Optimal. Leaf size=196 \[ -\frac{(-13 B+3 i A) \sqrt{\tan (c+d x)}}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{(7 B+3 i A) \sqrt{\tan (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{(-B+i A) \sqrt{\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}} \]
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Rubi [A] time = 0.592109, antiderivative size = 196, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.132, Rules used = {3595, 3596, 12, 3544, 205} \[ -\frac{(-13 B+3 i A) \sqrt{\tan (c+d x)}}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{(7 B+3 i A) \sqrt{\tan (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{(-B+i A) \sqrt{\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}} \]
Antiderivative was successfully verified.
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Rule 3595
Rule 3596
Rule 12
Rule 3544
Rule 205
Rubi steps
\begin{align*} \int \frac{\sqrt{\tan (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx &=\frac{(i A-B) \sqrt{\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac{\int \frac{\frac{1}{2} a (i A-B)-a (2 A-3 i B) \tan (c+d x)}{\sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}} \, dx}{5 a^2}\\ &=\frac{(i A-B) \sqrt{\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(3 i A+7 B) \sqrt{\tan (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}-\frac{\int \frac{\frac{1}{4} a^2 (9 i A+B)-\frac{1}{2} a^2 (3 A-7 i B) \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}} \, dx}{15 a^4}\\ &=\frac{(i A-B) \sqrt{\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(3 i A+7 B) \sqrt{\tan (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}-\frac{(3 i A-13 B) \sqrt{\tan (c+d x)}}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{\int \frac{15 a^3 (i A+B) \sqrt{a+i a \tan (c+d x)}}{8 \sqrt{\tan (c+d x)}} \, dx}{15 a^6}\\ &=\frac{(i A-B) \sqrt{\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(3 i A+7 B) \sqrt{\tan (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}-\frac{(3 i A-13 B) \sqrt{\tan (c+d x)}}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{(i A+B) \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx}{8 a^3}\\ &=\frac{(i A-B) \sqrt{\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(3 i A+7 B) \sqrt{\tan (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}-\frac{(3 i A-13 B) \sqrt{\tan (c+d x)}}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{(A-i B) \operatorname{Subst}\left (\int \frac{1}{-i a-2 a^2 x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{4 a d}\\ &=-\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{(i A-B) \sqrt{\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(3 i A+7 B) \sqrt{\tan (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}-\frac{(3 i A-13 B) \sqrt{\tan (c+d x)}}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}\\ \end{align*}
Mathematica [A] time = 5.70813, size = 215, normalized size = 1.1 \[ \frac{e^{-2 i (c+d x)} \sqrt{\tan (c+d x)} \sec ^2(c+d x) \left (\sqrt{-1+e^{2 i (c+d x)}} \left (-3 i A \left (3 e^{2 i (c+d x)}+e^{4 i (c+d x)}+1\right )-B \left (e^{2 i (c+d x)}+17 e^{4 i (c+d x)}-3\right )\right )+15 (B+i A) e^{5 i (c+d x)} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )\right )}{60 a^2 d \sqrt{-1+e^{2 i (c+d x)}} (\tan (c+d x)-i)^2 \sqrt{a+i a \tan (c+d x)}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.052, size = 1096, normalized size = 5.6 \begin{align*} \text{result 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 = 2.10997, size = 1357, normalized size = 6.92 \begin{align*} -\frac{{\left (15 \, \sqrt{\frac{1}{2}} a^{3} d \sqrt{\frac{i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (\frac{{\left (2 \, \sqrt{\frac{1}{2}} a^{3} d \sqrt{\frac{i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{2}{\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 i \, A + 4 \, B}\right ) - 15 \, \sqrt{\frac{1}{2}} a^{3} d \sqrt{\frac{i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (-\frac{{\left (2 \, \sqrt{\frac{1}{2}} a^{3} d \sqrt{\frac{i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt{2}{\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 i \, A + 4 \, B}\right ) - \sqrt{2}{\left ({\left (3 i \, A + 17 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (12 i \, A + 18 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (12 i \, A - 2 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i \, A - 3 \, B\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{120 \, a^{3} d} \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(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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