Integrand size = 16, antiderivative size = 41 \[ \int \frac {e^x (1+\sin (x))}{1-\cos (x)} \, dx=(-2+2 i) e^{(1+i) x} \operatorname {Hypergeometric2F1}\left (1-i,2,2-i,e^{i x}\right )+\frac {e^x \sin (x)}{1-\cos (x)} \]
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Time = 0.09 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {4551, 4549, 4528, 2225, 2283, 2326} \[ \int \frac {e^x (1+\sin (x))}{1-\cos (x)} \, dx=-4 i e^x \operatorname {Hypergeometric2F1}\left (-i,1,1-i,e^{i x}\right )+2 i e^x-\frac {e^x \sin (x)}{1-\cos (x)} \]
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Rule 2225
Rule 2283
Rule 2326
Rule 4528
Rule 4549
Rule 4551
Rubi steps \begin{align*} \text {integral}& = 2 \int \frac {e^x \sin (x)}{1-\cos (x)} \, dx+\int \frac {e^x (1-\sin (x))}{1-\cos (x)} \, dx \\ & = -\frac {e^x \sin (x)}{1-\cos (x)}+2 \int e^x \cot \left (\frac {x}{2}\right ) \, dx \\ & = -\frac {e^x \sin (x)}{1-\cos (x)}-2 i \int \left (-e^x-\frac {2 e^x}{-1+e^{i x}}\right ) \, dx \\ & = -\frac {e^x \sin (x)}{1-\cos (x)}+2 i \int e^x \, dx+4 i \int \frac {e^x}{-1+e^{i x}} \, dx \\ & = 2 i e^x-4 i e^x \operatorname {Hypergeometric2F1}\left (-i,1,1-i,e^{i x}\right )-\frac {e^x \sin (x)}{1-\cos (x)} \\ \end{align*}
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(100\) vs. \(2(41)=82\).
Time = 0.62 (sec) , antiderivative size = 100, normalized size of antiderivative = 2.44 \[ \int \frac {e^x (1+\sin (x))}{1-\cos (x)} \, dx=\frac {2 e^x \sin \left (\frac {x}{2}\right ) \left (\cos \left (\frac {x}{2}\right )+2 i \operatorname {Hypergeometric2F1}\left (-i,1,1-i,e^{i x}\right ) \sin \left (\frac {x}{2}\right )+(1+i) e^{i x} \operatorname {Hypergeometric2F1}\left (1,1-i,2-i,e^{i x}\right ) \sin \left (\frac {x}{2}\right )\right ) (1+\sin (x))}{(-1+\cos (x)) \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )^2} \]
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\[\int \frac {{\mathrm e}^{x} \left (\sin \left (x \right )+1\right )}{1-\cos \left (x \right )}d x\]
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\[ \int \frac {e^x (1+\sin (x))}{1-\cos (x)} \, dx=\int { -\frac {{\left (\sin \left (x\right ) + 1\right )} e^{x}}{\cos \left (x\right ) - 1} \,d x } \]
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\[ \int \frac {e^x (1+\sin (x))}{1-\cos (x)} \, dx=- \int \frac {e^{x}}{\cos {\left (x \right )} - 1}\, dx - \int \frac {e^{x} \sin {\left (x \right )}}{\cos {\left (x \right )} - 1}\, dx \]
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\[ \int \frac {e^x (1+\sin (x))}{1-\cos (x)} \, dx=\int { -\frac {{\left (\sin \left (x\right ) + 1\right )} e^{x}}{\cos \left (x\right ) - 1} \,d x } \]
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\[ \int \frac {e^x (1+\sin (x))}{1-\cos (x)} \, dx=\int { -\frac {{\left (\sin \left (x\right ) + 1\right )} e^{x}}{\cos \left (x\right ) - 1} \,d x } \]
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Timed out. \[ \int \frac {e^x (1+\sin (x))}{1-\cos (x)} \, dx=\int -\frac {{\mathrm {e}}^x\,\left (\sin \left (x\right )+1\right )}{\cos \left (x\right )-1} \,d x \]
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