Optimal. Leaf size=343 \[ -\frac {2 i (e+f x)^2}{3 a d}-\frac {2 i f (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{3 a d^2}+\frac {4 f (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{3 a d^2}+\frac {i f^2 \text {Li}_2\left (-i e^{i (c+d x)}\right )}{3 a d^3}-\frac {i f^2 \text {Li}_2\left (i e^{i (c+d x)}\right )}{3 a d^3}-\frac {2 i f^2 \text {Li}_2\left (-e^{2 i (c+d x)}\right )}{3 a d^3}-\frac {f^2 \sec (c+d x)}{3 a d^3}-\frac {f (e+f x) \sec ^2(c+d x)}{3 a d^2}-\frac {(e+f x)^2 \sec ^3(c+d x)}{3 a d}+\frac {f^2 \tan (c+d x)}{3 a d^3}+\frac {2 (e+f x)^2 \tan (c+d x)}{3 a d}+\frac {f (e+f x) \sec (c+d x) \tan (c+d x)}{3 a d^2}+\frac {(e+f x)^2 \sec ^2(c+d x) \tan (c+d x)}{3 a d} \]
[Out]
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Rubi [A]
time = 0.25, antiderivative size = 343, normalized size of antiderivative = 1.00, number of steps
used = 16, number of rules used = 12, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {4627, 4271,
3852, 8, 4269, 3800, 2221, 2317, 2438, 4494, 4270, 4266} \begin {gather*} \frac {i f^2 \text {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{3 a d^3}-\frac {i f^2 \text {PolyLog}\left (2,i e^{i (c+d x)}\right )}{3 a d^3}-\frac {2 i f^2 \text {PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{3 a d^3}-\frac {2 i f (e+f x) \text {ArcTan}\left (e^{i (c+d x)}\right )}{3 a d^2}+\frac {f^2 \tan (c+d x)}{3 a d^3}-\frac {f^2 \sec (c+d x)}{3 a d^3}+\frac {4 f (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{3 a d^2}-\frac {f (e+f x) \sec ^2(c+d x)}{3 a d^2}+\frac {f (e+f x) \tan (c+d x) \sec (c+d x)}{3 a d^2}+\frac {2 (e+f x)^2 \tan (c+d x)}{3 a d}-\frac {(e+f x)^2 \sec ^3(c+d x)}{3 a d}+\frac {(e+f x)^2 \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {2 i (e+f x)^2}{3 a d} \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
Rule 8
Rule 2221
Rule 2317
Rule 2438
Rule 3800
Rule 3852
Rule 4266
Rule 4269
Rule 4270
Rule 4271
Rule 4494
Rule 4627
Rubi steps
\begin {align*} \int \frac {(e+f x)^2 \sec ^2(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac {\int (e+f x)^2 \sec ^4(c+d x) \, dx}{a}-\frac {\int (e+f x)^2 \sec ^3(c+d x) \tan (c+d x) \, dx}{a}\\ &=-\frac {f (e+f x) \sec ^2(c+d x)}{3 a d^2}-\frac {(e+f x)^2 \sec ^3(c+d x)}{3 a d}+\frac {(e+f x)^2 \sec ^2(c+d x) \tan (c+d x)}{3 a d}+\frac {2 \int (e+f x)^2 \sec ^2(c+d x) \, dx}{3 a}+\frac {(2 f) \int (e+f x) \sec ^3(c+d x) \, dx}{3 a d}+\frac {f^2 \int \sec ^2(c+d x) \, dx}{3 a d^2}\\ &=-\frac {f^2 \sec (c+d x)}{3 a d^3}-\frac {f (e+f x) \sec ^2(c+d x)}{3 a d^2}-\frac {(e+f x)^2 \sec ^3(c+d x)}{3 a d}+\frac {2 (e+f x)^2 \tan (c+d x)}{3 a d}+\frac {f (e+f x) \sec (c+d x) \tan (c+d x)}{3 a d^2}+\frac {(e+f x)^2 \sec ^2(c+d x) \tan (c+d x)}{3 a d}+\frac {f \int (e+f x) \sec (c+d x) \, dx}{3 a d}-\frac {(4 f) \int (e+f x) \tan (c+d x) \, dx}{3 a d}-\frac {f^2 \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 a d^3}\\ &=-\frac {2 i (e+f x)^2}{3 a d}-\frac {2 i f (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{3 a d^2}-\frac {f^2 \sec (c+d x)}{3 a d^3}-\frac {f (e+f x) \sec ^2(c+d x)}{3 a d^2}-\frac {(e+f x)^2 \sec ^3(c+d x)}{3 a d}+\frac {f^2 \tan (c+d x)}{3 a d^3}+\frac {2 (e+f x)^2 \tan (c+d x)}{3 a d}+\frac {f (e+f x) \sec (c+d x) \tan (c+d x)}{3 a d^2}+\frac {(e+f x)^2 \sec ^2(c+d x) \tan (c+d x)}{3 a d}+\frac {(8 i f) \int \frac {e^{2 i (c+d x)} (e+f x)}{1+e^{2 i (c+d x)}} \, dx}{3 a d}-\frac {f^2 \int \log \left (1-i e^{i (c+d x)}\right ) \, dx}{3 a d^2}+\frac {f^2 \int \log \left (1+i e^{i (c+d x)}\right ) \, dx}{3 a d^2}\\ &=-\frac {2 i (e+f x)^2}{3 a d}-\frac {2 i f (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{3 a d^2}+\frac {4 f (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{3 a d^2}-\frac {f^2 \sec (c+d x)}{3 a d^3}-\frac {f (e+f x) \sec ^2(c+d x)}{3 a d^2}-\frac {(e+f x)^2 \sec ^3(c+d x)}{3 a d}+\frac {f^2 \tan (c+d x)}{3 a d^3}+\frac {2 (e+f x)^2 \tan (c+d x)}{3 a d}+\frac {f (e+f x) \sec (c+d x) \tan (c+d x)}{3 a d^2}+\frac {(e+f x)^2 \sec ^2(c+d x) \tan (c+d x)}{3 a d}+\frac {\left (i f^2\right ) \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i (c+d x)}\right )}{3 a d^3}-\frac {\left (i f^2\right ) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{i (c+d x)}\right )}{3 a d^3}-\frac {\left (4 f^2\right ) \int \log \left (1+e^{2 i (c+d x)}\right ) \, dx}{3 a d^2}\\ &=-\frac {2 i (e+f x)^2}{3 a d}-\frac {2 i f (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{3 a d^2}+\frac {4 f (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{3 a d^2}+\frac {i f^2 \text {Li}_2\left (-i e^{i (c+d x)}\right )}{3 a d^3}-\frac {i f^2 \text {Li}_2\left (i e^{i (c+d x)}\right )}{3 a d^3}-\frac {f^2 \sec (c+d x)}{3 a d^3}-\frac {f (e+f x) \sec ^2(c+d x)}{3 a d^2}-\frac {(e+f x)^2 \sec ^3(c+d x)}{3 a d}+\frac {f^2 \tan (c+d x)}{3 a d^3}+\frac {2 (e+f x)^2 \tan (c+d x)}{3 a d}+\frac {f (e+f x) \sec (c+d x) \tan (c+d x)}{3 a d^2}+\frac {(e+f x)^2 \sec ^2(c+d x) \tan (c+d x)}{3 a d}+\frac {\left (2 i f^2\right ) \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i (c+d x)}\right )}{3 a d^3}\\ &=-\frac {2 i (e+f x)^2}{3 a d}-\frac {2 i f (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{3 a d^2}+\frac {4 f (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{3 a d^2}+\frac {i f^2 \text {Li}_2\left (-i e^{i (c+d x)}\right )}{3 a d^3}-\frac {i f^2 \text {Li}_2\left (i e^{i (c+d x)}\right )}{3 a d^3}-\frac {2 i f^2 \text {Li}_2\left (-e^{2 i (c+d x)}\right )}{3 a d^3}-\frac {f^2 \sec (c+d x)}{3 a d^3}-\frac {f (e+f x) \sec ^2(c+d x)}{3 a d^2}-\frac {(e+f x)^2 \sec ^3(c+d x)}{3 a d}+\frac {f^2 \tan (c+d x)}{3 a d^3}+\frac {2 (e+f x)^2 \tan (c+d x)}{3 a d}+\frac {f (e+f x) \sec (c+d x) \tan (c+d x)}{3 a d^2}+\frac {(e+f x)^2 \sec ^2(c+d x) \tan (c+d x)}{3 a d}\\ \end {align*}
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Mathematica [A]
time = 5.16, size = 564, normalized size = 1.64 \begin {gather*} \frac {12 d f (e+f x) \log (1+i \cos (c+d x)-\sin (c+d x))+20 d f (e+f x) \log (1-i \cos (c+d x)+\sin (c+d x))-20 i f^2 \text {Li}_2(i \cos (c+d x)-\sin (c+d x))-12 i f^2 \text {Li}_2(-i \cos (c+d x)+\sin (c+d x))+\frac {6 d^2 f x (2 e+f x) (-i \cos (c)+\sin (c))}{\cos (c)+i (-1+\sin (c))}+\frac {10 d^2 f x (2 e+f x) (-i \cos (c)+\sin (c))}{\cos (c)+i (1+\sin (c))}+\frac {-2 f^2 \cos (c)-2 d f (e+f x) \cos (d x)+2 d^2 e^2 \cos (c+d x)+4 f^2 \cos (c+d x)+4 d^2 e f x \cos (c+d x)+2 d^2 f^2 x^2 \cos (c+d x)-2 d e f \cos (2 c+d x)-2 d f^2 x \cos (2 c+d x)-4 d^2 e^2 \cos (c+2 d x)-2 f^2 \cos (c+2 d x)-8 d^2 e f x \cos (c+2 d x)-4 d^2 f^2 x^2 \cos (c+2 d x)+8 d^2 e^2 \sin (d x)+2 f^2 \sin (d x)+16 d^2 e f x \sin (d x)+8 d^2 f^2 x^2 \sin (d x)+d^2 e^2 \sin (2 (c+d x))+2 f^2 \sin (2 (c+d x))+2 d^2 e f x \sin (2 (c+d x))+d^2 f^2 x^2 \sin (2 (c+d x))-2 f^2 \sin (2 c+d x)}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}}{12 a d^3} \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
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Maple [A]
time = 0.24, size = 573, normalized size = 1.67
method | result | size |
risch | \(-\frac {2 \left (i f^{2}+f^{2} {\mathrm e}^{3 i \left (d x +c \right )}+4 d^{2} f^{2} x^{2} {\mathrm e}^{i \left (d x +c \right )}+2 i d^{2} x^{2} f^{2}+i f^{2} {\mathrm e}^{2 i \left (d x +c \right )}+8 d^{2} e f x \,{\mathrm e}^{i \left (d x +c \right )}+i d \,f^{2} x \,{\mathrm e}^{i \left (d x +c \right )}+i d e f \,{\mathrm e}^{i \left (d x +c \right )}+i d \,f^{2} x \,{\mathrm e}^{3 i \left (d x +c \right )}+i d e f \,{\mathrm e}^{3 i \left (d x +c \right )}+4 i d^{2} e f x +f^{2} {\mathrm e}^{i \left (d x +c \right )}+4 d^{2} e^{2} {\mathrm e}^{i \left (d x +c \right )}+2 i d^{2} e^{2}\right )}{3 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{3} d^{3} a}+\frac {e f \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d^{2} a}+\frac {5 f \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) e}{3 a \,d^{2}}-\frac {8 f \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right ) e}{3 a \,d^{2}}-\frac {f^{2} c \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d^{3} a}-\frac {5 f^{2} c \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{3 a \,d^{3}}+\frac {8 f^{2} c \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{3 a \,d^{3}}-\frac {4 i f^{2} c^{2}}{3 a \,d^{3}}-\frac {8 i f^{2} c x}{3 a \,d^{2}}-\frac {5 i f^{2} \polylog \left (2, i {\mathrm e}^{i \left (d x +c \right )}\right )}{3 a \,d^{3}}+\frac {f^{2} \ln \left (1+i {\mathrm e}^{i \left (d x +c \right )}\right ) x}{d^{2} a}+\frac {f^{2} \ln \left (1+i {\mathrm e}^{i \left (d x +c \right )}\right ) c}{d^{3} a}-\frac {4 i f^{2} x^{2}}{3 a d}+\frac {5 f^{2} \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) x}{3 a \,d^{2}}+\frac {5 f^{2} \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) c}{3 a \,d^{3}}-\frac {i f^{2} \polylog \left (2, -i {\mathrm e}^{i \left (d x +c \right )}\right )}{d^{3} a}\) | \(573\) |
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than
twice the leaf count of optimal. 1328 vs. \(2 (290) = 580\).
time = 0.64, size = 1328, normalized size = 3.87 \begin {gather*} -\frac {8 \, d^{2} e^{2} + 4 \, f^{2} \cos \left (2 \, d x + 2 \, c\right ) + 4 i \, f^{2} \sin \left (2 \, d x + 2 \, c\right ) + 4 \, f^{2} - 10 \, {\left (d e f \cos \left (4 \, d x + 4 \, c\right ) + 2 i \, d e f \cos \left (3 \, d x + 3 \, c\right ) + 2 i \, d e f \cos \left (d x + c\right ) + i \, d e f \sin \left (4 \, d x + 4 \, c\right ) - 2 \, d e f \sin \left (3 \, d x + 3 \, c\right ) - 2 \, d e f \sin \left (d x + c\right ) - d e f\right )} \arctan \left (\sin \left (d x + c\right ) + 1, \cos \left (d x + c\right )\right ) - 6 \, {\left (d e f \cos \left (4 \, d x + 4 \, c\right ) + 2 i \, d e f \cos \left (3 \, d x + 3 \, c\right ) + 2 i \, d e f \cos \left (d x + c\right ) + i \, d e f \sin \left (4 \, d x + 4 \, c\right ) - 2 \, d e f \sin \left (3 \, d x + 3 \, c\right ) - 2 \, d e f \sin \left (d x + c\right ) - d e f\right )} \arctan \left (\sin \left (d x + c\right ) - 1, \cos \left (d x + c\right )\right ) + 10 \, {\left (d f^{2} x \cos \left (4 \, d x + 4 \, c\right ) + 2 i \, d f^{2} x \cos \left (3 \, d x + 3 \, c\right ) + 2 i \, d f^{2} x \cos \left (d x + c\right ) + i \, d f^{2} x \sin \left (4 \, d x + 4 \, c\right ) - 2 \, d f^{2} x \sin \left (3 \, d x + 3 \, c\right ) - 2 \, d f^{2} x \sin \left (d x + c\right ) - d f^{2} x\right )} \arctan \left (\cos \left (d x + c\right ), \sin \left (d x + c\right ) + 1\right ) - 6 \, {\left (d f^{2} x \cos \left (4 \, d x + 4 \, c\right ) + 2 i \, d f^{2} x \cos \left (3 \, d x + 3 \, c\right ) + 2 i \, d f^{2} x \cos \left (d x + c\right ) + i \, d f^{2} x \sin \left (4 \, d x + 4 \, c\right ) - 2 \, d f^{2} x \sin \left (3 \, d x + 3 \, c\right ) - 2 \, d f^{2} x \sin \left (d x + c\right ) - d f^{2} x\right )} \arctan \left (\cos \left (d x + c\right ), -\sin \left (d x + c\right ) + 1\right ) + 8 \, {\left (d^{2} f^{2} x^{2} + 2 \, d^{2} e f x\right )} \cos \left (4 \, d x + 4 \, c\right ) + 4 \, {\left (4 i \, d^{2} f^{2} x^{2} + d e f - i \, f^{2} + {\left (8 i \, d^{2} e f + d f^{2}\right )} x\right )} \cos \left (3 \, d x + 3 \, c\right ) + 4 \, {\left (-4 i \, d^{2} e^{2} + d f^{2} x + d e f - i \, f^{2}\right )} \cos \left (d x + c\right ) + 10 \, {\left (f^{2} \cos \left (4 \, d x + 4 \, c\right ) + 2 i \, f^{2} \cos \left (3 \, d x + 3 \, c\right ) + 2 i \, f^{2} \cos \left (d x + c\right ) + i \, f^{2} \sin \left (4 \, d x + 4 \, c\right ) - 2 \, f^{2} \sin \left (3 \, d x + 3 \, c\right ) - 2 \, f^{2} \sin \left (d x + c\right ) - f^{2}\right )} {\rm Li}_2\left (i \, e^{\left (i \, d x + i \, c\right )}\right ) + 6 \, {\left (f^{2} \cos \left (4 \, d x + 4 \, c\right ) + 2 i \, f^{2} \cos \left (3 \, d x + 3 \, c\right ) + 2 i \, f^{2} \cos \left (d x + c\right ) + i \, f^{2} \sin \left (4 \, d x + 4 \, c\right ) - 2 \, f^{2} \sin \left (3 \, d x + 3 \, c\right ) - 2 \, f^{2} \sin \left (d x + c\right ) - f^{2}\right )} {\rm Li}_2\left (-i \, e^{\left (i \, d x + i \, c\right )}\right ) + 5 \, {\left (-i \, d f^{2} x - i \, d e f + {\left (i \, d f^{2} x + i \, d e f\right )} \cos \left (4 \, d x + 4 \, c\right ) - 2 \, {\left (d f^{2} x + d e f\right )} \cos \left (3 \, d x + 3 \, c\right ) - 2 \, {\left (d f^{2} x + d e f\right )} \cos \left (d x + c\right ) - {\left (d f^{2} x + d e f\right )} \sin \left (4 \, d x + 4 \, c\right ) + 2 \, {\left (-i \, d f^{2} x - i \, d e f\right )} \sin \left (3 \, d x + 3 \, c\right ) + 2 \, {\left (-i \, d f^{2} x - i \, d e f\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 ) + 3 \, {\left (-i \, d f^{2} x - i \, d e f + {\left (i \, d f^{2} x + i \, d e f\right )} \cos \left (4 \, d x + 4 \, c\right ) - 2 \, {\left (d f^{2} x + d e f\right )} \cos \left (3 \, d x + 3 \, c\right ) - 2 \, {\left (d f^{2} x + d e f\right )} \cos \left (d x + c\right ) - {\left (d f^{2} x + d e f\right )} \sin \left (4 \, d x + 4 \, c\right ) + 2 \, {\left (-i \, d f^{2} x - i \, d e f\right )} \sin \left (3 \, d x + 3 \, c\right ) + 2 \, {\left (-i \, d f^{2} x - i \, d e f\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 ) + 8 \, {\left (i \, d^{2} f^{2} x^{2} + 2 i \, d^{2} e f x\right )} \sin \left (4 \, d x + 4 \, c\right ) - 4 \, {\left (4 \, d^{2} f^{2} x^{2} - i \, d e f - f^{2} + {\left (8 \, d^{2} e f - i \, d f^{2}\right )} x\right )} \sin \left (3 \, d x + 3 \, c\right ) + 4 \, {\left (4 \, d^{2} e^{2} + i \, d f^{2} x + i \, d e f + f^{2}\right )} \sin \left (d x + c\right )}{-6 i \, a d^{3} \cos \left (4 \, d x + 4 \, c\right ) + 12 \, a d^{3} \cos \left (3 \, d x + 3 \, c\right ) + 12 \, a d^{3} \cos \left (d x + c\right ) + 6 \, a d^{3} \sin \left (4 \, d x + 4 \, c\right ) + 12 i \, a d^{3} \sin \left (3 \, d x + 3 \, c\right ) + 12 i \, a d^{3} \sin \left (d x + c\right ) + 6 i \, a d^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice
the leaf count of optimal. 868 vs. \(2 (298) = 596\).
time = 0.44, size = 868, normalized size = 2.53 \begin {gather*} \frac {2 \, d^{2} f^{2} x^{2} + 4 \, d^{2} f x e - 2 \, {\left (2 \, d^{2} f^{2} x^{2} + 4 \, d^{2} f x e + 2 \, d^{2} e^{2} + f^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, d^{2} e^{2} - 2 \, {\left (d f^{2} x + d f e\right )} \cos \left (d x + c\right ) - 3 \, {\left (-i \, f^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - i \, f^{2} \cos \left (d x + c\right )\right )} {\rm Li}_2\left (i \, \cos \left (d x + c\right ) + \sin \left (d x + c\right )\right ) - 5 \, {\left (i \, f^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + i \, f^{2} \cos \left (d x + c\right )\right )} {\rm Li}_2\left (i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right ) - 3 \, {\left (i \, f^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + i \, f^{2} \cos \left (d x + c\right )\right )} {\rm Li}_2\left (-i \, \cos \left (d x + c\right ) + \sin \left (d x + c\right )\right ) - 5 \, {\left (-i \, f^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - i \, f^{2} \cos \left (d x + c\right )\right )} {\rm Li}_2\left (-i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right ) - 5 \, {\left ({\left (c f^{2} - d f e\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (c f^{2} - d f e\right )} \cos \left (d x + c\right )\right )} \log \left (\cos \left (d x + c\right ) + i \, \sin \left (d x + c\right ) + i\right ) - 3 \, {\left ({\left (c f^{2} - d f e\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (c f^{2} - d f e\right )} \cos \left (d x + c\right )\right )} \log \left (\cos \left (d x + c\right ) - i \, \sin \left (d x + c\right ) + i\right ) + 5 \, {\left ({\left (d f^{2} x + c f^{2}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (d f^{2} x + c f^{2}\right )} \cos \left (d x + c\right )\right )} \log \left (i \, \cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1\right ) + 3 \, {\left ({\left (d f^{2} x + c f^{2}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (d f^{2} x + c f^{2}\right )} \cos \left (d x + c\right )\right )} \log \left (i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right ) + 5 \, {\left ({\left (d f^{2} x + c f^{2}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (d f^{2} x + c f^{2}\right )} \cos \left (d x + c\right )\right )} \log \left (-i \, \cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1\right ) + 3 \, {\left ({\left (d f^{2} x + c f^{2}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (d f^{2} x + c f^{2}\right )} \cos \left (d x + c\right )\right )} \log \left (-i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right ) - 5 \, {\left ({\left (c f^{2} - d f e\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (c f^{2} - d f e\right )} \cos \left (d x + c\right )\right )} \log \left (-\cos \left (d x + c\right ) + i \, \sin \left (d x + c\right ) + i\right ) - 3 \, {\left ({\left (c f^{2} - d f e\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (c f^{2} - d f e\right )} \cos \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 (d^{2} f^{2} x^{2} + 2 \, d^{2} f x e + d^{2} e^{2}\right )} \sin \left (d x + c\right )}{6 \, {\left (a d^{3} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a d^{3} \cos \left (d x + c\right )\right )}} \end {gather*}
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 \begin {gather*} \frac {\int \frac {e^{2} \sec ^{2}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {f^{2} x^{2} \sec ^{2}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {2 e f x \sec ^{2}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F(-1)]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \text {Hanged} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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