Optimal. Leaf size=122 \[ \frac {(a+i a \tan (e+f x))^m \, _2F_1\left (1,m;m+1;\frac {1}{2} (i \tan (e+f x)+1)\right )}{2 f m (d+i c)}-\frac {d (a+i a \tan (e+f x))^m \, _2F_1\left (1,m;m+1;-\frac {d (i \tan (e+f x)+1)}{i c-d}\right )}{f m \left (c^2+d^2\right )} \]
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Rubi [A] time = 0.26, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3562, 3481, 68, 3599} \[ \frac {(a+i a \tan (e+f x))^m \, _2F_1\left (1,m;m+1;\frac {1}{2} (i \tan (e+f x)+1)\right )}{2 f m (d+i c)}-\frac {d (a+i a \tan (e+f x))^m \, _2F_1\left (1,m;m+1;-\frac {d (i \tan (e+f x)+1)}{i c-d}\right )}{f m \left (c^2+d^2\right )} \]
Antiderivative was successfully verified.
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Rule 68
Rule 3481
Rule 3562
Rule 3599
Rubi steps
\begin {align*} \int \frac {(a+i a \tan (e+f x))^m}{c+d \tan (e+f x)} \, dx &=\frac {\int (a+i a \tan (e+f x))^m \, dx}{c-i d}-\frac {d \int \frac {(a+i a \tan (e+f x))^m (i a+a \tan (e+f x))}{c+d \tan (e+f x)} \, dx}{a (c-i d)}\\ &=\frac {a \operatorname {Subst}\left (\int \frac {(a+x)^{-1+m}}{a-x} \, dx,x,i a \tan (e+f x)\right )}{(i c+d) f}+\frac {(a d) \operatorname {Subst}\left (\int \frac {(a+i a x)^{-1+m}}{c+d x} \, dx,x,\tan (e+f x)\right )}{(i c+d) f}\\ &=\frac {\, _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 (i c+d) f m}-\frac {d \, _2F_1\left (1,m;1+m;-\frac {d (1+i \tan (e+f x))}{i c-d}\right ) (a+i a \tan (e+f x))^m}{\left (c^2+d^2\right ) f m}\\ \end {align*}
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Mathematica [F] time = 14.32, size = 0, normalized size = 0.00 \[ \int \frac {(a+i a \tan (e+f x))^m}{c+d \tan (e+f x)} \, dx \]
Verification is Not applicable to the result.
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fricas [F] time = 0.73, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (\frac {2 \, a e^{\left (2 i \, f x + 2 i \, e\right )}}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{m} {\left (i \, e^{\left (2 i \, f x + 2 i \, e\right )} + i\right )}}{{\left (i \, c + d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c - d}, x\right ) \]
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 \[ \int \frac {{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{m}}{d \tan \left (f x + e\right ) + c}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 3.82, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +i a \tan \left (f x +e \right )\right )^{m}}{c +d \tan \left (f x +e \right )}\, dx \]
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 \[ \int \frac {{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{m}}{d \tan \left (f x + e\right ) + c}\,{d x} \]
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 \[ \int \frac {{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^m}{c+d\,\mathrm {tan}\left (e+f\,x\right )} \,d x \]
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 \[ \int \frac {\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{m}}{c + d \tan {\left (e + f x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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