\(\int \frac {1}{(b \sec (e+f x))^{5/2}} \, dx\) [446]

Optimal result

Integrand size = 12, antiderivative size = 72 \[ \int \frac {1}{(b \sec (e+f x))^{5/2}} \, dx=\frac {6 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 b^2 f \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}+\frac {2 \sin (e+f x)}{5 b f (b \sec (e+f x))^{3/2}} \]

[Out]

2/5*sin(f*x+e)/b/f/(b*sec(f*x+e))^(3/2)+6/5*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticE(sin(1/2*
f*x+1/2*e),2^(1/2))/b^2/f/cos(f*x+e)^(1/2)/(b*sec(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 72, 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 = {3854, 3856, 2719} \[ \int \frac {1}{(b \sec (e+f x))^{5/2}} \, dx=\frac {6 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 b^2 f \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}+\frac {2 \sin (e+f x)}{5 b f (b \sec (e+f x))^{3/2}} \]

[In]

Int[(b*Sec[e + f*x])^(-5/2),x]

[Out]

(6*EllipticE[(e + f*x)/2, 2])/(5*b^2*f*Sqrt[Cos[e + f*x]]*Sqrt[b*Sec[e + f*x]]) + (2*Sin[e + f*x])/(5*b*f*(b*S
ec[e + f*x])^(3/2))

Rule 2719

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

Rule 3854

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Csc[c + d*x])^(n + 1)/(b*d*n)), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[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 \sin (e+f x)}{5 b f (b \sec (e+f x))^{3/2}}+\frac {3 \int \frac {1}{\sqrt {b \sec (e+f x)}} \, dx}{5 b^2} \\ & = \frac {2 \sin (e+f x)}{5 b f (b \sec (e+f x))^{3/2}}+\frac {3 \int \sqrt {\cos (e+f x)} \, dx}{5 b^2 \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}} \\ & = \frac {6 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 b^2 f \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}+\frac {2 \sin (e+f x)}{5 b f (b \sec (e+f x))^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.83 \[ \int \frac {1}{(b \sec (e+f x))^{5/2}} \, dx=\frac {\sqrt {b \sec (e+f x)} \left (12 \sqrt {\cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )+\sin (e+f x)+\sin (3 (e+f x))\right )}{10 b^3 f} \]

[In]

Integrate[(b*Sec[e + f*x])^(-5/2),x]

[Out]

(Sqrt[b*Sec[e + f*x]]*(12*Sqrt[Cos[e + f*x]]*EllipticE[(e + f*x)/2, 2] + Sin[e + f*x] + Sin[3*(e + f*x)]))/(10
*b^3*f)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.84 (sec) , antiderivative size = 420, normalized size of antiderivative = 5.83

method	result	size
default	\(\frac {\frac {6 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, E\left (i \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right ), i\right ) \cos \left (f x +e \right )}{5}-\frac {6 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right ), i\right ) \cos \left (f x +e \right )}{5}+\frac {12 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, E\left (i \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right ), i\right )}{5}-\frac {12 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right ), i\right )}{5}+\frac {6 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, E\left (i \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right ), i\right ) \sec \left (f x +e \right )}{5}-\frac {6 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right ), i\right ) \sec \left (f x +e \right )}{5}+\frac {2 \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )}{5}+\frac {2 \sin \left (f x +e \right ) \cos \left (f x +e \right )}{5}+\frac {6 \sin \left (f x +e \right )}{5}}{f \left (\cos \left (f x +e \right )+1\right ) \sqrt {b \sec \left (f x +e \right )}\, b^{2}}\)	\(420\)

[In]

int(1/(b*sec(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/5/f/(cos(f*x+e)+1)/(b*sec(f*x+e))^(1/2)/b^2*(3*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*
EllipticE(I*(-cot(f*x+e)+csc(f*x+e)),I)*cos(f*x+e)-3*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)*cos(f*x+e)+6*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1)
)^(1/2)*EllipticE(I*(-cot(f*x+e)+csc(f*x+e)),I)-6*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)+3*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*Ellipt
icE(I*(-cot(f*x+e)+csc(f*x+e)),I)*sec(f*x+e)-3*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*El
lipticF(I*(-cot(f*x+e)+csc(f*x+e)),I)*sec(f*x+e)+sin(f*x+e)*cos(f*x+e)^2+sin(f*x+e)*cos(f*x+e)+3*sin(f*x+e))

Fricas [C] (verification not implemented)

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

Time = 0.11 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.32 \[ \int \frac {1}{(b \sec (e+f x))^{5/2}} \, dx=\frac {2 \, \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )^{2} \sin \left (f x + e\right ) + 3 i \, \sqrt {2} \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) - 3 i \, \sqrt {2} \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right )}{5 \, b^{3} f} \]

[In]

integrate(1/(b*sec(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/5*(2*sqrt(b/cos(f*x + e))*cos(f*x + e)^2*sin(f*x + e) + 3*I*sqrt(2)*sqrt(b)*weierstrassZeta(-4, 0, weierstra
ssPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e))) - 3*I*sqrt(2)*sqrt(b)*weierstrassZeta(-4, 0, weierstrassPInv
erse(-4, 0, cos(f*x + e) - I*sin(f*x + e))))/(b^3*f)

Sympy [F]

\[ \int \frac {1}{(b \sec (e+f x))^{5/2}} \, dx=\int \frac {1}{\left (b \sec {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate(1/(b*sec(f*x+e))**(5/2),x)

[Out]

Integral((b*sec(e + f*x))**(-5/2), x)

Maxima [F]

\[ \int \frac {1}{(b \sec (e+f x))^{5/2}} \, dx=\int { \frac {1}{\left (b \sec \left (f x + e\right )\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate(1/(b*sec(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*sec(f*x + e))^(-5/2), x)

Giac [F]

\[ \int \frac {1}{(b \sec (e+f x))^{5/2}} \, dx=\int { \frac {1}{\left (b \sec \left (f x + e\right )\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate(1/(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. \[ \int \frac {1}{(b \sec (e+f x))^{5/2}} \, dx=\int \frac {1}{{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \]

[In]

int(1/(b/cos(e + f*x))^(5/2),x)

[Out]

int(1/(b/cos(e + f*x))^(5/2), x)