\(\int (-\sec (e+f x))^n \sqrt {a+a \sec (e+f x)} \, dx\) [313]

Optimal result

Integrand size = 25, antiderivative size = 70 \[ \int (-\sec (e+f x))^n \sqrt {a+a \sec (e+f x)} \, dx=\frac {2 a \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1-n,\frac {3}{2},1-\sec (e+f x)\right ) (-\sec (e+f x))^n \sec ^{1-n}(e+f x) \sin (e+f x)}{f \sqrt {a+a \sec (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {3891, 69, 67} \[ \int (-\sec (e+f x))^n \sqrt {a+a \sec (e+f x)} \, dx=\frac {2 a \sin (e+f x) (-\sec (e+f x))^n \sec ^{1-n}(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1-n,\frac {3}{2},1-\sec (e+f x)\right )}{f \sqrt {a \sec (e+f x)+a}} \]

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

Rule 67

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

Rule 69

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 3891

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*} \text {integral}& = \frac {\left (a^2 \tan (e+f x)\right ) \text {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 ) \text {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 \operatorname {Hypergeometric2F1}\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*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.01 \[ \int (-\sec (e+f x))^n \sqrt {a+a \sec (e+f x)} \, dx=\frac {2 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1-n,\frac {3}{2},1-\sec (e+f x)\right ) (-\sec (e+f x))^n \sec ^{-n}(e+f x) \sqrt {a (1+\sec (e+f x))} \tan \left (\frac {1}{2} (e+f x)\right )}{f} \]

[In]

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

[Out]

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

Maple [F]

\[\int \left (-\sec \left (f x +e \right )\right )^{n} \sqrt {a +a \sec \left (f x +e \right )}d x\]

[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)

Fricas [F]

\[ \int (-\sec (e+f x))^n \sqrt {a+a \sec (e+f x)} \, dx=\int { \sqrt {a \sec \left (f x + e\right ) + a} \left (-\sec \left (f x + e\right )\right )^{n} \,d x } \]

[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)

Sympy [F]

\[ \int (-\sec (e+f x))^n \sqrt {a+a \sec (e+f x)} \, dx=\int \left (- \sec {\left (e + f x \right )}\right )^{n} \sqrt {a \left (\sec {\left (e + f x \right )} + 1\right )}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int (-\sec (e+f x))^n \sqrt {a+a \sec (e+f x)} \, dx=\int { \sqrt {a \sec \left (f x + e\right ) + a} \left (-\sec \left (f x + e\right )\right )^{n} \,d x } \]

[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)

Giac [F]

\[ \int (-\sec (e+f x))^n \sqrt {a+a \sec (e+f x)} \, dx=\int { \sqrt {a \sec \left (f x + e\right ) + a} \left (-\sec \left (f x + e\right )\right )^{n} \,d x } \]

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int (-\sec (e+f x))^n \sqrt {a+a \sec (e+f x)} \, dx=\int \sqrt {a+\frac {a}{\cos \left (e+f\,x\right )}}\,{\left (-\frac {1}{\cos \left (e+f\,x\right )}\right )}^n \,d x \]

[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)