Optimal. Leaf size=79 \[ -\frac {2 i f \text {Li}_2\left (i e^{i (c+d x)}\right )}{a d^2}+\frac {2 (e+f x) \log \left (1-i e^{i (c+d x)}\right )}{a d}-\frac {i (e+f x)^2}{2 a f} \]
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Rubi [A] time = 0.13, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4517, 2190, 2279, 2391} \[ -\frac {2 i f \text {PolyLog}\left (2,i e^{i (c+d x)}\right )}{a d^2}+\frac {2 (e+f x) \log \left (1-i e^{i (c+d x)}\right )}{a d}-\frac {i (e+f x)^2}{2 a f} \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2279
Rule 2391
Rule 4517
Rubi steps
\begin {align*} \int \frac {(e+f x) \cos (c+d x)}{a+a \sin (c+d x)} \, dx &=-\frac {i (e+f x)^2}{2 a f}+2 \int \frac {e^{i (c+d x)} (e+f x)}{a-i a e^{i (c+d x)}} \, dx\\ &=-\frac {i (e+f x)^2}{2 a f}+\frac {2 (e+f x) \log \left (1-i e^{i (c+d x)}\right )}{a d}-\frac {(2 f) \int \log \left (1-i e^{i (c+d x)}\right ) \, dx}{a d}\\ &=-\frac {i (e+f x)^2}{2 a f}+\frac {2 (e+f x) \log \left (1-i e^{i (c+d x)}\right )}{a d}+\frac {(2 i f) \operatorname {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i (c+d x)}\right )}{a d^2}\\ &=-\frac {i (e+f x)^2}{2 a f}+\frac {2 (e+f x) \log \left (1-i e^{i (c+d x)}\right )}{a d}-\frac {2 i f \text {Li}_2\left (i e^{i (c+d x)}\right )}{a d^2}\\ \end {align*}
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Mathematica [B] time = 0.57, size = 246, normalized size = 3.11 \[ \frac {-i c^2 f+4 d e \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )-4 i f \text {Li}_2\left (i e^{i (c+d x)}\right )-2 i c d f x+4 c f \log \left (1-i e^{i (c+d x)}\right )+4 \pi f \log \left (1+e^{-i (c+d x)}\right )+4 d f x \log \left (1-i e^{i (c+d x)}\right )+2 \pi f \log \left (1-i e^{i (c+d x)}\right )-2 \pi f \log \left (\sin \left (\frac {1}{4} (2 c+2 d x+\pi )\right )\right )-4 \pi f \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-4 c f \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+i \pi c f-i d^2 f x^2+i \pi d f x}{2 a d^2} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.46, size = 156, normalized size = 1.97 \[ \frac {-i \, f {\rm Li}_2\left (i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right ) + i \, f {\rm Li}_2\left (-i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right ) + {\left (d e - c f\right )} \log \left (\cos \left (d x + c\right ) + i \, \sin \left (d x + c\right ) + i\right ) + {\left (d f x + c f\right )} \log \left (i \, \cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1\right ) + {\left (d f x + c f\right )} \log \left (-i \, \cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1\right ) + {\left (d e - c f\right )} \log \left (-\cos \left (d x + c\right ) + i \, \sin \left (d x + c\right ) + i\right )}{a d^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (f x + e\right )} \cos \left (d x + c\right )}{a \sin \left (d x + c\right ) + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.23, size = 203, normalized size = 2.57 \[ -\frac {i f \,x^{2}}{2 a}+\frac {i e x}{a}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) e}{d a}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right ) e}{d a}-\frac {2 i f c x}{d a}-\frac {i f \,c^{2}}{d^{2} a}+\frac {2 f \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) x}{d a}+\frac {2 f \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) c}{d^{2} a}-\frac {2 i f \polylog \left (2, i {\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{2}}-\frac {2 f c \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d^{2} a}+\frac {2 f c \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{d^{2} a} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.95, size = 116, normalized size = 1.47 \[ \frac {-i \, d^{2} f x^{2} - 2 i \, d^{2} e x - 4 i \, d f x \arctan \left (\cos \left (d x + c\right ), \sin \left (d x + c\right ) + 1\right ) + 4 i \, d e \arctan \left (\sin \left (d x + c\right ) + 1, \cos \left (d x + c\right )\right ) - 4 i \, f {\rm Li}_2\left (i \, e^{\left (i \, d x + i \, c\right )}\right ) + 2 \, {\left (d f x + d e\right )} \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right )}{2 \, a d^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\cos \left (c+d\,x\right )\,\left (e+f\,x\right )}{a+a\,\sin \left (c+d\,x\right )} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {e \cos {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {f x \cos {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \]
Verification of antiderivative is not currently implemented for this CAS.
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