Optimal. Leaf size=273 \[ \frac{\left (c^2+2 i c d (1-n)-d^2 \left (2 n^2-4 n+1\right )\right ) (c+d \tan (e+f x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{c+d \tan (e+f x)}{c+i d}\right )}{8 a^2 f (n+1) (-d+i c)^3}+\frac{(c+d \tan (e+f x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{c+d \tan (e+f x)}{c-i d}\right )}{8 a^2 f (n+1) (d+i c)}+\frac{(-d (2-n)+i c) (c+d \tan (e+f x))^{n+1}}{4 a^2 f (c+i d)^2 (1+i \tan (e+f x))}-\frac{(c+d \tan (e+f x))^{n+1}}{4 f (-d+i c) (a+i a \tan (e+f x))^2} \]
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Rubi [A] time = 0.556523, antiderivative size = 273, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {3559, 3596, 3539, 3537, 68} \[ \frac{\left (c^2+2 i c d (1-n)-d^2 \left (2 n^2-4 n+1\right )\right ) (c+d \tan (e+f x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{c+d \tan (e+f x)}{c+i d}\right )}{8 a^2 f (n+1) (-d+i c)^3}+\frac{(c+d \tan (e+f x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{c+d \tan (e+f x)}{c-i d}\right )}{8 a^2 f (n+1) (d+i c)}+\frac{(-d (2-n)+i c) (c+d \tan (e+f x))^{n+1}}{4 a^2 f (c+i d)^2 (1+i \tan (e+f x))}-\frac{(c+d \tan (e+f x))^{n+1}}{4 f (-d+i c) (a+i a \tan (e+f x))^2} \]
Antiderivative was successfully verified.
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Rule 3559
Rule 3596
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))^2} \, dx &=-\frac{(c+d \tan (e+f x))^{1+n}}{4 (i c-d) f (a+i a \tan (e+f x))^2}-\frac{\int \frac{(c+d \tan (e+f x))^n (-a (2 i c-d (3-n))-i a d (1-n) \tan (e+f x))}{a+i a \tan (e+f x)} \, dx}{4 a^2 (i c-d)}\\ &=\frac{(i c-d (2-n)) (c+d \tan (e+f x))^{1+n}}{4 a^2 (c+i d)^2 f (1+i \tan (e+f x))}-\frac{(c+d \tan (e+f x))^{1+n}}{4 (i c-d) f (a+i a \tan (e+f x))^2}-\frac{\int (c+d \tan (e+f x))^n \left (-2 a^2 \left (c^2-d^2 (1-n)^2+i c d (2-n)\right )+2 a^2 d (c+i d (2-n)) n \tan (e+f x)\right ) \, dx}{8 a^4 (c+i d)^2}\\ &=\frac{(i c-d (2-n)) (c+d \tan (e+f x))^{1+n}}{4 a^2 (c+i d)^2 f (1+i \tan (e+f x))}-\frac{(c+d \tan (e+f x))^{1+n}}{4 (i c-d) f (a+i a \tan (e+f x))^2}+\frac{\int (1+i \tan (e+f x)) (c+d \tan (e+f x))^n \, dx}{8 a^2}+\frac{\left (c^2+2 i c d (1-n)-d^2 \left (1-4 n+2 n^2\right )\right ) \int (1-i \tan (e+f x)) (c+d \tan (e+f x))^n \, dx}{8 a^2 (c+i d)^2}\\ &=\frac{(i c-d (2-n)) (c+d \tan (e+f x))^{1+n}}{4 a^2 (c+i d)^2 f (1+i \tan (e+f x))}-\frac{(c+d \tan (e+f x))^{1+n}}{4 (i c-d) f (a+i a \tan (e+f x))^2}+\frac{i \operatorname{Subst}\left (\int \frac{(c-i d x)^n}{-1+x} \, dx,x,i \tan (e+f x)\right )}{8 a^2 f}+\frac{\left (2 c d (1-n)-i \left (c^2-d^2 \left (1-4 n+2 n^2\right )\right )\right ) \operatorname{Subst}\left (\int \frac{(c+i d x)^n}{-1+x} \, dx,x,-i \tan (e+f x)\right )}{8 a^2 (c+i d)^2 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}}{8 a^2 (i c+d) f (1+n)}+\frac{\left (c^2+2 i c d (1-n)-d^2 \left (1-4 n+2 n^2\right )\right ) \, _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}}{8 a^2 (i c-d)^3 f (1+n)}+\frac{(i c-d (2-n)) (c+d \tan (e+f x))^{1+n}}{4 a^2 (c+i d)^2 f (1+i \tan (e+f x))}-\frac{(c+d \tan (e+f x))^{1+n}}{4 (i c-d) f (a+i a \tan (e+f x))^2}\\ \end{align*}
Mathematica [F] time = 7.39494, size = 0, normalized size = 0. \[ \int \frac{(c+d \tan (e+f x))^n}{(a+i a \tan (e+f x))^2} \, dx \]
Verification is Not applicable to the result.
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Maple [F] time = 0.244, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( c+d\tan \left ( fx+e \right ) \right ) ^{n}}{ \left ( a+ia\tan \left ( fx+e \right ) \right ) ^{2}}}\, 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 (4 i \, f x + 4 i \, e\right )} + 2 \, e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{4 \, a^{2}}, 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}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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