Optimal. Leaf size=114 \[ -\frac {i (a+i a \tan (e+f x))^m (c+d \tan (e+f x))^n \left (\frac {c+d \tan (e+f x)}{c+i d}\right )^{-n} F_1\left (m;-n,1;m+1;-\frac {d (i \tan (e+f x)+1)}{i c-d},\frac {1}{2} (i \tan (e+f x)+1)\right )}{2 f m} \]
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Rubi [A] time = 0.17, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {3564, 137, 136} \[ -\frac {i (a+i a \tan (e+f x))^m (c+d \tan (e+f x))^n \left (\frac {c+d \tan (e+f x)}{c+i d}\right )^{-n} F_1\left (m;-n,1;m+1;-\frac {d (i \tan (e+f x)+1)}{i c-d},\frac {1}{2} (i \tan (e+f x)+1)\right )}{2 f m} \]
Antiderivative was successfully verified.
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Rule 136
Rule 137
Rule 3564
Rubi steps
\begin {align*} \int (a+i a \tan (e+f x))^m (c+d \tan (e+f x))^n \, dx &=\frac {\left (i a^2\right ) \operatorname {Subst}\left (\int \frac {(a+x)^{-1+m} \left (c-\frac {i d x}{a}\right )^n}{-a^2+a x} \, dx,x,i a \tan (e+f x)\right )}{f}\\ &=\frac {\left (i a^2 (c+d \tan (e+f x))^n \left (\frac {c+d \tan (e+f x)}{c+i d}\right )^{-n}\right ) \operatorname {Subst}\left (\int \frac {(a+x)^{-1+m} \left (\frac {c}{c+i d}-\frac {i d x}{a (c+i d)}\right )^n}{-a^2+a x} \, dx,x,i a \tan (e+f x)\right )}{f}\\ &=-\frac {i F_1\left (m;-n,1;1+m;-\frac {d (1+i \tan (e+f x))}{i c-d},\frac {1}{2} (1+i \tan (e+f x))\right ) (a+i a \tan (e+f x))^m (c+d \tan (e+f x))^n \left (\frac {c+d \tan (e+f x)}{c+i d}\right )^{-n}}{2 f m}\\ \end {align*}
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Mathematica [F] time = 7.87, size = 0, normalized size = 0.00 \[ \int (a+i a \tan (e+f x))^m (c+d \tan (e+f x))^n \, dx \]
Verification is Not applicable to the result.
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fricas [F] time = 1.14, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (\frac {2 \, a e^{\left (2 i \, f x + 2 i \, e\right )}}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{m} \left (\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{n}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{m} {\left (d \tan \left (f x + e\right ) + c\right )}^{n}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 2.64, size = 0, normalized size = 0.00 \[ \int \left (a +i a \tan \left (f x +e \right )\right )^{m} \left (c +d \tan \left (f x +e \right )\right )^{n}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{m} {\left (d \tan \left (f x + e\right ) + c\right )}^{n}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^m\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^n \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{m} \left (c + d \tan {\left (e + f x \right )}\right )^{n}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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