3.292 \(\int \frac{\sec ^n(e+f x)}{a+a \sec (e+f x)} \, dx\)

Optimal. Leaf size=174 \[ \frac{(1-n) \sin (e+f x) \sec ^{n-2}(e+f x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{2-n}{2},\frac{4-n}{2},\cos ^2(e+f x)\right )}{a f (2-n) \sqrt{\sin ^2(e+f x)}}-\frac{\sin (e+f x) \sec ^{n-1}(e+f x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1-n}{2},\frac{3-n}{2},\cos ^2(e+f x)\right )}{a f \sqrt{\sin ^2(e+f x)}}+\frac{\sin (e+f x) \sec ^n(e+f x)}{f (a \sec (e+f x)+a)} \]

[Out]

(Sec[e + f*x]^n*Sin[e + f*x])/(f*(a + a*Sec[e + f*x])) + ((1 - n)*Hypergeometric2F1[1/2, (2 - n)/2, (4 - n)/2,
Cos[e + f*x]^2]*Sec[e + f*x]^(-2 + n)*Sin[e + f*x])/(a*f*(2 - n)*Sqrt[Sin[e + f*x]^2]) - (Hypergeometric2F1[1
/2, (1 - n)/2, (3 - n)/2, Cos[e + f*x]^2]*Sec[e + f*x]^(-1 + n)*Sin[e + f*x])/(a*f*Sqrt[Sin[e + f*x]^2])

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Rubi [A]  time = 0.157535, antiderivative size = 174, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3820, 3787, 3772, 2643} \[ \frac{(1-n) \sin (e+f x) \sec ^{n-2}(e+f x) \, _2F_1\left (\frac{1}{2},\frac{2-n}{2};\frac{4-n}{2};\cos ^2(e+f x)\right )}{a f (2-n) \sqrt{\sin ^2(e+f x)}}-\frac{\sin (e+f x) \sec ^{n-1}(e+f x) \, _2F_1\left (\frac{1}{2},\frac{1-n}{2};\frac{3-n}{2};\cos ^2(e+f x)\right )}{a f \sqrt{\sin ^2(e+f x)}}+\frac{\sin (e+f x) \sec ^n(e+f x)}{f (a \sec (e+f x)+a)} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^n/(a + a*Sec[e + f*x]),x]

[Out]

(Sec[e + f*x]^n*Sin[e + f*x])/(f*(a + a*Sec[e + f*x])) + ((1 - n)*Hypergeometric2F1[1/2, (2 - n)/2, (4 - n)/2,
Cos[e + f*x]^2]*Sec[e + f*x]^(-2 + n)*Sin[e + f*x])/(a*f*(2 - n)*Sqrt[Sin[e + f*x]^2]) - (Hypergeometric2F1[1
/2, (1 - n)/2, (3 - n)/2, Cos[e + f*x]^2]*Sec[e + f*x]^(-1 + n)*Sin[e + f*x])/(a*f*Sqrt[Sin[e + f*x]^2])

Rule 3820

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[(b*d*Cot[e
+ f*x]*(d*Csc[e + f*x])^(n - 1))/(a*f*(a + b*Csc[e + f*x])), x] + Dist[(d*(n - 1))/(a*b), Int[(d*Csc[e + f*x]
)^(n - 1)*(a - b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0]

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3772

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x])^(n - 1)*((Sin[c + d*x]/b)^(n - 1)
*Int[1/(Sin[c + d*x]/b)^n, x]), x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rubi steps

\begin{align*} \int \frac{\sec ^n(e+f x)}{a+a \sec (e+f x)} \, dx &=\frac{\sec ^n(e+f x) \sin (e+f x)}{f (a+a \sec (e+f x))}-\frac{(1-n) \int \sec ^{-1+n}(e+f x) (a-a \sec (e+f x)) \, dx}{a^2}\\ &=\frac{\sec ^n(e+f x) \sin (e+f x)}{f (a+a \sec (e+f x))}-\frac{(1-n) \int \sec ^{-1+n}(e+f x) \, dx}{a}+\frac{(1-n) \int \sec ^n(e+f x) \, dx}{a}\\ &=\frac{\sec ^n(e+f x) \sin (e+f x)}{f (a+a \sec (e+f x))}-\frac{\left ((1-n) \cos ^n(e+f x) \sec ^n(e+f x)\right ) \int \cos ^{1-n}(e+f x) \, dx}{a}+\frac{\left ((1-n) \cos ^n(e+f x) \sec ^n(e+f x)\right ) \int \cos ^{-n}(e+f x) \, dx}{a}\\ &=\frac{\sec ^n(e+f x) \sin (e+f x)}{f (a+a \sec (e+f x))}+\frac{(1-n) \, _2F_1\left (\frac{1}{2},\frac{2-n}{2};\frac{4-n}{2};\cos ^2(e+f x)\right ) \sec ^{-2+n}(e+f x) \sin (e+f x)}{a f (2-n) \sqrt{\sin ^2(e+f x)}}-\frac{\, _2F_1\left (\frac{1}{2},\frac{1-n}{2};\frac{3-n}{2};\cos ^2(e+f x)\right ) \sec ^{-1+n}(e+f x) \sin (e+f x)}{a f \sqrt{\sin ^2(e+f x)}}\\ \end{align*}

Mathematica [F]  time = 0.946683, size = 0, normalized size = 0. \[ \int \frac{\sec ^n(e+f x)}{a+a \sec (e+f x)} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[Sec[e + f*x]^n/(a + a*Sec[e + f*x]),x]

[Out]

Integrate[Sec[e + f*x]^n/(a + a*Sec[e + f*x]), x]

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Maple [F]  time = 0.746, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( \sec \left ( fx+e \right ) \right ) ^{n}}{a+a\sec \left ( fx+e \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^n/(a+a*sec(f*x+e)),x)

[Out]

int(sec(f*x+e)^n/(a+a*sec(f*x+e)),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (f x + e\right )^{n}}{a \sec \left (f x + e\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^n/(a+a*sec(f*x+e)),x, algorithm="maxima")

[Out]

integrate(sec(f*x + e)^n/(a*sec(f*x + e) + a), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sec \left (f x + e\right )^{n}}{a \sec \left (f x + e\right ) + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^n/(a+a*sec(f*x+e)),x, algorithm="fricas")

[Out]

integral(sec(f*x + e)^n/(a*sec(f*x + e) + a), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\sec ^{n}{\left (e + f x \right )}}{\sec{\left (e + f x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)**n/(a+a*sec(f*x+e)),x)

[Out]

Integral(sec(e + f*x)**n/(sec(e + f*x) + 1), x)/a

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (f x + e\right )^{n}}{a \sec \left (f x + e\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^n/(a+a*sec(f*x+e)),x, algorithm="giac")

[Out]

integrate(sec(f*x + e)^n/(a*sec(f*x + e) + a), x)