Optimal. Leaf size=273 \[ \frac {\sqrt {2} \left (a^2+b^2 (m+1)\right ) \tan (e+f x) (a+b \sec (e+f x))^m \left (\frac {a+b \sec (e+f x)}{a+b}\right )^{-m} F_1\left (\frac {1}{2};\frac {1}{2},-m;\frac {3}{2};\frac {1}{2} (1-\sec (e+f x)),\frac {b (1-\sec (e+f x))}{a+b}\right )}{b^2 f (m+2) \sqrt {\sec (e+f x)+1}}-\frac {\sqrt {2} a (a+b) \tan (e+f x) (a+b \sec (e+f x))^m \left (\frac {a+b \sec (e+f x)}{a+b}\right )^{-m} F_1\left (\frac {1}{2};\frac {1}{2},-m-1;\frac {3}{2};\frac {1}{2} (1-\sec (e+f x)),\frac {b (1-\sec (e+f x))}{a+b}\right )}{b^2 f (m+2) \sqrt {\sec (e+f x)+1}}+\frac {\tan (e+f x) (a+b \sec (e+f x))^{m+1}}{b f (m+2)} \]
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Rubi [A] time = 0.35, antiderivative size = 273, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3840, 4007, 3834, 139, 138} \[ \frac {\sqrt {2} \left (a^2+b^2 (m+1)\right ) \tan (e+f x) (a+b \sec (e+f x))^m \left (\frac {a+b \sec (e+f x)}{a+b}\right )^{-m} F_1\left (\frac {1}{2};\frac {1}{2},-m;\frac {3}{2};\frac {1}{2} (1-\sec (e+f x)),\frac {b (1-\sec (e+f x))}{a+b}\right )}{b^2 f (m+2) \sqrt {\sec (e+f x)+1}}-\frac {\sqrt {2} a (a+b) \tan (e+f x) (a+b \sec (e+f x))^m \left (\frac {a+b \sec (e+f x)}{a+b}\right )^{-m} F_1\left (\frac {1}{2};\frac {1}{2},-m-1;\frac {3}{2};\frac {1}{2} (1-\sec (e+f x)),\frac {b (1-\sec (e+f x))}{a+b}\right )}{b^2 f (m+2) \sqrt {\sec (e+f x)+1}}+\frac {\tan (e+f x) (a+b \sec (e+f x))^{m+1}}{b f (m+2)} \]
Antiderivative was successfully verified.
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Rule 138
Rule 139
Rule 3834
Rule 3840
Rule 4007
Rubi steps
\begin {align*} \int \sec ^3(e+f x) (a+b \sec (e+f x))^m \, dx &=\frac {(a+b \sec (e+f x))^{1+m} \tan (e+f x)}{b f (2+m)}+\frac {\int \sec (e+f x) (b (1+m)-a \sec (e+f x)) (a+b \sec (e+f x))^m \, dx}{b (2+m)}\\ &=\frac {(a+b \sec (e+f x))^{1+m} \tan (e+f x)}{b f (2+m)}-\frac {a \int \sec (e+f x) (a+b \sec (e+f x))^{1+m} \, dx}{b^2 (2+m)}+\frac {\left (a^2+b^2 (1+m)\right ) \int \sec (e+f x) (a+b \sec (e+f x))^m \, dx}{b^2 (2+m)}\\ &=\frac {(a+b \sec (e+f x))^{1+m} \tan (e+f x)}{b f (2+m)}+\frac {(a \tan (e+f x)) \operatorname {Subst}\left (\int \frac {(a+b x)^{1+m}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sec (e+f x)\right )}{b^2 f (2+m) \sqrt {1-\sec (e+f x)} \sqrt {1+\sec (e+f x)}}-\frac {\left (\left (a^2+b^2 (1+m)\right ) \tan (e+f x)\right ) \operatorname {Subst}\left (\int \frac {(a+b x)^m}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sec (e+f x)\right )}{b^2 f (2+m) \sqrt {1-\sec (e+f x)} \sqrt {1+\sec (e+f x)}}\\ &=\frac {(a+b \sec (e+f x))^{1+m} \tan (e+f x)}{b f (2+m)}-\frac {\left (a (-a-b) (a+b \sec (e+f x))^m \left (-\frac {a+b \sec (e+f x)}{-a-b}\right )^{-m} \tan (e+f x)\right ) \operatorname {Subst}\left (\int \frac {\left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^{1+m}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sec (e+f x)\right )}{b^2 f (2+m) \sqrt {1-\sec (e+f x)} \sqrt {1+\sec (e+f x)}}-\frac {\left (\left (a^2+b^2 (1+m)\right ) (a+b \sec (e+f x))^m \left (-\frac {a+b \sec (e+f x)}{-a-b}\right )^{-m} \tan (e+f x)\right ) \operatorname {Subst}\left (\int \frac {\left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^m}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sec (e+f x)\right )}{b^2 f (2+m) \sqrt {1-\sec (e+f x)} \sqrt {1+\sec (e+f x)}}\\ &=\frac {(a+b \sec (e+f x))^{1+m} \tan (e+f x)}{b f (2+m)}-\frac {\sqrt {2} a (a+b) F_1\left (\frac {1}{2};\frac {1}{2},-1-m;\frac {3}{2};\frac {1}{2} (1-\sec (e+f x)),\frac {b (1-\sec (e+f x))}{a+b}\right ) (a+b \sec (e+f x))^m \left (\frac {a+b \sec (e+f x)}{a+b}\right )^{-m} \tan (e+f x)}{b^2 f (2+m) \sqrt {1+\sec (e+f x)}}+\frac {\sqrt {2} \left (a^2+b^2 (1+m)\right ) F_1\left (\frac {1}{2};\frac {1}{2},-m;\frac {3}{2};\frac {1}{2} (1-\sec (e+f x)),\frac {b (1-\sec (e+f x))}{a+b}\right ) (a+b \sec (e+f x))^m \left (\frac {a+b \sec (e+f x)}{a+b}\right )^{-m} \tan (e+f x)}{b^2 f (2+m) \sqrt {1+\sec (e+f x)}}\\ \end {align*}
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Mathematica [B] time = 26.77, size = 8899, normalized size = 32.60 \[ \text {Result too large to show} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{3}, x\right ) \]
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 \[ \int {\left (b \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{3}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.14, size = 0, normalized size = 0.00 \[ \int \left (\sec ^{3}\left (f x +e \right )\right ) \left (a +b \sec \left (f x +e \right )\right )^{m}\, dx \]
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 \[ \int {\left (b \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{3}\,{d x} \]
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 \[ \int \frac {{\left (a+\frac {b}{\cos \left (e+f\,x\right )}\right )}^m}{{\cos \left (e+f\,x\right )}^3} \,d x \]
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 \[ \int \left (a + b \sec {\left (e + f x \right )}\right )^{m} \sec ^{3}{\left (e + f x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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