Optimal. Leaf size=172 \[ \frac {i f \text {Li}_2\left (-i e^{i (c+d x)}\right )}{2 a d^2}-\frac {i f \text {Li}_2\left (i e^{i (c+d x)}\right )}{2 a d^2}+\frac {f \tan (c+d x)}{2 a d^2}-\frac {f \sec (c+d x)}{2 a d^2}-\frac {i (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{a d}-\frac {(e+f x) \sec ^2(c+d x)}{2 a d}+\frac {(e+f x) \tan (c+d x) \sec (c+d x)}{2 a d} \]
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Rubi [A] time = 0.14, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4531, 4185, 4181, 2279, 2391, 4409, 3767, 8} \[ \frac {i f \text {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{2 a d^2}-\frac {i f \text {PolyLog}\left (2,i e^{i (c+d x)}\right )}{2 a d^2}+\frac {f \tan (c+d x)}{2 a d^2}-\frac {f \sec (c+d x)}{2 a d^2}-\frac {i (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{a d}-\frac {(e+f x) \sec ^2(c+d x)}{2 a d}+\frac {(e+f x) \tan (c+d x) \sec (c+d x)}{2 a d} \]
Antiderivative was successfully verified.
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Rule 8
Rule 2279
Rule 2391
Rule 3767
Rule 4181
Rule 4185
Rule 4409
Rule 4531
Rubi steps
\begin {align*} \int \frac {(e+f x) \sec (c+d x)}{a+a \sin (c+d x)} \, dx &=\frac {\int (e+f x) \sec ^3(c+d x) \, dx}{a}-\frac {\int (e+f x) \sec ^2(c+d x) \tan (c+d x) \, dx}{a}\\ &=-\frac {f \sec (c+d x)}{2 a d^2}-\frac {(e+f x) \sec ^2(c+d x)}{2 a d}+\frac {(e+f x) \sec (c+d x) \tan (c+d x)}{2 a d}+\frac {\int (e+f x) \sec (c+d x) \, dx}{2 a}+\frac {f \int \sec ^2(c+d x) \, dx}{2 a d}\\ &=-\frac {i (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{a d}-\frac {f \sec (c+d x)}{2 a d^2}-\frac {(e+f x) \sec ^2(c+d x)}{2 a d}+\frac {(e+f x) \sec (c+d x) \tan (c+d x)}{2 a d}-\frac {f \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{2 a d^2}-\frac {f \int \log \left (1-i e^{i (c+d x)}\right ) \, dx}{2 a d}+\frac {f \int \log \left (1+i e^{i (c+d x)}\right ) \, dx}{2 a d}\\ &=-\frac {i (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{a d}-\frac {f \sec (c+d x)}{2 a d^2}-\frac {(e+f x) \sec ^2(c+d x)}{2 a d}+\frac {f \tan (c+d x)}{2 a d^2}+\frac {(e+f x) \sec (c+d x) \tan (c+d x)}{2 a d}+\frac {(i f) \operatorname {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i (c+d x)}\right )}{2 a d^2}-\frac {(i f) \operatorname {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{i (c+d x)}\right )}{2 a d^2}\\ &=-\frac {i (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{a d}+\frac {i f \text {Li}_2\left (-i e^{i (c+d x)}\right )}{2 a d^2}-\frac {i f \text {Li}_2\left (i e^{i (c+d x)}\right )}{2 a d^2}-\frac {f \sec (c+d x)}{2 a d^2}-\frac {(e+f x) \sec ^2(c+d x)}{2 a d}+\frac {f \tan (c+d x)}{2 a d^2}+\frac {(e+f x) \sec (c+d x) \tan (c+d x)}{2 a d}\\ \end {align*}
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Mathematica [B] time = 3.07, size = 655, normalized size = 3.81 \[ -\frac {(c+d x) (c f-d (2 e+f x)) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2+d e \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2 \left (2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+c+d x\right )+d e \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2 \left (-2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+c+d x\right )-\frac {f \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2 \left ((-1)^{3/4} (c+d x)^2+\frac {4 i \text {Li}_2\left (-i e^{i (c+d x)}\right )-3 i \pi (c+d x)-4 \pi \log \left (1+e^{-i (c+d x)}\right )+2 (-2 c-2 d x+\pi ) \log \left (1+i e^{i (c+d x)}\right )-2 \pi \log \left (\sin \left (\frac {1}{4} (2 c+2 d x-\pi )\right )\right )+4 \pi \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {2}}\right )}{\sqrt {2}}+\frac {f \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2 \left (\sqrt [4]{-1} (c+d x)^2+\frac {4 i \text {Li}_2\left (i e^{i (c+d x)}\right )-i \pi (c+d x)-4 \pi \log \left (1+e^{-i (c+d x)}\right )-2 (2 c+2 d x+\pi ) \log \left (1-i e^{i (c+d x)}\right )+2 \pi \log \left (\sin \left (\frac {1}{4} (2 c+2 d x+\pi )\right )\right )+4 \pi \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {2}}\right )}{\sqrt {2}}-4 f \sin \left (\frac {1}{2} (c+d x)\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )-c f \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2 \left (2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+c+d x\right )-c f \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2 \left (-2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+c+d x\right )+2 d (e+f x)}{4 a d^2 (\sin (c+d x)+1)} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.54, size = 508, normalized size = 2.95 \[ -\frac {2 \, d f x + 2 \, d e + 2 \, f \cos \left (d x + c\right ) - {\left (-i \, f \sin \left (d x + c\right ) - i \, f\right )} {\rm Li}_2\left (i \, \cos \left (d x + c\right ) + \sin \left (d x + c\right )\right ) - {\left (-i \, f \sin \left (d x + c\right ) - i \, f\right )} {\rm Li}_2\left (i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right ) - {\left (i \, f \sin \left (d x + c\right ) + i \, f\right )} {\rm Li}_2\left (-i \, \cos \left (d x + c\right ) + \sin \left (d x + c\right )\right ) - {\left (i \, f \sin \left (d x + c\right ) + i \, f\right )} {\rm Li}_2\left (-i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right ) - {\left (d e - c f + {\left (d e - c f\right )} \sin \left (d x + c\right )\right )} \log \left (\cos \left (d x + c\right ) + i \, \sin \left (d x + c\right ) + i\right ) + {\left (d e - c f + {\left (d e - c f\right )} \sin \left (d x + c\right )\right )} \log \left (\cos \left (d x + c\right ) - i \, \sin \left (d x + c\right ) + i\right ) - {\left (d f x + c f + {\left (d f x + c f\right )} \sin \left (d x + c\right )\right )} \log \left (i \, \cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1\right ) + {\left (d f x + c f + {\left (d f x + c f\right )} \sin \left (d x + c\right )\right )} \log \left (i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right ) - {\left (d f x + c f + {\left (d f x + c f\right )} \sin \left (d x + c\right )\right )} \log \left (-i \, \cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1\right ) + {\left (d f x + c f + {\left (d f x + c f\right )} \sin \left (d x + c\right )\right )} \log \left (-i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right ) - {\left (d e - c f + {\left (d e - c f\right )} \sin \left (d x + c\right )\right )} \log \left (-\cos \left (d x + c\right ) + i \, \sin \left (d x + c\right ) + i\right ) + {\left (d e - c f + {\left (d e - c f\right )} \sin \left (d x + c\right )\right )} \log \left (-\cos \left (d x + c\right ) - i \, \sin \left (d x + c\right ) + i\right )}{4 \, {\left (a d^{2} \sin \left (d x + c\right ) + a d^{2}\right )}} \]
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 \[ \int \frac {{\left (f x + e\right )} \sec \left (d x + c\right )}{a \sin \left (d x + c\right ) + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.34, size = 303, normalized size = 1.76 \[ -\frac {i \left (d f x \,{\mathrm e}^{i \left (d x +c \right )}+d e \,{\mathrm e}^{i \left (d x +c \right )}+f -i f \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2} a}-\frac {e \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) e}{2 d a}-\frac {f \ln \left (1+i {\mathrm e}^{i \left (d x +c \right )}\right ) x}{2 a d}-\frac {f \ln \left (1+i {\mathrm e}^{i \left (d x +c \right )}\right ) c}{2 a \,d^{2}}+\frac {i f \polylog \left (2, -i {\mathrm e}^{i \left (d x +c \right )}\right )}{2 a \,d^{2}}+\frac {f \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) x}{2 d a}+\frac {f \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) c}{2 d^{2} a}-\frac {i f \polylog \left (2, i {\mathrm e}^{i \left (d x +c \right )}\right )}{2 a \,d^{2}}+\frac {f c \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a \,d^{2}}-\frac {f c \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d^{2} a} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 1.39, size = 730, normalized size = 4.24 \[ \frac {{\left (2 \, d e \cos \left (2 \, d x + 2 \, c\right ) + 4 i \, d e \cos \left (d x + c\right ) + 2 i \, d e \sin \left (2 \, d x + 2 \, c\right ) - 4 \, d e \sin \left (d x + c\right ) - 2 \, d e\right )} \arctan \left (\sin \left (d x + c\right ) + 1, \cos \left (d x + c\right )\right ) - {\left (2 \, d e \cos \left (2 \, d x + 2 \, c\right ) + 4 i \, d e \cos \left (d x + c\right ) + 2 i \, d e \sin \left (2 \, d x + 2 \, c\right ) - 4 \, d e \sin \left (d x + c\right ) - 2 \, d e\right )} \arctan \left (\sin \left (d x + c\right ) - 1, \cos \left (d x + c\right )\right ) - {\left (2 \, d f x \cos \left (2 \, d x + 2 \, c\right ) + 4 i \, d f x \cos \left (d x + c\right ) + 2 i \, d f x \sin \left (2 \, d x + 2 \, c\right ) - 4 \, d f x \sin \left (d x + c\right ) - 2 \, d f x\right )} \arctan \left (\cos \left (d x + c\right ), \sin \left (d x + c\right ) + 1\right ) - {\left (2 \, d f x \cos \left (2 \, d x + 2 \, c\right ) + 4 i \, d f x \cos \left (d x + c\right ) + 2 i \, d f x \sin \left (2 \, d x + 2 \, c\right ) - 4 \, d f x \sin \left (d x + c\right ) - 2 \, d f x\right )} \arctan \left (\cos \left (d x + c\right ), -\sin \left (d x + c\right ) + 1\right ) - {\left (4 \, d f x + 4 \, d e - 4 i \, f\right )} \cos \left (d x + c\right ) - {\left (2 \, f \cos \left (2 \, d x + 2 \, c\right ) + 4 i \, f \cos \left (d x + c\right ) + 2 i \, f \sin \left (2 \, d x + 2 \, c\right ) - 4 \, f \sin \left (d x + c\right ) - 2 \, f\right )} {\rm Li}_2\left (i \, e^{\left (i \, d x + i \, c\right )}\right ) + {\left (2 \, f \cos \left (2 \, d x + 2 \, c\right ) + 4 i \, f \cos \left (d x + c\right ) + 2 i \, f \sin \left (2 \, d x + 2 \, c\right ) - 4 \, f \sin \left (d x + c\right ) - 2 \, f\right )} {\rm Li}_2\left (-i \, e^{\left (i \, d x + i \, c\right )}\right ) + {\left (i \, d f x + i \, d e + {\left (-i \, d f x - i \, d e\right )} \cos \left (2 \, d x + 2 \, c\right ) + 2 \, {\left (d f x + d e\right )} \cos \left (d x + c\right ) + {\left (d f x + d e\right )} \sin \left (2 \, d x + 2 \, c\right ) + {\left (2 i \, d f x + 2 i \, d e\right )} \sin \left (d x + c\right )\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 ) + {\left (-i \, d f x - i \, d e + {\left (i \, d f x + i \, d e\right )} \cos \left (2 \, d x + 2 \, c\right ) - 2 \, {\left (d f x + d e\right )} \cos \left (d x + c\right ) - {\left (d f x + d e\right )} \sin \left (2 \, d x + 2 \, c\right ) + {\left (-2 i \, d f x - 2 i \, d e\right )} \sin \left (d x + c\right )\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 ) + {\left (-4 i \, d f x - 4 i \, d e - 4 \, f\right )} \sin \left (d x + c\right ) - 4 \, f}{-4 i \, a d^{2} \cos \left (2 \, d x + 2 \, c\right ) + 8 \, a d^{2} \cos \left (d x + c\right ) + 4 \, a d^{2} \sin \left (2 \, d x + 2 \, c\right ) + 8 i \, a d^{2} \sin \left (d x + c\right ) + 4 i \, a d^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F(-1)] time = 0.00, size = -1, normalized size = -0.01 \[ \text {Hanged} \]
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 \[ \frac {\int \frac {e \sec {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {f x \sec {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \]
Verification of antiderivative is not currently implemented for this CAS.
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