Integrand size = 29, antiderivative size = 141 \[ \int \frac {\cosh ^3(c+d x) \sinh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {a^3 \left (a^2+b^2\right ) \log (a+b \sinh (c+d x))}{b^6 d}+\frac {a^2 \left (a^2+b^2\right ) \sinh (c+d x)}{b^5 d}-\frac {a \left (a^2+b^2\right ) \sinh ^2(c+d x)}{2 b^4 d}+\frac {\left (a^2+b^2\right ) \sinh ^3(c+d x)}{3 b^3 d}-\frac {a \sinh ^4(c+d x)}{4 b^2 d}+\frac {\sinh ^5(c+d x)}{5 b d} \]
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Time = 0.15 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2916, 12, 908} \[ \int \frac {\cosh ^3(c+d x) \sinh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {a^2 \left (a^2+b^2\right ) \sinh (c+d x)}{b^5 d}-\frac {a \left (a^2+b^2\right ) \sinh ^2(c+d x)}{2 b^4 d}+\frac {\left (a^2+b^2\right ) \sinh ^3(c+d x)}{3 b^3 d}-\frac {a^3 \left (a^2+b^2\right ) \log (a+b \sinh (c+d x))}{b^6 d}-\frac {a \sinh ^4(c+d x)}{4 b^2 d}+\frac {\sinh ^5(c+d x)}{5 b d} \]
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Rule 12
Rule 908
Rule 2916
Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int \frac {x^3 \left (-b^2-x^2\right )}{b^3 (a+x)} \, dx,x,b \sinh (c+d x)\right )}{b^3 d} \\ & = -\frac {\text {Subst}\left (\int \frac {x^3 \left (-b^2-x^2\right )}{a+x} \, dx,x,b \sinh (c+d x)\right )}{b^6 d} \\ & = -\frac {\text {Subst}\left (\int \left (-a^2 \left (a^2+b^2\right )+a \left (a^2+b^2\right ) x-\left (a^2+b^2\right ) x^2+a x^3-x^4+\frac {a^3 \left (a^2+b^2\right )}{a+x}\right ) \, dx,x,b \sinh (c+d x)\right )}{b^6 d} \\ & = -\frac {a^3 \left (a^2+b^2\right ) \log (a+b \sinh (c+d x))}{b^6 d}+\frac {a^2 \left (a^2+b^2\right ) \sinh (c+d x)}{b^5 d}-\frac {a \left (a^2+b^2\right ) \sinh ^2(c+d x)}{2 b^4 d}+\frac {\left (a^2+b^2\right ) \sinh ^3(c+d x)}{3 b^3 d}-\frac {a \sinh ^4(c+d x)}{4 b^2 d}+\frac {\sinh ^5(c+d x)}{5 b d} \\ \end{align*}
Time = 0.27 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.87 \[ \int \frac {\cosh ^3(c+d x) \sinh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {\frac {60 a^3 \left (a^2+b^2\right ) \log (a+b \sinh (c+d x))}{b^6}-\frac {60 a^2 \left (a^2+b^2\right ) \sinh (c+d x)}{b^5}+\frac {30 a \left (a^2+b^2\right ) \sinh ^2(c+d x)}{b^4}-\frac {20 \left (a^2+b^2\right ) \sinh ^3(c+d x)}{b^3}+\frac {15 a \sinh ^4(c+d x)}{b^2}-\frac {12 \sinh ^5(c+d x)}{b}}{60 d} \]
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Time = 99.05 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.02
method | result | size |
derivativedivides | \(\frac {\frac {\frac {\sinh \left (d x +c \right )^{5} b^{4}}{5}-\frac {a \sinh \left (d x +c \right )^{4} b^{3}}{4}+\frac {a^{2} b^{2} \sinh \left (d x +c \right )^{3}}{3}+\frac {b^{4} \sinh \left (d x +c \right )^{3}}{3}-\frac {a^{3} b \sinh \left (d x +c \right )^{2}}{2}-\frac {a \,b^{3} \sinh \left (d x +c \right )^{2}}{2}+a^{4} \sinh \left (d x +c \right )+a^{2} b^{2} \sinh \left (d x +c \right )}{b^{5}}-\frac {a^{3} \left (a^{2}+b^{2}\right ) \ln \left (a +b \sinh \left (d x +c \right )\right )}{b^{6}}}{d}\) | \(144\) |
default | \(\frac {\frac {\frac {\sinh \left (d x +c \right )^{5} b^{4}}{5}-\frac {a \sinh \left (d x +c \right )^{4} b^{3}}{4}+\frac {a^{2} b^{2} \sinh \left (d x +c \right )^{3}}{3}+\frac {b^{4} \sinh \left (d x +c \right )^{3}}{3}-\frac {a^{3} b \sinh \left (d x +c \right )^{2}}{2}-\frac {a \,b^{3} \sinh \left (d x +c \right )^{2}}{2}+a^{4} \sinh \left (d x +c \right )+a^{2} b^{2} \sinh \left (d x +c \right )}{b^{5}}-\frac {a^{3} \left (a^{2}+b^{2}\right ) \ln \left (a +b \sinh \left (d x +c \right )\right )}{b^{6}}}{d}\) | \(144\) |
risch | \(-\frac {{\mathrm e}^{-3 d x -3 c} a^{2}}{24 b^{3} d}+\frac {2 a^{5} c}{b^{6} d}-\frac {a^{5} \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 a \,{\mathrm e}^{d x +c}}{b}-1\right )}{b^{6} d}-\frac {a^{3} {\mathrm e}^{2 d x +2 c}}{8 b^{4} d}-\frac {{\mathrm e}^{-3 d x -3 c}}{96 b d}+\frac {{\mathrm e}^{d x +c} a^{4}}{2 b^{5} d}-\frac {{\mathrm e}^{-d x -c} a^{4}}{2 b^{5} d}-\frac {a^{3} {\mathrm e}^{-2 d x -2 c}}{8 b^{4} d}+\frac {{\mathrm e}^{-d x -c}}{16 b d}+\frac {a^{5} x}{b^{6}}-\frac {{\mathrm e}^{-5 d x -5 c}}{160 b d}+\frac {{\mathrm e}^{5 d x +5 c}}{160 b d}+\frac {{\mathrm e}^{3 d x +3 c}}{96 b d}-\frac {{\mathrm e}^{d x +c}}{16 b d}+\frac {2 a^{3} c}{b^{4} d}-\frac {a^{3} \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 a \,{\mathrm e}^{d x +c}}{b}-1\right )}{b^{4} d}-\frac {a \,{\mathrm e}^{4 d x +4 c}}{64 b^{2} d}-\frac {a \,{\mathrm e}^{-2 d x -2 c}}{16 b^{2} d}-\frac {a \,{\mathrm e}^{2 d x +2 c}}{16 b^{2} d}+\frac {a^{3} x}{b^{4}}-\frac {a \,{\mathrm e}^{-4 d x -4 c}}{64 b^{2} d}-\frac {3 \,{\mathrm e}^{-d x -c} a^{2}}{8 b^{3} d}+\frac {3 \,{\mathrm e}^{d x +c} a^{2}}{8 b^{3} d}+\frac {{\mathrm e}^{3 d x +3 c} a^{2}}{24 b^{3} d}\) | \(437\) |
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Leaf count of result is larger than twice the leaf count of optimal. 1660 vs. \(2 (133) = 266\).
Time = 0.28 (sec) , antiderivative size = 1660, normalized size of antiderivative = 11.77 \[ \int \frac {\cosh ^3(c+d x) \sinh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx=\text {Too large to display} \]
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Timed out. \[ \int \frac {\cosh ^3(c+d x) \sinh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx=\text {Timed out} \]
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Leaf count of result is larger than twice the leaf count of optimal. 300 vs. \(2 (133) = 266\).
Time = 0.22 (sec) , antiderivative size = 300, normalized size of antiderivative = 2.13 \[ \int \frac {\cosh ^3(c+d x) \sinh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {{\left (15 \, a b^{3} e^{\left (-d x - c\right )} - 6 \, b^{4} - 10 \, {\left (4 \, a^{2} b^{2} + b^{4}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 60 \, {\left (2 \, a^{3} b + a b^{3}\right )} e^{\left (-3 \, d x - 3 \, c\right )} - 60 \, {\left (8 \, a^{4} + 6 \, a^{2} b^{2} - b^{4}\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} e^{\left (5 \, d x + 5 \, c\right )}}{960 \, b^{5} d} - \frac {{\left (a^{5} + a^{3} b^{2}\right )} {\left (d x + c\right )}}{b^{6} d} - \frac {15 \, a b^{3} e^{\left (-4 \, d x - 4 \, c\right )} + 6 \, b^{4} e^{\left (-5 \, d x - 5 \, c\right )} + 60 \, {\left (8 \, a^{4} + 6 \, a^{2} b^{2} - b^{4}\right )} e^{\left (-d x - c\right )} + 60 \, {\left (2 \, a^{3} b + a b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, {\left (4 \, a^{2} b^{2} + b^{4}\right )} e^{\left (-3 \, d x - 3 \, c\right )}}{960 \, b^{5} d} - \frac {{\left (a^{5} + a^{3} b^{2}\right )} \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{b^{6} d} \]
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Time = 0.34 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.83 \[ \int \frac {\cosh ^3(c+d x) \sinh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {\frac {6 \, b^{4} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{5} - 15 \, a b^{3} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{4} + 40 \, a^{2} b^{2} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3} + 40 \, b^{4} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3} - 120 \, a^{3} b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} - 120 \, a b^{3} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 480 \, a^{4} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 480 \, a^{2} b^{2} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}}{b^{5}} - \frac {960 \, {\left (a^{5} + a^{3} b^{2}\right )} \log \left ({\left | b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, a \right |}\right )}{b^{6}}}{960 \, d} \]
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Time = 1.47 (sec) , antiderivative size = 307, normalized size of antiderivative = 2.18 \[ \int \frac {\cosh ^3(c+d x) \sinh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {{\mathrm {e}}^{5\,c+5\,d\,x}}{160\,b\,d}-\frac {{\mathrm {e}}^{-5\,c-5\,d\,x}}{160\,b\,d}-\frac {a\,{\mathrm {e}}^{-4\,c-4\,d\,x}}{64\,b^2\,d}-\frac {a\,{\mathrm {e}}^{4\,c+4\,d\,x}}{64\,b^2\,d}-\frac {\ln \left (2\,a\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-b+b\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}\right )\,\left (a^5+a^3\,b^2\right )}{b^6\,d}-\frac {{\mathrm {e}}^{-c-d\,x}\,\left (8\,a^4+6\,a^2\,b^2-b^4\right )}{16\,b^5\,d}+\frac {a^3\,x\,\left (a^2+b^2\right )}{b^6}-\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}\,\left (2\,a^3+a\,b^2\right )}{16\,b^4\,d}-\frac {{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (2\,a^3+a\,b^2\right )}{16\,b^4\,d}+\frac {{\mathrm {e}}^{c+d\,x}\,\left (8\,a^4+6\,a^2\,b^2-b^4\right )}{16\,b^5\,d}-\frac {{\mathrm {e}}^{-3\,c-3\,d\,x}\,\left (4\,a^2+b^2\right )}{96\,b^3\,d}+\frac {{\mathrm {e}}^{3\,c+3\,d\,x}\,\left (4\,a^2+b^2\right )}{96\,b^3\,d} \]
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