Optimal. Leaf size=104 \[ -\frac{2 (-1)^{3/4} \sqrt{a} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{(1+i) \sqrt{a} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d} \]
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Rubi [A] time = 0.269664, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3563, 3544, 205, 3599, 63, 217, 203} \[ -\frac{2 (-1)^{3/4} \sqrt{a} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{(1+i) \sqrt{a} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d} \]
Antiderivative was successfully verified.
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Rule 3563
Rule 3544
Rule 205
Rule 3599
Rule 63
Rule 217
Rule 203
Rubi steps
\begin{align*} \int \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)} \, dx &=-\left (i \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx\right )+\frac{\int \frac{\sqrt{a+i a \tan (c+d x)} (i a+a \tan (c+d x))}{\sqrt{\tan (c+d x)}} \, dx}{a}\\ &=\frac{(i a) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \sqrt{a+i a x}} \, dx,x,\tan (c+d x)\right )}{d}-\frac{\left (2 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{(2 i 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{(2 i 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{2 (-1)^{3/4} \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}\\ \end{align*}
Mathematica [B] time = 1.06521, size = 229, normalized size = 2.2 \[ \frac{i \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \left (\sqrt{2} \left (\log \left (-2 \sqrt{2} e^{i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}}-3 e^{2 i (c+d x)}+1\right )-\log \left (2 \sqrt{2} e^{i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}}-3 e^{2 i (c+d x)}+1\right )\right )-4 \log \left (\sqrt{-1+e^{2 i (c+d x)}}+e^{i (c+d x)}\right )\right ) \cos (c+d x) \sqrt{a+i a \tan (c+d x)}}{2 d \sqrt{-1+e^{2 i (c+d x)}}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.063, size = 347, normalized size = 3.3 \begin{align*} -{\frac{a}{2\,d \left ( -\tan \left ( dx+c \right ) +i \right ) }\sqrt{\tan \left ( dx+c \right ) }\sqrt{a \left ( 1+i\tan \left ( dx+c \right ) \right ) } \left ( 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}-\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}\tan \left ( dx+c \right ) +2\,i\sqrt{-ia}\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 ) \tan \left ( dx+c \right ) +2\,\sqrt{-ia}\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 ) \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} \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 = 2.5475, size = 1280, normalized size = 12.31 \begin{align*} -\frac{1}{2} \, \sqrt{\frac{4 i \, a}{d^{2}}} \log \left ({\left (\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 )} + i \, d \sqrt{\frac{4 i \, a}{d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) + \frac{1}{2} \, \sqrt{\frac{4 i \, a}{d^{2}}} \log \left ({\left (\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 )} - i \, d \sqrt{\frac{4 i \, a}{d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) + \frac{1}{2} \, \sqrt{\frac{2 i \, a}{d^{2}}} \log \left ({\left (\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 )} + i \, d \sqrt{\frac{2 i \, a}{d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) - \frac{1}{2} \, \sqrt{\frac{2 i \, a}{d^{2}}} \log \left ({\left (\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 )} - i \, d \sqrt{\frac{2 i \, a}{d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 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 \sqrt{a \left (i \tan{\left (c + d x \right )} + 1\right )} \sqrt{\tan{\left (c + d x \right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.20773, size = 153, normalized size = 1.47 \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 )}{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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