Integrand size = 16, antiderivative size = 144 \[ \int (c+d x)^m \sinh ^2(a+b x) \, dx=-\frac {(c+d x)^{1+m}}{2 d (1+m)}+\frac {2^{-3-m} e^{2 a-\frac {2 b c}{d}} (c+d x)^m \left (-\frac {b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {2 b (c+d x)}{d}\right )}{b}-\frac {2^{-3-m} e^{-2 a+\frac {2 b c}{d}} (c+d x)^m \left (\frac {b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 b (c+d x)}{d}\right )}{b} \]
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Time = 0.15 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {3393, 3388, 2212} \[ \int (c+d x)^m \sinh ^2(a+b x) \, dx=\frac {2^{-m-3} e^{2 a-\frac {2 b c}{d}} (c+d x)^m \left (-\frac {b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {2 b (c+d x)}{d}\right )}{b}-\frac {2^{-m-3} e^{\frac {2 b c}{d}-2 a} (c+d x)^m \left (\frac {b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {2 b (c+d x)}{d}\right )}{b}-\frac {(c+d x)^{m+1}}{2 d (m+1)} \]
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Rule 2212
Rule 3388
Rule 3393
Rubi steps \begin{align*} \text {integral}& = -\int \left (\frac {1}{2} (c+d x)^m-\frac {1}{2} (c+d x)^m \cosh (2 a+2 b x)\right ) \, dx \\ & = -\frac {(c+d x)^{1+m}}{2 d (1+m)}+\frac {1}{2} \int (c+d x)^m \cosh (2 a+2 b x) \, dx \\ & = -\frac {(c+d x)^{1+m}}{2 d (1+m)}+\frac {1}{4} \int e^{-i (2 i a+2 i b x)} (c+d x)^m \, dx+\frac {1}{4} \int e^{i (2 i a+2 i b x)} (c+d x)^m \, dx \\ & = -\frac {(c+d x)^{1+m}}{2 d (1+m)}+\frac {2^{-3-m} e^{2 a-\frac {2 b c}{d}} (c+d x)^m \left (-\frac {b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {2 b (c+d x)}{d}\right )}{b}-\frac {2^{-3-m} e^{-2 a+\frac {2 b c}{d}} (c+d x)^m \left (\frac {b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 b (c+d x)}{d}\right )}{b} \\ \end{align*}
Time = 0.12 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.91 \[ \int (c+d x)^m \sinh ^2(a+b x) \, dx=\frac {1}{8} (c+d x)^m \left (-\frac {4 (c+d x)}{d (1+m)}+\frac {2^{-m} e^{2 a-\frac {2 b c}{d}} \left (-\frac {b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {2 b (c+d x)}{d}\right )}{b}-\frac {2^{-m} e^{-2 a+\frac {2 b c}{d}} \left (\frac {b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 b (c+d x)}{d}\right )}{b}\right ) \]
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\[\int \left (d x +c \right )^{m} \sinh \left (b x +a \right )^{2}d x\]
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Time = 0.08 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.67 \[ \int (c+d x)^m \sinh ^2(a+b x) \, dx=-\frac {{\left (d m + d\right )} \cosh \left (\frac {d m \log \left (\frac {2 \, b}{d}\right ) - 2 \, b c + 2 \, a d}{d}\right ) \Gamma \left (m + 1, \frac {2 \, {\left (b d x + b c\right )}}{d}\right ) - {\left (d m + d\right )} \cosh \left (\frac {d m \log \left (-\frac {2 \, b}{d}\right ) + 2 \, b c - 2 \, a d}{d}\right ) \Gamma \left (m + 1, -\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) - {\left (d m + d\right )} \Gamma \left (m + 1, \frac {2 \, {\left (b d x + b c\right )}}{d}\right ) \sinh \left (\frac {d m \log \left (\frac {2 \, b}{d}\right ) - 2 \, b c + 2 \, a d}{d}\right ) + {\left (d m + d\right )} \Gamma \left (m + 1, -\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) \sinh \left (\frac {d m \log \left (-\frac {2 \, b}{d}\right ) + 2 \, b c - 2 \, a d}{d}\right ) + 4 \, {\left (b d x + b c\right )} \cosh \left (m \log \left (d x + c\right )\right ) + 4 \, {\left (b d x + b c\right )} \sinh \left (m \log \left (d x + c\right )\right )}{8 \, {\left (b d m + b d\right )}} \]
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\[ \int (c+d x)^m \sinh ^2(a+b x) \, dx=\int \left (c + d x\right )^{m} \sinh ^{2}{\left (a + b x \right )}\, dx \]
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Time = 0.07 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.71 \[ \int (c+d x)^m \sinh ^2(a+b x) \, dx=-\frac {{\left (d x + c\right )}^{m + 1} e^{\left (-2 \, a + \frac {2 \, b c}{d}\right )} E_{-m}\left (\frac {2 \, {\left (d x + c\right )} b}{d}\right )}{4 \, d} - \frac {{\left (d x + c\right )}^{m + 1} e^{\left (2 \, a - \frac {2 \, b c}{d}\right )} E_{-m}\left (-\frac {2 \, {\left (d x + c\right )} b}{d}\right )}{4 \, d} - \frac {{\left (d x + c\right )}^{m + 1}}{2 \, d {\left (m + 1\right )}} \]
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\[ \int (c+d x)^m \sinh ^2(a+b x) \, dx=\int { {\left (d x + c\right )}^{m} \sinh \left (b x + a\right )^{2} \,d x } \]
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Timed out. \[ \int (c+d x)^m \sinh ^2(a+b x) \, dx=\int {\mathrm {sinh}\left (a+b\,x\right )}^2\,{\left (c+d\,x\right )}^m \,d x \]
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