Optimal. Leaf size=154 \[ -\frac{2 i \sqrt{a+i a \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 \sqrt{a+i a \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{26 \sqrt{a+i a \tan (c+d x)}}{15 d \sqrt{\tan (c+d x)}}-\frac{(1+i) \sqrt{a} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d} \]
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Rubi [A] time = 0.413582, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {3561, 3598, 12, 3544, 205} \[ -\frac{2 i \sqrt{a+i a \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 \sqrt{a+i a \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{26 \sqrt{a+i a \tan (c+d x)}}{15 d \sqrt{\tan (c+d x)}}-\frac{(1+i) \sqrt{a} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d} \]
Antiderivative was successfully verified.
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Rule 3561
Rule 3598
Rule 12
Rule 3544
Rule 205
Rubi steps
\begin{align*} \int \frac{\sqrt{a+i a \tan (c+d x)}}{\tan ^{\frac{7}{2}}(c+d x)} \, dx &=-\frac{2 \sqrt{a+i a \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 \int \frac{\left (\frac{i a}{2}-2 a \tan (c+d x)\right ) \sqrt{a+i a \tan (c+d x)}}{\tan ^{\frac{5}{2}}(c+d x)} \, dx}{5 a}\\ &=-\frac{2 \sqrt{a+i a \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 i \sqrt{a+i a \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 \int \frac{\sqrt{a+i a \tan (c+d x)} \left (-\frac{13 a^2}{4}-\frac{1}{2} i a^2 \tan (c+d x)\right )}{\tan ^{\frac{3}{2}}(c+d x)} \, dx}{15 a^2}\\ &=-\frac{2 \sqrt{a+i a \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 i \sqrt{a+i a \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{26 \sqrt{a+i a \tan (c+d x)}}{15 d \sqrt{\tan (c+d x)}}+\frac{8 \int -\frac{15 i a^3 \sqrt{a+i a \tan (c+d x)}}{8 \sqrt{\tan (c+d x)}} \, dx}{15 a^3}\\ &=-\frac{2 \sqrt{a+i a \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 i \sqrt{a+i a \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{26 \sqrt{a+i a \tan (c+d x)}}{15 d \sqrt{\tan (c+d x)}}-i \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{2 \sqrt{a+i a \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 i \sqrt{a+i a \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{26 \sqrt{a+i a \tan (c+d x)}}{15 d \sqrt{\tan (c+d x)}}-\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 \sqrt{a+i a \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 i \sqrt{a+i a \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{26 \sqrt{a+i a \tan (c+d x)}}{15 d \sqrt{\tan (c+d x)}}\\ \end{align*}
Mathematica [F] time = 3.26072, size = 0, normalized size = 0. \[ \int \frac{\sqrt{a+i a \tan (c+d x)}}{\tan ^{\frac{7}{2}}(c+d x)} \, dx \]
Verification is Not applicable to the result.
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Maple [B] time = 0.06, size = 357, normalized size = 2.3 \begin{align*} -{\frac{1}{30\,d \left ( -\tan \left ( dx+c \right ) +i \right ) }\sqrt{a \left ( 1+i\tan \left ( dx+c \right ) \right ) } \left ( 15\,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 ) \left ( \tan \left ( dx+c \right ) \right ) ^{3}a-15\,\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 ) \left ( \tan \left ( dx+c \right ) \right ) ^{4}a+52\,\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) } \left ( \tan \left ( dx+c \right ) \right ) ^{3}-16\,\tan \left ( dx+c \right ) \sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{-ia}-56\,i \left ( \tan \left ( dx+c \right ) \right ) ^{2}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{-ia}+12\,i\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{-ia} \right ) \left ( \tan \left ( dx+c \right ) \right ) ^{-{\frac{5}{2}}}{\frac{1}{\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }}}{\frac{1}{\sqrt{-ia}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 2.49105, size = 1544, normalized size = 10.03 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.4031, size = 1251, normalized size = 8.12 \begin{align*} \frac{\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 (68 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 12 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 20 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 60 i\right )} e^{\left (i \, d x + i \, c\right )} + 15 \,{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \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 ) - 15 \,{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \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 )}{30 \,{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \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 = 1.34831, size = 178, normalized size = 1.16 \begin{align*} \frac{2 \, \sqrt{-2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 2 \, a^{2}} a^{4} \log \left (\sqrt{i \, a \tan \left (d x + c\right ) + a}\right )}{-\left (i + 1\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{5} + \left (6 i + 6\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a - \left (14 i + 14\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{2} + \left (16 i + 16\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{3} - \left (9 i + 9\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{4} + \left (2 i + 2\right ) \, a^{5}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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