Optimal. Leaf size=305 \[ -\frac{i \text{CosIntegral}\left (\frac{2 c f}{d}+2 f x\right ) \sin \left (2 e-\frac{2 c f}{d}\right )}{2 a^2 d}+\frac{i \text{CosIntegral}\left (\frac{4 c f}{d}+4 f x\right ) \sin \left (4 e-\frac{4 c f}{d}\right )}{4 a^2 d}-\frac{\text{CosIntegral}\left (\frac{2 c f}{d}+2 f x\right ) \cos \left (2 e-\frac{2 c f}{d}\right )}{2 a^2 d}+\frac{\text{CosIntegral}\left (\frac{4 c f}{d}+4 f x\right ) \cos \left (4 e-\frac{4 c f}{d}\right )}{4 a^2 d}+\frac{\sin \left (2 e-\frac{2 c f}{d}\right ) \text{Si}\left (2 x f+\frac{2 c f}{d}\right )}{2 a^2 d}-\frac{\sin \left (4 e-\frac{4 c f}{d}\right ) \text{Si}\left (4 x f+\frac{4 c f}{d}\right )}{4 a^2 d}-\frac{i \cos \left (2 e-\frac{2 c f}{d}\right ) \text{Si}\left (2 x f+\frac{2 c f}{d}\right )}{2 a^2 d}+\frac{i \cos \left (4 e-\frac{4 c f}{d}\right ) \text{Si}\left (4 x f+\frac{4 c f}{d}\right )}{4 a^2 d}+\frac{\log (c+d x)}{4 a^2 d} \]
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Rubi [A] time = 0.752218, antiderivative size = 305, normalized size of antiderivative = 1., 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 3728
Rule 3303
Rule 3299
Rule 3302
Rule 3312
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*}
Mathematica [A] time = 0.563459, size = 136, normalized size = 0.45 \[ \frac{-2 \left (\text{CosIntegral}\left (\frac{2 f (c+d x)}{d}\right )+i \text{Si}\left (\frac{2 f (c+d x)}{d}\right )\right ) \left (\cos \left (2 e-\frac{2 c f}{d}\right )+i \sin \left (2 e-\frac{2 c f}{d}\right )\right )+\left (\text{CosIntegral}\left (\frac{4 f (c+d x)}{d}\right )+i \text{Si}\left (\frac{4 f (c+d x)}{d}\right )\right ) \left (\cos \left (4 e-\frac{4 c f}{d}\right )+i \sin \left (4 e-\frac{4 c f}{d}\right )\right )+\log (c+d x)}{4 a^2 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.133, size = 382, normalized size = 1.3 \begin{align*}{\frac{-{\frac{i}{2}}}{{a}^{2}d}{\it Si} \left ( 2\,fx+2\,e+2\,{\frac{cf-de}{d}} \right ) \cos \left ( 2\,{\frac{cf-de}{d}} \right ) }+{\frac{{\frac{i}{2}}}{{a}^{2}d}{\it Ci} \left ( 2\,fx+2\,e+2\,{\frac{cf-de}{d}} \right ) \sin \left ( 2\,{\frac{cf-de}{d}} \right ) }+{\frac{{\frac{i}{4}}}{{a}^{2}d}{\it Si} \left ( 4\,fx+4\,e+4\,{\frac{cf-de}{d}} \right ) \cos \left ( 4\,{\frac{cf-de}{d}} \right ) }-{\frac{{\frac{i}{4}}}{{a}^{2}d}{\it Ci} \left ( 4\,fx+4\,e+4\,{\frac{cf-de}{d}} \right ) \sin \left ( 4\,{\frac{cf-de}{d}} \right ) }+{\frac{\ln \left ( \left ( fx+e \right ) d+cf-de \right ) }{4\,{a}^{2}d}}+{\frac{1}{4\,{a}^{2}d}{\it Si} \left ( 4\,fx+4\,e+4\,{\frac{cf-de}{d}} \right ) \sin \left ( 4\,{\frac{cf-de}{d}} \right ) }+{\frac{1}{4\,{a}^{2}d}{\it Ci} \left ( 4\,fx+4\,e+4\,{\frac{cf-de}{d}} \right ) \cos \left ( 4\,{\frac{cf-de}{d}} \right ) }-{\frac{1}{2\,{a}^{2}d}{\it Si} \left ( 2\,fx+2\,e+2\,{\frac{cf-de}{d}} \right ) \sin \left ( 2\,{\frac{cf-de}{d}} \right ) }-{\frac{1}{2\,{a}^{2}d}{\it Ci} \left ( 2\,fx+2\,e+2\,{\frac{cf-de}{d}} \right ) \cos \left ( 2\,{\frac{cf-de}{d}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.31033, size = 262, normalized size = 0.86 \begin{align*} -\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} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.68577, size = 197, normalized size = 0.65 \begin{align*} \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} \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 = 1.26621, size = 1332, normalized size = 4.37 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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