\(\int \frac {\csc ^4(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx\) [448]

Optimal result

Integrand size = 21, antiderivative size = 102 \[ \int \frac {\csc ^4(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\frac {\csc (e+f x)}{2 b f (b \sec (e+f x))^{3/2}}-\frac {\csc ^3(e+f x)}{3 b f (b \sec (e+f x))^{3/2}}+\frac {E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{2 b^2 f \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}} \] Output:

1/2*csc(f*x+e)/b/f/(b*sec(f*x+e))^(3/2)-1/3*csc(f*x+e)^3/b/f/(b*sec(f*x+e) 
)^(3/2)+1/2*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)
 

Mathematica [A] (verified)

Time = 0.59 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.77 \[ \int \frac {\csc ^4(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\frac {\left (-3+5 \csc ^2(e+f x)-2 \csc ^4(e+f x)+3 \sqrt {\cos (e+f x)} \csc (e+f x) E\left (\left .\frac {1}{2} (e+f x)\right |2\right )\right ) \sqrt {b \sec (e+f x)} \sin (e+f x)}{6 b^3 f} \] Input:

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

Output:

((-3 + 5*Csc[e + f*x]^2 - 2*Csc[e + f*x]^4 + 3*Sqrt[Cos[e + f*x]]*Csc[e + 
f*x]*EllipticE[(e + f*x)/2, 2])*Sqrt[b*Sec[e + f*x]]*Sin[e + f*x])/(6*b^3* 
f)
 

Rubi [A] (verified)

Time = 0.79 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 3103, 3042, 3105, 3042, 4258, 3042, 3119}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3103

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3105

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 3042

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

rule 3103

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

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 3119

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 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 = 2.61 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.70

method	result	size
default	\(\frac {\frac {i \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \operatorname {EllipticE}\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), i\right ) \left (-3-3 \sec \left (f x +e \right )\right )}{6}+\frac {i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \operatorname {EllipticF}\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), i\right ) \left (3+3 \sec \left (f x +e \right )\right )}{6}+\frac {\csc \left (f x +e \right )}{2}-\frac {\cot \left (f x +e \right ) \csc \left (f x +e \right )^{2}}{3}}{f \sqrt {b \sec \left (f x +e \right )}\, b^{2}}\)	\(173\)

Input:

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

Output:

1/f*(1/6*I*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*(1/(cos(f*x+e)+1))^(1/2)*Elli 
pticE(I*(cot(f*x+e)-csc(f*x+e)),I)*(-3-3*sec(f*x+e))+1/6*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+3*sec(f*x+e))+1/2*csc(f*x+e)-1/3*cot(f*x+e)*csc(f*x+e)^2)/(b*s 
ec(f*x+e))^(1/2)/b^2
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.59 \[ \int \frac {\csc ^4(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=-\frac {3 \, \sqrt {2} {\left (-i \, \cos \left (f x + e\right )^{2} + i\right )} \sqrt {b} \sin \left (f x + e\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + 3 \, \sqrt {2} {\left (i \, \cos \left (f x + e\right )^{2} - i\right )} \sqrt {b} \sin \left (f x + e\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) - 2 \, {\left (3 \, \cos \left (f x + e\right )^{4} - \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{12 \, {\left (b^{3} f \cos \left (f x + e\right )^{2} - b^{3} f\right )} \sin \left (f x + e\right )} \] Input:

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

Output:

-1/12*(3*sqrt(2)*(-I*cos(f*x + e)^2 + I)*sqrt(b)*sin(f*x + e)*weierstrassZ 
eta(-4, 0, weierstrassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e))) + 3* 
sqrt(2)*(I*cos(f*x + e)^2 - I)*sqrt(b)*sin(f*x + e)*weierstrassZeta(-4, 0, 
 weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e))) - 2*(3*cos(f*x 
 + e)^4 - cos(f*x + e)^2)*sqrt(b/cos(f*x + e)))/((b^3*f*cos(f*x + e)^2 - b 
^3*f)*sin(f*x + e))
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\csc ^4(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\int \frac {1}{{\sin \left (e+f\,x\right )}^4\,{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \] Input:

int(1/(sin(e + f*x)^4*(b/cos(e + f*x))^(5/2)),x)
 

Output:

int(1/(sin(e + f*x)^4*(b/cos(e + f*x))^(5/2)), x)
 

Reduce [F]

\[ \int \frac {\csc ^4(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\frac {\sqrt {b}\, \left (\int \frac {\sqrt {\sec \left (f x +e \right )}\, \csc \left (f x +e \right )^{4}}{\sec \left (f x +e \right )^{3}}d x \right )}{b^{3}} \] 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)**3,x))/b**3