Integrand size = 16, antiderivative size = 220 \[ \int (e x)^m \sinh ^3\left (a+b x^n\right ) \, dx=-\frac {3^{-\frac {1+m}{n}} e^{3 a} (e x)^{1+m} \left (-b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},-3 b x^n\right )}{8 e n}+\frac {3 e^a (e x)^{1+m} \left (-b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},-b x^n\right )}{8 e n}-\frac {3 e^{-a} (e x)^{1+m} \left (b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},b x^n\right )}{8 e n}+\frac {3^{-\frac {1+m}{n}} e^{-3 a} (e x)^{1+m} \left (b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},3 b x^n\right )}{8 e n} \]
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Time = 0.16 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {5470, 5468, 2250} \[ \int (e x)^m \sinh ^3\left (a+b x^n\right ) \, dx=-\frac {e^{3 a} 3^{-\frac {m+1}{n}} (e x)^{m+1} \left (-b x^n\right )^{-\frac {m+1}{n}} \Gamma \left (\frac {m+1}{n},-3 b x^n\right )}{8 e n}+\frac {3 e^a (e x)^{m+1} \left (-b x^n\right )^{-\frac {m+1}{n}} \Gamma \left (\frac {m+1}{n},-b x^n\right )}{8 e n}-\frac {3 e^{-a} (e x)^{m+1} \left (b x^n\right )^{-\frac {m+1}{n}} \Gamma \left (\frac {m+1}{n},b x^n\right )}{8 e n}+\frac {e^{-3 a} 3^{-\frac {m+1}{n}} (e x)^{m+1} \left (b x^n\right )^{-\frac {m+1}{n}} \Gamma \left (\frac {m+1}{n},3 b x^n\right )}{8 e n} \]
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Rule 2250
Rule 5468
Rule 5470
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {3}{4} (e x)^m \sinh \left (a+b x^n\right )+\frac {1}{4} (e x)^m \sinh \left (3 a+3 b x^n\right )\right ) \, dx \\ & = \frac {1}{4} \int (e x)^m \sinh \left (3 a+3 b x^n\right ) \, dx-\frac {3}{4} \int (e x)^m \sinh \left (a+b x^n\right ) \, dx \\ & = -\left (\frac {1}{8} \int e^{-3 a-3 b x^n} (e x)^m \, dx\right )+\frac {1}{8} \int e^{3 a+3 b x^n} (e x)^m \, dx+\frac {3}{8} \int e^{-a-b x^n} (e x)^m \, dx-\frac {3}{8} \int e^{a+b x^n} (e x)^m \, dx \\ & = -\frac {3^{-\frac {1+m}{n}} e^{3 a} (e x)^{1+m} \left (-b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},-3 b x^n\right )}{8 e n}+\frac {3 e^a (e x)^{1+m} \left (-b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},-b x^n\right )}{8 e n}-\frac {3 e^{-a} (e x)^{1+m} \left (b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},b x^n\right )}{8 e n}+\frac {3^{-\frac {1+m}{n}} e^{-3 a} (e x)^{1+m} \left (b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},3 b x^n\right )}{8 e n} \\ \end{align*}
Time = 0.24 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.84 \[ \int (e x)^m \sinh ^3\left (a+b x^n\right ) \, dx=\frac {3^{-\frac {1+m}{n}} e^{-3 a} x (e x)^m \left (-b^2 x^{2 n}\right )^{-\frac {1+m}{n}} \left (-e^{6 a} \left (b x^n\right )^{\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},-3 b x^n\right )+3^{\frac {1+m+n}{n}} e^{4 a} \left (b x^n\right )^{\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},-b x^n\right )+\left (-b x^n\right )^{\frac {1+m}{n}} \left (-3^{\frac {1+m+n}{n}} e^{2 a} \Gamma \left (\frac {1+m}{n},b x^n\right )+\Gamma \left (\frac {1+m}{n},3 b x^n\right )\right )\right )}{8 n} \]
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\[\int \left (e x \right )^{m} \sinh \left (a +b \,x^{n}\right )^{3}d x\]
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\[ \int (e x)^m \sinh ^3\left (a+b x^n\right ) \, dx=\int { \left (e x\right )^{m} \sinh \left (b x^{n} + a\right )^{3} \,d x } \]
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\[ \int (e x)^m \sinh ^3\left (a+b x^n\right ) \, dx=\int \left (e x\right )^{m} \sinh ^{3}{\left (a + b x^{n} \right )}\, dx \]
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\[ \int (e x)^m \sinh ^3\left (a+b x^n\right ) \, dx=\int { \left (e x\right )^{m} \sinh \left (b x^{n} + a\right )^{3} \,d x } \]
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\[ \int (e x)^m \sinh ^3\left (a+b x^n\right ) \, dx=\int { \left (e x\right )^{m} \sinh \left (b x^{n} + a\right )^{3} \,d x } \]
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Timed out. \[ \int (e x)^m \sinh ^3\left (a+b x^n\right ) \, dx=\int {\mathrm {sinh}\left (a+b\,x^n\right )}^3\,{\left (e\,x\right )}^m \,d x \]
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