Optimal. Leaf size=135 \[ -\frac{\sqrt [4]{-1} \sqrt{a} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{\sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{d}-\frac{(1-i) \sqrt{a} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d} \]
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Rubi [A] time = 0.369064, antiderivative size = 135, 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 = {3560, 21, 3555, 3544, 205, 3599, 63, 217, 203} \[ -\frac{\sqrt [4]{-1} \sqrt{a} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{\sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{d}-\frac{(1-i) \sqrt{a} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d} \]
Antiderivative was successfully verified.
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Rule 3560
Rule 21
Rule 3555
Rule 3544
Rule 205
Rule 3599
Rule 63
Rule 217
Rule 203
Rubi steps
\begin{align*} \int \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)} \, dx &=\frac{\sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{d}-\frac{\int \frac{\left (\frac{a}{2}+\frac{1}{2} i a \tan (c+d x)\right ) \sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx}{a}\\ &=\frac{\sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{d}-\frac{\int \frac{(a+i a \tan (c+d x))^{3/2}}{\sqrt{\tan (c+d x)}} \, dx}{2 a}\\ &=\frac{\sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{d}-\frac{i \int \frac{\sqrt{a+i a \tan (c+d x)} (i a+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{\sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{d}+\frac{a \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \sqrt{a+i a x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac{\left (2 i a^2\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{(1-i) \sqrt{a} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{\sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{d}+\frac{a \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+i a x^2}} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=-\frac{(1-i) \sqrt{a} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{\sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{d}+\frac{a \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 )}{d}\\ &=-\frac{\sqrt [4]{-1} \sqrt{a} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{(1-i) \sqrt{a} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{\sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{d}\\ \end{align*}
Mathematica [A] time = 3.61913, size = 249, normalized size = 1.84 \[ \frac{\sqrt{a+i a \tan (c+d x)} \left (8 \sqrt{\tan (c+d x)}+\frac{i e^{-i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}} \left (8 \log \left (\sqrt{-1+e^{2 i (c+d x)}}+e^{i (c+d x)}\right )+\sqrt{2} \left (\log \left (2 \sqrt{2} e^{i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}}-3 e^{2 i (c+d x)}+1\right )-\log \left (-2 \sqrt{2} e^{i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}}-3 e^{2 i (c+d x)}+1\right )\right )\right )}{\sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}}}\right )}{8 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.06, size = 228, normalized size = 1.7 \begin{align*}{\frac{1}{2\,d} \left ( \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}+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+2\,\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{-ia}\sqrt{ia} \right ) \sqrt{\tan \left ( dx+c \right ) }\sqrt{a \left ( 1+i\tan \left ( dx+c \right ) \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 \sqrt{i \, a \tan \left (d x + c\right ) + a} \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 = 2.44423, size = 1439, normalized size = 10.66 \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 = 1.2166, size = 161, normalized size = 1.19 \begin{align*} \frac{\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 ) \tan \left (d x + c\right )}{a \tan \left (d x + c\right ) + i \, a} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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