Optimal. Leaf size=168 \[ -\frac{i \sqrt{\tan (c+d x)}}{20 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{i \sqrt{\tan (c+d x)}}{10 a d (a+i a \tan (c+d x))^{3/2}}+\frac{i \sqrt{\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}} \]
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Rubi [A] time = 0.424374, antiderivative size = 168, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {3557, 3596, 12, 3544, 205} \[ -\frac{i \sqrt{\tan (c+d x)}}{20 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{i \sqrt{\tan (c+d x)}}{10 a d (a+i a \tan (c+d x))^{3/2}}+\frac{i \sqrt{\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}} \]
Antiderivative was successfully verified.
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Rule 3557
Rule 3596
Rule 12
Rule 3544
Rule 205
Rubi steps
\begin{align*} \int \frac{\sqrt{\tan (c+d x)}}{(a+i a \tan (c+d x))^{5/2}} \, dx &=\frac{i \sqrt{\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac{\int \frac{i a-4 a \tan (c+d x)}{\sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}} \, dx}{10 a^2}\\ &=\frac{i \sqrt{\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{i \sqrt{\tan (c+d x)}}{10 a d (a+i a \tan (c+d x))^{3/2}}-\frac{\int \frac{\frac{9 i a^2}{2}-3 a^2 \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}} \, dx}{30 a^4}\\ &=\frac{i \sqrt{\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{i \sqrt{\tan (c+d x)}}{10 a d (a+i a \tan (c+d x))^{3/2}}-\frac{i \sqrt{\tan (c+d x)}}{20 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{\int \frac{15 i a^3 \sqrt{a+i a \tan (c+d x)}}{4 \sqrt{\tan (c+d x)}} \, dx}{30 a^6}\\ &=\frac{i \sqrt{\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{i \sqrt{\tan (c+d x)}}{10 a d (a+i a \tan (c+d x))^{3/2}}-\frac{i \sqrt{\tan (c+d x)}}{20 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{i \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx}{8 a^3}\\ &=\frac{i \sqrt{\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{i \sqrt{\tan (c+d x)}}{10 a d (a+i a \tan (c+d x))^{3/2}}-\frac{i \sqrt{\tan (c+d x)}}{20 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{\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 ) \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 \sqrt{\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{i \sqrt{\tan (c+d x)}}{10 a d (a+i a \tan (c+d x))^{3/2}}-\frac{i \sqrt{\tan (c+d x)}}{20 a^2 d \sqrt{a+i a \tan (c+d x)}}\\ \end{align*}
Mathematica [A] time = 1.66907, size = 175, normalized size = 1.04 \[ \frac{i e^{-4 i (c+d x)} \sqrt{\tan (c+d x)} \left (\sqrt{-1+e^{2 i (c+d x)}} \left (3 e^{2 i (c+d x)}+e^{4 i (c+d x)}+1\right )-5 e^{5 i (c+d x)} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )\right )}{20 \sqrt{2} a^2 d \sqrt{-1+e^{2 i (c+d x)}} \sqrt{\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.05, size = 570, normalized size = 3.4 \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 = 2.92275, size = 1076, normalized size = 6.4 \begin{align*} \frac{{\left (10 \, a^{3} d \sqrt{\frac{i}{8 \, a^{5} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (\frac{1}{4} \,{\left (4 i \, a^{3} d \sqrt{\frac{i}{8 \, a^{5} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{2} \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}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 10 \, a^{3} d \sqrt{\frac{i}{8 \, a^{5} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (\frac{1}{4} \,{\left (-4 i \, a^{3} d \sqrt{\frac{i}{8 \, a^{5} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{2} \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}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt{2} \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}}{\left (i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 4 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 4 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{40 \, 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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