3.328 \(\int (d \sec (e+f x))^n \sqrt{a-a \sec (e+f x)} \, dx\)

Optimal. Leaf size=69 \[ \frac{2 a \tan (e+f x) (-\sec (e+f x))^{-n} (d \sec (e+f x))^n \text{Hypergeometric2F1}\left (\frac{1}{2},1-n,\frac{3}{2},\sec (e+f x)+1\right )}{f \sqrt{a-a \sec (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.0808413, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {3806, 67, 65} \[ \frac{2 a \tan (e+f x) (-\sec (e+f x))^{-n} (d \sec (e+f x))^n \, _2F_1\left (\frac{1}{2},1-n;\frac{3}{2};\sec (e+f x)+1\right )}{f \sqrt{a-a \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(d*Sec[e + f*x])^n*Sqrt[a - a*Sec[e + f*x]],x]

[Out]

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

Rule 3806

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

Rule 67

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[((-((b*c)/d))^IntPart[m]*(b*x)^FracPart[m])/
(-((d*x)/c))^FracPart[m], Int[(-((d*x)/c))^m*(c + d*x)^n, x], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !GtQ[c, 0] &&  !GtQ[-(d/(b*c)), 0]

Rule 65

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)*Hypergeometric2F1[-m, n +
1, n + 2, 1 + (d*x)/c])/(d*(n + 1)*(-(d/(b*c)))^m), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Inte
gerQ[m] || GtQ[-(d/(b*c)), 0])

Rubi steps

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

Mathematica [C]  time = 0.53145, size = 213, normalized size = 3.09 \[ \frac{2^{n-\frac{1}{2}} e^{\frac{1}{2} i (e+f (1-2 n) x)} \left (\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^{n-\frac{1}{2}} \csc \left (\frac{e}{2}+\frac{f x}{2}\right ) \sqrt{a-a \sec (e+f x)} \sec ^{-n-\frac{1}{2}}(e+f x) \left ((n+1) e^{i f n x} \text{Hypergeometric2F1}\left (1,\frac{1-n}{2},\frac{n+2}{2},-e^{2 i (e+f x)}\right )-n e^{i (e+f (n+1) x)} \text{Hypergeometric2F1}\left (1,1-\frac{n}{2},\frac{n+3}{2},-e^{2 i (e+f x)}\right )\right ) (d \sec (e+f x))^n}{f n (n+1)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(d*Sec[e + f*x])^n*Sqrt[a - a*Sec[e + f*x]],x]

[Out]

(2^(-1/2 + n)*E^((I/2)*(e + f*(1 - 2*n)*x))*(E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x))))^(-1/2 + n)*Csc[e/2 + (
f*x)/2]*(E^(I*f*n*x)*(1 + n)*Hypergeometric2F1[1, (1 - n)/2, (2 + n)/2, -E^((2*I)*(e + f*x))] - E^(I*(e + f*(1
+ n)*x))*n*Hypergeometric2F1[1, 1 - n/2, (3 + n)/2, -E^((2*I)*(e + f*x))])*Sec[e + f*x]^(-1/2 - n)*(d*Sec[e +
f*x])^n*Sqrt[a - a*Sec[e + f*x]])/(f*n*(1 + n))

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Maple [F]  time = 0.198, size = 0, normalized size = 0. \begin{align*} \int \left ( d\sec \left ( fx+e \right ) \right ) ^{n}\sqrt{a-a\sec \left ( fx+e \right ) }\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sec(f*x+e))^n*(a-a*sec(f*x+e))^(1/2),x)

[Out]

int((d*sec(f*x+e))^n*(a-a*sec(f*x+e))^(1/2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^n*(a-a*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(-a*sec(f*x + e) + a)*(d*sec(f*x + e))^n, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^n*(a-a*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(-a*sec(f*x + e) + a)*(d*sec(f*x + e))^n, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))**n*(a-a*sec(f*x+e))**(1/2),x)

[Out]

Integral((d*sec(e + f*x))**n*sqrt(-a*(sec(e + f*x) - 1)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^n*(a-a*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(-a*sec(f*x + e) + a)*(d*sec(f*x + e))^n, x)