Optimal. Leaf size=182 \[ -\frac{23 (-1)^{3/4} a^{5/2} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{4 d}-\frac{a^2 \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 d}+\frac{9 i a^2 \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 d}-\frac{(4+4 i) a^{5/2} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d} \]
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Rubi [A] time = 0.549187, antiderivative size = 182, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.321, Rules used = {3556, 3597, 3601, 3544, 205, 3599, 63, 217, 203} \[ -\frac{23 (-1)^{3/4} a^{5/2} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{4 d}-\frac{a^2 \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 d}+\frac{9 i a^2 \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 d}-\frac{(4+4 i) a^{5/2} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d} \]
Antiderivative was successfully verified.
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Rule 3556
Rule 3597
Rule 3601
Rule 3544
Rule 205
Rule 3599
Rule 63
Rule 217
Rule 203
Rubi steps
\begin{align*} \int \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{5/2} \, dx &=-\frac{a^2 \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 d}+\frac{1}{2} a \int \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)} \left (\frac{7 a}{2}+\frac{9}{2} i a \tan (c+d x)\right ) \, dx\\ &=\frac{9 i a^2 \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 d}-\frac{a^2 \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 d}+\frac{1}{2} \int \frac{\sqrt{a+i a \tan (c+d x)} \left (-\frac{9 i a^2}{4}+\frac{23}{4} a^2 \tan (c+d x)\right )}{\sqrt{\tan (c+d x)}} \, dx\\ &=\frac{9 i a^2 \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 d}-\frac{a^2 \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 d}+\frac{1}{8} (23 i a) \int \frac{(a-i a \tan (c+d x)) \sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx-\left (4 i a^2\right ) \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx\\ &=\frac{9 i a^2 \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 d}-\frac{a^2 \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 d}+\frac{\left (23 i a^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \sqrt{a+i a x}} \, dx,x,\tan (c+d x)\right )}{8 d}-\frac{\left (8 a^4\right ) \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 )}{d}\\ &=-\frac{(4+4 i) a^{5/2} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{9 i a^2 \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 d}-\frac{a^2 \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 d}+\frac{\left (23 i a^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+i a x^2}} \, dx,x,\sqrt{\tan (c+d x)}\right )}{4 d}\\ &=-\frac{(4+4 i) a^{5/2} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{9 i a^2 \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 d}-\frac{a^2 \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 d}+\frac{\left (23 i a^3\right ) \operatorname{Subst}\left (\int \frac{1}{1-i a x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{4 d}\\ &=-\frac{23 (-1)^{3/4} a^{5/2} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{4 d}-\frac{(4+4 i) a^{5/2} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{9 i a^2 \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 d}-\frac{a^2 \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 d}\\ \end{align*}
Mathematica [A] time = 3.05915, size = 192, normalized size = 1.05 \[ \frac{a^2 \sqrt{a+i a \tan (c+d x)} \left (\frac{e^{-i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}} \left (23 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{2} e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )-32 \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )\right )}{\sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}}}-2 \sqrt{\tan (c+d x)} (2 \tan (c+d x)-9 i)\right )}{8 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.04, size = 405, normalized size = 2.2 \begin{align*}{\frac{{a}^{2}}{8\,d}\sqrt{\tan \left ( dx+c \right ) }\sqrt{a \left ( 1+i\tan \left ( dx+c \right ) \right ) } \left ( 23\,i\ln \left ({\frac{1}{2} \left ( 2\,ia\tan \left ( dx+c \right ) +2\,\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{ia}+a \right ){\frac{1}{\sqrt{ia}}}} \right ) \sqrt{-ia}a+18\,i\sqrt{ia}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }-4\,\sqrt{ia}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\tan \left ( dx+c \right ) +32\,\ln \left ( 1/2\,{\frac{2\,ia\tan \left ( dx+c \right ) +2\,\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{ia}+a}{\sqrt{ia}}} \right ) a\sqrt{-ia}+8\,i\sqrt{ia}\sqrt{2}\ln \left ({\frac{1}{\tan \left ( dx+c \right ) +i} \left ( 2\,\sqrt{2}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }-ia+3\,a\tan \left ( dx+c \right ) \right ) } \right ) a+8\,\sqrt{ia}\sqrt{2}\ln \left ({\frac{2\,\sqrt{2}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }-ia+3\,a\tan \left ( dx+c \right ) }{\tan \left ( dx+c \right ) +i}} \right ) a \right ){\frac{1}{\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }}}{\frac{1}{\sqrt{-ia}}}{\frac{1}{\sqrt{ia}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \sqrt{\tan \left (d x + c\right )}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.52395, size = 1835, normalized size = 10.08 \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 [A] time = 1.28934, size = 173, normalized size = 0.95 \begin{align*} \frac{i \, \sqrt{-2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 2 \, a^{2}}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2}{\left (\frac{-i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + i \, a^{2}}{\sqrt{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{2} - 2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{3} + a^{4}}} + 1\right )} \log \left (\sqrt{i \, a \tan \left (d x + c\right ) + a}\right )}{i \, a \tan \left (d x + c\right ) - a} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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