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

Optimal. Leaf size=69 \[ \frac {2 a \sin (e+f x) (-\sec (e+f x))^{-n} \sec ^{n+1}(e+f x) \, _2F_1\left (\frac {1}{2},1-n;\frac {3}{2};\sec (e+f x)+1\right )}{f \sqrt {a-a \sec (e+f x)}} \]

[Out]

2*a*hypergeom([1/2, 1-n],[3/2],1+sec(f*x+e))*sec(f*x+e)^(1+n)*sin(f*x+e)/f/((-sec(f*x+e))^n)/(a-a*sec(f*x+e))^
(1/2)

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Rubi [A]  time = 0.08, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3806, 67, 65} \[ \frac {2 a \sin (e+f x) (-\sec (e+f x))^{-n} \sec ^{n+1}(e+f x) \, _2F_1\left (\frac {1}{2},1-n;\frac {3}{2};\sec (e+f x)+1\right )}{f \sqrt {a-a \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[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]]*Sec[e + f*x]^(1 + n)*Sin[e + f*x])/(f*(-Sec[e + f*x]
)^n*Sqrt[a - a*Sec[e + f*x]])

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])

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 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]

Rubi steps

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

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Mathematica [C]  time = 0.40, size = 185, normalized size = 2.68 \[ -\frac {2^n e^{\frac {1}{2} i (e+f (1-2 n) x)} \left (\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^n \cos (e+f x) \csc \left (\frac {1}{2} (e+f x)\right ) \sqrt {a-a \sec (e+f x)} \left (n e^{i (e+f (n+1) x)} \, _2F_1\left (1,1-\frac {n}{2};\frac {n+3}{2};-e^{2 i (e+f x)}\right )-(n+1) e^{i f n x} \, _2F_1\left (1,\frac {1-n}{2};\frac {n+2}{2};-e^{2 i (e+f x)}\right )\right )}{f n (n+1)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {-a \sec \left (f x + e\right ) + a} \sec \left (f x + e\right )^{n}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(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)*sec(f*x + e)^n, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {-a \sec \left (f x + e\right ) + a} \sec \left (f x + e\right )^{n}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(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)*sec(f*x + e)^n, x)

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maple [F]  time = 1.26, size = 0, normalized size = 0.00 \[ \int \left (\sec ^{n}\left (f x +e \right )\right ) \sqrt {a -a \sec \left (f x +e \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {-a \sec \left (f x + e\right ) + a} \sec \left (f x + e\right )^{n}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(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)*sec(f*x + e)^n, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \sqrt {a-\frac {a}{\cos \left (e+f\,x\right )}}\,{\left (\frac {1}{\cos \left (e+f\,x\right )}\right )}^n \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {- a \left (\sec {\left (e + f x \right )} - 1\right )} \sec ^{n}{\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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