Optimal. Leaf size=434 \[ -\frac{f \text{CosIntegral}\left (\frac{4 c f}{d}+4 f x\right ) \sin \left (4 e-\frac{4 c f}{d}\right )}{a^2 d^2}+\frac{f \text{CosIntegral}\left (\frac{2 c f}{d}+2 f x\right ) \sin \left (2 e-\frac{2 c f}{d}\right )}{a^2 d^2}-\frac{i f \text{CosIntegral}\left (\frac{2 c f}{d}+2 f x\right ) \cos \left (2 e-\frac{2 c f}{d}\right )}{a^2 d^2}+\frac{i f \text{CosIntegral}\left (\frac{4 c f}{d}+4 f x\right ) \cos \left (4 e-\frac{4 c f}{d}\right )}{a^2 d^2}+\frac{i f \sin \left (2 e-\frac{2 c f}{d}\right ) \text{Si}\left (2 x f+\frac{2 c f}{d}\right )}{a^2 d^2}-\frac{i f \sin \left (4 e-\frac{4 c f}{d}\right ) \text{Si}\left (4 x f+\frac{4 c f}{d}\right )}{a^2 d^2}+\frac{f \cos \left (2 e-\frac{2 c f}{d}\right ) \text{Si}\left (2 x f+\frac{2 c f}{d}\right )}{a^2 d^2}-\frac{f \cos \left (4 e-\frac{4 c f}{d}\right ) \text{Si}\left (4 x f+\frac{4 c f}{d}\right )}{a^2 d^2}+\frac{\sin ^2(2 e+2 f x)}{4 a^2 d (c+d x)}+\frac{i \sin (2 e+2 f x)}{2 a^2 d (c+d x)}-\frac{i \sin (4 e+4 f x)}{4 a^2 d (c+d x)}-\frac{\cos ^2(2 e+2 f x)}{4 a^2 d (c+d x)}+\frac{\cos (2 e+2 f x)}{2 a^2 d (c+d x)}-\frac{1}{4 a^2 d (c+d x)} \]
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Rubi [A] time = 0.680654, antiderivative size = 434, normalized size of antiderivative = 1., 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 3728
Rule 3297
Rule 3303
Rule 3299
Rule 3302
Rule 3313
Rule 12
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*}
Mathematica [A] time = 0.548107, size = 203, normalized size = 0.47 \[ \frac{4 f (c+d x) \left (\text{CosIntegral}\left (\frac{2 f (c+d x)}{d}\right )+i \text{Si}\left (\frac{2 f (c+d x)}{d}\right )\right ) \left (\sin \left (2 e-\frac{2 c f}{d}\right )-i \cos \left (2 e-\frac{2 c f}{d}\right )\right )+(c+d x) \left (\text{CosIntegral}\left (\frac{4 f (c+d x)}{d}\right )+i \text{Si}\left (\frac{4 f (c+d x)}{d}\right )\right ) \left (-4 f \sin \left (4 e-\frac{4 c f}{d}\right )+4 i f \cos \left (4 e-\frac{4 c f}{d}\right )\right )+2 d (\cos (2 (e+f x))+i \sin (2 (e+f x)))-d (\cos (4 (e+f x))+i \sin (4 (e+f x)))-d}{4 a^2 d^2 (c+d x)} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.125, size = 537, normalized size = 1.2 \begin{align*} -{\frac{1}{{a}^{2}f} \left ({\frac{i}{4}}{f}^{2} \left ( -2\,{\frac{\sin \left ( 2\,fx+2\,e \right ) }{ \left ( \left ( fx+e \right ) d+cf-de \right ) d}}+2\,{\frac{1}{d} \left ( 2\,{\frac{1}{d}{\it Si} \left ( 2\,fx+2\,e+2\,{\frac{cf-de}{d}} \right ) \sin \left ( 2\,{\frac{cf-de}{d}} \right ) }+2\,{\frac{1}{d}{\it Ci} \left ( 2\,fx+2\,e+2\,{\frac{cf-de}{d}} \right ) \cos \left ( 2\,{\frac{cf-de}{d}} \right ) } \right ) } \right ) -{\frac{i}{16}}{f}^{2} \left ( -4\,{\frac{\sin \left ( 4\,fx+4\,e \right ) }{ \left ( \left ( fx+e \right ) d+cf-de \right ) d}}+4\,{\frac{1}{d} \left ( 4\,{\frac{1}{d}{\it Si} \left ( 4\,fx+4\,e+4\,{\frac{cf-de}{d}} \right ) \sin \left ( 4\,{\frac{cf-de}{d}} \right ) }+4\,{\frac{1}{d}{\it Ci} \left ( 4\,fx+4\,e+4\,{\frac{cf-de}{d}} \right ) \cos \left ( 4\,{\frac{cf-de}{d}} \right ) } \right ) } \right ) +{\frac{{f}^{2}}{ \left ( 4\, \left ( fx+e \right ) d+4\,cf-4\,de \right ) d}}-{\frac{{f}^{2}}{16} \left ( -4\,{\frac{\cos \left ( 4\,fx+4\,e \right ) }{ \left ( \left ( fx+e \right ) d+cf-de \right ) d}}-4\,{\frac{1}{d} \left ( 4\,{\frac{1}{d}{\it Si} \left ( 4\,fx+4\,e+4\,{\frac{cf-de}{d}} \right ) \cos \left ( 4\,{\frac{cf-de}{d}} \right ) }-4\,{\frac{1}{d}{\it Ci} \left ( 4\,fx+4\,e+4\,{\frac{cf-de}{d}} \right ) \sin \left ( 4\,{\frac{cf-de}{d}} \right ) } \right ) } \right ) }+{\frac{{f}^{2}}{4} \left ( -2\,{\frac{\cos \left ( 2\,fx+2\,e \right ) }{ \left ( \left ( fx+e \right ) d+cf-de \right ) d}}-2\,{\frac{1}{d} \left ( 2\,{\frac{1}{d}{\it Si} \left ( 2\,fx+2\,e+2\,{\frac{cf-de}{d}} \right ) \cos \left ( 2\,{\frac{cf-de}{d}} \right ) }-2\,{\frac{1}{d}{\it Ci} \left ( 2\,fx+2\,e+2\,{\frac{cf-de}{d}} \right ) \sin \left ( 2\,{\frac{cf-de}{d}} \right ) } \right ) } \right ) } \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.50112, size = 289, normalized size = 0.67 \begin{align*} -\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} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.62667, size = 324, normalized size = 0.75 \begin{align*} \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 )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 2.24745, size = 1551, normalized size = 3.57 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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