Optimal. Leaf size=211 \[ \frac {2^{m+\frac {1}{2}} m \left (m^2+3 m+5\right ) \tan (e+f x) (\sec (e+f x)+1)^{-m-\frac {1}{2}} (a \sec (e+f x)+a)^m \, _2F_1\left (\frac {1}{2},\frac {1}{2}-m;\frac {3}{2};\frac {1}{2} (1-\sec (e+f x))\right )}{f (m+1) (m+2) (m+3)}+\frac {m \tan (e+f x) (a \sec (e+f x)+a)^{m+1}}{a f \left (m^2+5 m+6\right )}+\frac {\tan (e+f x) \sec ^2(e+f x) (a \sec (e+f x)+a)^m}{f (m+3)}+\frac {(m+4) \tan (e+f x) (a \sec (e+f x)+a)^m}{f (m+1) (m+2) (m+3)} \]
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Rubi [A] time = 0.35, antiderivative size = 211, normalized size of antiderivative = 1.00, 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 69
Rule 3824
Rule 3827
Rule 3828
Rule 4001
Rule 4010
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*}
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Mathematica [A] time = 1.36, size = 154, normalized size = 0.73 \[ \frac {\tan (e+f x) (\sec (e+f x)+1)^{-m-\frac {1}{2}} (a (\sec (e+f x)+1))^m \left (2^{m+\frac {3}{2}} m \left (m^2+3 m+5\right ) \, _2F_1\left (\frac {1}{2},-m-\frac {1}{2};\frac {3}{2};\frac {1}{2} (1-\sec (e+f x))\right )+\left (\left (2 m^2+5 m+2\right ) \sec ^2(e+f x)+(2 m+1) m \sec (e+f x)+m^2+m+4\right ) (\sec (e+f x)+1)^{m+\frac {1}{2}}\right )}{f (m+2) (m+3) (2 m+1)} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{4}, x\right ) \]
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 \[ \int {\left (a \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{4}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.16, size = 0, normalized size = 0.00 \[ \int \left (\sec ^{4}\left (f x +e \right )\right ) \left (a +a \sec \left (f x +e \right )\right )^{m}\, dx \]
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 \[ \int {\left (a \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{4}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^m}{{\cos \left (e+f\,x\right )}^4} \,d x \]
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 \[ \int \left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{m} \sec ^{4}{\left (e + f x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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