Optimal. Leaf size=349 \[ \frac{d \left (65 i c^2 d+15 c^3-117 c d^2+317 i d^3\right ) \sqrt{a+i a \tan (e+f x)}}{60 a^3 f (c-i d) (c+i d)^4 \sqrt{c+d \tan (e+f x)}}+\frac{15 c^2+70 i c d-151 d^2}{60 a^2 f (-d+i c)^3 \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{4 \sqrt{2} a^{5/2} f (c-i d)^{3/2}}+\frac{-17 d+5 i c}{30 a f (c+i d)^2 (a+i a \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}-\frac{1}{5 f (-d+i c) (a+i a \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}} \]
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Rubi [A] time = 1.24015, antiderivative size = 349, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {3559, 3596, 3598, 12, 3544, 208} \[ \frac{d \left (65 i c^2 d+15 c^3-117 c d^2+317 i d^3\right ) \sqrt{a+i a \tan (e+f x)}}{60 a^3 f (c-i d) (c+i d)^4 \sqrt{c+d \tan (e+f x)}}+\frac{15 c^2+70 i c d-151 d^2}{60 a^2 f (-d+i c)^3 \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{4 \sqrt{2} a^{5/2} f (c-i d)^{3/2}}+\frac{-17 d+5 i c}{30 a f (c+i d)^2 (a+i a \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}-\frac{1}{5 f (-d+i c) (a+i a \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}} \]
Antiderivative was successfully verified.
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Rule 3559
Rule 3596
Rule 3598
Rule 12
Rule 3544
Rule 208
Rubi steps
\begin{align*} \int \frac{1}{(a+i a \tan (e+f x))^{5/2} (c+d \tan (e+f x))^{3/2}} \, dx &=-\frac{1}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}}-\frac{\int \frac{-\frac{1}{2} a (5 i c-11 d)-3 i a d \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \, dx}{5 a^2 (i c-d)}\\ &=-\frac{1}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}}+\frac{5 i c-17 d}{30 a (c+i d)^2 f (a+i a \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}-\frac{\int \frac{-\frac{1}{4} a^2 \left (15 c^2+50 i c d-83 d^2\right )-a^2 (5 c+17 i d) d \tan (e+f x)}{\sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}} \, dx}{15 a^4 (c+i d)^2}\\ &=-\frac{1}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}}+\frac{5 i c-17 d}{30 a (c+i d)^2 f (a+i a \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}+\frac{15 c^2+70 i c d-151 d^2}{60 a^2 (i c-d)^3 f \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}-\frac{\int \frac{\sqrt{a+i a \tan (e+f x)} \left (\frac{1}{8} a^3 \left (15 i c^3-75 c^2 d-185 i c d^2+317 d^3\right )+\frac{1}{4} a^3 d \left (15 i c^2-70 c d-151 i d^2\right ) \tan (e+f x)\right )}{(c+d \tan (e+f x))^{3/2}} \, dx}{15 a^6 (i c-d)^3}\\ &=-\frac{1}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}}+\frac{5 i c-17 d}{30 a (c+i d)^2 f (a+i a \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}+\frac{15 c^2+70 i c d-151 d^2}{60 a^2 (i c-d)^3 f \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}+\frac{d \left (15 c^3+65 i c^2 d-117 c d^2+317 i d^3\right ) \sqrt{a+i a \tan (e+f x)}}{60 a^3 (c-i d) (c+i d)^4 f \sqrt{c+d \tan (e+f x)}}-\frac{2 \int \frac{15 i a^4 (c+i d)^4 \sqrt{a+i a \tan (e+f x)}}{16 \sqrt{c+d \tan (e+f x)}} \, dx}{15 a^7 (i c-d)^3 \left (c^2+d^2\right )}\\ &=-\frac{1}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}}+\frac{5 i c-17 d}{30 a (c+i d)^2 f (a+i a \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}+\frac{15 c^2+70 i c d-151 d^2}{60 a^2 (i c-d)^3 f \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}+\frac{d \left (15 c^3+65 i c^2 d-117 c d^2+317 i d^3\right ) \sqrt{a+i a \tan (e+f x)}}{60 a^3 (c-i d) (c+i d)^4 f \sqrt{c+d \tan (e+f x)}}+\frac{\int \frac{\sqrt{a+i a \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}} \, dx}{8 a^3 (c-i d)}\\ &=-\frac{1}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}}+\frac{5 i c-17 d}{30 a (c+i d)^2 f (a+i a \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}+\frac{15 c^2+70 i c d-151 d^2}{60 a^2 (i c-d)^3 f \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}+\frac{d \left (15 c^3+65 i c^2 d-117 c d^2+317 i d^3\right ) \sqrt{a+i a \tan (e+f x)}}{60 a^3 (c-i d) (c+i d)^4 f \sqrt{c+d \tan (e+f x)}}+\frac{\operatorname{Subst}\left (\int \frac{1}{a c-i a d-2 a^2 x^2} \, dx,x,\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{a+i a \tan (e+f x)}}\right )}{4 a (i c+d) f}\\ &=-\frac{i \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{4 \sqrt{2} a^{5/2} (c-i d)^{3/2} f}-\frac{1}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}}+\frac{5 i c-17 d}{30 a (c+i d)^2 f (a+i a \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}+\frac{15 c^2+70 i c d-151 d^2}{60 a^2 (i c-d)^3 f \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}+\frac{d \left (15 c^3+65 i c^2 d-117 c d^2+317 i d^3\right ) \sqrt{a+i a \tan (e+f x)}}{60 a^3 (c-i d) (c+i d)^4 f \sqrt{c+d \tan (e+f x)}}\\ \end{align*}
Mathematica [B] time = 8.9528, size = 788, normalized size = 2.26 \[ \frac{\sec ^3(e+f x) (\cos (f x)+i \sin (f x))^3 \left (\frac{\left (17 c^2+77 i c d-126 d^2\right ) \left (-\frac{\sin (e)}{60}+\frac{1}{60} i \cos (e)\right ) \cos (2 f x)}{(c+i d)^4}+\frac{\left (17 c^2+77 i c d-126 d^2\right ) \left (\frac{\cos (e)}{60}+\frac{1}{60} i \sin (e)\right ) \sin (2 f x)}{(c+i d)^4}+\frac{\left (\frac{1}{120} \cos (3 e)+\frac{1}{120} i \sin (3 e)\right ) \left (91 i c^2 d^2 \sin (e)-109 c^2 d^2 \cos (e)+23 c^3 d \sin (e)+91 i c^3 d \cos (e)+23 c^4 \cos (e)-109 c d^3 \sin (e)+223 i c d^3 \cos (e)+223 i d^4 \sin (e)+240 d^4 \cos (e)\right )}{(c-i d) (c+i d)^4 (-i c \cos (e)-i d \sin (e))}+\frac{2 \left (\frac{1}{2} i d^5 \sin (3 e-f x)-\frac{1}{2} i d^5 \sin (3 e+f x)+\frac{1}{2} d^5 \cos (3 e-f x)-\frac{1}{2} d^5 \cos (3 e+f x)\right )}{(c-i d) (c+i d)^4 (c \cos (e)+d \sin (e)) (c \cos (e+f x)+d \sin (e+f x))}+\frac{(7 c+16 i d) \left (\frac{\sin (e)}{60}+\frac{1}{60} i \cos (e)\right ) \cos (4 f x)}{(c+i d)^3}+\frac{\left (\frac{1}{40} \sin (3 e)+\frac{1}{40} i \cos (3 e)\right ) \cos (6 f x)}{(c+i d)^2}+\frac{(7 c+16 i d) \left (\frac{\cos (e)}{60}-\frac{1}{60} i \sin (e)\right ) \sin (4 f x)}{(c+i d)^3}+\frac{\left (\frac{1}{40} \cos (3 e)-\frac{1}{40} i \sin (3 e)\right ) \sin (6 f x)}{(c+i d)^2}\right ) \sqrt{\sec (e+f x) (c \cos (e+f x)+d \sin (e+f x))}}{f (a+i a \tan (e+f x))^{5/2}}-\frac{i e^{3 i e} \sqrt{e^{i f x}} \sec ^{\frac{5}{2}}(e+f x) (\cos (f x)+i \sin (f x))^{5/2} \log \left (2 \left (\sqrt{c-i d} e^{i (e+f x)}+\sqrt{1+e^{2 i (e+f x)}} \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}\right )\right )}{4 \sqrt{2} f (c-i d)^{3/2} \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \sqrt{1+e^{2 i (e+f x)}} (a+i a \tan (e+f x))^{5/2}} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 0.107, size = 7870, normalized size = 22.6 \begin{align*} \text{output too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.2659, size = 2824, normalized size = 8.09 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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