Optimal. Leaf size=129 \[ \frac{a (d \tan (e+f x))^{n+1} \text{Hypergeometric2F1}\left (1,\frac{n+1}{2},\frac{n+3}{2},-\tan ^2(e+f x)\right )}{d f (n+1)}+\frac{b \sec (e+f x) \cos ^2(e+f x)^{\frac{n+2}{2}} (d \tan (e+f x))^{n+1} \text{Hypergeometric2F1}\left (\frac{n+1}{2},\frac{n+2}{2},\frac{n+3}{2},\sin ^2(e+f x)\right )}{d f (n+1)} \]
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Rubi [A] time = 0.0852123, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3884, 3476, 364, 2617} \[ \frac{a (d \tan (e+f x))^{n+1} \, _2F_1\left (1,\frac{n+1}{2};\frac{n+3}{2};-\tan ^2(e+f x)\right )}{d f (n+1)}+\frac{b \sec (e+f x) \cos ^2(e+f x)^{\frac{n+2}{2}} (d \tan (e+f x))^{n+1} \, _2F_1\left (\frac{n+1}{2},\frac{n+2}{2};\frac{n+3}{2};\sin ^2(e+f x)\right )}{d f (n+1)} \]
Antiderivative was successfully verified.
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Rule 3884
Rule 3476
Rule 364
Rule 2617
Rubi steps
\begin{align*} \int (a+b \sec (e+f x)) (d \tan (e+f x))^n \, dx &=a \int (d \tan (e+f x))^n \, dx+b \int \sec (e+f x) (d \tan (e+f x))^n \, dx\\ &=\frac{b \cos ^2(e+f x)^{\frac{2+n}{2}} \, _2F_1\left (\frac{1+n}{2},\frac{2+n}{2};\frac{3+n}{2};\sin ^2(e+f x)\right ) \sec (e+f x) (d \tan (e+f x))^{1+n}}{d f (1+n)}+\frac{(a d) \operatorname{Subst}\left (\int \frac{x^n}{d^2+x^2} \, dx,x,d \tan (e+f x)\right )}{f}\\ &=\frac{a \, _2F_1\left (1,\frac{1+n}{2};\frac{3+n}{2};-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{1+n}}{d f (1+n)}+\frac{b \cos ^2(e+f x)^{\frac{2+n}{2}} \, _2F_1\left (\frac{1+n}{2},\frac{2+n}{2};\frac{3+n}{2};\sin ^2(e+f x)\right ) \sec (e+f x) (d \tan (e+f x))^{1+n}}{d f (1+n)}\\ \end{align*}
Mathematica [A] time = 0.675377, size = 106, normalized size = 0.82 \[ \frac{(d \tan (e+f x))^n \left (\frac{a \tan (e+f x) \text{Hypergeometric2F1}\left (1,\frac{n+1}{2},\frac{n+3}{2},-\tan ^2(e+f x)\right )}{n+1}+b \csc (e+f x) \left (-\tan ^2(e+f x)\right )^{\frac{1-n}{2}} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1-n}{2},\frac{3}{2},\sec ^2(e+f x)\right )\right )}{f} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.663, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\sec \left ( fx+e \right ) \right ) \left ( d\tan \left ( fx+e \right ) \right ) ^{n}\, 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 )} \left (d \tan \left (f x + e\right )\right )^{n}\,{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 )} \left (d \tan \left (f x + e\right )\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 (d \tan{\left (e + f x \right )}\right )^{n} \left (a + b \sec{\left (e + f x \right )}\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 )} \left (d \tan \left (f x + e\right )\right )^{n}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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