\(\int \frac {\csc ^4(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx\) [435]

Optimal result

Integrand size = 21, antiderivative size = 102 \[ \int \frac {\csc ^4(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\frac {\csc (e+f x)}{6 b f \sqrt {b \sec (e+f x)}}-\frac {\csc ^3(e+f x)}{3 b f \sqrt {b \sec (e+f x)}}-\frac {\sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {b \sec (e+f x)}}{6 b^2 f} \] Output:

1/6*csc(f*x+e)/b/f/(b*sec(f*x+e))^(1/2)-1/3*csc(f*x+e)^3/b/f/(b*sec(f*x+e) 
)^(1/2)-1/6*cos(f*x+e)^(1/2)*InverseJacobiAM(1/2*f*x+1/2*e,2^(1/2))*(b*sec 
(f*x+e))^(1/2)/b^2/f
 

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.61 \[ \int \frac {\csc ^4(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\frac {\csc (e+f x)-2 \csc ^3(e+f x)-\frac {\operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{\sqrt {\cos (e+f x)}}}{6 b f \sqrt {b \sec (e+f x)}} \] Input:

Integrate[Csc[e + f*x]^4/(b*Sec[e + f*x])^(3/2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.98, 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, 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.

\(\displaystyle \int \frac {\csc ^4(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\csc (e+f x)^4}{(b \sec (e+f x))^{3/2}}dx\)

\(\Big \downarrow \) 3103

\(\displaystyle -\frac {\int \csc ^2(e+f x) \sqrt {b \sec (e+f x)}dx}{6 b^2}-\frac {\csc ^3(e+f x)}{3 b f \sqrt {b \sec (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\int \csc (e+f x)^2 \sqrt {b \sec (e+f x)}dx}{6 b^2}-\frac {\csc ^3(e+f x)}{3 b f \sqrt {b \sec (e+f x)}}\)

\(\Big \downarrow \) 3105

\(\displaystyle -\frac {\frac {1}{2} \int \sqrt {b \sec (e+f x)}dx-\frac {b \csc (e+f x)}{f \sqrt {b \sec (e+f x)}}}{6 b^2}-\frac {\csc ^3(e+f x)}{3 b f \sqrt {b \sec (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {1}{2} \int \sqrt {b \csc \left (e+f x+\frac {\pi }{2}\right )}dx-\frac {b \csc (e+f x)}{f \sqrt {b \sec (e+f x)}}}{6 b^2}-\frac {\csc ^3(e+f x)}{3 b f \sqrt {b \sec (e+f x)}}\)

\(\Big \downarrow \) 4258

\(\displaystyle -\frac {\frac {1}{2} \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)}}dx-\frac {b \csc (e+f x)}{f \sqrt {b \sec (e+f x)}}}{6 b^2}-\frac {\csc ^3(e+f x)}{3 b f \sqrt {b \sec (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {1}{2} \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}}dx-\frac {b \csc (e+f x)}{f \sqrt {b \sec (e+f x)}}}{6 b^2}-\frac {\csc ^3(e+f x)}{3 b f \sqrt {b \sec (e+f x)}}\)

\(\Big \downarrow \) 3120

\(\displaystyle -\frac {\frac {\sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {b \sec (e+f x)}}{f}-\frac {b \csc (e+f x)}{f \sqrt {b \sec (e+f x)}}}{6 b^2}-\frac {\csc ^3(e+f x)}{3 b f \sqrt {b \sec (e+f x)}}\)

Input:

Int[Csc[e + f*x]^4/(b*Sec[e + f*x])^(3/2),x]
 

Output:

-1/3*Csc[e + f*x]^3/(b*f*Sqrt[b*Sec[e + f*x]]) - (-((b*Csc[e + f*x])/(f*Sq 
rt[b*Sec[e + f*x]])) + (Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[ 
b*Sec[e + f*x]])/f)/(6*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 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 = 2.20 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.04

method	result	size
default	\(\frac {\frac {i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \operatorname {EllipticF}\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), i\right ) \left (-1-\sec \left (f x +e \right )\right )}{6}+\frac {\left (-\cos \left (f x +e \right )^{2}-1\right ) \csc \left (f x +e \right )^{3}}{6}}{f b \sqrt {b \sec \left (f x +e \right )}}\)	\(106\)

Input:

int(csc(f*x+e)^4/(b*sec(f*x+e))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/f*(1/6*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*Elli 
pticF(I*(cot(f*x+e)-csc(f*x+e)),I)*(-1-sec(f*x+e))+1/6*(-cos(f*x+e)^2-1)*c 
sc(f*x+e)^3)/b/(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 = 148, normalized size of antiderivative = 1.45 \[ \int \frac {\csc ^4(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\frac {\sqrt {2} {\left (i \, \cos \left (f x + e\right )^{2} - i\right )} \sqrt {b} \sin \left (f x + e\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + \sqrt {2} {\left (-i \, \cos \left (f x + e\right )^{2} + i\right )} \sqrt {b} \sin \left (f x + e\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) + 2 \, {\left (\cos \left (f x + e\right )^{3} + \cos \left (f x + e\right )\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{12 \, {\left (b^{2} f \cos \left (f x + e\right )^{2} - b^{2} f\right )} \sin \left (f x + e\right )} \] Input:

integrate(csc(f*x+e)^4/(b*sec(f*x+e))^(3/2),x, algorithm="fricas")
 

Output:

1/12*(sqrt(2)*(I*cos(f*x + e)^2 - I)*sqrt(b)*sin(f*x + e)*weierstrassPInve 
rse(-4, 0, cos(f*x + e) + I*sin(f*x + e)) + sqrt(2)*(-I*cos(f*x + e)^2 + I 
)*sqrt(b)*sin(f*x + e)*weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x 
 + e)) + 2*(cos(f*x + e)^3 + cos(f*x + e))*sqrt(b/cos(f*x + e)))/((b^2*f*c 
os(f*x + e)^2 - b^2*f)*sin(f*x + e))
 

Sympy [F]

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

integrate(csc(f*x+e)**4/(b*sec(f*x+e))**(3/2),x)
 

Output:

Integral(csc(e + f*x)**4/(b*sec(e + f*x))**(3/2), x)
 

Maxima [F]

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

integrate(csc(f*x+e)^4/(b*sec(f*x+e))^(3/2),x, algorithm="maxima")
 

Output:

integrate(csc(f*x + e)^4/(b*sec(f*x + e))^(3/2), x)
 

Giac [F]

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

integrate(csc(f*x+e)^4/(b*sec(f*x+e))^(3/2),x, algorithm="giac")
 

Output:

integrate(csc(f*x + e)^4/(b*sec(f*x + e))^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\csc ^4(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\int \frac {1}{{\sin \left (e+f\,x\right )}^4\,{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \] Input:

int(1/(sin(e + f*x)^4*(b/cos(e + f*x))^(3/2)),x)
 

Output:

int(1/(sin(e + f*x)^4*(b/cos(e + f*x))^(3/2)), x)
 

Reduce [F]

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

int(csc(f*x+e)^4/(b*sec(f*x+e))^(3/2),x)
 

Output:

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