Optimal. Leaf size=205 \[ -\frac{39 \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{10 a^2 d}-\frac{151 (a+i a \tan (c+d x))^{3/2}}{30 a^3 d}+\frac{78 \sqrt{a+i a \tan (c+d x)}}{5 a^2 d}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{\tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{19 i \tan ^3(c+d x)}{6 a d \sqrt{a+i a \tan (c+d x)}} \]
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Rubi [A] time = 0.491853, antiderivative size = 205, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.269, Rules used = {3558, 3595, 3597, 3592, 3527, 3480, 206} \[ -\frac{39 \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{10 a^2 d}-\frac{151 (a+i a \tan (c+d x))^{3/2}}{30 a^3 d}+\frac{78 \sqrt{a+i a \tan (c+d x)}}{5 a^2 d}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{\tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{19 i \tan ^3(c+d x)}{6 a d \sqrt{a+i a \tan (c+d x)}} \]
Antiderivative was successfully verified.
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Rule 3558
Rule 3595
Rule 3597
Rule 3592
Rule 3527
Rule 3480
Rule 206
Rubi steps
\begin{align*} \int \frac{\tan ^5(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx &=-\frac{\tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac{\int \frac{\tan ^3(c+d x) \left (-4 a+\frac{11}{2} i a \tan (c+d x)\right )}{\sqrt{a+i a \tan (c+d x)}} \, dx}{3 a^2}\\ &=-\frac{\tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{19 i \tan ^3(c+d x)}{6 a d \sqrt{a+i a \tan (c+d x)}}+\frac{\int \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)} \left (-\frac{57 i a^2}{2}-\frac{117}{4} a^2 \tan (c+d x)\right ) \, dx}{3 a^4}\\ &=-\frac{\tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{19 i \tan ^3(c+d x)}{6 a d \sqrt{a+i a \tan (c+d x)}}-\frac{39 \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{10 a^2 d}+\frac{2 \int \tan (c+d x) \sqrt{a+i a \tan (c+d x)} \left (\frac{117 a^3}{2}-\frac{453}{8} i a^3 \tan (c+d x)\right ) \, dx}{15 a^5}\\ &=-\frac{\tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{19 i \tan ^3(c+d x)}{6 a d \sqrt{a+i a \tan (c+d x)}}-\frac{39 \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{10 a^2 d}-\frac{151 (a+i a \tan (c+d x))^{3/2}}{30 a^3 d}+\frac{2 \int \sqrt{a+i a \tan (c+d x)} \left (\frac{453 i a^3}{8}+\frac{117}{2} a^3 \tan (c+d x)\right ) \, dx}{15 a^5}\\ &=-\frac{\tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{19 i \tan ^3(c+d x)}{6 a d \sqrt{a+i a \tan (c+d x)}}+\frac{78 \sqrt{a+i a \tan (c+d x)}}{5 a^2 d}-\frac{39 \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{10 a^2 d}-\frac{151 (a+i a \tan (c+d x))^{3/2}}{30 a^3 d}-\frac{i \int \sqrt{a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=-\frac{\tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{19 i \tan ^3(c+d x)}{6 a d \sqrt{a+i a \tan (c+d x)}}+\frac{78 \sqrt{a+i a \tan (c+d x)}}{5 a^2 d}-\frac{39 \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{10 a^2 d}-\frac{151 (a+i a \tan (c+d x))^{3/2}}{30 a^3 d}-\frac{\operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+i a \tan (c+d x)}\right )}{2 a d}\\ &=-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{\tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{19 i \tan ^3(c+d x)}{6 a d \sqrt{a+i a \tan (c+d x)}}+\frac{78 \sqrt{a+i a \tan (c+d x)}}{5 a^2 d}-\frac{39 \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{10 a^2 d}-\frac{151 (a+i a \tan (c+d x))^{3/2}}{30 a^3 d}\\ \end{align*}
Mathematica [A] time = 1.71896, size = 149, normalized size = 0.73 \[ \frac{e^{-2 i (c+d x)} \left (115 e^{2 i (c+d x)}+855 e^{4 i (c+d x)}+1105 e^{6 i (c+d x)}+466 e^{8 i (c+d x)}-15 e^{3 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{5/2} \sinh ^{-1}\left (e^{i (c+d x)}\right )-5\right )}{30 a d \left (1+e^{2 i (c+d x)}\right )^3 \sqrt{a+i a \tan (c+d x)}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.023, size = 131, normalized size = 0.6 \begin{align*} 2\,{\frac{1}{d{a}^{4}} \left ( 1/5\, \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{5/2}- \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{3/2}a+4\,{a}^{2}\sqrt{a+ia\tan \left ( dx+c \right ) }-1/8\,{a}^{5/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a+ia\tan \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ) +9/4\,{\frac{{a}^{3}}{\sqrt{a+ia\tan \left ( dx+c \right ) }}}-1/6\,{\frac{{a}^{4}}{ \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{3/2}}} \right ) } \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: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.42575, size = 1123, normalized size = 5.48 \begin{align*} \frac{\sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (466 \, e^{\left (8 i \, d x + 8 i \, c\right )} + 1105 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 855 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 115 \, e^{\left (2 i \, d x + 2 i \, c\right )} - 5\right )} e^{\left (i \, d x + i \, c\right )} - 15 \, \sqrt{\frac{1}{2}}{\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{1}{a^{3} d^{2}}} \log \left ({\left (2 \, \sqrt{\frac{1}{2}} a^{2} d \sqrt{\frac{1}{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}}{\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 ) + 15 \, \sqrt{\frac{1}{2}}{\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{1}{a^{3} d^{2}}} \log \left (-{\left (2 \, \sqrt{\frac{1}{2}} a^{2} d \sqrt{\frac{1}{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}}{\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 )}{60 \,{\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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{5}{\left (c + d x \right )}}{\left (a \left (i \tan{\left (c + d x \right )} + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (d x + c\right )^{5}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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