Optimal. Leaf size=193 \[ \frac{(c+d \tan (e+f x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{c+d \tan (e+f x)}{c-i d}\right )}{4 a f (n+1) (d+i c)}+\frac{(i c+2 d n-d) (c+d \tan (e+f x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{c+d \tan (e+f x)}{c+i d}\right )}{4 a f (n+1) (c+i d)^2}-\frac{(c+d \tan (e+f x))^{n+1}}{2 f (-d+i c) (a+i a \tan (e+f x))} \]
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Rubi [A] time = 0.279009, antiderivative size = 193, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3552, 3539, 3537, 68} \[ \frac{(c+d \tan (e+f x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{c+d \tan (e+f x)}{c-i d}\right )}{4 a f (n+1) (d+i c)}+\frac{(i c+2 d n-d) (c+d \tan (e+f x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{c+d \tan (e+f x)}{c+i d}\right )}{4 a f (n+1) (c+i d)^2}-\frac{(c+d \tan (e+f x))^{n+1}}{2 f (-d+i c) (a+i a \tan (e+f x))} \]
Antiderivative was successfully verified.
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Rule 3552
Rule 3539
Rule 3537
Rule 68
Rubi steps
\begin{align*} \int \frac{(c+d \tan (e+f x))^n}{a+i a \tan (e+f x)} \, dx &=-\frac{(c+d \tan (e+f x))^{1+n}}{2 (i c-d) f (a+i a \tan (e+f x))}+\frac{\int (c+d \tan (e+f x))^n (a (i c-d (1-n))-i a d n \tan (e+f x)) \, dx}{2 a^2 (i c-d)}\\ &=-\frac{(c+d \tan (e+f x))^{1+n}}{2 (i c-d) f (a+i a \tan (e+f x))}+\frac{\int (1+i \tan (e+f x)) (c+d \tan (e+f x))^n \, dx}{4 a}+\frac{(c+i d (1-2 n)) \int (1-i \tan (e+f x)) (c+d \tan (e+f x))^n \, dx}{4 a (c+i d)}\\ &=-\frac{(c+d \tan (e+f x))^{1+n}}{2 (i c-d) f (a+i a \tan (e+f x))}+\frac{i \operatorname{Subst}\left (\int \frac{(c-i d x)^n}{-1+x} \, dx,x,i \tan (e+f x)\right )}{4 a f}-\frac{(i (c+i d (1-2 n))) \operatorname{Subst}\left (\int \frac{(c+i d x)^n}{-1+x} \, dx,x,-i \tan (e+f x)\right )}{4 a (c+i d) f}\\ &=\frac{\, _2F_1\left (1,1+n;2+n;\frac{c+d \tan (e+f x)}{c-i d}\right ) (c+d \tan (e+f x))^{1+n}}{4 a (i c+d) f (1+n)}+\frac{(i c-d (1-2 n)) \, _2F_1\left (1,1+n;2+n;\frac{c+d \tan (e+f x)}{c+i d}\right ) (c+d \tan (e+f x))^{1+n}}{4 a (c+i d)^2 f (1+n)}-\frac{(c+d \tan (e+f x))^{1+n}}{2 (i c-d) f (a+i a \tan (e+f x))}\\ \end{align*}
Mathematica [F] time = 24.987, size = 0, normalized size = 0. \[ \int \frac{(c+d \tan (e+f x))^n}{a+i a \tan (e+f x)} \, dx \]
Verification is Not applicable to the result.
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Maple [F] time = 0.327, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( c+d\tan \left ( fx+e \right ) \right ) ^{n}}{a+ia\tan \left ( fx+e \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{{\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}{\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )} e^{\left (-2 i \, f x - 2 i \, e\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 (d \tan \left (f x + e\right ) + c\right )}^{n}}{i \, a \tan \left (f x + e\right ) + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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