Optimal. Leaf size=79 \[ \frac{2 \sqrt{\tan (c+d x)} (1+i \tan (c+d x))^{-m} (a+i a \tan (c+d x))^m F_1\left (\frac{1}{2};1-m,1;\frac{3}{2};-i \tan (c+d x),i \tan (c+d x)\right )}{d} \]
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Rubi [A] time = 0.0994344, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3564, 130, 430, 429} \[ \frac{2 \sqrt{\tan (c+d x)} (1+i \tan (c+d x))^{-m} (a+i a \tan (c+d x))^m F_1\left (\frac{1}{2};1-m,1;\frac{3}{2};-i \tan (c+d x),i \tan (c+d x)\right )}{d} \]
Antiderivative was successfully verified.
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Rule 3564
Rule 130
Rule 430
Rule 429
Rubi steps
\begin{align*} \int \frac{(a+i a \tan (c+d x))^m}{\sqrt{\tan (c+d x)}} \, dx &=\frac{\left (i a^2\right ) \operatorname{Subst}\left (\int \frac{(a+x)^{-1+m}}{\sqrt{-\frac{i x}{a}} \left (-a^2+a x\right )} \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=-\frac{\left (2 a^3\right ) \operatorname{Subst}\left (\int \frac{\left (a+i a x^2\right )^{-1+m}}{-a^2+i a^2 x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=-\frac{\left (2 a^2 (1+i \tan (c+d x))^{-m} (a+i a \tan (c+d x))^m\right ) \operatorname{Subst}\left (\int \frac{\left (1+i x^2\right )^{-1+m}}{-a^2+i a^2 x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=\frac{2 F_1\left (\frac{1}{2};1-m,1;\frac{3}{2};-i \tan (c+d x),i \tan (c+d x)\right ) (1+i \tan (c+d x))^{-m} \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^m}{d}\\ \end{align*}
Mathematica [F] time = 5.234, size = 0, normalized size = 0. \[ \int \frac{(a+i a \tan (c+d x))^m}{\sqrt{\tan (c+d x)}} \, dx \]
Verification is Not applicable to the result.
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Maple [F] time = 0.23, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{m}{\frac{1}{\sqrt{\tan \left ( dx+c \right ) }}}}\, dx \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{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{m}}{\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 [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\left (\frac{2 \, a e^{\left (2 i \, d x + 2 i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{m} \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 (i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}, x\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 \frac{\left (a \left (i \tan{\left (c + d x \right )} + 1\right )\right )^{m}}{\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 [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{m}}{\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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