Optimal. Leaf size=104 \[ \frac {A (d+i c)+B (c+3 i d)}{4 a^2 f (1+i \tan (e+f x))}+\frac {x (A-i B) (c-i d)}{4 a^2}+\frac {(-B+i A) (c+i d)}{4 f (a+i a \tan (e+f x))^2} \]
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Rubi [A] time = 0.24, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3590, 3526, 8} \[ \frac {A (d+i c)+B (c+3 i d)}{4 a^2 f (1+i \tan (e+f x))}+\frac {x (A-i B) (c-i d)}{4 a^2}+\frac {(-B+i A) (c+i d)}{4 f (a+i a \tan (e+f x))^2} \]
Antiderivative was successfully verified.
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Rule 8
Rule 3526
Rule 3590
Rubi steps
\begin {align*} \int \frac {(A+B \tan (e+f x)) (c+d \tan (e+f x))}{(a+i a \tan (e+f x))^2} \, dx &=\frac {(i A-B) (c+i d)}{4 f (a+i a \tan (e+f x))^2}-\frac {i \int \frac {a (B (c+i d)+A (i c+d))+2 a B d \tan (e+f x)}{a+i a \tan (e+f x)} \, dx}{2 a^2}\\ &=\frac {B (c+3 i d)+A (i c+d)}{4 a^2 f (1+i \tan (e+f x))}+\frac {(i A-B) (c+i d)}{4 f (a+i a \tan (e+f x))^2}+\frac {((A-i B) (c-i d)) \int 1 \, dx}{4 a^2}\\ &=\frac {(A-i B) (c-i d) x}{4 a^2}+\frac {B (c+3 i d)+A (i c+d)}{4 a^2 f (1+i \tan (e+f x))}+\frac {(i A-B) (c+i d)}{4 f (a+i a \tan (e+f x))^2}\\ \end {align*}
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Mathematica [A] time = 1.89, size = 201, normalized size = 1.93 \[ -\frac {(A+B \tan (e+f x)) (c+d \tan (e+f x)) (\sin (2 (e+f x)) (A (4 i c f x+c+4 d f x+i d)+B (4 c f x+i c-4 i d f x-d))+\cos (2 (e+f x)) (A (c (4 f x+i)+d (-1-4 i f x))-B (4 i c f x+c+d (4 f x+i)))+4 i (A c+B d))}{16 a^2 f (\tan (e+f x)-i)^2 (A \cos (e+f x)+B \sin (e+f x)) (c \cos (e+f x)+d \sin (e+f x))} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.72, size = 84, normalized size = 0.81 \[ \frac {{\left ({\left (4 \, {\left (A - i \, B\right )} c + {\left (-4 i \, A - 4 \, B\right )} d\right )} f x e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (i \, A - B\right )} c - {\left (A + i \, B\right )} d + {\left (4 i \, A c + 4 i \, B d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{16 \, a^{2} f} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 1.74, size = 200, normalized size = 1.92 \[ -\frac {\frac {2 \, {\left (-i \, A c - B c - A d + i \, B d\right )} \log \left (-i \, \tan \left (f x + e\right ) + 1\right )}{a^{2}} + \frac {2 \, {\left (i \, A c + B c + A d - i \, B d\right )} \log \left (-i \, \tan \left (f x + e\right ) - 1\right )}{a^{2}} + \frac {-3 i \, A c \tan \left (f x + e\right )^{2} - 3 \, B c \tan \left (f x + e\right )^{2} - 3 \, A d \tan \left (f x + e\right )^{2} + 3 i \, B d \tan \left (f x + e\right )^{2} - 10 \, A c \tan \left (f x + e\right ) + 10 i \, B c \tan \left (f x + e\right ) + 10 i \, A d \tan \left (f x + e\right ) - 6 \, B d \tan \left (f x + e\right ) + 11 i \, A c + 3 \, B c + 3 \, A d + 5 i \, B d}{a^{2} {\left (\tan \left (f x + e\right ) - i\right )}^{2}}}{16 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.26, size = 338, normalized size = 3.25 \[ \frac {A \ln \left (\tan \left (f x +e \right )+i\right ) d}{8 f \,a^{2}}+\frac {B \ln \left (\tan \left (f x +e \right )+i\right ) c}{8 f \,a^{2}}-\frac {i B \ln \left (\tan \left (f x +e \right )+i\right ) d}{8 f \,a^{2}}-\frac {i A c}{4 f \,a^{2} \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {i A d}{4 f \,a^{2} \left (\tan \left (f x +e \right )-i\right )}+\frac {i \ln \left (\tan \left (f x +e \right )-i\right ) B d}{8 f \,a^{2}}+\frac {c A}{4 f \,a^{2} \left (\tan \left (f x +e \right )-i\right )}+\frac {3 B d}{4 f \,a^{2} \left (\tan \left (f x +e \right )-i\right )}+\frac {A d}{4 f \,a^{2} \left (\tan \left (f x +e \right )-i\right )^{2}}+\frac {B c}{4 f \,a^{2} \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {i B c}{4 f \,a^{2} \left (\tan \left (f x +e \right )-i\right )}+\frac {i A \ln \left (\tan \left (f x +e \right )+i\right ) c}{8 f \,a^{2}}+\frac {i B d}{4 f \,a^{2} \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {i \ln \left (\tan \left (f x +e \right )-i\right ) A c}{8 f \,a^{2}}-\frac {\ln \left (\tan \left (f x +e \right )-i\right ) A d}{8 f \,a^{2}}-\frac {\ln \left (\tan \left (f x +e \right )-i\right ) B c}{8 f \,a^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 9.50, size = 159, normalized size = 1.53 \[ -\frac {B\,d\,f\,x-A\,c\,f\,x+A\,d\,f\,x\,1{}\mathrm {i}+B\,c\,f\,x\,1{}\mathrm {i}}{4\,a^2\,f}+\frac {\left (A\,c+3\,B\,d-A\,d\,1{}\mathrm {i}-B\,c\,1{}\mathrm {i}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^3+\left (2\,A\,d+2\,B\,c+B\,d\,4{}\mathrm {i}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^2+\left (3\,A\,c+B\,d+A\,d\,1{}\mathrm {i}+B\,c\,1{}\mathrm {i}\right )\,\mathrm {tan}\left (e+f\,x\right )+A\,c\,2{}\mathrm {i}+B\,d\,2{}\mathrm {i}}{f\,\left (4\,a^2\,{\mathrm {tan}\left (e+f\,x\right )}^4+8\,a^2\,{\mathrm {tan}\left (e+f\,x\right )}^2+4\,a^2\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.55, size = 298, normalized size = 2.87 \[ \begin {cases} \frac {\left (\left (16 i A a^{2} c f e^{4 i e} + 16 i B a^{2} d f e^{4 i e}\right ) e^{- 2 i f x} + \left (4 i A a^{2} c f e^{2 i e} - 4 A a^{2} d f e^{2 i e} - 4 B a^{2} c f e^{2 i e} - 4 i B a^{2} d f e^{2 i e}\right ) e^{- 4 i f x}\right ) e^{- 6 i e}}{64 a^{4} f^{2}} & \text {for}\: 64 a^{4} f^{2} e^{6 i e} \neq 0 \\x \left (- \frac {A c - i A d - i B c - B d}{4 a^{2}} + \frac {\left (A c e^{4 i e} + 2 A c e^{2 i e} + A c - i A d e^{4 i e} + i A d - i B c e^{4 i e} + i B c - B d e^{4 i e} + 2 B d e^{2 i e} - B d\right ) e^{- 4 i e}}{4 a^{2}}\right ) & \text {otherwise} \end {cases} - \frac {x \left (- A c + i A d + i B c + B d\right )}{4 a^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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