Optimal. Leaf size=78 \[ \frac{d (a+i a \tan (e+f x))^m}{f m}-\frac{(d+i c) (a+i a \tan (e+f x))^m \, _2F_1\left (1,m;m+1;\frac{1}{2} (i \tan (e+f x)+1)\right )}{2 f m} \]
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Rubi [A] time = 0.0697399, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {3527, 3481, 68} \[ \frac{d (a+i a \tan (e+f x))^m}{f m}-\frac{(d+i c) (a+i a \tan (e+f x))^m \, _2F_1\left (1,m;m+1;\frac{1}{2} (i \tan (e+f x)+1)\right )}{2 f m} \]
Antiderivative was successfully verified.
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Rule 3527
Rule 3481
Rule 68
Rubi steps
\begin{align*} \int (a+i a \tan (e+f x))^m (c+d \tan (e+f x)) \, dx &=\frac{d (a+i a \tan (e+f x))^m}{f m}-(-c+i d) \int (a+i a \tan (e+f x))^m \, dx\\ &=\frac{d (a+i a \tan (e+f x))^m}{f m}-\frac{(a (i c+d)) \operatorname{Subst}\left (\int \frac{(a+x)^{-1+m}}{a-x} \, dx,x,i a \tan (e+f x)\right )}{f}\\ &=\frac{d (a+i a \tan (e+f x))^m}{f m}-\frac{(i c+d) \, _2F_1\left (1,m;1+m;\frac{1}{2} (1+i \tan (e+f x))\right ) (a+i a \tan (e+f x))^m}{2 f m}\\ \end{align*}
Mathematica [A] time = 5.78361, size = 152, normalized size = 1.95 \[ \frac{2^{m-1} \left (e^{i f x}\right )^m \left (\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^m \sec ^{-m}(e+f x) (\cos (f x)+i \sin (f x))^{-m} (a+i a \tan (e+f x))^m \left ((m+1) (d-i c)-i m (c-i d) e^{2 i (e+f x)} \, _2F_1\left (1,1;m+2;-e^{2 i (e+f x)}\right )\right )}{f m (m+1)} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.771, size = 0, normalized size = 0. \begin{align*} \int \left ( a+ia\tan \left ( fx+e \right ) \right ) ^{m} \left ( c+d\tan \left ( fx+e \right ) \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{\left (d \tan \left (f x + e\right ) + c\right )}{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{m}\,{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 ({\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d\right )} \left (\frac{2 \, a e^{\left (2 i \, f x + 2 i \, e\right )}}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{m}}{e^{\left (2 i \, f x + 2 i \, e\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 \left (a \left (i \tan{\left (e + f x \right )} + 1\right )\right )^{m} \left (c + d \tan{\left (e + f x \right )}\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{\left (d \tan \left (f x + e\right ) + c\right )}{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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