\(\int (d \sec (e+f x))^n \sqrt {1+\sec (e+f x)} \, dx\) [304]

Optimal result

Integrand size = 23, antiderivative size = 64 \[ \int (d \sec (e+f x))^n \sqrt {1+\sec (e+f x)} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (\frac {1}{2},n,1+n,\sec (e+f x)\right ) (d \sec (e+f x))^n \tan (e+f x)}{f n \sqrt {1-\sec (e+f x)} \sqrt {1+\sec (e+f x)}} \]

[Out]

-hypergeom([1/2, n],[1+n],sec(f*x+e))*(d*sec(f*x+e))^n*tan(f*x+e)/f/n/(1-sec(f*x+e))^(1/2)/(1+sec(f*x+e))^(1/2
)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {3891, 66} \[ \int (d \sec (e+f x))^n \sqrt {1+\sec (e+f x)} \, dx=-\frac {\tan (e+f x) (d \sec (e+f x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},n,n+1,\sec (e+f x)\right )}{f n \sqrt {1-\sec (e+f x)} \sqrt {\sec (e+f x)+1}} \]

[In]

Int[(d*Sec[e + f*x])^n*Sqrt[1 + Sec[e + f*x]],x]

[Out]

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

Rule 66

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x)^(m + 1)/(b*(m + 1)))*Hypergeometr
ic2F1[-n, m + 1, m + 2, (-d)*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 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 {(d \tan (e+f x)) \text {Subst}\left (\int \frac {(d x)^{-1+n}}{\sqrt {1-x}} \, dx,x,\sec (e+f x)\right )}{f \sqrt {1-\sec (e+f x)} \sqrt {1+\sec (e+f x)}} \\ & = -\frac {\operatorname {Hypergeometric2F1}\left (\frac {1}{2},n,1+n,\sec (e+f x)\right ) (d \sec (e+f x))^n \tan (e+f x)}{f n \sqrt {1-\sec (e+f x)} \sqrt {1+\sec (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.05 \[ \int (d \sec (e+f x))^n \sqrt {1+\sec (e+f x)} \, dx=\frac {2 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1-n,\frac {3}{2},1-\sec (e+f x)\right ) \sec ^{1-n}(e+f x) (d \sec (e+f x))^n \sin (e+f x)}{f \sqrt {1+\sec (e+f x)}} \]

[In]

Integrate[(d*Sec[e + f*x])^n*Sqrt[1 + Sec[e + f*x]],x]

[Out]

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

Maple [F]

\[\int \left (d \sec \left (f x +e \right )\right )^{n} \sqrt {\sec \left (f x +e \right )+1}d x\]

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral((d*sec(f*x + e))^n*sqrt(sec(f*x + e) + 1), x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

integrate((d*sec(f*x + e))^n*sqrt(sec(f*x + e) + 1), x)

Giac [F]

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

[In]

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

[Out]

integrate((d*sec(f*x + e))^n*sqrt(sec(f*x + e) + 1), x)

Mupad [F(-1)]

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

[In]

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

[Out]

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