\(\int (b \sec (e+f x))^{5/2} \, dx\) [407]

Optimal result

Integrand size = 12, antiderivative size = 70 \[ \int (b \sec (e+f x))^{5/2} \, dx=\frac {2 b^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {b \sec (e+f x)}}{3 f}+\frac {2 b (b \sec (e+f x))^{3/2} \sin (e+f x)}{3 f} \]

[Out]

2/3*b*(b*sec(f*x+e))^(3/2)*sin(f*x+e)/f+2/3*b^2*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticF(sin(
1/2*f*x+1/2*e),2^(1/2))*cos(f*x+e)^(1/2)*(b*sec(f*x+e))^(1/2)/f

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3853, 3856, 2720} \[ \int (b \sec (e+f x))^{5/2} \, dx=\frac {2 b^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {b \sec (e+f x)}}{3 f}+\frac {2 b \sin (e+f x) (b \sec (e+f x))^{3/2}}{3 f} \]

[In]

Int[(b*Sec[e + f*x])^(5/2),x]

[Out]

(2*b^2*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[b*Sec[e + f*x]])/(3*f) + (2*b*(b*Sec[e + f*x])^(3/2)*
Sin[e + f*x])/(3*f)

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rubi steps \begin{align*} \text {integral}& = \frac {2 b (b \sec (e+f x))^{3/2} \sin (e+f x)}{3 f}+\frac {1}{3} b^2 \int \sqrt {b \sec (e+f x)} \, dx \\ & = \frac {2 b (b \sec (e+f x))^{3/2} \sin (e+f x)}{3 f}+\frac {1}{3} \left (b^2 \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}\right ) \int \frac {1}{\sqrt {\cos (e+f x)}} \, dx \\ & = \frac {2 b^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {b \sec (e+f x)}}{3 f}+\frac {2 b (b \sec (e+f x))^{3/2} \sin (e+f x)}{3 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.73 \[ \int (b \sec (e+f x))^{5/2} \, dx=\frac {2 b^2 \sqrt {b \sec (e+f x)} \left (\sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )+\tan (e+f x)\right )}{3 f} \]

[In]

Integrate[(b*Sec[e + f*x])^(5/2),x]

[Out]

(2*b^2*Sqrt[b*Sec[e + f*x]]*(Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2] + Tan[e + f*x]))/(3*f)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.70 (sec) , antiderivative size = 144, normalized size of antiderivative = 2.06

method	result	size
default	\(-\frac {2 \sqrt {b \sec \left (f x +e \right )}\, b^{2} \left (i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right ), i\right ) \cos \left (f x +e \right )+i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right ), i\right )-\tan \left (f x +e \right )\right )}{3 f}\)	\(144\)

[In]

int((b*sec(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-2/3/f*(b*sec(f*x+e))^(1/2)*b^2*(I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(-co
t(f*x+e)+csc(f*x+e)),I)*cos(f*x+e)+I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(-
cot(f*x+e)+csc(f*x+e)),I)-tan(f*x+e))

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.10 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.44 \[ \int (b \sec (e+f x))^{5/2} \, dx=\frac {-i \, \sqrt {2} b^{\frac {5}{2}} \cos \left (f x + e\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + i \, \sqrt {2} b^{\frac {5}{2}} \cos \left (f x + e\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) + 2 \, b^{2} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{3 \, f \cos \left (f x + e\right )} \]

[In]

integrate((b*sec(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/3*(-I*sqrt(2)*b^(5/2)*cos(f*x + e)*weierstrassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e)) + I*sqrt(2)*b^(
5/2)*cos(f*x + e)*weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e)) + 2*b^2*sqrt(b/cos(f*x + e))*sin(f
*x + e))/(f*cos(f*x + e))

Sympy [F]

\[ \int (b \sec (e+f x))^{5/2} \, dx=\int \left (b \sec {\left (e + f x \right )}\right )^{\frac {5}{2}}\, dx \]

[In]

integrate((b*sec(f*x+e))**(5/2),x)

[Out]

Integral((b*sec(e + f*x))**(5/2), x)

Maxima [F]

\[ \int (b \sec (e+f x))^{5/2} \, dx=\int { \left (b \sec \left (f x + e\right )\right )^{\frac {5}{2}} \,d x } \]

[In]

integrate((b*sec(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*sec(f*x + e))^(5/2), x)

Giac [F]

\[ \int (b \sec (e+f x))^{5/2} \, dx=\int { \left (b \sec \left (f x + e\right )\right )^{\frac {5}{2}} \,d x } \]

[In]

integrate((b*sec(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e))^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int (b \sec (e+f x))^{5/2} \, dx=\int {\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{5/2} \,d x \]

[In]

int((b/cos(e + f*x))^(5/2),x)

[Out]

int((b/cos(e + f*x))^(5/2), x)