Optimal. Leaf size=86 \[ -\frac{i 2^{-\frac{m}{2}-1} (1+i \tan (c+d x))^{m/2} (e \cos (c+d x))^m \, _2F_1\left (-\frac{m}{2},\frac{m+4}{2};1-\frac{m}{2};\frac{1}{2} (1-i \tan (c+d x))\right )}{a d m} \]
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Rubi [A] time = 0.238043, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {3515, 3505, 3523, 70, 69} \[ -\frac{i 2^{-\frac{m}{2}-1} (1+i \tan (c+d x))^{m/2} (e \cos (c+d x))^m \, _2F_1\left (-\frac{m}{2},\frac{m+4}{2};1-\frac{m}{2};\frac{1}{2} (1-i \tan (c+d x))\right )}{a d m} \]
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}{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}}{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))^{-1-\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)^{-2-\frac{m}{2}} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\left (2^{-2-\frac{m}{2}} (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 )^{m/2}\right ) \operatorname{Subst}\left (\int \left (\frac{1}{2}+\frac{i x}{2}\right )^{-2-\frac{m}{2}} (a-i a x)^{-1-\frac{m}{2}} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac{i 2^{-1-\frac{m}{2}} (e \cos (c+d x))^m \, _2F_1\left (-\frac{m}{2},\frac{4+m}{2};1-\frac{m}{2};\frac{1}{2} (1-i \tan (c+d x))\right ) (1+i \tan (c+d x))^{m/2}}{a d m}\\ \end{align*}
Mathematica [A] time = 59.1586, size = 147, normalized size = 1.71 \[ \frac{2^{-m-1} e^{-i (c+2 d x)} \left (1+e^{2 i (c+d x)}\right )^2 \left (e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )\right )^m (\cos (d x)+i \sin (d x)) \, _2F_1\left (1,\frac{m+2}{2};-\frac{m}{2};-e^{2 i (c+d x)}\right ) \cos ^{-m-1}(c+d x) (e \cos (c+d x))^m}{a d (m+2) (\tan (c+d x)-i)} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.868, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( e\cos \left ( dx+c \right ) \right ) ^{m}}{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(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \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{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}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (-2 i \, d x - 2 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(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \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}}{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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