Optimal. Leaf size=201 \[ -\frac{7 \sqrt{a+i a \tan (c+d x)}}{2 a^2 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{13 i \sqrt{a+i a \tan (c+d x)}}{2 a^2 d \sqrt{\tan (c+d x)}}-\frac{\left (\frac{1}{4}-\frac{i}{4}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac{5}{2 a d \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}+\frac{1}{3 d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \]
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Rubi [A] time = 0.57659, antiderivative size = 201, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {3559, 3596, 3598, 12, 3544, 205} \[ -\frac{7 \sqrt{a+i a \tan (c+d x)}}{2 a^2 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{13 i \sqrt{a+i a \tan (c+d x)}}{2 a^2 d \sqrt{\tan (c+d x)}}-\frac{\left (\frac{1}{4}-\frac{i}{4}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac{5}{2 a d \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}+\frac{1}{3 d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \]
Antiderivative was successfully verified.
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Rule 3559
Rule 3596
Rule 3598
Rule 12
Rule 3544
Rule 205
Rubi steps
\begin{align*} \int \frac{1}{\tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx &=\frac{1}{3 d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac{\int \frac{\frac{9 a}{2}-3 i a \tan (c+d x)}{\tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}} \, dx}{3 a^2}\\ &=\frac{1}{3 d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac{5}{2 a d \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}+\frac{\int \frac{\sqrt{a+i a \tan (c+d x)} \left (\frac{63 a^2}{4}-15 i a^2 \tan (c+d x)\right )}{\tan ^{\frac{5}{2}}(c+d x)} \, dx}{3 a^4}\\ &=\frac{1}{3 d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac{5}{2 a d \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{7 \sqrt{a+i a \tan (c+d x)}}{2 a^2 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \int \frac{\sqrt{a+i a \tan (c+d x)} \left (-\frac{117 i a^3}{8}-\frac{63}{4} a^3 \tan (c+d x)\right )}{\tan ^{\frac{3}{2}}(c+d x)} \, dx}{9 a^5}\\ &=\frac{1}{3 d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac{5}{2 a d \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{7 \sqrt{a+i a \tan (c+d x)}}{2 a^2 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{13 i \sqrt{a+i a \tan (c+d x)}}{2 a^2 d \sqrt{\tan (c+d x)}}+\frac{4 \int -\frac{9 a^4 \sqrt{a+i a \tan (c+d x)}}{16 \sqrt{\tan (c+d x)}} \, dx}{9 a^6}\\ &=\frac{1}{3 d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac{5}{2 a d \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{7 \sqrt{a+i a \tan (c+d x)}}{2 a^2 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{13 i \sqrt{a+i a \tan (c+d x)}}{2 a^2 d \sqrt{\tan (c+d x)}}-\frac{\int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx}{4 a^2}\\ &=\frac{1}{3 d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac{5}{2 a d \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{7 \sqrt{a+i a \tan (c+d x)}}{2 a^2 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{13 i \sqrt{a+i a \tan (c+d x)}}{2 a^2 d \sqrt{\tan (c+d x)}}+\frac{i \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 )}{2 d}\\ &=-\frac{\left (\frac{1}{4}-\frac{i}{4}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac{1}{3 d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac{5}{2 a d \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{7 \sqrt{a+i a \tan (c+d x)}}{2 a^2 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{13 i \sqrt{a+i a \tan (c+d x)}}{2 a^2 d \sqrt{\tan (c+d x)}}\\ \end{align*}
Mathematica [A] time = 1.18988, size = 186, normalized size = 0.93 \[ \frac{i e^{-4 i (c+d x)} \sqrt{\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (18 e^{2 i (c+d x)}-87 e^{4 i (c+d x)}+52 e^{6 i (c+d x)}+3 e^{3 i (c+d x)} \left (-1+e^{2 i (c+d x)}\right )^{3/2} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )+1\right )}{6 \sqrt{2} a^2 d \left (-1+e^{2 i (c+d x)}\right ) \sqrt{\tan (c+d x)}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.048, size = 545, normalized size = 2.7 \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 = 3.04096, size = 1346, normalized size = 6.7 \begin{align*} -\frac{\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 (52 \, e^{\left (8 i \, d x + 8 i \, c\right )} - 35 \, e^{\left (6 i \, d x + 6 i \, c\right )} - 69 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 19 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (i \, d x + i \, c\right )} + 3 \,{\left (a^{2} d e^{\left (8 i \, d x + 8 i \, c\right )} - 2 \, a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \sqrt{-\frac{i}{2 \, a^{3} d^{2}}} \log \left (\frac{1}{4} \,{\left (2 \, a^{2} d \sqrt{-\frac{i}{2 \, a^{3} 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 ) - 3 \,{\left (a^{2} d e^{\left (8 i \, d x + 8 i \, c\right )} - 2 \, a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \sqrt{-\frac{i}{2 \, a^{3} d^{2}}} \log \left (-\frac{1}{4} \,{\left (2 \, a^{2} d \sqrt{-\frac{i}{2 \, a^{3} 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 )}{12 \,{\left (a^{2} d e^{\left (8 i \, d x + 8 i \, c\right )} - 2 \, a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )}} \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 [A] time = 1.31453, size = 174, normalized size = 0.87 \begin{align*} \frac{2 \, \sqrt{-2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 2 \, a^{2}} a^{3} \log \left (\sqrt{i \, a \tan \left (d x + c\right ) + a}\right )}{-\left (i - 1\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{6} + \left (5 i - 5\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{5} a - \left (9 i - 9\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a^{2} + \left (7 i - 7\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{3} - \left (2 i - 2\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{4}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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