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 = 0.41, antiderivative size = 227, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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 64
Rule 103
Rule 151
Rule 156
Rule 848
Rule 6677
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*}
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Mathematica [F] time = 6.32, size = 0, normalized size = 0.00 \[ \int \frac {\left (c (d \tan (e+f x))^p\right )^n}{(a+i a \tan (e+f x))^2} \, dx \]
Verification is Not applicable to the result.
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fricas [F] time = 0.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )} e^{\left (n p \log \left (\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right ) - 4 i \, f x + n \log \relax (c) - 4 i \, e\right )}}{4 \, a^{2}}, x\right ) \]
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 \[ \int \frac {\left (\left (d \tan \left (f x + e\right )\right )^{p} c\right )^{n}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F(-1)] time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c \left (d \tan \left (f x +e \right )\right )^{p}\right )^{n}}{\left (a +i a \tan \left (f x +e \right )\right )^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^p\right )}^n}{{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2} \,d x \]
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 \[ - \frac {\int \frac {\left (c \left (d \tan {\left (e + f x \right )}\right )^{p}\right )^{n}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\, dx}{a^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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