Optimal. Leaf size=296 \[ -\frac{a \left (a^2 (n p+1)+3 b^2 n p\right ) \sin (e+f x) \cos (e+f x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{2} (1-n p),\frac{1}{2} (3-n p),\cos ^2(e+f x)\right ) \left (c (d \sec (e+f x))^p\right )^n}{f \left (1-n^2 p^2\right ) \sqrt{\sin ^2(e+f x)}}+\frac{b \left (3 a^2 (n p+2)+b^2 (n p+1)\right ) \sin (e+f x) \text{Hypergeometric2F1}\left (\frac{1}{2},-\frac{n p}{2},\frac{1}{2} (2-n p),\cos ^2(e+f x)\right ) \left (c (d \sec (e+f x))^p\right )^n}{f n p (n p+2) \sqrt{\sin ^2(e+f x)}}+\frac{a b^2 (2 n p+5) \tan (e+f x) \left (c (d \sec (e+f x))^p\right )^n}{f (n p+1) (n p+2)}+\frac{b^2 \tan (e+f x) (a+b \sec (e+f x)) \left (c (d \sec (e+f x))^p\right )^n}{f (n p+2)} \]
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Rubi [A] time = 0.514726, antiderivative size = 296, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {3948, 3842, 4047, 3772, 2643, 4046} \[ -\frac{a \left (a^2 (n p+1)+3 b^2 n p\right ) \sin (e+f x) \cos (e+f x) \, _2F_1\left (\frac{1}{2},\frac{1}{2} (1-n p);\frac{1}{2} (3-n p);\cos ^2(e+f x)\right ) \left (c (d \sec (e+f x))^p\right )^n}{f \left (1-n^2 p^2\right ) \sqrt{\sin ^2(e+f x)}}+\frac{b \left (3 a^2 (n p+2)+b^2 (n p+1)\right ) \sin (e+f x) \, _2F_1\left (\frac{1}{2},-\frac{n p}{2};\frac{1}{2} (2-n p);\cos ^2(e+f x)\right ) \left (c (d \sec (e+f x))^p\right )^n}{f n p (n p+2) \sqrt{\sin ^2(e+f x)}}+\frac{a b^2 (2 n p+5) \tan (e+f x) \left (c (d \sec (e+f x))^p\right )^n}{f (n p+1) (n p+2)}+\frac{b^2 \tan (e+f x) (a+b \sec (e+f x)) \left (c (d \sec (e+f x))^p\right )^n}{f (n p+2)} \]
Antiderivative was successfully verified.
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Rule 3948
Rule 3842
Rule 4047
Rule 3772
Rule 2643
Rule 4046
Rubi steps
\begin{align*} \int \left (c (d \sec (e+f x))^p\right )^n (a+b \sec (e+f x))^3 \, dx &=\left ((d \sec (e+f x))^{-n p} \left (c (d \sec (e+f x))^p\right )^n\right ) \int (d \sec (e+f x))^{n p} (a+b \sec (e+f x))^3 \, dx\\ &=\frac{b^2 \left (c (d \sec (e+f x))^p\right )^n (a+b \sec (e+f x)) \tan (e+f x)}{f (2+n p)}+\frac{\left ((d \sec (e+f x))^{-n p} \left (c (d \sec (e+f x))^p\right )^n\right ) \int (d \sec (e+f x))^{n p} \left (a d \left (b^2 n p+a^2 (2+n p)\right )+b d \left (b^2 (1+n p)+3 a^2 (2+n p)\right ) \sec (e+f x)+a b^2 d (5+2 n p) \sec ^2(e+f x)\right ) \, dx}{d (2+n p)}\\ &=\frac{b^2 \left (c (d \sec (e+f x))^p\right )^n (a+b \sec (e+f x)) \tan (e+f x)}{f (2+n p)}+\frac{\left ((d \sec (e+f x))^{-n p} \left (c (d \sec (e+f x))^p\right )^n\right ) \int (d \sec (e+f x))^{n p} \left (a d \left (b^2 n p+a^2 (2+n p)\right )+a b^2 d (5+2 n p) \sec ^2(e+f x)\right ) \, dx}{d (2+n p)}+\frac{\left (b \left (b^2 (1+n p)+3 a^2 (2+n p)\right ) (d \sec (e+f x))^{-n p} \left (c (d \sec (e+f x))^p\right )^n\right ) \int (d \sec (e+f x))^{1+n p} \, dx}{d (2+n p)}\\ &=\frac{a b^2 (5+2 n p) \left (c (d \sec (e+f x))^p\right )^n \tan (e+f x)}{f (1+n p) (2+n p)}+\frac{b^2 \left (c (d \sec (e+f x))^p\right )^n (a+b \sec (e+f x)) \tan (e+f x)}{f (2+n p)}+\frac{\left (b \left (b^2 (1+n p)+3 a^2 (2+n p)\right ) \left (\frac{\cos (e+f x)}{d}\right )^{n p} \left (c (d \sec (e+f x))^p\right )^n\right ) \int \left (\frac{\cos (e+f x)}{d}\right )^{-1-n p} \, dx}{d (2+n p)}+\frac{\left (a \left (3 b^2 n p+a^2 (1+n p)\right ) (d \sec (e+f x))^{-n p} \left (c (d \sec (e+f x))^p\right )^n\right ) \int (d \sec (e+f x))^{n p} \, dx}{1+n p}\\ &=\frac{b \left (b^2 (1+n p)+3 a^2 (2+n p)\right ) \, _2F_1\left (\frac{1}{2},-\frac{n p}{2};\frac{1}{2} (2-n p);\cos ^2(e+f x)\right ) \left (c (d \sec (e+f x))^p\right )^n \sin (e+f x)}{f n p (2+n p) \sqrt{\sin ^2(e+f x)}}+\frac{a b^2 (5+2 n p) \left (c (d \sec (e+f x))^p\right )^n \tan (e+f x)}{f (1+n p) (2+n p)}+\frac{b^2 \left (c (d \sec (e+f x))^p\right )^n (a+b \sec (e+f x)) \tan (e+f x)}{f (2+n p)}+\frac{\left (a \left (3 b^2 n p+a^2 (1+n p)\right ) \left (\frac{\cos (e+f x)}{d}\right )^{n p} \left (c (d \sec (e+f x))^p\right )^n\right ) \int \left (\frac{\cos (e+f x)}{d}\right )^{-n p} \, dx}{1+n p}\\ &=\frac{b \left (b^2 (1+n p)+3 a^2 (2+n p)\right ) \, _2F_1\left (\frac{1}{2},-\frac{n p}{2};\frac{1}{2} (2-n p);\cos ^2(e+f x)\right ) \left (c (d \sec (e+f x))^p\right )^n \sin (e+f x)}{f n p (2+n p) \sqrt{\sin ^2(e+f x)}}-\frac{a \left (3 b^2 n p+a^2 (1+n p)\right ) \cos (e+f x) \, _2F_1\left (\frac{1}{2},\frac{1}{2} (1-n p);\frac{1}{2} (3-n p);\cos ^2(e+f x)\right ) \left (c (d \sec (e+f x))^p\right )^n \sin (e+f x)}{f \left (1-n^2 p^2\right ) \sqrt{\sin ^2(e+f x)}}+\frac{a b^2 (5+2 n p) \left (c (d \sec (e+f x))^p\right )^n \tan (e+f x)}{f (1+n p) (2+n p)}+\frac{b^2 \left (c (d \sec (e+f x))^p\right )^n (a+b \sec (e+f x)) \tan (e+f x)}{f (2+n p)}\\ \end{align*}
Mathematica [A] time = 1.05196, size = 278, normalized size = 0.94 \[ -\frac{\left (-\tan ^2(e+f x)\right )^{3/2} \csc ^3(e+f x) \left (c (d \sec (e+f x))^p\right )^n \left (b n p \left ((n p+2) \left (3 a^2 (n p+3) \cos ^2(e+f x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{2} (n p+1),\frac{1}{2} (n p+3),\sec ^2(e+f x)\right )+b^2 (n p+1) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{2} (n p+3),\frac{1}{2} (n p+5),\sec ^2(e+f x)\right )\right )+3 a b \left (n^2 p^2+4 n p+3\right ) \cos (e+f x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{n p}{2}+1,\frac{n p}{2}+2,\sec ^2(e+f x)\right )\right )+a^3 \left (n^3 p^3+6 n^2 p^2+11 n p+6\right ) \cos ^3(e+f x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{n p}{2},\frac{n p}{2}+1,\sec ^2(e+f x)\right )\right )}{f n p (n p+1) (n p+2) (n p+3)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.163, size = 0, normalized size = 0. \begin{align*} \int \left ( c \left ( d\sec \left ( fx+e \right ) \right ) ^{p} \right ) ^{n} \left ( a+b\sec \left ( fx+e \right ) \right ) ^{3}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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 ({\left (b^{3} \sec \left (f x + e\right )^{3} + 3 \, a b^{2} \sec \left (f x + e\right )^{2} + 3 \, a^{2} b \sec \left (f x + e\right ) + a^{3}\right )} \left (\left (d \sec \left (f x + e\right )\right )^{p} c\right )^{n}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c \left (d \sec{\left (e + f x \right )}\right )^{p}\right )^{n} \left (a + b \sec{\left (e + f x \right )}\right )^{3}\, dx \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{\left (b \sec \left (f x + e\right ) + a\right )}^{3} \left (\left (d \sec \left (f x + e\right )\right )^{p} c\right )^{n}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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