Optimal. Leaf size=212 \[ \frac {a^3 b^3 \text {ArcTan}\left (\frac {b \cosh (x)+a \sinh (x)}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2}}+\frac {a^3 b^2 \cosh (x)}{\left (a^2-b^2\right )^3}-\frac {a b^2 \cosh ^3(x)}{3 \left (a^2-b^2\right )^2}-\frac {a \cosh ^3(x)}{3 \left (a^2-b^2\right )}+\frac {a \cosh ^5(x)}{5 \left (a^2-b^2\right )}-\frac {a^2 b^3 \sinh (x)}{\left (a^2-b^2\right )^3}+\frac {a^2 b \sinh ^3(x)}{3 \left (a^2-b^2\right )^2}-\frac {b \sinh ^3(x)}{3 \left (a^2-b^2\right )}-\frac {b \sinh ^5(x)}{5 \left (a^2-b^2\right )} \]
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Rubi [A]
time = 0.30, antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps
used = 17, number of rules used = 9, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {3188, 2644,
14, 2645, 30, 2717, 2718, 3153, 212} \begin {gather*} -\frac {b \sinh ^5(x)}{5 \left (a^2-b^2\right )}-\frac {b \sinh ^3(x)}{3 \left (a^2-b^2\right )}+\frac {a^2 b \sinh ^3(x)}{3 \left (a^2-b^2\right )^2}+\frac {a \cosh ^5(x)}{5 \left (a^2-b^2\right )}-\frac {a \cosh ^3(x)}{3 \left (a^2-b^2\right )}-\frac {a b^2 \cosh ^3(x)}{3 \left (a^2-b^2\right )^2}-\frac {a^2 b^3 \sinh (x)}{\left (a^2-b^2\right )^3}+\frac {a^3 b^3 \text {ArcTan}\left (\frac {a \sinh (x)+b \cosh (x)}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2}}+\frac {a^3 b^2 \cosh (x)}{\left (a^2-b^2\right )^3} \end {gather*}
Antiderivative was successfully verified.
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Rule 14
Rule 30
Rule 212
Rule 2644
Rule 2645
Rule 2717
Rule 2718
Rule 3153
Rule 3188
Rubi steps
\begin {align*} \int \frac {\cosh ^3(x) \sinh ^3(x)}{a \cosh (x)+b \sinh (x)} \, dx &=\frac {a \int \cosh ^2(x) \sinh ^3(x) \, dx}{a^2-b^2}-\frac {b \int \cosh ^3(x) \sinh ^2(x) \, dx}{a^2-b^2}+\frac {(a b) \int \frac {\cosh ^2(x) \sinh ^2(x)}{a \cosh (x)+b \sinh (x)} \, dx}{a^2-b^2}\\ &=\frac {\left (a^2 b\right ) \int \cosh (x) \sinh ^2(x) \, dx}{\left (a^2-b^2\right )^2}-\frac {\left (a b^2\right ) \int \cosh ^2(x) \sinh (x) \, dx}{\left (a^2-b^2\right )^2}+\frac {\left (a^2 b^2\right ) \int \frac {\cosh (x) \sinh (x)}{a \cosh (x)+b \sinh (x)} \, dx}{\left (a^2-b^2\right )^2}-\frac {a \text {Subst}\left (\int x^2 \left (1-x^2\right ) \, dx,x,\cosh (x)\right )}{a^2-b^2}-\frac {(i b) \text {Subst}\left (\int x^2 \left (1-x^2\right ) \, dx,x,i \sinh (x)\right )}{a^2-b^2}\\ &=\frac {\left (a^3 b^2\right ) \int \sinh (x) \, dx}{\left (a^2-b^2\right )^3}-\frac {\left (a^2 b^3\right ) \int \cosh (x) \, dx}{\left (a^2-b^2\right )^3}+\frac {\left (a^3 b^3\right ) \int \frac {1}{a \cosh (x)+b \sinh (x)} \, dx}{\left (a^2-b^2\right )^3}+\frac {\left (i a^2 b\right ) \text {Subst}\left (\int x^2 \, dx,x,i \sinh (x)\right )}{\left (a^2-b^2\right )^2}-\frac {\left (a b^2\right ) \text {Subst}\left (\int x^2 \, dx,x,\cosh (x)\right )}{\left (a^2-b^2\right )^2}-\frac {a \text {Subst}\left (\int \left (x^2-x^4\right ) \, dx,x,\cosh (x)\right )}{a^2-b^2}-\frac {(i b) \text {Subst}\left (\int \left (x^2-x^4\right ) \, dx,x,i \sinh (x)\right )}{a^2-b^2}\\ &=\frac {a^3 b^2 \cosh (x)}{\left (a^2-b^2\right )^3}-\frac {a b^2 \cosh ^3(x)}{3 \left (a^2-b^2\right )^2}-\frac {a \cosh ^3(x)}{3 \left (a^2-b^2\right )}+\frac {a \cosh ^5(x)}{5 \left (a^2-b^2\right )}-\frac {a^2 b^3 \sinh (x)}{\left (a^2-b^2\right )^3}+\frac {a^2 b \sinh ^3(x)}{3 \left (a^2-b^2\right )^2}-\frac {b \sinh ^3(x)}{3 \left (a^2-b^2\right )}-\frac {b \sinh ^5(x)}{5 \left (a^2-b^2\right )}+\frac {\left (i a^3 b^3\right ) \text {Subst}\left (\int \frac {1}{a^2-b^2-x^2} \, dx,x,-i b \cosh (x)-i a \sinh (x)\right )}{\left (a^2-b^2\right )^3}\\ &=\frac {a^3 b^3 \tan ^{-1}\left (\frac {b \cosh (x)+a \sinh (x)}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2}}+\frac {a^3 b^2 \cosh (x)}{\left (a^2-b^2\right )^3}-\frac {a b^2 \cosh ^3(x)}{3 \left (a^2-b^2\right )^2}-\frac {a \cosh ^3(x)}{3 \left (a^2-b^2\right )}+\frac {a \cosh ^5(x)}{5 \left (a^2-b^2\right )}-\frac {a^2 b^3 \sinh (x)}{\left (a^2-b^2\right )^3}+\frac {a^2 b \sinh ^3(x)}{3 \left (a^2-b^2\right )^2}-\frac {b \sinh ^3(x)}{3 \left (a^2-b^2\right )}-\frac {b \sinh ^5(x)}{5 \left (a^2-b^2\right )}\\ \end {align*}
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Mathematica [A]
time = 1.52, size = 325, normalized size = 1.53 \begin {gather*} \frac {1}{32} \left (\frac {4 a b \left (3 a^4+10 a^2 b^2+3 b^4\right ) \text {ArcTan}\left (\frac {b+a \tanh \left (\frac {x}{2}\right )}{\sqrt {a-b} \sqrt {a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2}}+\frac {2 a \left (a^4+10 a^2 b^2+5 b^4\right ) \cosh (x)}{(a-b)^3 (a+b)^3}-\frac {2 a \left (a^2+3 b^2\right ) \cosh (3 x)}{3 (a-b)^2 (a+b)^2}+\frac {2 a \cosh (5 x)}{5 (a-b) (a+b)}+\frac {2 b \left (5 a^4+10 a^2 b^2+b^4\right ) \sinh (x)}{(-a+b)^3 (a+b)^3}-3 \left (\frac {4 a b \text {ArcTan}\left (\frac {b+a \tanh \left (\frac {x}{2}\right )}{\sqrt {a-b} \sqrt {a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2}}+\frac {2 a \cosh (x)}{a^2-b^2}+\frac {2 b \sinh (x)}{-a^2+b^2}\right )+\frac {2 b \left (3 a^2+b^2\right ) \sinh (3 x)}{3 (a-b)^2 (a+b)^2}-\frac {2 b \sinh (5 x)}{5 (a-b) (a+b)}\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 1.16, size = 266, normalized size = 1.25
method | result | size |
default | \(-\frac {16}{5 \left (16 a +16 b \right ) \left (\tanh \left (\frac {x}{2}\right )-1\right )^{5}}-\frac {4}{\left (8 a +8 b \right ) \left (\tanh \left (\frac {x}{2}\right )-1\right )^{4}}-\frac {a +3 b}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {5 a +7 b}{12 \left (a +b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}+\frac {a \left (a +3 b \right )}{8 \left (a +b \right )^{3} \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {4}{\left (8 a -8 b \right ) \left (\tanh \left (\frac {x}{2}\right )+1\right )^{4}}+\frac {16}{5 \left (16 a -16 b \right ) \left (\tanh \left (\frac {x}{2}\right )+1\right )^{5}}-\frac {-5 a +7 b}{12 \left (a -b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}-\frac {a -3 b}{8 \left (a -b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {a \left (a -3 b \right )}{8 \left (a -b \right )^{3} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {2 a^{3} b^{3} \arctan \left (\frac {2 a \tanh \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3} \sqrt {a^{2}-b^{2}}}\) | \(266\) |
risch | \(\frac {{\mathrm e}^{5 x}}{160 a +160 b}-\frac {{\mathrm e}^{3 x} a}{96 \left (a +b \right )^{2}}+\frac {{\mathrm e}^{3 x} b}{96 \left (a +b \right )^{2}}-\frac {{\mathrm e}^{x} a^{2}}{16 \left (a +b \right )^{3}}-\frac {{\mathrm e}^{x} a b}{4 \left (a +b \right )^{3}}-\frac {{\mathrm e}^{x} b^{2}}{16 \left (a +b \right )^{3}}-\frac {{\mathrm e}^{-x} a^{2}}{16 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {{\mathrm e}^{-x} a b}{4 a^{3}-12 a^{2} b +12 a \,b^{2}-4 b^{3}}-\frac {{\mathrm e}^{-x} b^{2}}{16 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {{\mathrm e}^{-3 x} a}{96 \left (a^{2}-2 a b +b^{2}\right )}-\frac {{\mathrm e}^{-3 x} b}{96 \left (a^{2}-2 a b +b^{2}\right )}+\frac {{\mathrm e}^{-5 x}}{160 a -160 b}-\frac {b^{3} a^{3} \ln \left ({\mathrm e}^{x}-\frac {a -b}{\sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3}}+\frac {b^{3} a^{3} \ln \left ({\mathrm e}^{x}+\frac {a -b}{\sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3}}\) | \(324\) |
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. 2440 vs.
\(2 (196) = 392\).
time = 0.46, size = 4935, normalized size = 23.28 \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(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.41, size = 325, normalized size = 1.53 \begin {gather*} \frac {2 \, a^{3} b^{3} \arctan \left (\frac {a e^{x} + b e^{x}}{\sqrt {a^{2} - b^{2}}}\right )}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt {a^{2} - b^{2}}} - \frac {{\left (30 \, a^{2} e^{\left (4 \, x\right )} - 120 \, a b e^{\left (4 \, x\right )} + 30 \, b^{2} e^{\left (4 \, x\right )} + 5 \, a^{2} e^{\left (2 \, x\right )} - 5 \, b^{2} e^{\left (2 \, x\right )} - 3 \, a^{2} + 6 \, a b - 3 \, b^{2}\right )} e^{\left (-5 \, x\right )}}{480 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}} + \frac {3 \, a^{4} e^{\left (5 \, x\right )} + 12 \, a^{3} b e^{\left (5 \, x\right )} + 18 \, a^{2} b^{2} e^{\left (5 \, x\right )} + 12 \, a b^{3} e^{\left (5 \, x\right )} + 3 \, b^{4} e^{\left (5 \, x\right )} - 5 \, a^{4} e^{\left (3 \, x\right )} - 10 \, a^{3} b e^{\left (3 \, x\right )} + 10 \, a b^{3} e^{\left (3 \, x\right )} + 5 \, b^{4} e^{\left (3 \, x\right )} - 30 \, a^{4} e^{x} - 180 \, a^{3} b e^{x} - 300 \, a^{2} b^{2} e^{x} - 180 \, a b^{3} e^{x} - 30 \, b^{4} e^{x}}{480 \, {\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 2.07, size = 371, normalized size = 1.75 \begin {gather*} \frac {{\mathrm {e}}^{-5\,x}}{160\,a-160\,b}+\frac {{\mathrm {e}}^{5\,x}}{160\,a+160\,b}+\frac {2\,\mathrm {atan}\left (\frac {a^3\,b^3\,{\mathrm {e}}^x\,\sqrt {a^{14}-7\,a^{12}\,b^2+21\,a^{10}\,b^4-35\,a^8\,b^6+35\,a^6\,b^8-21\,a^4\,b^{10}+7\,a^2\,b^{12}-b^{14}}}{a^7\,\sqrt {a^6\,b^6}+b^7\,\sqrt {a^6\,b^6}-3\,a^2\,b^5\,\sqrt {a^6\,b^6}+3\,a^3\,b^4\,\sqrt {a^6\,b^6}+3\,a^4\,b^3\,\sqrt {a^6\,b^6}-3\,a^5\,b^2\,\sqrt {a^6\,b^6}-a\,b^6\,\sqrt {a^6\,b^6}-a^6\,b\,\sqrt {a^6\,b^6}}\right )\,\sqrt {a^6\,b^6}}{\sqrt {a^{14}-7\,a^{12}\,b^2+21\,a^{10}\,b^4-35\,a^8\,b^6+35\,a^6\,b^8-21\,a^4\,b^{10}+7\,a^2\,b^{12}-b^{14}}}-\frac {{\mathrm {e}}^{-x}\,\left (a^2-4\,a\,b+b^2\right )}{16\,{\left (a-b\right )}^3}-\frac {{\mathrm {e}}^{-3\,x}\,\left (a+b\right )}{96\,{\left (a-b\right )}^2}-\frac {{\mathrm {e}}^{3\,x}\,\left (a-b\right )}{96\,{\left (a+b\right )}^2}-\frac {{\mathrm {e}}^x\,\left (a^2+4\,a\,b+b^2\right )}{16\,{\left (a+b\right )}^3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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