Optimal. Leaf size=436 \[ -\frac{f \text{CosIntegral}\left (\frac{4 c f}{d}+4 f x\right ) \sin \left (4 e-\frac{4 c f}{d}\right )}{a^2 d^2}-\frac{f \text{CosIntegral}\left (\frac{2 c f}{d}+2 f x\right ) \sin \left (2 e-\frac{2 c f}{d}\right )}{a^2 d^2}-\frac{i f \text{CosIntegral}\left (\frac{2 c f}{d}+2 f x\right ) \cos \left (2 e-\frac{2 c f}{d}\right )}{a^2 d^2}-\frac{i f \text{CosIntegral}\left (\frac{4 c f}{d}+4 f x\right ) \cos \left (4 e-\frac{4 c f}{d}\right )}{a^2 d^2}+\frac{i f \sin \left (2 e-\frac{2 c f}{d}\right ) \text{Si}\left (2 x f+\frac{2 c f}{d}\right )}{a^2 d^2}+\frac{i f \sin \left (4 e-\frac{4 c f}{d}\right ) \text{Si}\left (4 x f+\frac{4 c f}{d}\right )}{a^2 d^2}-\frac{f \cos \left (2 e-\frac{2 c f}{d}\right ) \text{Si}\left (2 x f+\frac{2 c f}{d}\right )}{a^2 d^2}-\frac{f \cos \left (4 e-\frac{4 c f}{d}\right ) \text{Si}\left (4 x f+\frac{4 c f}{d}\right )}{a^2 d^2}+\frac{\sin ^2(2 e+2 f x)}{4 a^2 d (c+d x)}+\frac{i \sin (2 e+2 f x)}{2 a^2 d (c+d x)}+\frac{i \sin (4 e+4 f x)}{4 a^2 d (c+d x)}-\frac{\cos ^2(2 e+2 f x)}{4 a^2 d (c+d x)}-\frac{\cos (2 e+2 f x)}{2 a^2 d (c+d x)}-\frac{1}{4 a^2 d (c+d x)} \]
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Rubi [A] time = 0.738799, antiderivative size = 436, 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 \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*}
Mathematica [A] time = 1.37489, size = 467, normalized size = 1.07 \[ -\frac{\left (\cos \left (2 \left (f \left (x-\frac{c}{d}\right )+e\right )\right )-i \sin \left (2 \left (f \left (x-\frac{c}{d}\right )+e\right )\right )\right ) \left (4 f (c+d x) \text{CosIntegral}\left (\frac{4 f (c+d x)}{d}\right ) \left (\sin \left (2 e-\frac{2 f (c+d x)}{d}\right )+i \cos \left (2 e-\frac{2 f (c+d x)}{d}\right )\right )+4 i f (c+d x) (\cos (2 f x)+i \sin (2 f x)) \text{CosIntegral}\left (\frac{2 f (c+d x)}{d}\right )-4 i c f \text{Si}\left (\frac{4 f (c+d x)}{d}\right ) \sin \left (2 e-\frac{2 f (c+d x)}{d}\right )-4 i d f x \text{Si}\left (\frac{4 f (c+d x)}{d}\right ) \sin \left (2 e-\frac{2 f (c+d x)}{d}\right )+4 c f \text{Si}\left (\frac{4 f (c+d x)}{d}\right ) \cos \left (2 e-\frac{2 f (c+d x)}{d}\right )+4 d f x \text{Si}\left (\frac{4 f (c+d x)}{d}\right ) \cos \left (2 e-\frac{2 f (c+d x)}{d}\right )+i d \sin \left (2 \left (f \left (x-\frac{c}{d}\right )+e\right )\right )-i d \sin \left (2 \left (f \left (\frac{c}{d}+x\right )+e\right )\right )+d \cos \left (2 \left (f \left (x-\frac{c}{d}\right )+e\right )\right )+d \cos \left (2 \left (f \left (\frac{c}{d}+x\right )+e\right )\right )+4 i c f \sin (2 f x) \text{Si}\left (\frac{2 f (c+d x)}{d}\right )+4 i d f x \sin (2 f x) \text{Si}\left (\frac{2 f (c+d x)}{d}\right )+4 c f \cos (2 f x) \text{Si}\left (\frac{2 f (c+d x)}{d}\right )+4 d f x \cos (2 f x) \text{Si}\left (\frac{2 f (c+d x)}{d}\right )-2 i d \sin \left (\frac{2 c f}{d}\right )+2 d \cos \left (\frac{2 c f}{d}\right )\right )}{4 a^2 d^2 (c+d x)} \]
Warning: Unable to verify antiderivative.
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Maple [A] time = 0.277, size = 175, normalized size = 0.4 \begin{align*} -{\frac{1}{4\,{a}^{2}d \left ( dx+c \right ) }}-{\frac{f{{\rm e}^{-4\,i \left ( fx+e \right ) }}}{4\,{a}^{2} \left ( dfx+cf \right ) d}}+{\frac{if}{{a}^{2}{d}^{2}}{{\rm e}^{{\frac{4\,i \left ( cf-de \right ) }{d}}}}{\it Ei} \left ( 1,4\,ifx+4\,ie+{\frac{4\,i \left ( cf-de \right ) }{d}} \right ) }-{\frac{f{{\rm e}^{-2\,i \left ( fx+e \right ) }}}{2\,{a}^{2} \left ( dfx+cf \right ) d}}+{\frac{if}{{a}^{2}{d}^{2}}{{\rm e}^{{\frac{2\,i \left ( cf-de \right ) }{d}}}}{\it Ei} \left ( 1,2\,ifx+2\,ie+{\frac{2\,i \left ( cf-de \right ) }{d}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.50772, size = 284, normalized size = 0.65 \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 = 1.62919, size = 362, normalized size = 0.83 \begin{align*} \frac{{\left ({\left ({\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 )} +{\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 )} - 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{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 = 1.99281, size = 1551, normalized size = 3.56 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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