Optimal. Leaf size=211 \[ \frac{2^{m+\frac{1}{2}} m \left (m^2+3 m+5\right ) \tan (e+f x) (\sec (e+f x)+1)^{-m-\frac{1}{2}} (a \sec (e+f x)+a)^m \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{2}-m,\frac{3}{2},\frac{1}{2} (1-\sec (e+f x))\right )}{f (m+1) (m+2) (m+3)}+\frac{m \tan (e+f x) (a \sec (e+f x)+a)^{m+1}}{a f \left (m^2+5 m+6\right )}+\frac{\tan (e+f x) \sec ^2(e+f x) (a \sec (e+f x)+a)^m}{f (m+3)}+\frac{(m+4) \tan (e+f x) (a \sec (e+f x)+a)^m}{f (m+1) (m+2) (m+3)} \]
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Rubi [A] time = 0.352669, antiderivative size = 211, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3824, 4010, 4001, 3828, 3827, 69} \[ \frac{2^{m+\frac{1}{2}} m \left (m^2+3 m+5\right ) \tan (e+f x) (\sec (e+f x)+1)^{-m-\frac{1}{2}} (a \sec (e+f x)+a)^m \, _2F_1\left (\frac{1}{2},\frac{1}{2}-m;\frac{3}{2};\frac{1}{2} (1-\sec (e+f x))\right )}{f (m+1) (m+2) (m+3)}+\frac{m \tan (e+f x) (a \sec (e+f x)+a)^{m+1}}{a f \left (m^2+5 m+6\right )}+\frac{\tan (e+f x) \sec ^2(e+f x) (a \sec (e+f x)+a)^m}{f (m+3)}+\frac{(m+4) \tan (e+f x) (a \sec (e+f x)+a)^m}{f (m+1) (m+2) (m+3)} \]
Antiderivative was successfully verified.
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Rule 3824
Rule 4010
Rule 4001
Rule 3828
Rule 3827
Rule 69
Rubi steps
\begin{align*} \int \sec ^4(e+f x) (a+a \sec (e+f x))^m \, dx &=\frac{\sec ^2(e+f x) (a+a \sec (e+f x))^m \tan (e+f x)}{f (3+m)}+\frac{\int \sec ^2(e+f x) (a+a \sec (e+f x))^m (2 a+a m \sec (e+f x)) \, dx}{a (3+m)}\\ &=\frac{\sec ^2(e+f x) (a+a \sec (e+f x))^m \tan (e+f x)}{f (3+m)}+\frac{m (a+a \sec (e+f x))^{1+m} \tan (e+f x)}{a f \left (6+5 m+m^2\right )}+\frac{\int \sec (e+f x) (a+a \sec (e+f x))^m \left (a^2 m (1+m)+a^2 (4+m) \sec (e+f x)\right ) \, dx}{a^2 (2+m) (3+m)}\\ &=\frac{(4+m) (a+a \sec (e+f x))^m \tan (e+f x)}{f (1+m) (2+m) (3+m)}+\frac{\sec ^2(e+f x) (a+a \sec (e+f x))^m \tan (e+f x)}{f (3+m)}+\frac{m (a+a \sec (e+f x))^{1+m} \tan (e+f x)}{a f \left (6+5 m+m^2\right )}+\frac{\left (m \left (5+3 m+m^2\right )\right ) \int \sec (e+f x) (a+a \sec (e+f x))^m \, dx}{(1+m) (2+m) (3+m)}\\ &=\frac{(4+m) (a+a \sec (e+f x))^m \tan (e+f x)}{f (1+m) (2+m) (3+m)}+\frac{\sec ^2(e+f x) (a+a \sec (e+f x))^m \tan (e+f x)}{f (3+m)}+\frac{m (a+a \sec (e+f x))^{1+m} \tan (e+f x)}{a f \left (6+5 m+m^2\right )}+\frac{\left (m \left (5+3 m+m^2\right ) (1+\sec (e+f x))^{-m} (a+a \sec (e+f x))^m\right ) \int \sec (e+f x) (1+\sec (e+f x))^m \, dx}{(1+m) (2+m) (3+m)}\\ &=\frac{(4+m) (a+a \sec (e+f x))^m \tan (e+f x)}{f (1+m) (2+m) (3+m)}+\frac{\sec ^2(e+f x) (a+a \sec (e+f x))^m \tan (e+f x)}{f (3+m)}+\frac{m (a+a \sec (e+f x))^{1+m} \tan (e+f x)}{a f \left (6+5 m+m^2\right )}-\frac{\left (m \left (5+3 m+m^2\right ) (1+\sec (e+f x))^{-\frac{1}{2}-m} (a+a \sec (e+f x))^m \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(1+x)^{-\frac{1}{2}+m}}{\sqrt{1-x}} \, dx,x,\sec (e+f x)\right )}{f (1+m) (2+m) (3+m) \sqrt{1-\sec (e+f x)}}\\ &=\frac{(4+m) (a+a \sec (e+f x))^m \tan (e+f x)}{f (1+m) (2+m) (3+m)}+\frac{\sec ^2(e+f x) (a+a \sec (e+f x))^m \tan (e+f x)}{f (3+m)}+\frac{2^{\frac{1}{2}+m} m \left (5+3 m+m^2\right ) \, _2F_1\left (\frac{1}{2},\frac{1}{2}-m;\frac{3}{2};\frac{1}{2} (1-\sec (e+f x))\right ) (1+\sec (e+f x))^{-\frac{1}{2}-m} (a+a \sec (e+f x))^m \tan (e+f x)}{f (1+m) (2+m) (3+m)}+\frac{m (a+a \sec (e+f x))^{1+m} \tan (e+f x)}{a f \left (6+5 m+m^2\right )}\\ \end{align*}
Mathematica [A] time = 1.27768, size = 154, normalized size = 0.73 \[ \frac{\tan (e+f x) (\sec (e+f x)+1)^{-m-\frac{1}{2}} (a (\sec (e+f x)+1))^m \left (2^{m+\frac{3}{2}} m \left (m^2+3 m+5\right ) \text{Hypergeometric2F1}\left (\frac{1}{2},-m-\frac{1}{2},\frac{3}{2},\frac{1}{2} (1-\sec (e+f x))\right )+\left (\left (2 m^2+5 m+2\right ) \sec ^2(e+f x)+(2 m+1) m \sec (e+f x)+m^2+m+4\right ) (\sec (e+f x)+1)^{m+\frac{1}{2}}\right )}{f (m+2) (m+3) (2 m+1)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.31, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( fx+e \right ) \right ) ^{4} \left ( a+a\sec \left ( fx+e \right ) \right ) ^{m}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{4}\,{d x} \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 (a \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{4}, 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 (a \left (\sec{\left (e + f x \right )} + 1\right )\right )^{m} \sec ^{4}{\left (e + f x \right )}\, 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 (a \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{4}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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