Optimal. Leaf size=434 \[ -\frac {f \text {Ci}\left (4 x f+\frac {4 c f}{d}\right ) \sin \left (4 e-\frac {4 c f}{d}\right )}{a^2 d^2}+\frac {f \text {Ci}\left (2 x f+\frac {2 c f}{d}\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{a^2 d^2}-\frac {i f \text {Ci}\left (2 x f+\frac {2 c f}{d}\right ) \cos \left (2 e-\frac {2 c f}{d}\right )}{a^2 d^2}+\frac {i f \text {Ci}\left (4 x f+\frac {4 c f}{d}\right ) \cos \left (4 e-\frac {4 c f}{d}\right )}{a^2 d^2}+\frac {i f \sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (2 x f+\frac {2 c f}{d}\right )}{a^2 d^2}-\frac {i f \sin \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (4 x f+\frac {4 c f}{d}\right )}{a^2 d^2}+\frac {f \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (2 x f+\frac {2 c f}{d}\right )}{a^2 d^2}-\frac {f \cos \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (4 x f+\frac {4 c f}{d}\right )}{a^2 d^2}+\frac {\sin ^2(2 e+2 f x)}{4 a^2 d (c+d x)}+\frac {i \sin (2 e+2 f x)}{2 a^2 d (c+d x)}-\frac {i \sin (4 e+4 f x)}{4 a^2 d (c+d x)}-\frac {\cos ^2(2 e+2 f x)}{4 a^2 d (c+d x)}+\frac {\cos (2 e+2 f x)}{2 a^2 d (c+d x)}-\frac {1}{4 a^2 d (c+d x)} \]
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Rubi [A] time = 0.68, antiderivative size = 434, 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 = {3728, 3297, 3303, 3299, 3302, 3313, 12} \[ -\frac {f \text {CosIntegral}\left (\frac {4 c f}{d}+4 f x\right ) \sin \left (4 e-\frac {4 c f}{d}\right )}{a^2 d^2}+\frac {f \text {CosIntegral}\left (\frac {2 c f}{d}+2 f x\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{a^2 d^2}-\frac {i f \text {CosIntegral}\left (\frac {2 c f}{d}+2 f x\right ) \cos \left (2 e-\frac {2 c f}{d}\right )}{a^2 d^2}+\frac {i f \text {CosIntegral}\left (\frac {4 c f}{d}+4 f x\right ) \cos \left (4 e-\frac {4 c f}{d}\right )}{a^2 d^2}+\frac {i f \sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (2 x f+\frac {2 c f}{d}\right )}{a^2 d^2}-\frac {i f \sin \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (4 x f+\frac {4 c f}{d}\right )}{a^2 d^2}+\frac {f \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (2 x f+\frac {2 c f}{d}\right )}{a^2 d^2}-\frac {f \cos \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (4 x f+\frac {4 c f}{d}\right )}{a^2 d^2}+\frac {\sin ^2(2 e+2 f x)}{4 a^2 d (c+d x)}+\frac {i \sin (2 e+2 f x)}{2 a^2 d (c+d x)}-\frac {i \sin (4 e+4 f x)}{4 a^2 d (c+d x)}-\frac {\cos ^2(2 e+2 f x)}{4 a^2 d (c+d x)}+\frac {\cos (2 e+2 f x)}{2 a^2 d (c+d x)}-\frac {1}{4 a^2 d (c+d x)} \]
Antiderivative was successfully verified.
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Rule 12
Rule 3297
Rule 3299
Rule 3302
Rule 3303
Rule 3313
Rule 3728
Rubi steps
\begin {align*} \int \frac {1}{(c+d x)^2 (a+i a \cot (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 = 0.59, size = 203, normalized size = 0.47 \[ \frac {4 f (c+d x) \left (\text {Ci}\left (\frac {2 f (c+d x)}{d}\right )+i \text {Si}\left (\frac {2 f (c+d x)}{d}\right )\right ) \left (\sin \left (2 e-\frac {2 c f}{d}\right )-i \cos \left (2 e-\frac {2 c f}{d}\right )\right )+(c+d x) \left (\text {Ci}\left (\frac {4 f (c+d x)}{d}\right )+i \text {Si}\left (\frac {4 f (c+d x)}{d}\right )\right ) \left (-4 f \sin \left (4 e-\frac {4 c f}{d}\right )+4 i f \cos \left (4 e-\frac {4 c f}{d}\right )\right )+2 d (\cos (2 (e+f x))+i \sin (2 (e+f x)))-d (\cos (4 (e+f x))+i \sin (4 (e+f x)))-d}{4 a^2 d^2 (c+d x)} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.60, size = 129, normalized size = 0.30 \[ \frac {{\left (4 i \, d f x + 4 i \, c f\right )} {\rm Ei}\left (\frac {4 i \, d f x + 4 i \, c f}{d}\right ) e^{\left (\frac {4 i \, d e - 4 i \, c f}{d}\right )} + {\left (-4 i \, d f x - 4 i \, c f\right )} {\rm Ei}\left (\frac {2 i \, d f x + 2 i \, c f}{d}\right ) e^{\left (\frac {2 i \, d e - 2 i \, c f}{d}\right )} - d e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, d e^{\left (2 i \, f x + 2 i \, e\right )} - d}{4 \, {\left (a^{2} d^{3} x + a^{2} c d^{2}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 64.09, size = 2369, normalized size = 5.46 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 1.43, size = 537, normalized size = 1.24 \[ -\frac {\frac {i f^{2} \left (-\frac {2 \sin \left (2 f x +2 e \right )}{\left (\left (f x +e \right ) d +c f -d e \right ) d}+\frac {\frac {4 \Si \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \sin \left (\frac {2 c f -2 d e}{d}\right )}{d}+\frac {4 \Ci \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \cos \left (\frac {2 c f -2 d e}{d}\right )}{d}}{d}\right )}{4}-\frac {i f^{2} \left (-\frac {4 \sin \left (4 f x +4 e \right )}{\left (\left (f x +e \right ) d +c f -d e \right ) d}+\frac {\frac {16 \Si \left (4 f x +4 e +\frac {4 c f -4 d e}{d}\right ) \sin \left (\frac {4 c f -4 d e}{d}\right )}{d}+\frac {16 \Ci \left (4 f x +4 e +\frac {4 c f -4 d e}{d}\right ) \cos \left (\frac {4 c f -4 d e}{d}\right )}{d}}{d}\right )}{16}+\frac {f^{2}}{4 \left (\left (f x +e \right ) d +c f -d e \right ) d}-\frac {f^{2} \left (-\frac {4 \cos \left (4 f x +4 e \right )}{\left (\left (f x +e \right ) d +c f -d e \right ) d}-\frac {4 \left (\frac {4 \Si \left (4 f x +4 e +\frac {4 c f -4 d e}{d}\right ) \cos \left (\frac {4 c f -4 d e}{d}\right )}{d}-\frac {4 \Ci \left (4 f x +4 e +\frac {4 c f -4 d e}{d}\right ) \sin \left (\frac {4 c f -4 d e}{d}\right )}{d}\right )}{d}\right )}{16}+\frac {f^{2} \left (-\frac {2 \cos \left (2 f x +2 e \right )}{\left (\left (f x +e \right ) d +c f -d e \right ) d}-\frac {2 \left (\frac {2 \Si \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \cos \left (\frac {2 c f -2 d e}{d}\right )}{d}-\frac {2 \Ci \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \sin \left (\frac {2 c f -2 d e}{d}\right )}{d}\right )}{d}\right )}{4}}{a^{2} f} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.81, size = 214, normalized size = 0.49 \[ -\frac {64 \, f^{2} \cos \left (-\frac {4 \, {\left (d e - c f\right )}}{d}\right ) E_{2}\left (-\frac {4 i \, {\left (f x + e\right )} d - 4 i \, d e + 4 i \, c f}{d}\right ) - 128 \, f^{2} \cos \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right ) E_{2}\left (-\frac {2 i \, {\left (f x + e\right )} d - 2 i \, d e + 2 i \, c f}{d}\right ) + 128 i \, f^{2} E_{2}\left (-\frac {2 i \, {\left (f x + e\right )} d - 2 i \, d e + 2 i \, c f}{d}\right ) \sin \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right ) - 64 i \, f^{2} E_{2}\left (-\frac {4 i \, {\left (f x + e\right )} d - 4 i \, d e + 4 i \, c f}{d}\right ) \sin \left (-\frac {4 \, {\left (d e - c f\right )}}{d}\right ) + 64 \, f^{2}}{256 \, {\left ({\left (f x + e\right )} a^{2} d^{2} - a^{2} d^{2} e + a^{2} c d f\right )} f} \]
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 \[ \int \frac {1}{{\left (a+a\,\mathrm {cot}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2\,{\left (c+d\,x\right )}^2} \,d x \]
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 \[ - \frac {\int \frac {1}{c^{2} \cot ^{2}{\left (e + f x \right )} - 2 i c^{2} \cot {\left (e + f x \right )} - c^{2} + 2 c d x \cot ^{2}{\left (e + f x \right )} - 4 i c d x \cot {\left (e + f x \right )} - 2 c d x + d^{2} x^{2} \cot ^{2}{\left (e + f x \right )} - 2 i d^{2} x^{2} \cot {\left (e + f x \right )} - d^{2} x^{2}}\, dx}{a^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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