Optimal. Leaf size=436 \[ -\frac {1}{4 a^2 d (c+d x)}-\frac {\cos (2 e+2 f x)}{2 a^2 d (c+d x)}-\frac {\cos ^2(2 e+2 f x)}{4 a^2 d (c+d x)}-\frac {i f \cos \left (2 e-\frac {2 c f}{d}\right ) \text {CosIntegral}\left (\frac {2 c f}{d}+2 f x\right )}{a^2 d^2}-\frac {i f \cos \left (4 e-\frac {4 c f}{d}\right ) \text {CosIntegral}\left (\frac {4 c f}{d}+4 f x\right )}{a^2 d^2}-\frac {f \text {CosIntegral}\left (\frac {4 c f}{d}+4 f x\right ) \sin \left (4 e-\frac {4 c f}{d}\right )}{a^2 d^2}-\frac {f \text {CosIntegral}\left (\frac {2 c f}{d}+2 f x\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{a^2 d^2}+\frac {i \sin (2 e+2 f x)}{2 a^2 d (c+d x)}+\frac {\sin ^2(2 e+2 f x)}{4 a^2 d (c+d x)}+\frac {i \sin (4 e+4 f x)}{4 a^2 d (c+d x)}-\frac {f \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 c f}{d}+2 f x\right )}{a^2 d^2}+\frac {i f \sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 c f}{d}+2 f x\right )}{a^2 d^2}-\frac {f \cos \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (\frac {4 c f}{d}+4 f x\right )}{a^2 d^2}+\frac {i f \sin \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (\frac {4 c f}{d}+4 f x\right )}{a^2 d^2} \]
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Rubi [A]
time = 0.51, antiderivative size = 436, normalized size of antiderivative = 1.00, number of steps
used = 24, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3809, 3378,
3384, 3380, 3383, 3394, 12} \begin {gather*} -\frac {f \text {CosIntegral}\left (\frac {4 c f}{d}+4 f x\right ) \sin \left (4 e-\frac {4 c f}{d}\right )}{a^2 d^2}-\frac {f \text {CosIntegral}\left (\frac {2 c f}{d}+2 f x\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{a^2 d^2}-\frac {i f \text {CosIntegral}\left (\frac {2 c f}{d}+2 f x\right ) \cos \left (2 e-\frac {2 c f}{d}\right )}{a^2 d^2}-\frac {i f \text {CosIntegral}\left (\frac {4 c f}{d}+4 f x\right ) \cos \left (4 e-\frac {4 c f}{d}\right )}{a^2 d^2}+\frac {i f \sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (2 x f+\frac {2 c f}{d}\right )}{a^2 d^2}+\frac {i f \sin \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (4 x f+\frac {4 c f}{d}\right )}{a^2 d^2}-\frac {f \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (2 x f+\frac {2 c f}{d}\right )}{a^2 d^2}-\frac {f \cos \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (4 x f+\frac {4 c f}{d}\right )}{a^2 d^2}+\frac {\sin ^2(2 e+2 f x)}{4 a^2 d (c+d x)}+\frac {i \sin (2 e+2 f x)}{2 a^2 d (c+d x)}+\frac {i \sin (4 e+4 f x)}{4 a^2 d (c+d x)}-\frac {\cos ^2(2 e+2 f x)}{4 a^2 d (c+d x)}-\frac {\cos (2 e+2 f x)}{2 a^2 d (c+d x)}-\frac {1}{4 a^2 d (c+d x)} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 3378
Rule 3380
Rule 3383
Rule 3384
Rule 3394
Rule 3809
Rubi steps
\begin {align*} \int \frac {1}{(c+d x)^2 (a+i a \tan (e+f x))^2} \, dx &=\int \left (\frac {1}{4 a^2 (c+d x)^2}+\frac {\cos (2 e+2 f x)}{2 a^2 (c+d x)^2}+\frac {\cos ^2(2 e+2 f x)}{4 a^2 (c+d x)^2}-\frac {i \sin (2 e+2 f x)}{2 a^2 (c+d x)^2}-\frac {\sin ^2(2 e+2 f x)}{4 a^2 (c+d x)^2}-\frac {i \sin (4 e+4 f x)}{4 a^2 (c+d x)^2}\right ) \, dx\\ &=-\frac {1}{4 a^2 d (c+d x)}-\frac {i \int \frac {\sin (4 e+4 f x)}{(c+d x)^2} \, dx}{4 a^2}-\frac {i \int \frac {\sin (2 e+2 f x)}{(c+d x)^2} \, dx}{2 a^2}+\frac {\int \frac {\cos ^2(2 e+2 f x)}{(c+d x)^2} \, dx}{4 a^2}-\frac {\int \frac {\sin ^2(2 e+2 f x)}{(c+d x)^2} \, dx}{4 a^2}+\frac {\int \frac {\cos (2 e+2 f x)}{(c+d x)^2} \, dx}{2 a^2}\\ &=-\frac {1}{4 a^2 d (c+d x)}-\frac {\cos (2 e+2 f x)}{2 a^2 d (c+d x)}-\frac {\cos ^2(2 e+2 f x)}{4 a^2 d (c+d x)}+\frac {i \sin (2 e+2 f x)}{2 a^2 d (c+d x)}+\frac {\sin ^2(2 e+2 f x)}{4 a^2 d (c+d x)}+\frac {i \sin (4 e+4 f x)}{4 a^2 d (c+d x)}-\frac {(i f) \int \frac {\cos (2 e+2 f x)}{c+d x} \, dx}{a^2 d}-\frac {(i f) \int \frac {\cos (4 e+4 f x)}{c+d x} \, dx}{a^2 d}-\frac {f \int \frac {\sin (2 e+2 f x)}{c+d x} \, dx}{a^2 d}+\frac {f \int -\frac {\sin (4 e+4 f x)}{2 (c+d x)} \, dx}{a^2 d}-\frac {f \int \frac {\sin (4 e+4 f x)}{2 (c+d x)} \, dx}{a^2 d}\\ &=-\frac {1}{4 a^2 d (c+d x)}-\frac {\cos (2 e+2 f x)}{2 a^2 d (c+d x)}-\frac {\cos ^2(2 e+2 f x)}{4 a^2 d (c+d x)}+\frac {i \sin (2 e+2 f x)}{2 a^2 d (c+d x)}+\frac {\sin ^2(2 e+2 f x)}{4 a^2 d (c+d x)}+\frac {i \sin (4 e+4 f x)}{4 a^2 d (c+d x)}-2 \frac {f \int \frac {\sin (4 e+4 f x)}{c+d x} \, dx}{2 a^2 d}-\frac {\left (i f \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}{a^2 d}-\frac {\left (i f \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}{a^2 d}-\frac {\left (f \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}{a^2 d}+\frac {\left (i f \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}{a^2 d}+\frac {\left (i f \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}{a^2 d}-\frac {\left (f \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}{a^2 d}\\ &=-\frac {1}{4 a^2 d (c+d x)}-\frac {\cos (2 e+2 f x)}{2 a^2 d (c+d x)}-\frac {\cos ^2(2 e+2 f x)}{4 a^2 d (c+d x)}-\frac {i f \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Ci}\left (\frac {2 c f}{d}+2 f x\right )}{a^2 d^2}-\frac {i f \cos \left (4 e-\frac {4 c f}{d}\right ) \text {Ci}\left (\frac {4 c f}{d}+4 f x\right )}{a^2 d^2}-\frac {f \text {Ci}\left (\frac {2 c f}{d}+2 f x\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{a^2 d^2}+\frac {i \sin (2 e+2 f x)}{2 a^2 d (c+d x)}+\frac {\sin ^2(2 e+2 f x)}{4 a^2 d (c+d x)}+\frac {i \sin (4 e+4 f x)}{4 a^2 d (c+d x)}-\frac {f \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 c f}{d}+2 f x\right )}{a^2 d^2}+\frac {i f \sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 c f}{d}+2 f x\right )}{a^2 d^2}+\frac {i f \sin \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (\frac {4 c f}{d}+4 f x\right )}{a^2 d^2}-2 \left (\frac {\left (f \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}{2 a^2 d}+\frac {\left (f \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}{2 a^2 d}\right )\\ &=-\frac {1}{4 a^2 d (c+d x)}-\frac {\cos (2 e+2 f x)}{2 a^2 d (c+d x)}-\frac {\cos ^2(2 e+2 f x)}{4 a^2 d (c+d x)}-\frac {i f \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Ci}\left (\frac {2 c f}{d}+2 f x\right )}{a^2 d^2}-\frac {i f \cos \left (4 e-\frac {4 c f}{d}\right ) \text {Ci}\left (\frac {4 c f}{d}+4 f x\right )}{a^2 d^2}-\frac {f \text {Ci}\left (\frac {2 c f}{d}+2 f x\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{a^2 d^2}+\frac {i \sin (2 e+2 f x)}{2 a^2 d (c+d x)}+\frac {\sin ^2(2 e+2 f x)}{4 a^2 d (c+d x)}+\frac {i \sin (4 e+4 f x)}{4 a^2 d (c+d x)}-\frac {f \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 c f}{d}+2 f x\right )}{a^2 d^2}+\frac {i f \sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 c f}{d}+2 f x\right )}{a^2 d^2}+\frac {i f \sin \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (\frac {4 c f}{d}+4 f x\right )}{a^2 d^2}-2 \left (\frac {f \text {Ci}\left (\frac {4 c f}{d}+4 f x\right ) \sin \left (4 e-\frac {4 c f}{d}\right )}{2 a^2 d^2}+\frac {f \cos \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (\frac {4 c f}{d}+4 f x\right )}{2 a^2 d^2}\right )\\ \end {align*}
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Mathematica [A]
time = 1.73, size = 467, normalized size = 1.07 \begin {gather*} -\frac {\left (\cos \left (2 \left (e+f \left (-\frac {c}{d}+x\right )\right )\right )-i \sin \left (2 \left (e+f \left (-\frac {c}{d}+x\right )\right )\right )\right ) \left (2 d \cos \left (\frac {2 c f}{d}\right )+d \cos \left (2 \left (e+f \left (-\frac {c}{d}+x\right )\right )\right )+d \cos \left (2 \left (e+f \left (\frac {c}{d}+x\right )\right )\right )-2 i d \sin \left (\frac {2 c f}{d}\right )+4 i f (c+d x) \text {CosIntegral}\left (\frac {2 f (c+d x)}{d}\right ) (\cos (2 f x)+i \sin (2 f x))+i d \sin \left (2 \left (e+f \left (-\frac {c}{d}+x\right )\right )\right )-i d \sin \left (2 \left (e+f \left (\frac {c}{d}+x\right )\right )\right )+4 f (c+d x) \text {CosIntegral}\left (\frac {4 f (c+d x)}{d}\right ) \left (i \cos \left (2 e-\frac {2 f (c+d x)}{d}\right )+\sin \left (2 e-\frac {2 f (c+d x)}{d}\right )\right )+4 c f \cos (2 f x) \text {Si}\left (\frac {2 f (c+d x)}{d}\right )+4 d f x \cos (2 f x) \text {Si}\left (\frac {2 f (c+d x)}{d}\right )+4 i c f \sin (2 f x) \text {Si}\left (\frac {2 f (c+d x)}{d}\right )+4 i d f x \sin (2 f x) \text {Si}\left (\frac {2 f (c+d x)}{d}\right )+4 c f \cos \left (2 e-\frac {2 f (c+d x)}{d}\right ) \text {Si}\left (\frac {4 f (c+d x)}{d}\right )+4 d f x \cos \left (2 e-\frac {2 f (c+d x)}{d}\right ) \text {Si}\left (\frac {4 f (c+d x)}{d}\right )-4 i c f \sin \left (2 e-\frac {2 f (c+d x)}{d}\right ) \text {Si}\left (\frac {4 f (c+d x)}{d}\right )-4 i d f x \sin \left (2 e-\frac {2 f (c+d x)}{d}\right ) \text {Si}\left (\frac {4 f (c+d x)}{d}\right )\right )}{4 a^2 d^2 (c+d x)} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.48, size = 536, normalized size = 1.23
method | result | size |
risch | \(-\frac {1}{4 a^{2} d \left (d x +c \right )}-\frac {f \,{\mathrm e}^{-4 i \left (f x +e \right )}}{4 a^{2} \left (d x f +c f \right ) d}+\frac {i f \,{\mathrm e}^{\frac {4 i \left (c f -d e \right )}{d}} \expIntegral \left (1, 4 i f x +4 i e +\frac {4 i \left (c f -d e \right )}{d}\right )}{a^{2} d^{2}}-\frac {f \,{\mathrm e}^{-2 i \left (f x +e \right )}}{2 a^{2} \left (d x f +c f \right ) d}+\frac {i f \,{\mathrm e}^{\frac {2 i \left (c f -d e \right )}{d}} \expIntegral \left (1, 2 i f x +2 i e +\frac {2 i \left (c f -d e \right )}{d}\right )}{a^{2} d^{2}}\) | \(175\) |
default | \(\frac {-\frac {i f^{2} \left (-\frac {2 \sin \left (2 f x +2 e \right )}{\left (c f -d e +d \left (f x +e \right )\right ) d}+\frac {\frac {4 \sinIntegral \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \sin \left (\frac {2 c f -2 d e}{d}\right )}{d}+\frac {4 \cosineIntegral \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \cos \left (\frac {2 c f -2 d e}{d}\right )}{d}}{d}\right )}{4}-\frac {i f^{2} \left (-\frac {4 \sin \left (4 f x +4 e \right )}{\left (c f -d e +d \left (f x +e \right )\right ) d}+\frac {\frac {16 \sinIntegral \left (4 f x +4 e +\frac {4 c f -4 d e}{d}\right ) \sin \left (\frac {4 c f -4 d e}{d}\right )}{d}+\frac {16 \cosineIntegral \left (4 f x +4 e +\frac {4 c f -4 d e}{d}\right ) \cos \left (\frac {4 c f -4 d e}{d}\right )}{d}}{d}\right )}{16}+\frac {f^{2} \left (-\frac {4 \cos \left (4 f x +4 e \right )}{\left (c f -d e +d \left (f x +e \right )\right ) d}-\frac {4 \left (\frac {4 \sinIntegral \left (4 f x +4 e +\frac {4 c f -4 d e}{d}\right ) \cos \left (\frac {4 c f -4 d e}{d}\right )}{d}-\frac {4 \cosineIntegral \left (4 f x +4 e +\frac {4 c f -4 d e}{d}\right ) \sin \left (\frac {4 c f -4 d e}{d}\right )}{d}\right )}{d}\right )}{16}+\frac {f^{2} \left (-\frac {2 \cos \left (2 f x +2 e \right )}{\left (c f -d e +d \left (f x +e \right )\right ) d}-\frac {2 \left (\frac {2 \sinIntegral \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \cos \left (\frac {2 c f -2 d e}{d}\right )}{d}-\frac {2 \cosineIntegral \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \sin \left (\frac {2 c f -2 d e}{d}\right )}{d}\right )}{d}\right )}{4}-\frac {f^{2}}{4 \left (c f -d e +d \left (f x +e \right )\right ) d}}{a^{2} f}\) | \(536\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.44, size = 225, normalized size = 0.52 \begin {gather*} -\frac {2 \, f^{2} \cos \left (\frac {2 \, {\left (c f - d e\right )}}{d}\right ) E_{2}\left (-\frac {2 \, {\left (-i \, {\left (f x + e\right )} d - i \, c f + i \, d e\right )}}{d}\right ) + f^{2} \cos \left (\frac {4 \, {\left (c f - d e\right )}}{d}\right ) E_{2}\left (-\frac {4 \, {\left (-i \, {\left (f x + e\right )} d - i \, c f + i \, d e\right )}}{d}\right ) + i \, f^{2} E_{2}\left (-\frac {4 \, {\left (-i \, {\left (f x + e\right )} d - i \, c f + i \, d e\right )}}{d}\right ) \sin \left (\frac {4 \, {\left (c f - d e\right )}}{d}\right ) + 2 i \, f^{2} E_{2}\left (-\frac {2 \, {\left (-i \, {\left (f x + e\right )} d - i \, c f + i \, d e\right )}}{d}\right ) \sin \left (\frac {2 \, {\left (c f - d e\right )}}{d}\right ) + f^{2}}{4 \, {\left ({\left (f x + e\right )} a^{2} d^{2} + a^{2} c d f - a^{2} d^{2} e\right )} f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.38, size = 147, normalized size = 0.34 \begin {gather*} -\frac {{\left ({\left (4 \, {\left (i \, d f x + i \, c f\right )} {\rm Ei}\left (-\frac {2 \, {\left (i \, d f x + i \, c f\right )}}{d}\right ) e^{\left (-\frac {2 \, {\left (-i \, c f + i \, d e\right )}}{d}\right )} + 4 \, {\left (i \, d f x + i \, c f\right )} {\rm Ei}\left (-\frac {4 \, {\left (i \, d f x + i \, c f\right )}}{d}\right ) e^{\left (-\frac {4 \, {\left (-i \, c f + i \, d e\right )}}{d}\right )} + d\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, d e^{\left (2 i \, f x + 2 i \, e\right )} + d\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{4 \, {\left (a^{2} d^{3} x + a^{2} c d^{2}\right )}} \end {gather*}
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 \begin {gather*} - \frac {\int \frac {1}{c^{2} \tan ^{2}{\left (e + f x \right )} - 2 i c^{2} \tan {\left (e + f x \right )} - c^{2} + 2 c d x \tan ^{2}{\left (e + f x \right )} - 4 i c d x \tan {\left (e + f x \right )} - 2 c d x + d^{2} x^{2} \tan ^{2}{\left (e + f x \right )} - 2 i d^{2} x^{2} \tan {\left (e + f x \right )} - d^{2} x^{2}}\, dx}{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. 2135 vs. \(2 (423) = 846\).
time = 13.09, size = 2135, normalized size = 4.90 \begin {gather*} \text {Too large to display} \end {gather*}
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 \begin {gather*} \int \frac {1}{{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2\,{\left (c+d\,x\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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