Optimal. Leaf size=91 \[ \frac {(c-i d)^2 x}{4 a^2}+\frac {(c+i d) (i c+3 d)}{4 a^2 f (1+i \tan (e+f x))}+\frac {i (c+i d)^2}{4 f (a+i a \tan (e+f x))^2} \]
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Rubi [A]
time = 0.10, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps
used = 3, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {3621, 3607, 8}
\begin {gather*} \frac {(c+i d) (3 d+i c)}{4 a^2 f (1+i \tan (e+f x))}+\frac {x (c-i d)^2}{4 a^2}+\frac {i (c+i d)^2}{4 f (a+i a \tan (e+f x))^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 8
Rule 3607
Rule 3621
Rubi steps
\begin {align*} \int \frac {(c+d \tan (e+f x))^2}{(a+i a \tan (e+f x))^2} \, dx &=\frac {i (c+i d)^2}{4 f (a+i a \tan (e+f x))^2}+\frac {\int \frac {a \left (c^2-2 i c d+d^2\right )-2 i a d^2 \tan (e+f x)}{a+i a \tan (e+f x)} \, dx}{2 a^2}\\ &=\frac {(c+i d) (i c+3 d)}{4 a^2 f (1+i \tan (e+f x))}+\frac {i (c+i d)^2}{4 f (a+i a \tan (e+f x))^2}+\frac {(c-i d)^2 \int 1 \, dx}{4 a^2}\\ &=\frac {(c-i d)^2 x}{4 a^2}+\frac {(c+i d) (i c+3 d)}{4 a^2 f (1+i \tan (e+f x))}+\frac {i (c+i d)^2}{4 f (a+i a \tan (e+f x))^2}\\ \end {align*}
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Mathematica [A]
time = 1.07, size = 134, normalized size = 1.47 \begin {gather*} -\frac {\sec ^2(e+f x) \left (4 i \left (c^2+d^2\right )+\left (c d (-2-8 i f x)+c^2 (i+4 f x)-d^2 (i+4 f x)\right ) \cos (2 (e+f x))+\left (d^2 (-1-4 i f x)+c^2 (1+4 i f x)+2 c d (i+4 f x)\right ) \sin (2 (e+f x))\right )}{16 a^2 f (-i+\tan (e+f x))^2} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.21, size = 123, normalized size = 1.35
method | result | size |
derivativedivides | \(\frac {\left (\frac {1}{8} i d^{2}-\frac {1}{8} i c^{2}-\frac {1}{4} c d \right ) \ln \left (\tan \left (f x +e \right )-i\right )-\frac {-c d +\frac {1}{2} i c^{2}-\frac {1}{2} i d^{2}}{2 \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {\frac {1}{2} i c d -\frac {1}{4} c^{2}-\frac {3}{4} d^{2}}{\tan \left (f x +e \right )-i}-\frac {i \left (2 i c d -c^{2}+d^{2}\right ) \ln \left (\tan \left (f x +e \right )+i\right )}{8}}{f \,a^{2}}\) | \(123\) |
default | \(\frac {\left (\frac {1}{8} i d^{2}-\frac {1}{8} i c^{2}-\frac {1}{4} c d \right ) \ln \left (\tan \left (f x +e \right )-i\right )-\frac {-c d +\frac {1}{2} i c^{2}-\frac {1}{2} i d^{2}}{2 \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {\frac {1}{2} i c d -\frac {1}{4} c^{2}-\frac {3}{4} d^{2}}{\tan \left (f x +e \right )-i}-\frac {i \left (2 i c d -c^{2}+d^{2}\right ) \ln \left (\tan \left (f x +e \right )+i\right )}{8}}{f \,a^{2}}\) | \(123\) |
risch | \(-\frac {i x c d}{2 a^{2}}+\frac {x \,c^{2}}{4 a^{2}}-\frac {x \,d^{2}}{4 a^{2}}+\frac {i {\mathrm e}^{-2 i \left (f x +e \right )} c^{2}}{4 a^{2} f}+\frac {i {\mathrm e}^{-2 i \left (f x +e \right )} d^{2}}{4 a^{2} f}-\frac {{\mathrm e}^{-4 i \left (f x +e \right )} c d}{8 a^{2} f}+\frac {i {\mathrm e}^{-4 i \left (f x +e \right )} c^{2}}{16 a^{2} f}-\frac {i {\mathrm e}^{-4 i \left (f x +e \right )} d^{2}}{16 a^{2} f}\) | \(132\) |
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 = 0.73, size = 83, normalized size = 0.91 \begin {gather*} \frac {{\left (4 \, {\left (c^{2} - 2 i \, c d - d^{2}\right )} f x e^{\left (4 i \, f x + 4 i \, e\right )} + i \, c^{2} - 2 \, c d - i \, d^{2} - 4 \, {\left (-i \, c^{2} - i \, d^{2}\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} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.24, size = 258, normalized size = 2.84 \begin {gather*} \begin {cases} \frac {\left (\left (16 i a^{2} c^{2} f e^{4 i e} + 16 i a^{2} d^{2} f e^{4 i e}\right ) e^{- 2 i f x} + \left (4 i a^{2} c^{2} f e^{2 i e} - 8 a^{2} c d f e^{2 i e} - 4 i a^{2} d^{2} f e^{2 i e}\right ) e^{- 4 i f x}\right ) e^{- 6 i e}}{64 a^{4} f^{2}} & \text {for}\: a^{4} f^{2} e^{6 i e} \neq 0 \\x \left (- \frac {c^{2} - 2 i c d - d^{2}}{4 a^{2}} + \frac {\left (c^{2} e^{4 i e} + 2 c^{2} e^{2 i e} + c^{2} - 2 i c d e^{4 i e} + 2 i c d - d^{2} e^{4 i e} + 2 d^{2} e^{2 i e} - d^{2}\right ) e^{- 4 i e}}{4 a^{2}}\right ) & \text {otherwise} \end {cases} + \frac {x \left (c^{2} - 2 i c d - d^{2}\right )}{4 a^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] Both result and optimal contain complex but leaf count of result is larger than twice
the leaf count of optimal. 176 vs. \(2 (73) = 146\).
time = 0.64, size = 176, normalized size = 1.93 \begin {gather*} -\frac {\frac {2 \, {\left (-i \, c^{2} - 2 \, c d + i \, d^{2}\right )} \log \left (-i \, \tan \left (f x + e\right ) + 1\right )}{a^{2}} + \frac {2 \, {\left (i \, c^{2} + 2 \, c d - i \, d^{2}\right )} \log \left (-i \, \tan \left (f x + e\right ) - 1\right )}{a^{2}} + \frac {-3 i \, c^{2} \tan \left (f x + e\right )^{2} - 6 \, c d \tan \left (f x + e\right )^{2} + 3 i \, d^{2} \tan \left (f x + e\right )^{2} - 10 \, c^{2} \tan \left (f x + e\right ) + 20 i \, c d \tan \left (f x + e\right ) - 6 \, d^{2} \tan \left (f x + e\right ) + 11 i \, c^{2} + 6 \, c d + 5 i \, d^{2}}{a^{2} {\left (\tan \left (f x + e\right ) - i\right )}^{2}}}{16 \, f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 5.24, size = 93, normalized size = 1.02 \begin {gather*} -\frac {x\,{\left (d+c\,1{}\mathrm {i}\right )}^2}{4\,a^2}+\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (\frac {c\,d}{2\,a^2}+\frac {c^2\,1{}\mathrm {i}}{4\,a^2}+\frac {d^2\,3{}\mathrm {i}}{4\,a^2}\right )+\frac {c^2}{2\,a^2}+\frac {d^2}{2\,a^2}}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (e+f\,x\right )-\mathrm {i}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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