Integrand size = 14, antiderivative size = 85 \[ \int x^{2+m} \sinh ^2(a+b x) \, dx=-\frac {x^{3+m}}{2 (3+m)}+\frac {2^{-5-m} e^{2 a} x^m (-b x)^{-m} \Gamma (3+m,-2 b x)}{b^3}-\frac {2^{-5-m} e^{-2 a} x^m (b x)^{-m} \Gamma (3+m,2 b x)}{b^3} \]
[Out]
Time = 0.10 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3393, 3388, 2212} \[ \int x^{2+m} \sinh ^2(a+b x) \, dx=\frac {e^{2 a} 2^{-m-5} x^m (-b x)^{-m} \Gamma (m+3,-2 b x)}{b^3}-\frac {e^{-2 a} 2^{-m-5} x^m (b x)^{-m} \Gamma (m+3,2 b x)}{b^3}-\frac {x^{m+3}}{2 (m+3)} \]
[In]
[Out]
Rule 2212
Rule 3388
Rule 3393
Rubi steps \begin{align*} \text {integral}& = -\int \left (\frac {x^{2+m}}{2}-\frac {1}{2} x^{2+m} \cosh (2 a+2 b x)\right ) \, dx \\ & = -\frac {x^{3+m}}{2 (3+m)}+\frac {1}{2} \int x^{2+m} \cosh (2 a+2 b x) \, dx \\ & = -\frac {x^{3+m}}{2 (3+m)}+\frac {1}{4} \int e^{-i (2 i a+2 i b x)} x^{2+m} \, dx+\frac {1}{4} \int e^{i (2 i a+2 i b x)} x^{2+m} \, dx \\ & = -\frac {x^{3+m}}{2 (3+m)}+\frac {2^{-5-m} e^{2 a} x^m (-b x)^{-m} \Gamma (3+m,-2 b x)}{b^3}-\frac {2^{-5-m} e^{-2 a} x^m (b x)^{-m} \Gamma (3+m,2 b x)}{b^3} \\ \end{align*}
Time = 0.09 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.92 \[ \int x^{2+m} \sinh ^2(a+b x) \, dx=\frac {1}{32} x^m \left (-\frac {16 x^3}{3+m}+\frac {2^{-m} e^{2 a} (-b x)^{-m} \Gamma (3+m,-2 b x)}{b^3}-\frac {2^{-m} e^{-2 a} (b x)^{-m} \Gamma (3+m,2 b x)}{b^3}\right ) \]
[In]
[Out]
\[\int x^{2+m} \sinh \left (b x +a \right )^{2}d x\]
[In]
[Out]
none
Time = 0.07 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.60 \[ \int x^{2+m} \sinh ^2(a+b x) \, dx=-\frac {4 \, b x \cosh \left ({\left (m + 2\right )} \log \left (x\right )\right ) + {\left (m + 3\right )} \cosh \left ({\left (m + 2\right )} \log \left (2 \, b\right ) + 2 \, a\right ) \Gamma \left (m + 3, 2 \, b x\right ) - {\left (m + 3\right )} \cosh \left ({\left (m + 2\right )} \log \left (-2 \, b\right ) - 2 \, a\right ) \Gamma \left (m + 3, -2 \, b x\right ) - {\left (m + 3\right )} \Gamma \left (m + 3, 2 \, b x\right ) \sinh \left ({\left (m + 2\right )} \log \left (2 \, b\right ) + 2 \, a\right ) + {\left (m + 3\right )} \Gamma \left (m + 3, -2 \, b x\right ) \sinh \left ({\left (m + 2\right )} \log \left (-2 \, b\right ) - 2 \, a\right ) + 4 \, b x \sinh \left ({\left (m + 2\right )} \log \left (x\right )\right )}{8 \, {\left (b m + 3 \, b\right )}} \]
[In]
[Out]
\[ \int x^{2+m} \sinh ^2(a+b x) \, dx=\int x^{m + 2} \sinh ^{2}{\left (a + b x \right )}\, dx \]
[In]
[Out]
none
Time = 0.08 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.84 \[ \int x^{2+m} \sinh ^2(a+b x) \, dx=-\frac {1}{4} \, \left (2 \, b x\right )^{-m - 3} x^{m + 3} e^{\left (-2 \, a\right )} \Gamma \left (m + 3, 2 \, b x\right ) - \frac {1}{4} \, \left (-2 \, b x\right )^{-m - 3} x^{m + 3} e^{\left (2 \, a\right )} \Gamma \left (m + 3, -2 \, b x\right ) - \frac {x^{m + 3}}{2 \, {\left (m + 3\right )}} \]
[In]
[Out]
\[ \int x^{2+m} \sinh ^2(a+b x) \, dx=\int { x^{m + 2} \sinh \left (b x + a\right )^{2} \,d x } \]
[In]
[Out]
Timed out. \[ \int x^{2+m} \sinh ^2(a+b x) \, dx=\int x^{m+2}\,{\mathrm {sinh}\left (a+b\,x\right )}^2 \,d x \]
[In]
[Out]