\(\int \csc ^4(e+f x) (b \sec (e+f x))^{5/2} \, dx\) [409]

Optimal result

Integrand size = 21, antiderivative size = 123 \[ \int \csc ^4(e+f x) (b \sec (e+f x))^{5/2} \, dx=-\frac {5 b^3 \csc (e+f x)}{2 f \sqrt {b \sec (e+f x)}}+\frac {5 b^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {b \sec (e+f x)}}{2 f}+\frac {b \csc (e+f x) (b \sec (e+f x))^{3/2}}{f}-\frac {b \csc ^3(e+f x) (b \sec (e+f x))^{3/2}}{3 f} \] Output:

-5/2*b^3*csc(f*x+e)/f/(b*sec(f*x+e))^(1/2)+5/2*b^2*cos(f*x+e)^(1/2)*Invers 
eJacobiAM(1/2*f*x+1/2*e,2^(1/2))*(b*sec(f*x+e))^(1/2)/f+b*csc(f*x+e)*(b*se 
c(f*x+e))^(3/2)/f-1/3*b*csc(f*x+e)^3*(b*sec(f*x+e))^(3/2)/f
 

Mathematica [A] (verified)

Time = 0.79 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.64 \[ \int \csc ^4(e+f x) (b \sec (e+f x))^{5/2} \, dx=\frac {b \left (4-\cot ^2(e+f x) \left (11+2 \csc ^2(e+f x)\right )+15 \cos ^{\frac {3}{2}}(e+f x) \csc (e+f x) \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )\right ) (b \sec (e+f x))^{3/2} \sin (e+f x)}{6 f} \] Input:

Integrate[Csc[e + f*x]^4*(b*Sec[e + f*x])^(5/2),x]
 

Output:

(b*(4 - Cot[e + f*x]^2*(11 + 2*Csc[e + f*x]^2) + 15*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])/ 
(6*f)
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.05, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 3105, 3042, 3106, 3042, 3105, 3042, 4258, 3042, 3120}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \csc ^4(e+f x) (b \sec (e+f x))^{5/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \csc (e+f x)^4 (b \sec (e+f x))^{5/2}dx\)

\(\Big \downarrow \) 3105

\(\displaystyle \frac {3}{2} \int \csc ^2(e+f x) (b \sec (e+f x))^{5/2}dx-\frac {b \csc ^3(e+f x) (b \sec (e+f x))^{3/2}}{3 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3}{2} \int \csc (e+f x)^2 (b \sec (e+f x))^{5/2}dx-\frac {b \csc ^3(e+f x) (b \sec (e+f x))^{3/2}}{3 f}\)

\(\Big \downarrow \) 3106

\(\displaystyle \frac {3}{2} \left (\frac {5}{3} b^2 \int \csc ^2(e+f x) \sqrt {b \sec (e+f x)}dx+\frac {2 b \csc (e+f x) (b \sec (e+f x))^{3/2}}{3 f}\right )-\frac {b \csc ^3(e+f x) (b \sec (e+f x))^{3/2}}{3 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3}{2} \left (\frac {5}{3} b^2 \int \csc (e+f x)^2 \sqrt {b \sec (e+f x)}dx+\frac {2 b \csc (e+f x) (b \sec (e+f x))^{3/2}}{3 f}\right )-\frac {b \csc ^3(e+f x) (b \sec (e+f x))^{3/2}}{3 f}\)

\(\Big \downarrow \) 3105

\(\displaystyle \frac {3}{2} \left (\frac {5}{3} b^2 \left (\frac {1}{2} \int \sqrt {b \sec (e+f x)}dx-\frac {b \csc (e+f x)}{f \sqrt {b \sec (e+f x)}}\right )+\frac {2 b \csc (e+f x) (b \sec (e+f x))^{3/2}}{3 f}\right )-\frac {b \csc ^3(e+f x) (b \sec (e+f x))^{3/2}}{3 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3}{2} \left (\frac {5}{3} b^2 \left (\frac {1}{2} \int \sqrt {b \csc \left (e+f x+\frac {\pi }{2}\right )}dx-\frac {b \csc (e+f x)}{f \sqrt {b \sec (e+f x)}}\right )+\frac {2 b \csc (e+f x) (b \sec (e+f x))^{3/2}}{3 f}\right )-\frac {b \csc ^3(e+f x) (b \sec (e+f x))^{3/2}}{3 f}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {3}{2} \left (\frac {5}{3} b^2 \left (\frac {1}{2} \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)}}dx-\frac {b \csc (e+f x)}{f \sqrt {b \sec (e+f x)}}\right )+\frac {2 b \csc (e+f x) (b \sec (e+f x))^{3/2}}{3 f}\right )-\frac {b \csc ^3(e+f x) (b \sec (e+f x))^{3/2}}{3 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3}{2} \left (\frac {5}{3} b^2 \left (\frac {1}{2} \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}}dx-\frac {b \csc (e+f x)}{f \sqrt {b \sec (e+f x)}}\right )+\frac {2 b \csc (e+f x) (b \sec (e+f x))^{3/2}}{3 f}\right )-\frac {b \csc ^3(e+f x) (b \sec (e+f x))^{3/2}}{3 f}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {3}{2} \left (\frac {5}{3} b^2 \left (\frac {\sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {b \sec (e+f x)}}{f}-\frac {b \csc (e+f x)}{f \sqrt {b \sec (e+f x)}}\right )+\frac {2 b \csc (e+f x) (b \sec (e+f x))^{3/2}}{3 f}\right )-\frac {b \csc ^3(e+f x) (b \sec (e+f x))^{3/2}}{3 f}\)

Input:

Int[Csc[e + f*x]^4*(b*Sec[e + f*x])^(5/2),x]
 

Output:

-1/3*(b*Csc[e + f*x]^3*(b*Sec[e + f*x])^(3/2))/f + (3*((2*b*Csc[e + f*x]*( 
b*Sec[e + f*x])^(3/2))/(3*f) + (5*b^2*(-((b*Csc[e + f*x])/(f*Sqrt[b*Sec[e 
+ f*x]])) + (Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[b*Sec[e + f 
*x]])/f))/3))/2
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3105

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*(m - 1))), x] + Simp[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 3106

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] + Simp[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 3120

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 4258

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(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]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 53.49 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.01

method	result	size
default	\(\frac {\left (\frac {2 \sec \left (f x +e \right ) \csc \left (f x +e \right )^{3}}{3}+\frac {5 \cot \left (f x +e \right )^{3}}{2}-\frac {7 \cot \left (f x +e \right ) \csc \left (f x +e \right )^{2}}{2}+\frac {5 i \left (\cos \left (f x +e \right )+1\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \operatorname {EllipticF}\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), i\right )}{2}\right ) b^{2} \sqrt {b \sec \left (f x +e \right )}}{f}\)	\(124\)

Input:

int(csc(f*x+e)^4*(b*sec(f*x+e))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/f*(2/3*sec(f*x+e)*csc(f*x+e)^3+5/2*cot(f*x+e)^3-7/2*cot(f*x+e)*csc(f*x+e 
)^2+5/2*I*EllipticF(I*(cot(f*x+e)-csc(f*x+e)),I)*(1/(cos(f*x+e)+1))^(1/2)* 
(cos(f*x+e)+1)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2))*b^2*(b*sec(f*x+e))^(1/2)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.57 \[ \int \csc ^4(e+f x) (b \sec (e+f x))^{5/2} \, dx=-\frac {15 \, \sqrt {2} {\left (i \, b^{2} \cos \left (f x + e\right )^{3} - i \, b^{2} \cos \left (f x + e\right )\right )} \sqrt {b} \sin \left (f x + e\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + 15 \, \sqrt {2} {\left (-i \, b^{2} \cos \left (f x + e\right )^{3} + i \, b^{2} \cos \left (f x + e\right )\right )} \sqrt {b} \sin \left (f x + e\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) + 2 \, {\left (15 \, b^{2} \cos \left (f x + e\right )^{4} - 21 \, b^{2} \cos \left (f x + e\right )^{2} + 4 \, b^{2}\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{12 \, {\left (f \cos \left (f x + e\right )^{3} - f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )} \] Input:

integrate(csc(f*x+e)^4*(b*sec(f*x+e))^(5/2),x, algorithm="fricas")
 

Output:

-1/12*(15*sqrt(2)*(I*b^2*cos(f*x + e)^3 - I*b^2*cos(f*x + e))*sqrt(b)*sin( 
f*x + e)*weierstrassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e)) + 15*sq 
rt(2)*(-I*b^2*cos(f*x + e)^3 + I*b^2*cos(f*x + e))*sqrt(b)*sin(f*x + e)*we 
ierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e)) + 2*(15*b^2*cos(f* 
x + e)^4 - 21*b^2*cos(f*x + e)^2 + 4*b^2)*sqrt(b/cos(f*x + e)))/((f*cos(f* 
x + e)^3 - f*cos(f*x + e))*sin(f*x + e))
 

Sympy [F(-1)]

Timed out. \[ \int \csc ^4(e+f x) (b \sec (e+f x))^{5/2} \, dx=\text {Timed out} \] Input:

integrate(csc(f*x+e)**4*(b*sec(f*x+e))**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \csc ^4(e+f x) (b \sec (e+f x))^{5/2} \, dx=\int { \left (b \sec \left (f x + e\right )\right )^{\frac {5}{2}} \csc \left (f x + e\right )^{4} \,d x } \] Input:

integrate(csc(f*x+e)^4*(b*sec(f*x+e))^(5/2),x, algorithm="maxima")
 

Output:

integrate((b*sec(f*x + e))^(5/2)*csc(f*x + e)^4, x)
 

Giac [F]

\[ \int \csc ^4(e+f x) (b \sec (e+f x))^{5/2} \, dx=\int { \left (b \sec \left (f x + e\right )\right )^{\frac {5}{2}} \csc \left (f x + e\right )^{4} \,d x } \] Input:

integrate(csc(f*x+e)^4*(b*sec(f*x+e))^(5/2),x, algorithm="giac")
 

Output:

integrate((b*sec(f*x + e))^(5/2)*csc(f*x + e)^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int \csc ^4(e+f x) (b \sec (e+f x))^{5/2} \, dx=\int \frac {{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{5/2}}{{\sin \left (e+f\,x\right )}^4} \,d x \] Input:

int((b/cos(e + f*x))^(5/2)/sin(e + f*x)^4,x)
 

Output:

int((b/cos(e + f*x))^(5/2)/sin(e + f*x)^4, x)
 

Reduce [F]

\[ \int \csc ^4(e+f x) (b \sec (e+f x))^{5/2} \, dx=\sqrt {b}\, \left (\int \sqrt {\sec \left (f x +e \right )}\, \csc \left (f x +e \right )^{4} \sec \left (f x +e \right )^{2}d x \right ) b^{2} \] Input:

int(csc(f*x+e)^4*(b*sec(f*x+e))^(5/2),x)
 

Output:

sqrt(b)*int(sqrt(sec(e + f*x))*csc(e + f*x)**4*sec(e + f*x)**2,x)*b**2