Optimal. Leaf size=105 \[ -\frac{i 2^{n-\frac{m}{2}} (a+i a \tan (c+d x))^n (e \cos (c+d x))^m (1+i \tan (c+d x))^{\frac{1}{2} (m-2 n)} \, _2F_1\left (-\frac{m}{2},\frac{1}{2} (m-2 n+2);1-\frac{m}{2};\frac{1}{2} (1-i \tan (c+d x))\right )}{d m} \]
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Rubi [A] time = 0.230718, antiderivative size = 105, 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} \[ -\frac{i 2^{n-\frac{m}{2}} (a+i a \tan (c+d x))^n (e \cos (c+d x))^m (1+i \tan (c+d x))^{\frac{1}{2} (m-2 n)} \, _2F_1\left (-\frac{m}{2},\frac{1}{2} (m-2 n+2);1-\frac{m}{2};\frac{1}{2} (1-i \tan (c+d x))\right )}{d m} \]
Antiderivative was successfully verified.
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Rule 3515
Rule 3505
Rule 3523
Rule 70
Rule 69
Rubi steps
\begin{align*} \int (e \cos (c+d x))^m (a+i a \tan (c+d x))^n \, dx &=\left ((e \cos (c+d x))^m (e \sec (c+d x))^m\right ) \int (e \sec (c+d x))^{-m} (a+i a \tan (c+d x))^n \, 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{m}{2}+n} \, 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)^{-1-\frac{m}{2}+n} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\left (2^{-1-\frac{m}{2}+n} a (e \cos (c+d x))^m (a-i a \tan (c+d x))^{m/2} (a+i a \tan (c+d x))^n \left (\frac{a+i a \tan (c+d x)}{a}\right )^{\frac{m}{2}-n}\right ) \operatorname{Subst}\left (\int \left (\frac{1}{2}+\frac{i x}{2}\right )^{-1-\frac{m}{2}+n} (a-i a x)^{-1-\frac{m}{2}} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac{i 2^{-\frac{m}{2}+n} (e \cos (c+d x))^m \, _2F_1\left (-\frac{m}{2},\frac{1}{2} (2+m-2 n);1-\frac{m}{2};\frac{1}{2} (1-i \tan (c+d x))\right ) (1+i \tan (c+d x))^{\frac{1}{2} (m-2 n)} (a+i a \tan (c+d x))^n}{d m}\\ \end{align*}
Mathematica [A] time = 13.2095, size = 192, normalized size = 1.83 \[ \frac{i 2^{n-m} \left (1+e^{2 i (c+d x)}\right ) \left (e^{i d x}\right )^n \left (e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )\right )^m \left (\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^n \cos ^{-m}(c+d x) \sec ^{-n}(c+d x) (\cos (d x)+i \sin (d x))^{-n} (a+i a \tan (c+d x))^n (e \cos (c+d x))^m \, _2F_1\left (1,\frac{m+2}{2};-\frac{m}{2}+n+1;-e^{2 i (c+d x)}\right )}{d (m-2 n)} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.725, size = 0, normalized size = 0. \begin{align*} \int \left ( e\cos \left ( dx+c \right ) \right ) ^{m} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{n}\, 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 \left (e \cos \left (d x + c\right )\right )^{m}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n}\,{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 (\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 (\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 )^{n}, x\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 [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e \cos \left (d x + c\right )\right )^{m}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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