{3} n^{3} + 12 \, a^{2} b^{3} n^{2} + 8 \, a^{2} b^{3} n - \pi ^{2} {\left (b^{3} n^{3} + 3 \, b^{3} n^{2} + 2 \, b^{3} n\right )}\right )} x^{3} + 6 \, {\left (4 \, a^{3} b^{2} n^{2} + 4 \, a^{3} b^{2} n - 3 \, \pi ^{2} {\left (a b^{2} n^{2} + a b^{2} n\right )}\right )} x^{2} - 3 \, {\left (\pi ^{4} b n - 24 \, \pi ^{2} a^{2} b n + 16 \, a^{4} b n\right )} x\right )} {\left (b^{2} x^{2} + 2 \, a b x + \frac {1}{4} \, \pi ^{2} + a^{2}\right )}^{\frac {1}{2} \, n} \cos \left (2 \, n \arctan \left (-\frac {2 \, b x}{\pi } - \frac {2 \, a}{\pi } + \frac {\sqrt {4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}}}{\pi }\right )\right ) - {\left (2 \, \pi {\left (b^{4} n^{4} + 6 \, b^{4} n^{3} + 11 \, b^{4} n^{2} + 6 \, b^{4} n\right )} x^{4} + 3 \, \pi ^{5} - 120 \, \pi ^{3} a^{2} + 240 \, \pi a^{4} - 16 \, \pi {\left (a b^{3} n^{3} + 3 \, a b^{3} n^{2} + 2 \, a b^{3} n\right )} x^{3} - 6 \, {\left (\pi ^{3} {\left (b^{2} n^{2} + b^{2} n\right )} - 12 \, \pi {\left (a^{2} b^{2} n^{2} + a^{2} b^{2} n\right )}\right )} x^{2} + 48 \, {\left (\pi ^{3} a b n - 4 \, \pi a^{3} b n\right )} x\right )} {\left (b^{2} x^{2} + 2 \, a b x + \frac {1}{4} \, \pi ^{2} + a^{2}\right )}^{\frac {1}{2} \, n} \sin \left (2 \, n \arctan \left (-\frac {2 \, b x}{\pi } - \frac {2 \, a}{\pi } + \frac {\sqrt {4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}}}{\pi }\right )\right )}{4 \, {\left (b^{5} n^{5} + 15 \, b^{5} n^{4} + 85 \, b^{5} n^{3} + 225 \, b^{5} n^{2} + 274 \, b^{5} n + 120 \, b^{5}\right )}} \]
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 \[ \int x^{4} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{n}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 33.06, size = 504228, normalized size = 3055.93 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [C] time = 0.54, size = 380, normalized size = 2.30 \[ \frac {{\left (4 \, {\left (n^{4} + 10 \, n^{3} + 35 \, n^{2} + 50 \, n + 24\right )} b^{5} x^{5} - 3 i \, \pi ^{5} + 30 \, \pi ^{4} a + 120 i \, \pi ^{3} a^{2} - 240 \, \pi ^{2} a^{3} - 240 i \, \pi a^{4} + 96 \, a^{5} - 2 \, {\left (i \, \pi {\left (n^{4} + 6 \, n^{3} + 11 \, n^{2} + 6 \, n\right )} b^{4} - 2 \, {\left (n^{4} + 6 \, n^{3} + 11 \, n^{2} + 6 \, n\right )} a b^{4}\right )} x^{4} + {\left (4 \, \pi ^{2} {\left (n^{3} + 3 \, n^{2} + 2 \, n\right )} b^{3} + 16 i \, \pi {\left (n^{3} + 3 \, n^{2} + 2 \, n\right )} a b^{3} - 16 \, {\left (n^{3} + 3 \, n^{2} + 2 \, n\right )} a^{2} b^{3}\right )} x^{3} + {\left (6 i \, \pi ^{3} {\left (n^{2} + n\right )} b^{2} - 36 \, \pi ^{2} {\left (n^{2} + n\right )} a b^{2} - 72 i \, \pi {\left (n^{2} + n\right )} a^{2} b^{2} + 48 \, {\left (n^{2} + n\right )} a^{3} b^{2}\right )} x^{2} - {\left (6 \, \pi ^{4} b n + 48 i \, \pi ^{3} a b n - 144 \, \pi ^{2} a^{2} b n - 192 i \, \pi a^{3} b n + 96 \, a^{4} b n\right )} x\right )} {\left (\cosh \left (-n \log \left (-i \, \pi + 2 \, b x + 2 \, a\right )\right ) - \sinh \left (-n \log \left (-i \, \pi + 2 \, b x + 2 \, a\right )\right )\right )}}{{\left (2^{n + 2} n^{5} + 15 \cdot 2^{n + 2} n^{4} + 85 \cdot 2^{n + 2} n^{3} + 225 \cdot 2^{n + 2} n^{2} + 137 \cdot 2^{n + 3} n + 15 \cdot 2^{n + 5}\right )} b^{5}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.19, size = 546, normalized size = 3.31 \[ -{\left (\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}{2}-\frac {\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}{2}\right )}^n\,\left (\frac {3\,{\left (\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^5}{4\,b^5\,\left (n^5+15\,n^4+85\,n^3+225\,n^2+274\,n+120\right )}-\frac {x^5\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )}{n^5+15\,n^4+85\,n^3+225\,n^2+274\,n+120}+\frac {3\,n\,x\,{\left (\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^4}{2\,b^4\,\left (n^5+15\,n^4+85\,n^3+225\,n^2+274\,n+120\right )}+\frac {n\,x^4\,\left (\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )\,\left (n^3+6\,n^2+11\,n+6\right )}{2\,b\,\left (n^5+15\,n^4+85\,n^3+225\,n^2+274\,n+120\right )}+\frac {3\,n\,x^2\,\left (n+1\right )\,{\left (\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^3}{2\,b^3\,\left (n^5+15\,n^4+85\,n^3+225\,n^2+274\,n+120\right )}+\frac {n\,x^3\,{\left (\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^2\,\left (n^2+3\,n+2\right )}{b^2\,\left (n^5+15\,n^4+85\,n^3+225\,n^2+274\,n+120\right )}\right ) \]
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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