Optimal. Leaf size=112 \[ \frac {(A-i B) x}{8 a^3}+\frac {i A-B}{6 f (a+i a \tan (e+f x))^3}+\frac {i A+B}{8 a f (a+i a \tan (e+f x))^2}+\frac {i A+B}{8 f \left (a^3+i a^3 \tan (e+f x)\right )} \]
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Rubi [A]
time = 0.06, 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 = {3607, 3560, 8}
\begin {gather*} \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} \end {gather*}
Antiderivative was successfully verified.
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Rule 8
Rule 3560
Rule 3607
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 = 0.47, size = 150, normalized size = 1.34 \begin {gather*} \frac {\sec ^3(e+f x) ((-27 A+3 i B) \cos (e+f x)+2 (-A-i B+6 i A f x+6 B f x) \cos (3 (e+f x))-9 i A \sin (e+f x)-9 B \sin (e+f x)+2 i A \sin (3 (e+f x))-2 B \sin (3 (e+f x))-12 A f x \sin (3 (e+f x))+12 i B f x \sin (3 (e+f x)))}{96 a^3 f (-i+\tan (e+f x))^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.22, size = 110, normalized size = 0.98
method | result | size |
derivativedivides | \(\frac {\left (-\frac {i A}{16}-\frac {B}{16}\right ) \ln \left (-i+\tan \left (f x +e \right )\right )-\frac {-\frac {A}{8}+\frac {i B}{8}}{-i+\tan \left (f x +e \right )}-\frac {\frac {i A}{4}+\frac {B}{4}}{2 \left (-i+\tan \left (f x +e \right )\right )^{2}}-\frac {\frac {A}{2}+\frac {i B}{2}}{3 \left (-i+\tan \left (f x +e \right )\right )^{3}}+\frac {i \left (-i B +A \right ) \ln \left (i+\tan \left (f x +e \right )\right )}{16}}{f \,a^{3}}\) | \(110\) |
default | \(\frac {\left (-\frac {i A}{16}-\frac {B}{16}\right ) \ln \left (-i+\tan \left (f x +e \right )\right )-\frac {-\frac {A}{8}+\frac {i B}{8}}{-i+\tan \left (f x +e \right )}-\frac {\frac {i A}{4}+\frac {B}{4}}{2 \left (-i+\tan \left (f x +e \right )\right )^{2}}-\frac {\frac {A}{2}+\frac {i B}{2}}{3 \left (-i+\tan \left (f x +e \right )\right )^{3}}+\frac {i \left (-i B +A \right ) \ln \left (i+\tan \left (f x +e \right )\right )}{16}}{f \,a^{3}}\) | \(110\) |
risch | \(-\frac {i x B}{8 a^{3}}+\frac {x A}{8 a^{3}}+\frac {{\mathrm e}^{-2 i \left (f x +e \right )} B}{16 a^{3} f}+\frac {3 i {\mathrm e}^{-2 i \left (f x +e \right )} A}{16 a^{3} f}-\frac {{\mathrm e}^{-4 i \left (f x +e \right )} B}{32 a^{3} f}+\frac {3 i {\mathrm e}^{-4 i \left (f x +e \right )} A}{32 a^{3} f}-\frac {{\mathrm e}^{-6 i \left (f x +e \right )} B}{48 a^{3} f}+\frac {i {\mathrm e}^{-6 i \left (f x +e \right )} A}{48 a^{3} f}\) | \(128\) |
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 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 6.15, size = 80, normalized size = 0.71 \begin {gather*} \frac {{\left (12 \, {\left (A - i \, B\right )} f x e^{\left (6 i \, f x + 6 i \, e\right )} - 6 \, {\left (-3 i \, A - B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} - 3 \, {\left (-3 i \, A + 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} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.27, size = 258, normalized size = 2.30 \begin {gather*} \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}\: 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}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.70, size = 140, normalized size = 1.25 \begin {gather*} -\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} \end {gather*}
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 \begin {gather*} -\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} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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