Optimal. Leaf size=134 \[ -\frac {b^4 \tanh ^{-1}\left (\frac {(b+a \coth (x)) \sinh (x)}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac {a b^2 \cosh (x)}{\left (a^2-b^2\right )^2}-\frac {a \cosh (x)}{a^2-b^2}+\frac {a \cosh ^3(x)}{3 \left (a^2-b^2\right )}-\frac {b^3 \sinh (x)}{\left (a^2-b^2\right )^2}-\frac {b \sinh ^3(x)}{3 \left (a^2-b^2\right )} \]
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Rubi [A]
time = 0.18, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps
used = 9, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {3592, 3567,
2713, 2718, 3590, 212} \begin {gather*} -\frac {b \sinh ^3(x)}{3 \left (a^2-b^2\right )}+\frac {a \cosh ^3(x)}{3 \left (a^2-b^2\right )}+\frac {a b^2 \cosh (x)}{\left (a^2-b^2\right )^2}-\frac {a \cosh (x)}{a^2-b^2}-\frac {b^4 \tanh ^{-1}\left (\frac {\sinh (x) (a \coth (x)+b)}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-\frac {b^3 \sinh (x)}{\left (a^2-b^2\right )^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 212
Rule 2713
Rule 2718
Rule 3567
Rule 3590
Rule 3592
Rubi steps
\begin {align*} \int \frac {\sinh ^3(x)}{a+b \coth (x)} \, dx &=\frac {\int (a-b \coth (x)) \sinh ^3(x) \, dx}{a^2-b^2}+\frac {b^2 \int \frac {\sinh (x)}{a+b \coth (x)} \, dx}{a^2-b^2}\\ &=-\frac {b \sinh ^3(x)}{3 \left (a^2-b^2\right )}+\frac {b^2 \int (a-b \coth (x)) \sinh (x) \, dx}{\left (a^2-b^2\right )^2}+\frac {b^4 \int \frac {\text {csch}(x)}{a+b \coth (x)} \, dx}{\left (a^2-b^2\right )^2}+\frac {a \int \sinh ^3(x) \, dx}{a^2-b^2}\\ &=-\frac {b^3 \sinh (x)}{\left (a^2-b^2\right )^2}-\frac {b \sinh ^3(x)}{3 \left (a^2-b^2\right )}+\frac {\left (a b^2\right ) \int \sinh (x) \, dx}{\left (a^2-b^2\right )^2}-\frac {b^4 \text {Subst}\left (\int \frac {1}{a^2-b^2-x^2} \, dx,x,i (-i b-i a \coth (x)) \sinh (x)\right )}{\left (a^2-b^2\right )^2}-\frac {a \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cosh (x)\right )}{a^2-b^2}\\ &=-\frac {b^4 \tanh ^{-1}\left (\frac {(b+a \coth (x)) \sinh (x)}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac {a b^2 \cosh (x)}{\left (a^2-b^2\right )^2}-\frac {a \cosh (x)}{a^2-b^2}+\frac {a \cosh ^3(x)}{3 \left (a^2-b^2\right )}-\frac {b^3 \sinh (x)}{\left (a^2-b^2\right )^2}-\frac {b \sinh ^3(x)}{3 \left (a^2-b^2\right )}\\ \end {align*}
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Mathematica [A]
time = 0.55, size = 171, normalized size = 1.28 \begin {gather*} \frac {24 b^4 \sqrt {a+b} \text {ArcTan}\left (\frac {a+b \tanh \left (\frac {x}{2}\right )}{\sqrt {-a+b} \sqrt {a+b}}\right )-3 a \sqrt {-a+b} \left (3 a^3+3 a^2 b-7 a b^2-7 b^3\right ) \cosh (x)-a (-a+b)^{3/2} (a+b)^2 \cosh (3 x)+3 b \sqrt {-a+b} \left (a^3+a^2 b-5 a b^2-5 b^3\right ) \sinh (x)+b (-a+b)^{3/2} (a+b)^2 \sinh (3 x)}{12 (-a+b)^{5/2} (a+b)^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.70, size = 175, normalized size = 1.31
method | result | size |
risch | \(\frac {{\mathrm e}^{3 x}}{24 a +24 b}-\frac {3 \,{\mathrm e}^{x} a}{8 \left (a +b \right )^{2}}-\frac {5 \,{\mathrm e}^{x} b}{8 \left (a +b \right )^{2}}-\frac {3 \,{\mathrm e}^{-x} a}{8 \left (a -b \right )^{2}}+\frac {5 \,{\mathrm e}^{-x} b}{8 \left (a -b \right )^{2}}+\frac {{\mathrm e}^{-3 x}}{24 a -24 b}+\frac {b^{4} \ln \left ({\mathrm e}^{x}-\frac {a -b}{\sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2}}-\frac {b^{4} \ln \left ({\mathrm e}^{x}+\frac {a -b}{\sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2}}\) | \(172\) |
default | \(\frac {2 b^{4} \arctan \left (\frac {2 \tanh \left (\frac {x}{2}\right ) b +2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {-a^{2}+b^{2}}}-\frac {16}{\left (32 a -32 b \right ) \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {32}{3 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3} \left (32 a -32 b \right )}-\frac {-2 b +a}{2 \left (a -b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {32}{3 \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3} \left (32 a +32 b \right )}-\frac {16}{\left (32 a +32 b \right ) \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {-2 b -a}{2 \left (a +b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}\) | \(175\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \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. 902 vs.
\(2 (126) = 252\).
time = 0.40, size = 1859, normalized size = 13.87 \begin {gather*} \text {Too large to display} \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 \frac {\sinh ^{3}{\left (x \right )}}{a + b \coth {\left (x \right )}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.41, size = 163, normalized size = 1.22 \begin {gather*} -\frac {2 \, b^{4} \arctan \left (-\frac {a e^{x} + b e^{x}}{\sqrt {-a^{2} + b^{2}}}\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {{\left (9 \, a e^{\left (2 \, x\right )} - 15 \, b e^{\left (2 \, x\right )} - a + b\right )} e^{\left (-3 \, x\right )}}{24 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} + \frac {a^{2} e^{\left (3 \, x\right )} + 2 \, a b e^{\left (3 \, x\right )} + b^{2} e^{\left (3 \, x\right )} - 9 \, a^{2} e^{x} - 24 \, a b e^{x} - 15 \, b^{2} e^{x}}{24 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.86, size = 172, normalized size = 1.28 \begin {gather*} \frac {{\mathrm {e}}^{-3\,x}}{24\,a-24\,b}+\frac {{\mathrm {e}}^{3\,x}}{24\,a+24\,b}-\frac {{\mathrm {e}}^{-x}\,\left (3\,a-5\,b\right )}{8\,{\left (a-b\right )}^2}-\frac {{\mathrm {e}}^x\,\left (3\,a+5\,b\right )}{8\,{\left (a+b\right )}^2}-\frac {b^4\,\ln \left (2\,a^3\,b-2\,a\,b^3+a^4-b^4+{\mathrm {e}}^x\,{\left (a+b\right )}^{7/2}\,\sqrt {a-b}\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}+\frac {b^4\,\ln \left (2\,a\,b^3-2\,a^3\,b-a^4+b^4+{\mathrm {e}}^x\,{\left (a+b\right )}^{7/2}\,\sqrt {a-b}\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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