Optimal. Leaf size=198 \[ \frac{23 i \sqrt{a+i a \tan (c+d x)}}{15 a d \tan ^{\frac{3}{2}}(c+d x)}-\frac{7 \sqrt{a+i a \tan (c+d x)}}{5 a d \tan ^{\frac{5}{2}}(c+d x)}+\frac{1}{d \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}+\frac{61 \sqrt{a+i a \tan (c+d x)}}{15 a d \sqrt{\tan (c+d x)}}-\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{\sqrt{a} d} \]
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Rubi [A] time = 0.559027, antiderivative size = 198, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {3559, 3598, 12, 3544, 205} \[ \frac{23 i \sqrt{a+i a \tan (c+d x)}}{15 a d \tan ^{\frac{3}{2}}(c+d x)}-\frac{7 \sqrt{a+i a \tan (c+d x)}}{5 a d \tan ^{\frac{5}{2}}(c+d x)}+\frac{1}{d \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}+\frac{61 \sqrt{a+i a \tan (c+d x)}}{15 a d \sqrt{\tan (c+d x)}}-\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{\sqrt{a} d} \]
Antiderivative was successfully verified.
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Rule 3559
Rule 3598
Rule 12
Rule 3544
Rule 205
Rubi steps
\begin{align*} \int \frac{1}{\tan ^{\frac{7}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}} \, dx &=\frac{1}{d \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}+\frac{\int \frac{\sqrt{a+i a \tan (c+d x)} \left (\frac{7 a}{2}-3 i a \tan (c+d x)\right )}{\tan ^{\frac{7}{2}}(c+d x)} \, dx}{a^2}\\ &=\frac{1}{d \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{7 \sqrt{a+i a \tan (c+d x)}}{5 a d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 \int \frac{\sqrt{a+i a \tan (c+d x)} \left (-\frac{23 i a^2}{4}-7 a^2 \tan (c+d x)\right )}{\tan ^{\frac{5}{2}}(c+d x)} \, dx}{5 a^3}\\ &=\frac{1}{d \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{7 \sqrt{a+i a \tan (c+d x)}}{5 a d \tan ^{\frac{5}{2}}(c+d x)}+\frac{23 i \sqrt{a+i a \tan (c+d x)}}{15 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 \int \frac{\sqrt{a+i a \tan (c+d x)} \left (-\frac{61 a^3}{8}+\frac{23}{4} i a^3 \tan (c+d x)\right )}{\tan ^{\frac{3}{2}}(c+d x)} \, dx}{15 a^4}\\ &=\frac{1}{d \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{7 \sqrt{a+i a \tan (c+d x)}}{5 a d \tan ^{\frac{5}{2}}(c+d x)}+\frac{23 i \sqrt{a+i a \tan (c+d x)}}{15 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{61 \sqrt{a+i a \tan (c+d x)}}{15 a d \sqrt{\tan (c+d x)}}+\frac{8 \int -\frac{15 i a^4 \sqrt{a+i a \tan (c+d x)}}{16 \sqrt{\tan (c+d x)}} \, dx}{15 a^5}\\ &=\frac{1}{d \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{7 \sqrt{a+i a \tan (c+d x)}}{5 a d \tan ^{\frac{5}{2}}(c+d x)}+\frac{23 i \sqrt{a+i a \tan (c+d x)}}{15 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{61 \sqrt{a+i a \tan (c+d x)}}{15 a d \sqrt{\tan (c+d x)}}-\frac{i \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx}{2 a}\\ &=\frac{1}{d \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{7 \sqrt{a+i a \tan (c+d x)}}{5 a d \tan ^{\frac{5}{2}}(c+d x)}+\frac{23 i \sqrt{a+i a \tan (c+d x)}}{15 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{61 \sqrt{a+i a \tan (c+d x)}}{15 a d \sqrt{\tan (c+d x)}}-\frac{a \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{\left (\frac{1}{2}+\frac{i}{2}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{\sqrt{a} d}+\frac{1}{d \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{7 \sqrt{a+i a \tan (c+d x)}}{5 a d \tan ^{\frac{5}{2}}(c+d x)}+\frac{23 i \sqrt{a+i a \tan (c+d x)}}{15 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{61 \sqrt{a+i a \tan (c+d x)}}{15 a d \sqrt{\tan (c+d x)}}\\ \end{align*}
Mathematica [A] time = 1.28967, size = 191, normalized size = 0.96 \[ \frac{i \sqrt{\tan (c+d x)} \left (\sqrt{-1+e^{2 i (c+d x)}} \left (165 e^{2 i (c+d x)}-205 e^{4 i (c+d x)}+103 e^{6 i (c+d x)}-15\right )-15 e^{i (c+d x)} \left (-1+e^{2 i (c+d x)}\right )^3 \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )\right )}{15 \sqrt{2} d \left (-1+e^{2 i (c+d x)}\right )^{7/2} \sqrt{\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.073, size = 473, normalized size = 2.4 \begin{align*}{\frac{1}{60\,ad \left ( -\tan \left ( dx+c \right ) +i \right ) ^{2}}\sqrt{a \left ( 1+i\tan \left ( dx+c \right ) \right ) } \left ( 30\,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 ) ^{4}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 ) ^{5}a-396\,i\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{-ia} \left ( \tan \left ( dx+c \right ) \right ) ^{3}+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 ) ^{3}a+244\,\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{-ia} \left ( \tan \left ( dx+c \right ) \right ) ^{4}+16\,i\tan \left ( dx+c \right ) \sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{-ia}-144\,\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{-ia} \left ( \tan \left ( dx+c \right ) \right ) ^{2}+24\,\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 [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{i \, a \tan \left (d x + c\right ) + a} \tan \left (d x + c\right )^{\frac{7}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 3.09587, size = 1423, normalized size = 7.19 \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 (206 i \, e^{\left (8 i \, d x + 8 i \, c\right )} - 204 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 80 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 300 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 30 i\right )} e^{\left (i \, d x + i \, c\right )} + 15 \,{\left (a d e^{\left (8 i \, d x + 8 i \, c\right )} - 3 \, a d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} - a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt{\frac{2 i}{a d^{2}}} \log \left (\frac{1}{4} \,{\left (i \, a d \sqrt{\frac{2 i}{a d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + \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 )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 15 \,{\left (a d e^{\left (8 i \, d x + 8 i \, c\right )} - 3 \, a d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} - a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt{\frac{2 i}{a d^{2}}} \log \left (\frac{1}{4} \,{\left (-i \, a d \sqrt{\frac{2 i}{a d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + \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 )}\right )} e^{\left (-i \, d x - i \, c\right )}\right )}{60 \,{\left (a d e^{\left (8 i \, d x + 8 i \, c\right )} - 3 \, a d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} - a d 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(-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 = 1.31121, size = 196, normalized size = 0.99 \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 )}^{6} + \left (6 i + 6\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{5} a - \left (14 i + 14\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a^{2} + \left (16 i + 16\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{3} - \left (9 i + 9\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{4} + \left (2 i + 2\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{5}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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