Optimal. Leaf size=104 \[ -\frac{i 2^{-\frac{m}{2}-\frac{1}{2}} (1+i \tan (c+d x))^{\frac{m+1}{2}} (e \cos (c+d x))^m \, _2F_1\left (-\frac{m}{2},\frac{m+3}{2};1-\frac{m}{2};\frac{1}{2} (1-i \tan (c+d x))\right )}{d m \sqrt{a+i a \tan (c+d x)}} \]
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Rubi [A] time = 0.290182, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {3515, 3505, 3523, 70, 69} \[ -\frac{i 2^{-\frac{m}{2}-\frac{1}{2}} (1+i \tan (c+d x))^{\frac{m+1}{2}} (e \cos (c+d x))^m \, _2F_1\left (-\frac{m}{2},\frac{m+3}{2};1-\frac{m}{2};\frac{1}{2} (1-i \tan (c+d x))\right )}{d m \sqrt{a+i a \tan (c+d x)}} \]
Antiderivative was successfully verified.
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Rule 3515
Rule 3505
Rule 3523
Rule 70
Rule 69
Rubi steps
\begin{align*} \int \frac{(e \cos (c+d x))^m}{\sqrt{a+i a \tan (c+d x)}} \, dx &=\left ((e \cos (c+d x))^m (e \sec (c+d x))^m\right ) \int \frac{(e \sec (c+d x))^{-m}}{\sqrt{a+i a \tan (c+d x)}} \, dx\\ &=\left ((e \cos (c+d x))^m (a-i a \tan (c+d x))^{m/2} (a+i a \tan (c+d x))^{m/2}\right ) \int (a-i a \tan (c+d x))^{-m/2} (a+i a \tan (c+d x))^{-\frac{1}{2}-\frac{m}{2}} \, dx\\ &=\frac{\left (a^2 (e \cos (c+d x))^m (a-i a \tan (c+d x))^{m/2} (a+i a \tan (c+d x))^{m/2}\right ) \operatorname{Subst}\left (\int (a-i a x)^{-1-\frac{m}{2}} (a+i a x)^{-\frac{3}{2}-\frac{m}{2}} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\left (2^{-\frac{3}{2}-\frac{m}{2}} a (e \cos (c+d x))^m (a-i a \tan (c+d x))^{m/2} \left (\frac{a+i a \tan (c+d x)}{a}\right )^{\frac{1}{2}+\frac{m}{2}}\right ) \operatorname{Subst}\left (\int \left (\frac{1}{2}+\frac{i x}{2}\right )^{-\frac{3}{2}-\frac{m}{2}} (a-i a x)^{-1-\frac{m}{2}} \, dx,x,\tan (c+d x)\right )}{d \sqrt{a+i a \tan (c+d x)}}\\ &=-\frac{i 2^{-\frac{1}{2}-\frac{m}{2}} (e \cos (c+d x))^m \, _2F_1\left (-\frac{m}{2},\frac{3+m}{2};1-\frac{m}{2};\frac{1}{2} (1-i \tan (c+d x))\right ) (1+i \tan (c+d x))^{\frac{1+m}{2}}}{d m \sqrt{a+i a \tan (c+d x)}}\\ \end{align*}
Mathematica [A] time = 14.8944, size = 143, normalized size = 1.38 \[ \frac{i 4^{-m} \left (1+e^{2 i (c+d x)}\right ) \left (e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )\right )^m \left (e e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )\right )^m \, _2F_1\left (1,\frac{m+2}{2};\frac{1-m}{2};-e^{2 i (c+d x)}\right ) \cos ^{-m}(c+d x)}{d (m+1) \sqrt{a+i a \tan (c+d x)}} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.363, size = 0, normalized size = 0. \begin{align*} \int{ \left ( e\cos \left ( dx+c \right ) \right ) ^{m}{\frac{1}{\sqrt{a+ia\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 (e \cos \left (d x + c\right )\right )^{m}}{\sqrt{i \, a \tan \left (d x + c\right ) + a}}\,{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{\sqrt{2} \left (\frac{1}{2} \,{\left (e e^{\left (2 i \, d x + 2 i \, c\right )} + e\right )} e^{\left (-i \, d x - i \, c\right )}\right )^{m} \sqrt{\frac{a}{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 )}}{2 \, a}, 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 (e \cos{\left (c + d x \right )}\right )^{m}}{\sqrt{a \left (i \tan{\left (c + d x \right )} + 1\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 (e \cos \left (d x + c\right )\right )^{m}}{\sqrt{i \, a \tan \left (d x + c\right ) + a}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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