Integrand size = 18, antiderivative size = 40 \[ \int e^{2 (a+b x)} \sinh ^2(a+b x) \, dx=-\frac {e^{2 a+2 b x}}{4 b}+\frac {e^{4 a+4 b x}}{16 b}+\frac {x}{4} \]
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Time = 0.02 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2320, 12, 272, 45} \[ \int e^{2 (a+b x)} \sinh ^2(a+b x) \, dx=-\frac {e^{2 a+2 b x}}{4 b}+\frac {e^{4 a+4 b x}}{16 b}+\frac {x}{4} \]
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Rule 12
Rule 45
Rule 272
Rule 2320
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (1-x^2\right )^2}{4 x} \, dx,x,e^{a+b x}\right )}{b} \\ & = \frac {\text {Subst}\left (\int \frac {\left (1-x^2\right )^2}{x} \, dx,x,e^{a+b x}\right )}{4 b} \\ & = \frac {\text {Subst}\left (\int \frac {(1-x)^2}{x} \, dx,x,e^{2 a+2 b x}\right )}{8 b} \\ & = \frac {\text {Subst}\left (\int \left (-2+\frac {1}{x}+x\right ) \, dx,x,e^{2 a+2 b x}\right )}{8 b} \\ & = -\frac {e^{2 a+2 b x}}{4 b}+\frac {e^{4 a+4 b x}}{16 b}+\frac {x}{4} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.80 \[ \int e^{2 (a+b x)} \sinh ^2(a+b x) \, dx=\frac {-4 e^{2 (a+b x)}+e^{4 (a+b x)}+4 b x}{16 b} \]
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Time = 0.15 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.82
method | result | size |
risch | \(-\frac {{\mathrm e}^{2 b x +2 a}}{4 b}+\frac {{\mathrm e}^{4 b x +4 a}}{16 b}+\frac {x}{4}\) | \(33\) |
default | \(\frac {x}{4}-\frac {\sinh \left (2 b x +2 a \right )}{4 b}+\frac {\sinh \left (4 b x +4 a \right )}{16 b}-\frac {\cosh \left (2 b x +2 a \right )}{4 b}+\frac {\cosh \left (4 b x +4 a \right )}{16 b}\) | \(61\) |
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Leaf count of result is larger than twice the leaf count of optimal. 92 vs. \(2 (32) = 64\).
Time = 0.27 (sec) , antiderivative size = 92, normalized size of antiderivative = 2.30 \[ \int e^{2 (a+b x)} \sinh ^2(a+b x) \, dx=\frac {{\left (4 \, b x + 1\right )} \cosh \left (b x + a\right )^{2} - 2 \, {\left (4 \, b x - 1\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + {\left (4 \, b x + 1\right )} \sinh \left (b x + a\right )^{2} - 4}{16 \, {\left (b \cosh \left (b x + a\right )^{2} - 2 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b \sinh \left (b x + a\right )^{2}\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 139 vs. \(2 (29) = 58\).
Time = 0.44 (sec) , antiderivative size = 139, normalized size of antiderivative = 3.48 \[ \int e^{2 (a+b x)} \sinh ^2(a+b x) \, dx=\begin {cases} \frac {x e^{2 a} e^{2 b x} \sinh ^{2}{\left (a + b x \right )}}{4} - \frac {x e^{2 a} e^{2 b x} \sinh {\left (a + b x \right )} \cosh {\left (a + b x \right )}}{2} + \frac {x e^{2 a} e^{2 b x} \cosh ^{2}{\left (a + b x \right )}}{4} + \frac {e^{2 a} e^{2 b x} \sinh ^{2}{\left (a + b x \right )}}{2 b} - \frac {e^{2 a} e^{2 b x} \sinh {\left (a + b x \right )} \cosh {\left (a + b x \right )}}{4 b} & \text {for}\: b \neq 0 \\x e^{2 a} \sinh ^{2}{\left (a \right )} & \text {otherwise} \end {cases} \]
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Time = 0.19 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.92 \[ \int e^{2 (a+b x)} \sinh ^2(a+b x) \, dx=\frac {1}{4} \, x - \frac {{\left (4 \, e^{\left (-2 \, b x - 2 \, a\right )} - 1\right )} e^{\left (4 \, b x + 4 \, a\right )}}{16 \, b} + \frac {a}{4 \, b} \]
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Time = 0.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.80 \[ \int e^{2 (a+b x)} \sinh ^2(a+b x) \, dx=\frac {1}{4} \, x + \frac {e^{\left (4 \, b x + 4 \, a\right )}}{16 \, b} - \frac {e^{\left (2 \, b x + 2 \, a\right )}}{4 \, b} \]
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Time = 2.37 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.80 \[ \int e^{2 (a+b x)} \sinh ^2(a+b x) \, dx=\frac {x}{4}-\frac {\frac {{\mathrm {e}}^{2\,a+2\,b\,x}}{4}-\frac {{\mathrm {e}}^{4\,a+4\,b\,x}}{16}}{b} \]
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