Optimal. Leaf size=112 \[ \frac {B+i A}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {x (A-i B)}{8 a^3}+\frac {-B+i A}{6 f (a+i a \tan (e+f x))^3}+\frac {B+i A}{8 a f (a+i a \tan (e+f x))^2} \]
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Rubi [A] time = 0.09, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {3526, 3479, 8} \[ \frac {B+i A}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {x (A-i B)}{8 a^3}+\frac {-B+i A}{6 f (a+i a \tan (e+f x))^3}+\frac {B+i A}{8 a f (a+i a \tan (e+f x))^2} \]
Antiderivative was successfully verified.
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Rule 8
Rule 3479
Rule 3526
Rubi steps
\begin {align*} \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3} \, dx &=\frac {i A-B}{6 f (a+i a \tan (e+f x))^3}+\frac {(A-i B) \int \frac {1}{(a+i a \tan (e+f x))^2} \, dx}{2 a}\\ &=\frac {i A-B}{6 f (a+i a \tan (e+f x))^3}+\frac {i A+B}{8 a f (a+i a \tan (e+f x))^2}+\frac {(A-i B) \int \frac {1}{a+i a \tan (e+f x)} \, dx}{4 a^2}\\ &=\frac {i A-B}{6 f (a+i a \tan (e+f x))^3}+\frac {i A+B}{8 a f (a+i a \tan (e+f x))^2}+\frac {i A+B}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {(A-i B) \int 1 \, dx}{8 a^3}\\ &=\frac {(A-i B) x}{8 a^3}+\frac {i A-B}{6 f (a+i a \tan (e+f x))^3}+\frac {i A+B}{8 a f (a+i a \tan (e+f x))^2}+\frac {i A+B}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}\\ \end {align*}
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Mathematica [A] time = 0.79, size = 150, normalized size = 1.34 \[ \frac {\sec ^3(e+f x) ((-27 A+3 i B) \cos (e+f x)+2 (6 i A f x-A+6 B f x-i B) \cos (3 (e+f x))-9 i A \sin (e+f x)+2 i A \sin (3 (e+f x))-12 A f x \sin (3 (e+f x))-9 B \sin (e+f x)-2 B \sin (3 (e+f x))+12 i B f x \sin (3 (e+f x)))}{96 a^3 f (\tan (e+f x)-i)^3} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.82, size = 76, normalized size = 0.68 \[ \frac {{\left (12 \, {\left (A - i \, B\right )} f x e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (18 i \, A + 6 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (9 i \, A - 3 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i \, A - 2 \, B\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{96 \, a^{3} f} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 2.04, size = 140, normalized size = 1.25 \[ -\frac {\frac {6 \, {\left (i \, A + B\right )} \log \left (\tan \left (f x + e\right ) - i\right )}{a^{3}} + \frac {6 \, {\left (-i \, A - B\right )} \log \left (i \, \tan \left (f x + e\right ) - 1\right )}{a^{3}} + \frac {-11 i \, A \tan \left (f x + e\right )^{3} - 11 \, B \tan \left (f x + e\right )^{3} - 45 \, A \tan \left (f x + e\right )^{2} + 45 i \, B \tan \left (f x + e\right )^{2} + 69 i \, A \tan \left (f x + e\right ) + 69 \, B \tan \left (f x + e\right ) + 51 \, A - 19 i \, B}{a^{3} {\left (\tan \left (f x + e\right ) - i\right )}^{3}}}{96 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.25, size = 203, normalized size = 1.81 \[ \frac {B \ln \left (\tan \left (f x +e \right )+i\right )}{16 f \,a^{3}}+\frac {i A \ln \left (\tan \left (f x +e \right )+i\right )}{16 f \,a^{3}}-\frac {A}{6 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{3}}-\frac {i B}{6 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{3}}+\frac {A}{8 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )}-\frac {i B}{8 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )}-\frac {i A}{8 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {B}{8 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {i \ln \left (\tan \left (f x +e \right )-i\right ) A}{16 f \,a^{3}}-\frac {\ln \left (\tan \left (f x +e \right )-i\right ) B}{16 f \,a^{3}} \]
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 = 9.13, size = 111, normalized size = 0.99 \[ -\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (\frac {B}{8\,a^3}+\frac {A\,1{}\mathrm {i}}{8\,a^3}\right )-\frac {A\,5{}\mathrm {i}}{12\,a^3}-\frac {B}{12\,a^3}+\mathrm {tan}\left (e+f\,x\right )\,\left (\frac {3\,A}{8\,a^3}-\frac {B\,3{}\mathrm {i}}{8\,a^3}\right )}{f\,\left (-{\mathrm {tan}\left (e+f\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (e+f\,x\right )}^2+\mathrm {tan}\left (e+f\,x\right )\,3{}\mathrm {i}+1\right )}-\frac {x\,\left (B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{8\,a^3} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.51, size = 264, normalized size = 2.36 \[ \begin {cases} - \frac {\left (\left (- 512 i A a^{6} f^{2} e^{6 i e} + 512 B a^{6} f^{2} e^{6 i e}\right ) e^{- 6 i f x} + \left (- 2304 i A a^{6} f^{2} e^{8 i e} + 768 B a^{6} f^{2} e^{8 i e}\right ) e^{- 4 i f x} + \left (- 4608 i A a^{6} f^{2} e^{10 i e} - 1536 B a^{6} f^{2} e^{10 i e}\right ) e^{- 2 i f x}\right ) e^{- 12 i e}}{24576 a^{9} f^{3}} & \text {for}\: 24576 a^{9} f^{3} e^{12 i e} \neq 0 \\x \left (- \frac {A - i B}{8 a^{3}} + \frac {\left (A e^{6 i e} + 3 A e^{4 i e} + 3 A e^{2 i e} + A - i B e^{6 i e} - i B e^{4 i e} + i B e^{2 i e} + i B\right ) e^{- 6 i e}}{8 a^{3}}\right ) & \text {otherwise} \end {cases} - \frac {x \left (- A + i B\right )}{8 a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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