Optimal. Leaf size=171 \[ \frac {\left (a^2-b^2\right ) \tan (e+f x) \, _2F_1\left (1,\frac {1}{2} (n p+1);\frac {1}{2} (n p+3);-\tan ^2(e+f x)\right ) \left (c (d \tan (e+f x))^p\right )^n}{f (n p+1)}+\frac {2 a b \tan ^2(e+f x) \, _2F_1\left (1,\frac {1}{2} (n p+2);\frac {1}{2} (n p+4);-\tan ^2(e+f x)\right ) \left (c (d \tan (e+f x))^p\right )^n}{f (n p+2)}+\frac {b^2 \tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{f (n p+1)} \]
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Rubi [A] time = 0.32, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {6677, 1802, 808, 364} \[ \frac {\left (a^2-b^2\right ) \tan (e+f x) \, _2F_1\left (1,\frac {1}{2} (n p+1);\frac {1}{2} (n p+3);-\tan ^2(e+f x)\right ) \left (c (d \tan (e+f x))^p\right )^n}{f (n p+1)}+\frac {2 a b \tan ^2(e+f x) \, _2F_1\left (1,\frac {1}{2} (n p+2);\frac {1}{2} (n p+4);-\tan ^2(e+f x)\right ) \left (c (d \tan (e+f x))^p\right )^n}{f (n p+2)}+\frac {b^2 \tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{f (n p+1)} \]
Antiderivative was successfully verified.
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Rule 364
Rule 808
Rule 1802
Rule 6677
Rubi steps
\begin {align*} \int \left (c (d \tan (e+f x))^p\right )^n (a+b \tan (e+f x))^2 \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (c (d x)^p\right )^n (a+b x)^2}{1+x^2} \, 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+b x)^2}{1+x^2} \, 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 \left (b^2 (d x)^{n p}+\frac {(d x)^{n p} \left (a^2-b^2+2 a b x\right )}{1+x^2}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {b^2 \tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{f (1+n p)}+\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 (a^2-b^2+2 a b x\right )}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {b^2 \tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{f (1+n p)}+\frac {\left (\left (a^2-b^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}}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}+\frac {\left (2 a b (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)^{1+n p}}{1+x^2} \, dx,x,\tan (e+f x)\right )}{d f}\\ &=\frac {b^2 \tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{f (1+n p)}+\frac {\left (a^2-b^2\right ) \, _2F_1\left (1,\frac {1}{2} (1+n p);\frac {1}{2} (3+n p);-\tan ^2(e+f x)\right ) \tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{f (1+n p)}+\frac {2 a b \, _2F_1\left (1,\frac {1}{2} (2+n p);\frac {1}{2} (4+n p);-\tan ^2(e+f x)\right ) \tan ^2(e+f x) \left (c (d \tan (e+f x))^p\right )^n}{f (2+n p)}\\ \end {align*}
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Mathematica [A] time = 0.57, size = 136, normalized size = 0.80 \[ \frac {\tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n \left (\left (a^2-b^2\right ) (n p+2) \, _2F_1\left (1,\frac {1}{2} (n p+1);\frac {1}{2} (n p+3);-\tan ^2(e+f x)\right )+b \left (2 a (n p+1) \tan (e+f x) \, _2F_1\left (1,\frac {n p}{2}+1;\frac {n p}{2}+2;-\tan ^2(e+f x)\right )+b (n p+2)\right )\right )}{f (n p+1) (n p+2)} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b^{2} \tan \left (f x + e\right )^{2} + 2 \, a b \tan \left (f x + e\right ) + a^{2}\right )} \left (\left (d \tan \left (f x + e\right )\right )^{p} c\right )^{n}, x\right ) \]
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 \[ \int {\left (b \tan \left (f x + e\right ) + a\right )}^{2} \left (\left (d \tan \left (f x + e\right )\right )^{p} c\right )^{n}\,{d x} \]
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 \[ \int \left (c \left (d \tan \left (f x +e \right )\right )^{p}\right )^{n} \left (a +b \tan \left (f x +e \right )\right )^{2}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \tan \left (f x + e\right ) + a\right )}^{2} \left (\left (d \tan \left (f x + e\right )\right )^{p} c\right )^{n}\,{d x} \]
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 \[ \int {\left (c\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^p\right )}^n\,{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2 \,d x \]
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 \[ \int \left (c \left (d \tan {\left (e + f x \right )}\right )^{p}\right )^{n} \left (a + b \tan {\left (e + f x \right )}\right )^{2}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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