\(\int \csc ^2(e+f x) (b \sec (e+f x))^{5/2} \, dx\) [408]

Optimal result

Integrand size = 21, antiderivative size = 98 \[ \int \csc ^2(e+f x) (b \sec (e+f x))^{5/2} \, dx=-\frac {5 b^3 \csc (e+f x)}{3 f \sqrt {b \sec (e+f x)}}+\frac {5 b^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {b \sec (e+f x)}}{3 f}+\frac {2 b \csc (e+f x) (b \sec (e+f x))^{3/2}}{3 f} \]

[Out]

2/3*b*csc(f*x+e)*(b*sec(f*x+e))^(3/2)/f-5/3*b^3*csc(f*x+e)/f/(b*sec(f*x+e))^(1/2)+5/3*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 = 0.07 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2706, 2705, 3856, 2720} \[ \int \csc ^2(e+f x) (b \sec (e+f x))^{5/2} \, dx=-\frac {5 b^3 \csc (e+f x)}{3 f \sqrt {b \sec (e+f x)}}+\frac {5 b^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {b \sec (e+f x)}}{3 f}+\frac {2 b \csc (e+f x) (b \sec (e+f x))^{3/2}}{3 f} \]

[In]

Int[Csc[e + f*x]^2*(b*Sec[e + f*x])^(5/2),x]

[Out]

(-5*b^3*Csc[e + f*x])/(3*f*Sqrt[b*Sec[e + f*x]]) + (5*b^2*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[b*
Sec[e + f*x]])/(3*f) + (2*b*Csc[e + f*x]*(b*Sec[e + f*x])^(3/2))/(3*f)

Rule 2705

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[(-a)*b*(a*Cs
c[e + f*x])^(m - 1)*((b*Sec[e + f*x])^(n - 1)/(f*(m - 1))), x] + Dist[a^2*((m + n - 2)/(m - 1)), Int[(a*Csc[e
+ f*x])^(m - 2)*(b*Sec[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && IntegersQ[2*m, 2*n] &&
!GtQ[n, m]

Rule 2706

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*b*(a*Csc[e
+ f*x])^(m - 1)*((b*Sec[e + f*x])^(n - 1)/(f*(n - 1))), x] + Dist[b^2*((m + n - 2)/(n - 1)), Int[(a*Csc[e + f
*x])^m*(b*Sec[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && IntegersQ[2*m, 2*n]

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 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 \csc (e+f x) (b \sec (e+f x))^{3/2}}{3 f}+\frac {1}{3} \left (5 b^2\right ) \int \csc ^2(e+f x) \sqrt {b \sec (e+f x)} \, dx \\ & = -\frac {5 b^3 \csc (e+f x)}{3 f \sqrt {b \sec (e+f x)}}+\frac {2 b \csc (e+f x) (b \sec (e+f x))^{3/2}}{3 f}+\frac {1}{6} \left (5 b^2\right ) \int \sqrt {b \sec (e+f x)} \, dx \\ & = -\frac {5 b^3 \csc (e+f x)}{3 f \sqrt {b \sec (e+f x)}}+\frac {2 b \csc (e+f x) (b \sec (e+f x))^{3/2}}{3 f}+\frac {1}{6} \left (5 b^2 \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}\right ) \int \frac {1}{\sqrt {\cos (e+f x)}} \, dx \\ & = -\frac {5 b^3 \csc (e+f x)}{3 f \sqrt {b \sec (e+f x)}}+\frac {5 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 \csc (e+f x) (b \sec (e+f x))^{3/2}}{3 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.68 \[ \int \csc ^2(e+f x) (b \sec (e+f x))^{5/2} \, dx=\frac {b \left (2-3 \cot ^2(e+f x)+5 \cos ^{\frac {3}{2}}(e+f x) \csc (e+f x) \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )\right ) (b \sec (e+f x))^{3/2} \sin (e+f x)}{3 f} \]

[In]

Integrate[Csc[e + f*x]^2*(b*Sec[e + f*x])^(5/2),x]

[Out]

(b*(2 - 3*Cot[e + f*x]^2 + 5*Cos[e + f*x]^(3/2)*Csc[e + f*x]*EllipticF[(e + f*x)/2, 2])*(b*Sec[e + f*x])^(3/2)
*Sin[e + f*x])/(3*f)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 7.19 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.62

method	result	size
default	\(-\frac {i b^{2} \sqrt {b \sec \left (f x +e \right )}\, \left (5 \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right ), i\right ) \cos \left (f x +e \right )+5 \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right ), i\right )-5 i \cot \left (f x +e \right )+2 i \csc \left (f x +e \right ) \sec \left (f x +e \right )\right )}{3 f}\)	\(159\)

[In]

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

[Out]

-1/3*I/f*b^2*(b*sec(f*x+e))^(1/2)*(5*(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)+5*(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)-5*I*cot(f*x+e)+2*I*csc(f*x+e)*sec(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 = 131, normalized size of antiderivative = 1.34 \[ \int \csc ^2(e+f x) (b \sec (e+f x))^{5/2} \, dx=\frac {-5 i \, \sqrt {2} b^{\frac {5}{2}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + 5 i \, \sqrt {2} b^{\frac {5}{2}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) - 2 \, {\left (5 \, b^{2} \cos \left (f x + e\right )^{2} - 2 \, b^{2}\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{6 \, f \cos \left (f x + e\right ) \sin \left (f x + e\right )} \]

[In]

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

[Out]

1/6*(-5*I*sqrt(2)*b^(5/2)*cos(f*x + e)*sin(f*x + e)*weierstrassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e))
+ 5*I*sqrt(2)*b^(5/2)*cos(f*x + e)*sin(f*x + e)*weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e)) - 2*
(5*b^2*cos(f*x + e)^2 - 2*b^2)*sqrt(b/cos(f*x + e)))/(f*cos(f*x + e)*sin(f*x + e))

Sympy [F(-1)]

Timed out. \[ \int \csc ^2(e+f x) (b \sec (e+f x))^{5/2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int((b/cos(e + f*x))^(5/2)/sin(e + f*x)^2,x)

[Out]

int((b/cos(e + f*x))^(5/2)/sin(e + f*x)^2, x)