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 = 0.348737, antiderivative size = 273, normalized size of antiderivative = 1., 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 3840
Rule 4007
Rule 3834
Rule 139
Rule 138
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*}
Mathematica [B] time = 26.2263, size = 8908, normalized size = 32.63 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.275, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( fx+e \right ) \right ) ^{3} \left ( a+b\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 (b \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{3}\,{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 (b \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{3}, 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 + b \sec{\left (e + f x \right )}\right )^{m} \sec ^{3}{\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 (b \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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