Optimal. Leaf size=305 \[ -\frac {i \text {Ci}\left (2 x f+\frac {2 c f}{d}\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{2 a^2 d}+\frac {i \text {Ci}\left (4 x f+\frac {4 c f}{d}\right ) \sin \left (4 e-\frac {4 c f}{d}\right )}{4 a^2 d}-\frac {\text {Ci}\left (2 x f+\frac {2 c f}{d}\right ) \cos \left (2 e-\frac {2 c f}{d}\right )}{2 a^2 d}+\frac {\text {Ci}\left (4 x f+\frac {4 c f}{d}\right ) \cos \left (4 e-\frac {4 c f}{d}\right )}{4 a^2 d}+\frac {\sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (2 x f+\frac {2 c f}{d}\right )}{2 a^2 d}-\frac {\sin \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (4 x f+\frac {4 c f}{d}\right )}{4 a^2 d}-\frac {i \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (2 x f+\frac {2 c f}{d}\right )}{2 a^2 d}+\frac {i \cos \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (4 x f+\frac {4 c f}{d}\right )}{4 a^2 d}+\frac {\log (c+d x)}{4 a^2 d} \]
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Rubi [A] time = 0.75, antiderivative size = 305, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3728, 3303, 3299, 3302, 3312} \[ -\frac {i \text {CosIntegral}\left (\frac {2 c f}{d}+2 f x\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{2 a^2 d}+\frac {i \text {CosIntegral}\left (\frac {4 c f}{d}+4 f x\right ) \sin \left (4 e-\frac {4 c f}{d}\right )}{4 a^2 d}-\frac {\text {CosIntegral}\left (\frac {2 c f}{d}+2 f x\right ) \cos \left (2 e-\frac {2 c f}{d}\right )}{2 a^2 d}+\frac {\text {CosIntegral}\left (\frac {4 c f}{d}+4 f x\right ) \cos \left (4 e-\frac {4 c f}{d}\right )}{4 a^2 d}+\frac {\sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (2 x f+\frac {2 c f}{d}\right )}{2 a^2 d}-\frac {\sin \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (4 x f+\frac {4 c f}{d}\right )}{4 a^2 d}-\frac {i \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (2 x f+\frac {2 c f}{d}\right )}{2 a^2 d}+\frac {i \cos \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (4 x f+\frac {4 c f}{d}\right )}{4 a^2 d}+\frac {\log (c+d x)}{4 a^2 d} \]
Antiderivative was successfully verified.
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Rule 3299
Rule 3302
Rule 3303
Rule 3312
Rule 3728
Rubi steps
\begin {align*} \int \frac {1}{(c+d x) (a+i a \cot (e+f x))^2} \, dx &=\int \left (\frac {1}{4 a^2 (c+d x)}-\frac {\cos (2 e+2 f x)}{2 a^2 (c+d x)}+\frac {\cos ^2(2 e+2 f x)}{4 a^2 (c+d x)}-\frac {i \sin (2 e+2 f x)}{2 a^2 (c+d x)}-\frac {\sin ^2(2 e+2 f x)}{4 a^2 (c+d x)}+\frac {i \sin (4 e+4 f x)}{4 a^2 (c+d x)}\right ) \, dx\\ &=\frac {\log (c+d x)}{4 a^2 d}+\frac {i \int \frac {\sin (4 e+4 f x)}{c+d x} \, dx}{4 a^2}-\frac {i \int \frac {\sin (2 e+2 f x)}{c+d x} \, dx}{2 a^2}+\frac {\int \frac {\cos ^2(2 e+2 f x)}{c+d x} \, dx}{4 a^2}-\frac {\int \frac {\sin ^2(2 e+2 f x)}{c+d x} \, dx}{4 a^2}-\frac {\int \frac {\cos (2 e+2 f x)}{c+d x} \, dx}{2 a^2}\\ &=\frac {\log (c+d x)}{4 a^2 d}-\frac {\int \left (\frac {1}{2 (c+d x)}-\frac {\cos (4 e+4 f x)}{2 (c+d x)}\right ) \, dx}{4 a^2}+\frac {\int \left (\frac {1}{2 (c+d x)}+\frac {\cos (4 e+4 f x)}{2 (c+d x)}\right ) \, dx}{4 a^2}+\frac {\left (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}{4 a^2}-\frac {\left (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}{2 a^2}-\frac {\cos \left (2 e-\frac {2 c f}{d}\right ) \int \frac {\cos \left (\frac {2 c f}{d}+2 f x\right )}{c+d x} \, dx}{2 a^2}+\frac {\left (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}{4 a^2}-\frac {\left (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}{2 a^2}+\frac {\sin \left (2 e-\frac {2 c f}{d}\right ) \int \frac {\sin \left (\frac {2 c f}{d}+2 f x\right )}{c+d x} \, dx}{2 a^2}\\ &=-\frac {\cos \left (2 e-\frac {2 c f}{d}\right ) \text {Ci}\left (\frac {2 c f}{d}+2 f x\right )}{2 a^2 d}+\frac {\log (c+d x)}{4 a^2 d}+\frac {i \text {Ci}\left (\frac {4 c f}{d}+4 f x\right ) \sin \left (4 e-\frac {4 c f}{d}\right )}{4 a^2 d}-\frac {i \text {Ci}\left (\frac {2 c f}{d}+2 f x\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{2 a^2 d}-\frac {i \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 c f}{d}+2 f x\right )}{2 a^2 d}+\frac {\sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 c f}{d}+2 f x\right )}{2 a^2 d}+\frac {i \cos \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (\frac {4 c f}{d}+4 f x\right )}{4 a^2 d}+2 \frac {\int \frac {\cos (4 e+4 f x)}{c+d x} \, dx}{8 a^2}\\ &=-\frac {\cos \left (2 e-\frac {2 c f}{d}\right ) \text {Ci}\left (\frac {2 c f}{d}+2 f x\right )}{2 a^2 d}+\frac {\log (c+d x)}{4 a^2 d}+\frac {i \text {Ci}\left (\frac {4 c f}{d}+4 f x\right ) \sin \left (4 e-\frac {4 c f}{d}\right )}{4 a^2 d}-\frac {i \text {Ci}\left (\frac {2 c f}{d}+2 f x\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{2 a^2 d}-\frac {i \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 c f}{d}+2 f x\right )}{2 a^2 d}+\frac {\sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 c f}{d}+2 f x\right )}{2 a^2 d}+\frac {i \cos \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (\frac {4 c f}{d}+4 f x\right )}{4 a^2 d}+2 \left (\frac {\cos \left (4 e-\frac {4 c f}{d}\right ) \int \frac {\cos \left (\frac {4 c f}{d}+4 f x\right )}{c+d x} \, dx}{8 a^2}-\frac {\sin \left (4 e-\frac {4 c f}{d}\right ) \int \frac {\sin \left (\frac {4 c f}{d}+4 f x\right )}{c+d x} \, dx}{8 a^2}\right )\\ &=-\frac {\cos \left (2 e-\frac {2 c f}{d}\right ) \text {Ci}\left (\frac {2 c f}{d}+2 f x\right )}{2 a^2 d}+\frac {\log (c+d x)}{4 a^2 d}+\frac {i \text {Ci}\left (\frac {4 c f}{d}+4 f x\right ) \sin \left (4 e-\frac {4 c f}{d}\right )}{4 a^2 d}-\frac {i \text {Ci}\left (\frac {2 c f}{d}+2 f x\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{2 a^2 d}-\frac {i \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 c f}{d}+2 f x\right )}{2 a^2 d}+\frac {\sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 c f}{d}+2 f x\right )}{2 a^2 d}+\frac {i \cos \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (\frac {4 c f}{d}+4 f x\right )}{4 a^2 d}+2 \left (\frac {\cos \left (4 e-\frac {4 c f}{d}\right ) \text {Ci}\left (\frac {4 c f}{d}+4 f x\right )}{8 a^2 d}-\frac {\sin \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (\frac {4 c f}{d}+4 f x\right )}{8 a^2 d}\right )\\ \end {align*}
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Mathematica [A] time = 0.61, size = 136, normalized size = 0.45 \[ \frac {-2 \left (\text {Ci}\left (\frac {2 f (c+d x)}{d}\right )+i \text {Si}\left (\frac {2 f (c+d x)}{d}\right )\right ) \left (\cos \left (2 e-\frac {2 c f}{d}\right )+i \sin \left (2 e-\frac {2 c f}{d}\right )\right )+\left (\text {Ci}\left (\frac {4 f (c+d x)}{d}\right )+i \text {Si}\left (\frac {4 f (c+d x)}{d}\right )\right ) \left (\cos \left (4 e-\frac {4 c f}{d}\right )+i \sin \left (4 e-\frac {4 c f}{d}\right )\right )+\log (c+d x)}{4 a^2 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.88, size = 80, normalized size = 0.26 \[ \frac {{\rm Ei}\left (\frac {4 i \, d f x + 4 i \, c f}{d}\right ) e^{\left (\frac {4 i \, d e - 4 i \, c f}{d}\right )} - 2 \, {\rm Ei}\left (\frac {2 i \, d f x + 2 i \, c f}{d}\right ) e^{\left (\frac {2 i \, d e - 2 i \, c f}{d}\right )} + \log \left (\frac {d x + c}{d}\right )}{4 \, a^{2} d} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 1.83, size = 987, normalized size = 3.24 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 1.42, size = 382, normalized size = 1.25 \[ -\frac {i \Si \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \cos \left (\frac {2 c f -2 d e}{d}\right )}{2 a^{2} d}+\frac {i \Ci \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \sin \left (\frac {2 c f -2 d e}{d}\right )}{2 a^{2} d}+\frac {i \Si \left (4 f x +4 e +\frac {4 c f -4 d e}{d}\right ) \cos \left (\frac {4 c f -4 d e}{d}\right )}{4 a^{2} d}-\frac {i \Ci \left (4 f x +4 e +\frac {4 c f -4 d e}{d}\right ) \sin \left (\frac {4 c f -4 d e}{d}\right )}{4 a^{2} d}+\frac {\ln \left (\left (f x +e \right ) d +c f -d e \right )}{4 a^{2} d}+\frac {\Si \left (4 f x +4 e +\frac {4 c f -4 d e}{d}\right ) \sin \left (\frac {4 c f -4 d e}{d}\right )}{4 a^{2} d}+\frac {\Ci \left (4 f x +4 e +\frac {4 c f -4 d e}{d}\right ) \cos \left (\frac {4 c f -4 d e}{d}\right )}{4 a^{2} d}-\frac {\Si \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \sin \left (\frac {2 c f -2 d e}{d}\right )}{2 a^{2} d}-\frac {\Ci \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \cos \left (\frac {2 c f -2 d e}{d}\right )}{2 a^{2} d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.96, size = 194, normalized size = 0.64 \[ -\frac {f \cos \left (-\frac {4 \, {\left (d e - c f\right )}}{d}\right ) E_{1}\left (-\frac {4 i \, {\left (f x + e\right )} d - 4 i \, d e + 4 i \, c f}{d}\right ) - 2 \, f \cos \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right ) E_{1}\left (-\frac {2 i \, {\left (f x + e\right )} d - 2 i \, d e + 2 i \, c f}{d}\right ) + 2 i \, f E_{1}\left (-\frac {2 i \, {\left (f x + e\right )} d - 2 i \, d e + 2 i \, c f}{d}\right ) \sin \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right ) - i \, f E_{1}\left (-\frac {4 i \, {\left (f x + e\right )} d - 4 i \, d e + 4 i \, c f}{d}\right ) \sin \left (-\frac {4 \, {\left (d e - c f\right )}}{d}\right ) - f \log \left ({\left (f x + e\right )} d - d e + c f\right )}{4 \, a^{2} d f} \]
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 \[ \int \frac {1}{{\left (a+a\,\mathrm {cot}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2\,\left (c+d\,x\right )} \,d x \]
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 \[ - \frac {\int \frac {1}{c \cot ^{2}{\left (e + f x \right )} - 2 i c \cot {\left (e + f x \right )} - c + d x \cot ^{2}{\left (e + f x \right )} - 2 i d x \cot {\left (e + f x \right )} - d x}\, dx}{a^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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