Optimal. Leaf size=142 \[ -\frac {b \left (3 a^2+b^2\right ) (a+b \coth (c+d x))^2}{2 d}-\frac {4 a b^2 \left (a^2+b^2\right ) \coth (c+d x)}{d}+\frac {b \left (5 a^4+10 a^2 b^2+b^4\right ) \log (\sinh (c+d x))}{d}+a x \left (a^4+10 a^2 b^2+5 b^4\right )-\frac {b (a+b \coth (c+d x))^4}{4 d}-\frac {2 a b (a+b \coth (c+d x))^3}{3 d} \]
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Rubi [A] time = 0.21, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3482, 3528, 3525, 3475} \[ -\frac {b \left (3 a^2+b^2\right ) (a+b \coth (c+d x))^2}{2 d}-\frac {4 a b^2 \left (a^2+b^2\right ) \coth (c+d x)}{d}+\frac {b \left (10 a^2 b^2+5 a^4+b^4\right ) \log (\sinh (c+d x))}{d}+a x \left (10 a^2 b^2+a^4+5 b^4\right )-\frac {b (a+b \coth (c+d x))^4}{4 d}-\frac {2 a b (a+b \coth (c+d x))^3}{3 d} \]
Antiderivative was successfully verified.
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Rule 3475
Rule 3482
Rule 3525
Rule 3528
Rubi steps
\begin {align*} \int (a+b \coth (c+d x))^5 \, dx &=-\frac {b (a+b \coth (c+d x))^4}{4 d}+\int (a+b \coth (c+d x))^3 \left (a^2+b^2+2 a b \coth (c+d x)\right ) \, dx\\ &=-\frac {2 a b (a+b \coth (c+d x))^3}{3 d}-\frac {b (a+b \coth (c+d x))^4}{4 d}+\int (a+b \coth (c+d x))^2 \left (a \left (a^2+3 b^2\right )+b \left (3 a^2+b^2\right ) \coth (c+d x)\right ) \, dx\\ &=-\frac {b \left (3 a^2+b^2\right ) (a+b \coth (c+d x))^2}{2 d}-\frac {2 a b (a+b \coth (c+d x))^3}{3 d}-\frac {b (a+b \coth (c+d x))^4}{4 d}+\int (a+b \coth (c+d x)) \left (a^4+6 a^2 b^2+b^4+4 a b \left (a^2+b^2\right ) \coth (c+d x)\right ) \, dx\\ &=a \left (a^4+10 a^2 b^2+5 b^4\right ) x-\frac {4 a b^2 \left (a^2+b^2\right ) \coth (c+d x)}{d}-\frac {b \left (3 a^2+b^2\right ) (a+b \coth (c+d x))^2}{2 d}-\frac {2 a b (a+b \coth (c+d x))^3}{3 d}-\frac {b (a+b \coth (c+d x))^4}{4 d}+\left (b \left (5 a^4+10 a^2 b^2+b^4\right )\right ) \int \coth (c+d x) \, dx\\ &=a \left (a^4+10 a^2 b^2+5 b^4\right ) x-\frac {4 a b^2 \left (a^2+b^2\right ) \coth (c+d x)}{d}-\frac {b \left (3 a^2+b^2\right ) (a+b \coth (c+d x))^2}{2 d}-\frac {2 a b (a+b \coth (c+d x))^3}{3 d}-\frac {b (a+b \coth (c+d x))^4}{4 d}+\frac {b \left (5 a^4+10 a^2 b^2+b^4\right ) \log (\sinh (c+d x))}{d}\\ \end {align*}
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Mathematica [A] time = 0.82, size = 141, normalized size = 0.99 \[ -\frac {60 a b^2 \left (2 a^2+b^2\right ) \coth (c+d x)+6 b^3 \left (10 a^2+b^2\right ) \coth ^2(c+d x)-12 b \left (5 a^4+10 a^2 b^2+b^4\right ) \log (\tanh (c+d x))+20 a b^4 \coth ^3(c+d x)-6 (a-b)^5 \log (\tanh (c+d x)+1)+6 (a+b)^5 \log (1-\tanh (c+d x))+3 b^5 \coth ^4(c+d x)}{12 d} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.46, size = 2748, normalized size = 19.35 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.17, size = 226, normalized size = 1.59 \[ \frac {3 \, {\left (a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}\right )} {\left (d x + c\right )} + 3 \, {\left (5 \, a^{4} b + 10 \, a^{2} b^{3} + b^{5}\right )} \log \left ({\left | e^{\left (2 \, d x + 2 \, c\right )} - 1 \right |}\right ) + \frac {4 \, {\left (15 \, a^{3} b^{2} + 10 \, a b^{4} - 3 \, {\left (5 \, a^{3} b^{2} + 5 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} e^{\left (6 \, d x + 6 \, c\right )} + 3 \, {\left (15 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 10 \, a b^{4} + b^{5}\right )} e^{\left (4 \, d x + 4 \, c\right )} - {\left (45 \, a^{3} b^{2} + 15 \, a^{2} b^{3} + 25 \, a b^{4} + 3 \, b^{5}\right )} e^{\left (2 \, d x + 2 \, c\right )}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{4}}}{3 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.02, size = 322, normalized size = 2.27 \[ \frac {\ln \left (\coth \left (d x +c \right )+1\right ) a^{5}}{2 d}-\frac {5 \ln \left (\coth \left (d x +c \right )+1\right ) a^{4} b}{2 d}+\frac {5 \ln \left (\coth \left (d x +c \right )+1\right ) a^{3} b^{2}}{d}-\frac {5 \ln \left (\coth \left (d x +c \right )+1\right ) a^{2} b^{3}}{d}+\frac {5 \ln \left (\coth \left (d x +c \right )+1\right ) a \,b^{4}}{2 d}-\frac {\ln \left (\coth \left (d x +c \right )+1\right ) b^{5}}{2 d}-\frac {\left (\coth ^{2}\left (d x +c \right )\right ) b^{5}}{2 d}-\frac {5 \left (\coth ^{3}\left (d x +c \right )\right ) a \,b^{4}}{3 d}-\frac {5 \left (\coth ^{2}\left (d x +c \right )\right ) a^{2} b^{3}}{d}-\frac {5 a \,b^{4} \coth \left (d x +c \right )}{d}-\frac {10 a^{3} b^{2} \coth \left (d x +c \right )}{d}-\frac {\ln \left (\coth \left (d x +c \right )-1\right ) a^{5}}{2 d}-\frac {5 \ln \left (\coth \left (d x +c \right )-1\right ) a^{4} b}{2 d}-\frac {5 \ln \left (\coth \left (d x +c \right )-1\right ) a^{3} b^{2}}{d}-\frac {5 \ln \left (\coth \left (d x +c \right )-1\right ) a^{2} b^{3}}{d}-\frac {5 \ln \left (\coth \left (d x +c \right )-1\right ) a \,b^{4}}{2 d}-\frac {\ln \left (\coth \left (d x +c \right )-1\right ) b^{5}}{2 d}-\frac {b^{5} \left (\coth ^{4}\left (d x +c \right )\right )}{4 d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.33, size = 348, normalized size = 2.45 \[ \frac {5}{3} \, a b^{4} {\left (3 \, x + \frac {3 \, c}{d} - \frac {4 \, {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} - 2\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}}\right )} + b^{5} {\left (x + \frac {c}{d} + \frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{d} + \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{d} + \frac {4 \, {\left (e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )}\right )}}{d {\left (4 \, e^{\left (-2 \, d x - 2 \, c\right )} - 6 \, e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, e^{\left (-6 \, d x - 6 \, c\right )} - e^{\left (-8 \, d x - 8 \, c\right )} - 1\right )}}\right )} + 10 \, a^{2} b^{3} {\left (x + \frac {c}{d} + \frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{d} + \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{d} + \frac {2 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )}}\right )} + 10 \, a^{3} b^{2} {\left (x + \frac {c}{d} + \frac {2}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}}\right )} + a^{5} x + \frac {5 \, a^{4} b \log \left (\sinh \left (d x + c\right )\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.28, size = 244, normalized size = 1.72 \[ x\,{\left (a-b\right )}^5-\frac {4\,\left (5\,a^3\,b^2+5\,a^2\,b^3+5\,a\,b^4+b^5\right )}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )}+\frac {\ln \left ({\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}-1\right )\,\left (5\,a^4\,b+10\,a^2\,b^3+b^5\right )}{d}-\frac {4\,\left (5\,a^2\,b^3+5\,a\,b^4+2\,b^5\right )}{d\,\left ({\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {4\,b^5}{d\,\left (6\,{\mathrm {e}}^{4\,c+4\,d\,x}-4\,{\mathrm {e}}^{2\,c+2\,d\,x}-4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1\right )}-\frac {8\,\left (3\,b^5+5\,a\,b^4\right )}{3\,d\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}-3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}-1\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 13.42, size = 325, normalized size = 2.29 \[ \begin {cases} a^{5} x + \tilde {\infty } a^{4} b x + \tilde {\infty } a^{3} b^{2} x + \tilde {\infty } a^{2} b^{3} x + \tilde {\infty } a b^{4} x + \tilde {\infty } b^{5} x & \text {for}\: c = \log {\left (- e^{- d x} \right )} \vee c = \log {\left (e^{- d x} \right )} \\x \left (a + b \coth {\relax (c )}\right )^{5} & \text {for}\: d = 0 \\a^{5} x + 5 a^{4} b x - \frac {5 a^{4} b \log {\left (\tanh {\left (c + d x \right )} + 1 \right )}}{d} + \frac {5 a^{4} b \log {\left (\tanh {\left (c + d x \right )} \right )}}{d} + 10 a^{3} b^{2} x - \frac {10 a^{3} b^{2}}{d \tanh {\left (c + d x \right )}} + 10 a^{2} b^{3} x - \frac {10 a^{2} b^{3} \log {\left (\tanh {\left (c + d x \right )} + 1 \right )}}{d} + \frac {10 a^{2} b^{3} \log {\left (\tanh {\left (c + d x \right )} \right )}}{d} - \frac {5 a^{2} b^{3}}{d \tanh ^{2}{\left (c + d x \right )}} + 5 a b^{4} x - \frac {5 a b^{4}}{d \tanh {\left (c + d x \right )}} - \frac {5 a b^{4}}{3 d \tanh ^{3}{\left (c + d x \right )}} + b^{5} x - \frac {b^{5} \log {\left (\tanh {\left (c + d x \right )} + 1 \right )}}{d} + \frac {b^{5} \log {\left (\tanh {\left (c + d x \right )} \right )}}{d} - \frac {b^{5}}{2 d \tanh ^{2}{\left (c + d x \right )}} - \frac {b^{5}}{4 d \tanh ^{4}{\left (c + d x \right )}} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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