Optimal. Leaf size=121 \[ \frac {x^3 \coth ^{-1}(\tanh (a+b x))^{1+n}}{b (1+n)}-\frac {3 x^2 \coth ^{-1}(\tanh (a+b x))^{2+n}}{b^2 (1+n) (2+n)}+\frac {6 x \coth ^{-1}(\tanh (a+b x))^{3+n}}{b^3 (1+n) (2+n) (3+n)}-\frac {6 \coth ^{-1}(\tanh (a+b x))^{4+n}}{b^4 (1+n) (2+n) (3+n) (4+n)} \]
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Rubi [A]
time = 0.06, 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 = {2199, 2188, 30}
\begin {gather*} -\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)} \end {gather*}
Antiderivative was successfully verified.
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Rule 30
Rule 2188
Rule 2199
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 \text {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 = 0.06, size = 106, normalized size = 0.88 \begin {gather*} \frac {\coth ^{-1}(\tanh (a+b x))^{1+n} \left (b^3 \left (24+26 n+9 n^2+n^3\right ) x^3-3 b^2 \left (12+7 n+n^2\right ) x^2 \coth ^{-1}(\tanh (a+b x))+6 b (4+n) x \coth ^{-1}(\tanh (a+b x))^2-6 \coth ^{-1}(\tanh (a+b x))^3\right )}{b^4 (1+n) (2+n) (3+n) (4+n)} \end {gather*}
Antiderivative was successfully verified.
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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order
3.
time = 11.51, size = 129477, normalized size = 1070.06
method | result | size |
risch | \(\text {Expression too large to display}\) | \(129477\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [C] Result contains complex when optimal does not.
time = 0.56, size = 257, normalized size = 2.12 \begin {gather*} \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} - 4 \, {\left (i \, \pi {\left (n^{3} + 3 \, n^{2} + 2 \, n\right )} b^{3} - 2 \, {\left (n^{3} + 3 \, n^{2} + 2 \, n\right )} a b^{3}\right )} x^{3} + 6 \, {\left (\pi ^{2} {\left (n^{2} + n\right )} b^{2} + 4 i \, \pi {\left (n^{2} + n\right )} a b^{2} - 4 \, {\left (n^{2} + n\right )} a^{2} b^{2}\right )} x^{2} - 6 \, {\left (-i \, \pi ^{3} b n + 6 \, \pi ^{2} a b n + 12 i \, \pi a^{2} b n - 8 \, 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}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 411 vs.
\(2 (121) = 242\).
time = 0.40, size = 411, normalized size = 3.40 \begin {gather*} \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 )}} \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*} \begin {cases} \frac {x^{4} \operatorname {acoth}^{n}{\left (\tanh {\left (a \right )} \right )}}{4} & \text {for}\: b = 0 \\- \frac {x^{3}}{3 b \operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )}} - \frac {x^{2}}{2 b^{2} \operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}} - \frac {x}{b^{3} \operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )}} + \frac {\log {\left (\operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )} \right )}}{b^{4}} & \text {for}\: n = -4 \\\int \frac {x^{3}}{\operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx & \text {for}\: n = -3 \\\int \frac {x^{3}}{\operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx & \text {for}\: n = -2 \\\int \frac {x^{3}}{\operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx & \text {for}\: n = -1 \\\frac {b^{3} n^{3} x^{3} \operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )} \operatorname {acoth}^{n}{\left (\tanh {\left (a + b x \right )} \right )}}{b^{4} n^{4} + 10 b^{4} n^{3} + 35 b^{4} n^{2} + 50 b^{4} n + 24 b^{4}} + \frac {9 b^{3} n^{2} x^{3} \operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )} \operatorname {acoth}^{n}{\left (\tanh {\left (a + b x \right )} \right )}}{b^{4} n^{4} + 10 b^{4} n^{3} + 35 b^{4} n^{2} + 50 b^{4} n + 24 b^{4}} + \frac {26 b^{3} n x^{3} \operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )} \operatorname {acoth}^{n}{\left (\tanh {\left (a + b x \right )} \right )}}{b^{4} n^{4} + 10 b^{4} n^{3} + 35 b^{4} n^{2} + 50 b^{4} n + 24 b^{4}} + \frac {24 b^{3} x^{3} \operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )} \operatorname {acoth}^{n}{\left (\tanh {\left (a + b x \right )} \right )}}{b^{4} n^{4} + 10 b^{4} n^{3} + 35 b^{4} n^{2} + 50 b^{4} n + 24 b^{4}} - \frac {3 b^{2} n^{2} x^{2} \operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )} \operatorname {acoth}^{n}{\left (\tanh {\left (a + b x \right )} \right )}}{b^{4} n^{4} + 10 b^{4} n^{3} + 35 b^{4} n^{2} + 50 b^{4} n + 24 b^{4}} - \frac {21 b^{2} n x^{2} \operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )} \operatorname {acoth}^{n}{\left (\tanh {\left (a + b x \right )} \right )}}{b^{4} n^{4} + 10 b^{4} n^{3} + 35 b^{4} n^{2} + 50 b^{4} n + 24 b^{4}} - \frac {36 b^{2} x^{2} \operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )} \operatorname {acoth}^{n}{\left (\tanh {\left (a + b x \right )} \right )}}{b^{4} n^{4} + 10 b^{4} n^{3} + 35 b^{4} n^{2} + 50 b^{4} n + 24 b^{4}} + \frac {6 b n x \operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )} \operatorname {acoth}^{n}{\left (\tanh {\left (a + b x \right )} \right )}}{b^{4} n^{4} + 10 b^{4} n^{3} + 35 b^{4} n^{2} + 50 b^{4} n + 24 b^{4}} + \frac {24 b x \operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )} \operatorname {acoth}^{n}{\left (\tanh {\left (a + b x \right )} \right )}}{b^{4} n^{4} + 10 b^{4} n^{3} + 35 b^{4} n^{2} + 50 b^{4} n + 24 b^{4}} - \frac {6 \operatorname {acoth}^{4}{\left (\tanh {\left (a + b x \right )} \right )} \operatorname {acoth}^{n}{\left (\tanh {\left (a + b x \right )} \right )}}{b^{4} n^{4} + 10 b^{4} n^{3} + 35 b^{4} n^{2} + 50 b^{4} n + 24 b^{4}} & \text {otherwise} \end {cases} \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 [B]
time = 1.50, size = 418, normalized size = 3.45 \begin {gather*} -{\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 ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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