Optimal. Leaf size=449 \[ -\frac {3 i \text {Ci}\left (2 x f+\frac {2 c f}{d}\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{8 a^3 d}-\frac {i \text {Ci}\left (6 x f+\frac {6 c f}{d}\right ) \sin \left (6 e-\frac {6 c f}{d}\right )}{8 a^3 d}-\frac {3 i \text {Ci}\left (4 x f+\frac {4 c f}{d}\right ) \sin \left (4 e-\frac {4 c f}{d}\right )}{8 a^3 d}+\frac {3 \text {Ci}\left (2 x f+\frac {2 c f}{d}\right ) \cos \left (2 e-\frac {2 c f}{d}\right )}{8 a^3 d}+\frac {3 \text {Ci}\left (4 x f+\frac {4 c f}{d}\right ) \cos \left (4 e-\frac {4 c f}{d}\right )}{8 a^3 d}+\frac {\text {Ci}\left (6 x f+\frac {6 c f}{d}\right ) \cos \left (6 e-\frac {6 c f}{d}\right )}{8 a^3 d}-\frac {3 \sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (2 x f+\frac {2 c f}{d}\right )}{8 a^3 d}-\frac {3 \sin \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (4 x f+\frac {4 c f}{d}\right )}{8 a^3 d}-\frac {\sin \left (6 e-\frac {6 c f}{d}\right ) \text {Si}\left (6 x f+\frac {6 c f}{d}\right )}{8 a^3 d}-\frac {3 i \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (2 x f+\frac {2 c f}{d}\right )}{8 a^3 d}-\frac {3 i \cos \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (4 x f+\frac {4 c f}{d}\right )}{8 a^3 d}-\frac {i \cos \left (6 e-\frac {6 c f}{d}\right ) \text {Si}\left (6 x f+\frac {6 c f}{d}\right )}{8 a^3 d}+\frac {\log (c+d x)}{8 a^3 d} \]
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Rubi [A] time = 1.78, antiderivative size = 449, normalized size of antiderivative = 1.00, number of steps used = 53, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3728, 3303, 3299, 3302, 3312, 4406, 4428} \[ -\frac {3 i \text {CosIntegral}\left (\frac {2 c f}{d}+2 f x\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{8 a^3 d}-\frac {i \text {CosIntegral}\left (\frac {6 c f}{d}+6 f x\right ) \sin \left (6 e-\frac {6 c f}{d}\right )}{8 a^3 d}-\frac {3 i \text {CosIntegral}\left (\frac {4 c f}{d}+4 f x\right ) \sin \left (4 e-\frac {4 c f}{d}\right )}{8 a^3 d}+\frac {3 \text {CosIntegral}\left (\frac {2 c f}{d}+2 f x\right ) \cos \left (2 e-\frac {2 c f}{d}\right )}{8 a^3 d}+\frac {3 \text {CosIntegral}\left (\frac {4 c f}{d}+4 f x\right ) \cos \left (4 e-\frac {4 c f}{d}\right )}{8 a^3 d}+\frac {\text {CosIntegral}\left (\frac {6 c f}{d}+6 f x\right ) \cos \left (6 e-\frac {6 c f}{d}\right )}{8 a^3 d}-\frac {3 \sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (2 x f+\frac {2 c f}{d}\right )}{8 a^3 d}-\frac {3 \sin \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (4 x f+\frac {4 c f}{d}\right )}{8 a^3 d}-\frac {\sin \left (6 e-\frac {6 c f}{d}\right ) \text {Si}\left (6 x f+\frac {6 c f}{d}\right )}{8 a^3 d}-\frac {3 i \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (2 x f+\frac {2 c f}{d}\right )}{8 a^3 d}-\frac {3 i \cos \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (4 x f+\frac {4 c f}{d}\right )}{8 a^3 d}-\frac {i \cos \left (6 e-\frac {6 c f}{d}\right ) \text {Si}\left (6 x f+\frac {6 c f}{d}\right )}{8 a^3 d}+\frac {\log (c+d x)}{8 a^3 d} \]
Antiderivative was successfully verified.
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Rule 3299
Rule 3302
Rule 3303
Rule 3312
Rule 3728
Rule 4406
Rule 4428
Rubi steps
\begin {align*} \int \frac {1}{(c+d x) (a+i a \tan (e+f x))^3} \, dx &=\int \left (\frac {1}{8 a^3 (c+d x)}+\frac {3 \cos (2 e+2 f x)}{8 a^3 (c+d x)}+\frac {3 \cos ^2(2 e+2 f x)}{8 a^3 (c+d x)}+\frac {\cos ^3(2 e+2 f x)}{8 a^3 (c+d x)}-\frac {3 i \sin (2 e+2 f x)}{8 a^3 (c+d x)}-\frac {3 i \cos ^2(2 e+2 f x) \sin (2 e+2 f x)}{8 a^3 (c+d x)}-\frac {3 \sin ^2(2 e+2 f x)}{8 a^3 (c+d x)}+\frac {i \sin ^3(2 e+2 f x)}{8 a^3 (c+d x)}-\frac {3 i \sin (4 e+4 f x)}{8 a^3 (c+d x)}-\frac {3 \sin (2 e+2 f x) \sin (4 e+4 f x)}{16 a^3 (c+d x)}\right ) \, dx\\ &=\frac {\log (c+d x)}{8 a^3 d}+\frac {i \int \frac {\sin ^3(2 e+2 f x)}{c+d x} \, dx}{8 a^3}-\frac {(3 i) \int \frac {\sin (2 e+2 f x)}{c+d x} \, dx}{8 a^3}-\frac {(3 i) \int \frac {\cos ^2(2 e+2 f x) \sin (2 e+2 f x)}{c+d x} \, dx}{8 a^3}-\frac {(3 i) \int \frac {\sin (4 e+4 f x)}{c+d x} \, dx}{8 a^3}+\frac {\int \frac {\cos ^3(2 e+2 f x)}{c+d x} \, dx}{8 a^3}-\frac {3 \int \frac {\sin (2 e+2 f x) \sin (4 e+4 f x)}{c+d x} \, dx}{16 a^3}+\frac {3 \int \frac {\cos (2 e+2 f x)}{c+d x} \, dx}{8 a^3}+\frac {3 \int \frac {\cos ^2(2 e+2 f x)}{c+d x} \, dx}{8 a^3}-\frac {3 \int \frac {\sin ^2(2 e+2 f x)}{c+d x} \, dx}{8 a^3}\\ &=\frac {\log (c+d x)}{8 a^3 d}+\frac {i \int \left (\frac {3 \sin (2 e+2 f x)}{4 (c+d x)}-\frac {\sin (6 e+6 f x)}{4 (c+d x)}\right ) \, dx}{8 a^3}-\frac {(3 i) \int \left (\frac {\sin (2 e+2 f x)}{4 (c+d x)}+\frac {\sin (6 e+6 f x)}{4 (c+d x)}\right ) \, dx}{8 a^3}+\frac {\int \left (\frac {3 \cos (2 e+2 f x)}{4 (c+d x)}+\frac {\cos (6 e+6 f x)}{4 (c+d x)}\right ) \, dx}{8 a^3}-\frac {3 \int \left (\frac {\cos (2 e+2 f x)}{2 (c+d x)}-\frac {\cos (6 e+6 f x)}{2 (c+d x)}\right ) \, dx}{16 a^3}-\frac {3 \int \left (\frac {1}{2 (c+d x)}-\frac {\cos (4 e+4 f x)}{2 (c+d x)}\right ) \, dx}{8 a^3}+\frac {3 \int \left (\frac {1}{2 (c+d x)}+\frac {\cos (4 e+4 f x)}{2 (c+d x)}\right ) \, dx}{8 a^3}-\frac {\left (3 i \cos \left (4 e-\frac {4 c f}{d}\right )\right ) \int \frac {\sin \left (\frac {4 c f}{d}+4 f x\right )}{c+d x} \, dx}{8 a^3}-\frac {\left (3 i \cos \left (2 e-\frac {2 c f}{d}\right )\right ) \int \frac {\sin \left (\frac {2 c f}{d}+2 f x\right )}{c+d x} \, dx}{8 a^3}+\frac {\left (3 \cos \left (2 e-\frac {2 c f}{d}\right )\right ) \int \frac {\cos \left (\frac {2 c f}{d}+2 f x\right )}{c+d x} \, dx}{8 a^3}-\frac {\left (3 i \sin \left (4 e-\frac {4 c f}{d}\right )\right ) \int \frac {\cos \left (\frac {4 c f}{d}+4 f x\right )}{c+d x} \, dx}{8 a^3}-\frac {\left (3 i \sin \left (2 e-\frac {2 c f}{d}\right )\right ) \int \frac {\cos \left (\frac {2 c f}{d}+2 f x\right )}{c+d x} \, dx}{8 a^3}-\frac {\left (3 \sin \left (2 e-\frac {2 c f}{d}\right )\right ) \int \frac {\sin \left (\frac {2 c f}{d}+2 f x\right )}{c+d x} \, dx}{8 a^3}\\ &=\frac {3 \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Ci}\left (\frac {2 c f}{d}+2 f x\right )}{8 a^3 d}+\frac {\log (c+d x)}{8 a^3 d}-\frac {3 i \text {Ci}\left (\frac {4 c f}{d}+4 f x\right ) \sin \left (4 e-\frac {4 c f}{d}\right )}{8 a^3 d}-\frac {3 i \text {Ci}\left (\frac {2 c f}{d}+2 f x\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{8 a^3 d}-\frac {3 i \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 c f}{d}+2 f x\right )}{8 a^3 d}-\frac {3 \sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 c f}{d}+2 f x\right )}{8 a^3 d}-\frac {3 i \cos \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (\frac {4 c f}{d}+4 f x\right )}{8 a^3 d}-\frac {i \int \frac {\sin (6 e+6 f x)}{c+d x} \, dx}{32 a^3}-\frac {(3 i) \int \frac {\sin (6 e+6 f x)}{c+d x} \, dx}{32 a^3}+\frac {\int \frac {\cos (6 e+6 f x)}{c+d x} \, dx}{32 a^3}+\frac {3 \int \frac {\cos (6 e+6 f x)}{c+d x} \, dx}{32 a^3}+2 \frac {3 \int \frac {\cos (4 e+4 f x)}{c+d x} \, dx}{16 a^3}\\ &=\frac {3 \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Ci}\left (\frac {2 c f}{d}+2 f x\right )}{8 a^3 d}+\frac {\log (c+d x)}{8 a^3 d}-\frac {3 i \text {Ci}\left (\frac {4 c f}{d}+4 f x\right ) \sin \left (4 e-\frac {4 c f}{d}\right )}{8 a^3 d}-\frac {3 i \text {Ci}\left (\frac {2 c f}{d}+2 f x\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{8 a^3 d}-\frac {3 i \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 c f}{d}+2 f x\right )}{8 a^3 d}-\frac {3 \sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 c f}{d}+2 f x\right )}{8 a^3 d}-\frac {3 i \cos \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (\frac {4 c f}{d}+4 f x\right )}{8 a^3 d}-\frac {\left (i \cos \left (6 e-\frac {6 c f}{d}\right )\right ) \int \frac {\sin \left (\frac {6 c f}{d}+6 f x\right )}{c+d x} \, dx}{32 a^3}-\frac {\left (3 i \cos \left (6 e-\frac {6 c f}{d}\right )\right ) \int \frac {\sin \left (\frac {6 c f}{d}+6 f x\right )}{c+d x} \, dx}{32 a^3}+\frac {\cos \left (6 e-\frac {6 c f}{d}\right ) \int \frac {\cos \left (\frac {6 c f}{d}+6 f x\right )}{c+d x} \, dx}{32 a^3}+\frac {\left (3 \cos \left (6 e-\frac {6 c f}{d}\right )\right ) \int \frac {\cos \left (\frac {6 c f}{d}+6 f x\right )}{c+d x} \, dx}{32 a^3}-\frac {\left (i \sin \left (6 e-\frac {6 c f}{d}\right )\right ) \int \frac {\cos \left (\frac {6 c f}{d}+6 f x\right )}{c+d x} \, dx}{32 a^3}-\frac {\left (3 i \sin \left (6 e-\frac {6 c f}{d}\right )\right ) \int \frac {\cos \left (\frac {6 c f}{d}+6 f x\right )}{c+d x} \, dx}{32 a^3}-\frac {\sin \left (6 e-\frac {6 c f}{d}\right ) \int \frac {\sin \left (\frac {6 c f}{d}+6 f x\right )}{c+d x} \, dx}{32 a^3}-\frac {\left (3 \sin \left (6 e-\frac {6 c f}{d}\right )\right ) \int \frac {\sin \left (\frac {6 c f}{d}+6 f x\right )}{c+d x} \, dx}{32 a^3}+2 \left (\frac {\left (3 \cos \left (4 e-\frac {4 c f}{d}\right )\right ) \int \frac {\cos \left (\frac {4 c f}{d}+4 f x\right )}{c+d x} \, dx}{16 a^3}-\frac {\left (3 \sin \left (4 e-\frac {4 c f}{d}\right )\right ) \int \frac {\sin \left (\frac {4 c f}{d}+4 f x\right )}{c+d x} \, dx}{16 a^3}\right )\\ &=\frac {3 \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Ci}\left (\frac {2 c f}{d}+2 f x\right )}{8 a^3 d}+\frac {\cos \left (6 e-\frac {6 c f}{d}\right ) \text {Ci}\left (\frac {6 c f}{d}+6 f x\right )}{8 a^3 d}+\frac {\log (c+d x)}{8 a^3 d}-\frac {i \text {Ci}\left (\frac {6 c f}{d}+6 f x\right ) \sin \left (6 e-\frac {6 c f}{d}\right )}{8 a^3 d}-\frac {3 i \text {Ci}\left (\frac {4 c f}{d}+4 f x\right ) \sin \left (4 e-\frac {4 c f}{d}\right )}{8 a^3 d}-\frac {3 i \text {Ci}\left (\frac {2 c f}{d}+2 f x\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{8 a^3 d}-\frac {3 i \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 c f}{d}+2 f x\right )}{8 a^3 d}-\frac {3 \sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 c f}{d}+2 f x\right )}{8 a^3 d}-\frac {3 i \cos \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (\frac {4 c f}{d}+4 f x\right )}{8 a^3 d}+2 \left (\frac {3 \cos \left (4 e-\frac {4 c f}{d}\right ) \text {Ci}\left (\frac {4 c f}{d}+4 f x\right )}{16 a^3 d}-\frac {3 \sin \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (\frac {4 c f}{d}+4 f x\right )}{16 a^3 d}\right )-\frac {i \cos \left (6 e-\frac {6 c f}{d}\right ) \text {Si}\left (\frac {6 c f}{d}+6 f x\right )}{8 a^3 d}-\frac {\sin \left (6 e-\frac {6 c f}{d}\right ) \text {Si}\left (\frac {6 c f}{d}+6 f x\right )}{8 a^3 d}\\ \end {align*}
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Mathematica [A] time = 0.80, size = 336, normalized size = 0.75 \[ \frac {\sec ^3(e+f x) (\cos (f x)+i \sin (f x))^3 \left (\left (\cos \left (e-\frac {4 c f}{d}\right )-i \sin \left (e-\frac {4 c f}{d}\right )\right ) \left (-i \text {Ci}\left (\frac {6 f (c+d x)}{d}\right ) \sin \left (2 e-\frac {2 c f}{d}\right )+\text {Ci}\left (\frac {6 f (c+d x)}{d}\right ) \cos \left (2 e-\frac {2 c f}{d}\right )+3 \text {Ci}\left (\frac {2 f (c+d x)}{d}\right ) \left (\cos \left (2 e-\frac {2 c f}{d}\right )+i \sin \left (2 e-\frac {2 c f}{d}\right )\right )+3 \text {Ci}\left (\frac {4 f (c+d x)}{d}\right )+3 \sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 f (c+d x)}{d}\right )-\sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {6 f (c+d x)}{d}\right )-3 i \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 f (c+d x)}{d}\right )-i \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {6 f (c+d x)}{d}\right )-3 i \text {Si}\left (\frac {4 f (c+d x)}{d}\right )\right )+i \sin (3 e) \log (f (c+d x))+\cos (3 e) \log (f (c+d x))\right )}{8 d (a+i a \tan (e+f x))^3} \]
Warning: Unable to verify antiderivative.
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fricas [A] time = 0.44, size = 111, normalized size = 0.25 \[ \frac {3 \, {\rm Ei}\left (\frac {-2 i \, d f x - 2 i \, c f}{d}\right ) e^{\left (\frac {-2 i \, d e + 2 i \, c f}{d}\right )} + 3 \, {\rm Ei}\left (\frac {-4 i \, d f x - 4 i \, c f}{d}\right ) e^{\left (\frac {-4 i \, d e + 4 i \, c f}{d}\right )} + {\rm Ei}\left (\frac {-6 i \, d f x - 6 i \, c f}{d}\right ) e^{\left (\frac {-6 i \, d e + 6 i \, c f}{d}\right )} + \log \left (\frac {d x + c}{d}\right )}{8 \, a^{3} d} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 1.46, size = 846, normalized size = 1.88 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.62, size = 560, normalized size = 1.25 \[ -\frac {3 i \Si \left (4 f x +4 e +\frac {4 c f -4 d e}{d}\right ) \cos \left (\frac {4 c f -4 d e}{d}\right )}{8 a^{3} d}+\frac {3 i \Ci \left (4 f x +4 e +\frac {4 c f -4 d e}{d}\right ) \sin \left (\frac {4 c f -4 d e}{d}\right )}{8 a^{3} d}-\frac {3 i \Si \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \cos \left (\frac {2 c f -2 d e}{d}\right )}{8 a^{3} d}+\frac {3 i \Ci \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \sin \left (\frac {2 c f -2 d e}{d}\right )}{8 a^{3} d}-\frac {i \Si \left (6 f x +6 e +\frac {6 c f -6 d e}{d}\right ) \cos \left (\frac {6 c f -6 d e}{d}\right )}{8 a^{3} d}+\frac {i \Ci \left (6 f x +6 e +\frac {6 c f -6 d e}{d}\right ) \sin \left (\frac {6 c f -6 d e}{d}\right )}{8 a^{3} d}+\frac {\Si \left (6 f x +6 e +\frac {6 c f -6 d e}{d}\right ) \sin \left (\frac {6 c f -6 d e}{d}\right )}{8 a^{3} d}+\frac {\Ci \left (6 f x +6 e +\frac {6 c f -6 d e}{d}\right ) \cos \left (\frac {6 c f -6 d e}{d}\right )}{8 a^{3} d}+\frac {3 \Si \left (4 f x +4 e +\frac {4 c f -4 d e}{d}\right ) \sin \left (\frac {4 c f -4 d e}{d}\right )}{8 a^{3} d}+\frac {3 \Ci \left (4 f x +4 e +\frac {4 c f -4 d e}{d}\right ) \cos \left (\frac {4 c f -4 d e}{d}\right )}{8 a^{3} d}+\frac {3 \Si \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \sin \left (\frac {2 c f -2 d e}{d}\right )}{8 a^{3} d}+\frac {3 \Ci \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \cos \left (\frac {2 c f -2 d e}{d}\right )}{8 a^{3} d}+\frac {\ln \left (\left (f x +e \right ) d +c f -d e \right )}{8 a^{3} d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 3.02, size = 270, normalized size = 0.60 \[ -\frac {f \cos \left (-\frac {6 \, {\left (d e - c f\right )}}{d}\right ) E_{1}\left (\frac {6 i \, {\left (f x + e\right )} d - 6 i \, d e + 6 i \, c f}{d}\right ) + 3 \, f \cos \left (-\frac {4 \, {\left (d e - c f\right )}}{d}\right ) E_{1}\left (\frac {4 i \, {\left (f x + e\right )} d - 4 i \, d e + 4 i \, c f}{d}\right ) + 3 \, f \cos \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right ) E_{1}\left (\frac {2 i \, {\left (f x + e\right )} d - 2 i \, d e + 2 i \, c f}{d}\right ) + 3 i \, f E_{1}\left (\frac {2 i \, {\left (f x + e\right )} d - 2 i \, d e + 2 i \, c f}{d}\right ) \sin \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right ) + 3 i \, f E_{1}\left (\frac {4 i \, {\left (f x + e\right )} d - 4 i \, d e + 4 i \, c f}{d}\right ) \sin \left (-\frac {4 \, {\left (d e - c f\right )}}{d}\right ) + i \, f E_{1}\left (\frac {6 i \, {\left (f x + e\right )} d - 6 i \, d e + 6 i \, c f}{d}\right ) \sin \left (-\frac {6 \, {\left (d e - c f\right )}}{d}\right ) - f \log \left ({\left (f x + e\right )} d - d e + c f\right )}{8 \, a^{3} d f} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3\,\left (c+d\,x\right )} \,d x \]
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 \[ \frac {i \int \frac {1}{c \tan ^{3}{\left (e + f x \right )} - 3 i c \tan ^{2}{\left (e + f x \right )} - 3 c \tan {\left (e + f x \right )} + i c + d x \tan ^{3}{\left (e + f x \right )} - 3 i d x \tan ^{2}{\left (e + f x \right )} - 3 d x \tan {\left (e + f x \right )} + i d x}\, dx}{a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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