Optimal. Leaf size=227 \[ \frac{\left (2 n^2 p^2-4 n p+1\right ) \tan (e+f x) \, _2F_1(1,n p+1;n p+2;-i \tan (e+f x)) \left (c (d \tan (e+f x))^p\right )^n}{8 a^2 f (n p+1)}+\frac{\tan (e+f x) \, _2F_1(1,n p+1;n p+2;i \tan (e+f x)) \left (c (d \tan (e+f x))^p\right )^n}{8 a^2 f (n p+1)}+\frac{(2-n p) \tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{4 a^2 f (1+i \tan (e+f x))}+\frac{\tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{4 a^2 f (1+i \tan (e+f x))^2} \]
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Rubi [A] time = 0.414228, antiderivative size = 227, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {6677, 848, 103, 151, 156, 64} \[ \frac{\left (2 n^2 p^2-4 n p+1\right ) \tan (e+f x) \, _2F_1(1,n p+1;n p+2;-i \tan (e+f x)) \left (c (d \tan (e+f x))^p\right )^n}{8 a^2 f (n p+1)}+\frac{\tan (e+f x) \, _2F_1(1,n p+1;n p+2;i \tan (e+f x)) \left (c (d \tan (e+f x))^p\right )^n}{8 a^2 f (n p+1)}+\frac{(2-n p) \tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{4 a^2 f (1+i \tan (e+f x))}+\frac{\tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{4 a^2 f (1+i \tan (e+f x))^2} \]
Antiderivative was successfully verified.
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Rule 6677
Rule 848
Rule 103
Rule 151
Rule 156
Rule 64
Rubi steps
\begin{align*} \int \frac{\left (c (d \tan (e+f x))^p\right )^n}{(a+i a \tan (e+f x))^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (c (d x)^p\right )^n}{(a+i a x)^2 \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\left ((d \tan (e+f x))^{-n p} \left (c (d \tan (e+f x))^p\right )^n\right ) \operatorname{Subst}\left (\int \frac{(d x)^{n p}}{(a+i a x)^2 \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\left ((d \tan (e+f x))^{-n p} \left (c (d \tan (e+f x))^p\right )^n\right ) \operatorname{Subst}\left (\int \frac{(d x)^{n p}}{\left (\frac{1}{a}-\frac{i x}{a}\right ) (a+i a x)^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{4 a^2 f (1+i \tan (e+f x))^2}-\frac{\left (i (d \tan (e+f x))^{-n p} \left (c (d \tan (e+f x))^p\right )^n\right ) \operatorname{Subst}\left (\int \frac{(d x)^{n p} (i d (3-n p)+d (1-n p) x)}{\left (\frac{1}{a}-\frac{i x}{a}\right ) (a+i a x)^2} \, dx,x,\tan (e+f x)\right )}{4 a d f}\\ &=\frac{\tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{4 a^2 f (1+i \tan (e+f x))^2}+\frac{(2-n p) \tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{4 a^2 f (1+i \tan (e+f x))}-\frac{\left ((d \tan (e+f x))^{-n p} \left (c (d \tan (e+f x))^p\right )^n\right ) \operatorname{Subst}\left (\int \frac{(d x)^{n p} \left (-2 d^2 (1-n p)^2-2 i d^2 n p (2-n p) x\right )}{\left (\frac{1}{a}-\frac{i x}{a}\right ) (a+i a x)} \, dx,x,\tan (e+f x)\right )}{8 a^2 d^2 f}\\ &=\frac{\tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{4 a^2 f (1+i \tan (e+f x))^2}+\frac{(2-n p) \tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{4 a^2 f (1+i \tan (e+f x))}+\frac{\left ((d \tan (e+f x))^{-n p} \left (c (d \tan (e+f x))^p\right )^n\right ) \operatorname{Subst}\left (\int \frac{(d x)^{n p}}{\frac{1}{a}-\frac{i x}{a}} \, dx,x,\tan (e+f x)\right )}{8 a^3 f}+\frac{\left (\left (1-4 n p+2 n^2 p^2\right ) (d \tan (e+f x))^{-n p} \left (c (d \tan (e+f x))^p\right )^n\right ) \operatorname{Subst}\left (\int \frac{(d x)^{n p}}{a+i a x} \, dx,x,\tan (e+f x)\right )}{8 a f}\\ &=\frac{\left (1-4 n p+2 n^2 p^2\right ) \, _2F_1(1,1+n p;2+n p;-i \tan (e+f x)) \tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{8 a^2 f (1+n p)}+\frac{\, _2F_1(1,1+n p;2+n p;i \tan (e+f x)) \tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{8 a^2 f (1+n p)}+\frac{\tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{4 a^2 f (1+i \tan (e+f x))^2}+\frac{(2-n p) \tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{4 a^2 f (1+i \tan (e+f x))}\\ \end{align*}
Mathematica [F] time = 5.97575, size = 0, normalized size = 0. \[ \int \frac{\left (c (d \tan (e+f x))^p\right )^n}{(a+i a \tan (e+f x))^2} \, dx \]
Verification is Not applicable to the result.
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Maple [F] time = 5.925, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( c \left ( d\tan \left ( fx+e \right ) \right ) ^{p} \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 (c \left (\frac{-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{p}\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 (\left (d \tan \left (f x + e\right )\right )^{p} 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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