Optimal. Leaf size=217 \[ -\frac{a^2 \tan ^{\frac{5}{2}}(c+d x)}{2 d \sqrt{a+i a \tan (c+d x)}}+\frac{i a^2 \tan ^{\frac{3}{2}}(c+d x)}{2 d \sqrt{a+i a \tan (c+d x)}}-\frac{11 \sqrt [4]{-1} a^{3/2} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{4 d}-\frac{(2-2 i) a^{3/2} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{5 a \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 d} \]
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Rubi [A] time = 0.714239, antiderivative size = 217, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 10, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {3556, 3595, 3597, 3601, 3544, 205, 3599, 63, 217, 203} \[ -\frac{a^2 \tan ^{\frac{5}{2}}(c+d x)}{2 d \sqrt{a+i a \tan (c+d x)}}+\frac{i a^2 \tan ^{\frac{3}{2}}(c+d x)}{2 d \sqrt{a+i a \tan (c+d x)}}-\frac{11 \sqrt [4]{-1} a^{3/2} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{4 d}-\frac{(2-2 i) a^{3/2} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{5 a \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 d} \]
Antiderivative was successfully verified.
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Rule 3556
Rule 3595
Rule 3597
Rule 3601
Rule 3544
Rule 205
Rule 3599
Rule 63
Rule 217
Rule 203
Rubi steps
\begin{align*} \int \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx &=-\frac{a^2 \tan ^{\frac{5}{2}}(c+d x)}{2 d \sqrt{a+i a \tan (c+d x)}}+\frac{1}{2} a \int \frac{\tan ^{\frac{3}{2}}(c+d x) \left (\frac{9 a}{2}+\frac{7}{2} i a \tan (c+d x)\right )}{\sqrt{a+i a \tan (c+d x)}} \, dx\\ &=\frac{i a^2 \tan ^{\frac{3}{2}}(c+d x)}{2 d \sqrt{a+i a \tan (c+d x)}}-\frac{a^2 \tan ^{\frac{5}{2}}(c+d x)}{2 d \sqrt{a+i a \tan (c+d x)}}-\frac{\int \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)} \left (\frac{3 i a^2}{2}-\frac{5}{2} a^2 \tan (c+d x)\right ) \, dx}{2 a}\\ &=\frac{i a^2 \tan ^{\frac{3}{2}}(c+d x)}{2 d \sqrt{a+i a \tan (c+d x)}}-\frac{a^2 \tan ^{\frac{5}{2}}(c+d x)}{2 d \sqrt{a+i a \tan (c+d x)}}+\frac{5 a \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 d}-\frac{\int \frac{\sqrt{a+i a \tan (c+d x)} \left (\frac{5 a^3}{4}+\frac{11}{4} i a^3 \tan (c+d x)\right )}{\sqrt{\tan (c+d x)}} \, dx}{2 a^2}\\ &=\frac{i a^2 \tan ^{\frac{3}{2}}(c+d x)}{2 d \sqrt{a+i a \tan (c+d x)}}-\frac{a^2 \tan ^{\frac{5}{2}}(c+d x)}{2 d \sqrt{a+i a \tan (c+d x)}}+\frac{5 a \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 d}+\frac{11}{8} \int \frac{(a-i a \tan (c+d x)) \sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx-(2 a) \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx\\ &=\frac{i a^2 \tan ^{\frac{3}{2}}(c+d x)}{2 d \sqrt{a+i a \tan (c+d x)}}-\frac{a^2 \tan ^{\frac{5}{2}}(c+d x)}{2 d \sqrt{a+i a \tan (c+d x)}}+\frac{5 a \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 d}+\frac{\left (11 a^2\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 (4 i a^3\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{(2-2 i) a^{3/2} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{i a^2 \tan ^{\frac{3}{2}}(c+d x)}{2 d \sqrt{a+i a \tan (c+d x)}}-\frac{a^2 \tan ^{\frac{5}{2}}(c+d x)}{2 d \sqrt{a+i a \tan (c+d x)}}+\frac{5 a \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 d}+\frac{\left (11 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+i a x^2}} \, dx,x,\sqrt{\tan (c+d x)}\right )}{4 d}\\ &=-\frac{(2-2 i) a^{3/2} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{i a^2 \tan ^{\frac{3}{2}}(c+d x)}{2 d \sqrt{a+i a \tan (c+d x)}}-\frac{a^2 \tan ^{\frac{5}{2}}(c+d x)}{2 d \sqrt{a+i a \tan (c+d x)}}+\frac{5 a \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 d}+\frac{\left (11 a^2\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{11 \sqrt [4]{-1} a^{3/2} \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{(2-2 i) a^{3/2} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{i a^2 \tan ^{\frac{3}{2}}(c+d x)}{2 d \sqrt{a+i a \tan (c+d x)}}-\frac{a^2 \tan ^{\frac{5}{2}}(c+d x)}{2 d \sqrt{a+i a \tan (c+d x)}}+\frac{5 a \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 d}\\ \end{align*}
Mathematica [A] time = 3.01684, size = 234, normalized size = 1.08 \[ \frac{a (5+2 i \tan (c+d x)) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 d}+\frac{i a e^{-i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}} \sqrt{\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (16 \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )-11 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{2} e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )\right )}{4 \sqrt{2} d \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.038, size = 403, normalized size = 1.9 \begin{align*} -{\frac{a}{8\,d}\sqrt{\tan \left ( dx+c \right ) }\sqrt{a \left ( 1+i\tan \left ( dx+c \right ) \right ) } \left ( 4\,i\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 ) \sqrt{ia}a-4\,i\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 ) -4\,\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+16\,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 ) a\sqrt{-ia}-11\,\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}-10\,\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{-ia}\sqrt{ia} \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{3}{2}} \tan \left (d x + c\right )^{\frac{3}{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.45974, size = 1785, normalized size = 8.23 \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 = 1.22207, size = 205, normalized size = 0.94 \begin{align*} \frac{{\left (-2 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} + 2 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a\right )} \sqrt{-2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 2 \, a^{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 )}{2 \,{\left ({\left (a \tan \left (d x + c\right ) - i \, a\right )} a + 2 i \, a^{2}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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