\(\int \frac {\csc (e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx\) [1402]

Optimal result

Integrand size = 31, antiderivative size = 527 \[ \int \frac {\csc (e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx=-\frac {\arctan \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f g^{5/2}}+\frac {b^{7/2} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a \left (-a^2+b^2\right )^{7/4} f g^{5/2}}-\frac {\text {arctanh}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f g^{5/2}}+\frac {b^{7/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a \left (-a^2+b^2\right )^{7/4} f g^{5/2}}+\frac {2}{3 a f g (g \cos (e+f x))^{3/2}}-\frac {2 b \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 \left (a^2-b^2\right ) f g^2 \sqrt {g \cos (e+f x)}}+\frac {b^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{\left (a^2-b^2\right ) \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) f g^2 \sqrt {g \cos (e+f x)}}+\frac {b^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{\left (a^2-b^2\right ) \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) f g^2 \sqrt {g \cos (e+f x)}}+\frac {2 b (b-a \sin (e+f x))}{3 a \left (a^2-b^2\right ) f g (g \cos (e+f x))^{3/2}} \] Output:

-arctan((g*cos(f*x+e))^(1/2)/g^(1/2))/a/f/g^(5/2)+b^(7/2)*arctan(b^(1/2)*( 
g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/a/(-a^2+b^2)^(7/4)/f/g^(5/2) 
-arctanh((g*cos(f*x+e))^(1/2)/g^(1/2))/a/f/g^(5/2)+b^(7/2)*arctanh(b^(1/2) 
*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/a/(-a^2+b^2)^(7/4)/f/g^(5/ 
2)+2/3/a/f/g/(g*cos(f*x+e))^(3/2)-2/3*b*cos(f*x+e)^(1/2)*InverseJacobiAM(1 
/2*f*x+1/2*e,2^(1/2))/(a^2-b^2)/f/g^2/(g*cos(f*x+e))^(1/2)+b^3*cos(f*x+e)^ 
(1/2)*EllipticPi(sin(1/2*f*x+1/2*e),2*b/(b-(-a^2+b^2)^(1/2)),2^(1/2))/(a^2 
-b^2)/(a^2-b*(b-(-a^2+b^2)^(1/2)))/f/g^2/(g*cos(f*x+e))^(1/2)+b^3*cos(f*x+ 
e)^(1/2)*EllipticPi(sin(1/2*f*x+1/2*e),2*b/(b+(-a^2+b^2)^(1/2)),2^(1/2))/( 
a^2-b^2)/(a^2-b*(b+(-a^2+b^2)^(1/2)))/f/g^2/(g*cos(f*x+e))^(1/2)+2/3*b*(b- 
a*sin(f*x+e))/a/(a^2-b^2)/f/g/(g*cos(f*x+e))^(3/2)
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 30.27 (sec) , antiderivative size = 2136, normalized size of antiderivative = 4.05 \[ \int \frac {\csc (e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx=\text {Result too large to show} \] Input:

Integrate[Csc[e + f*x]/((g*Cos[e + f*x])^(5/2)*(a + b*Sin[e + f*x])),x]
 

Output:

(Cos[e + f*x]^(5/2)*((-8*a*b*(a + b*Sqrt[1 - Cos[e + f*x]^2])*((5*a*(a^2 - 
 b^2)*AppellF1[1/4, 1/2, 1, 5/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^ 
2 + b^2)]*Sqrt[Cos[e + f*x]])/(Sqrt[1 - Cos[e + f*x]^2]*(5*(a^2 - b^2)*App 
ellF1[1/4, 1/2, 1, 5/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)] 
 - 2*(2*b^2*AppellF1[5/4, 1/2, 2, 9/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2 
)/(-a^2 + b^2)] + (-a^2 + b^2)*AppellF1[5/4, 3/2, 1, 9/4, Cos[e + f*x]^2, 
(b^2*Cos[e + f*x]^2)/(-a^2 + b^2)])*Cos[e + f*x]^2)*(a^2 + b^2*(-1 + Cos[e 
 + f*x]^2))) - ((1/8 - I/8)*Sqrt[b]*(2*ArcTan[1 - ((1 + I)*Sqrt[b]*Sqrt[Co 
s[e + f*x]])/(-a^2 + b^2)^(1/4)] - 2*ArcTan[1 + ((1 + I)*Sqrt[b]*Sqrt[Cos[ 
e + f*x]])/(-a^2 + b^2)^(1/4)] + Log[Sqrt[-a^2 + b^2] - (1 + I)*Sqrt[b]*(- 
a^2 + b^2)^(1/4)*Sqrt[Cos[e + f*x]] + I*b*Cos[e + f*x]] - Log[Sqrt[-a^2 + 
b^2] + (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Cos[e + f*x]] + I*b*Cos[e + 
 f*x]]))/(-a^2 + b^2)^(3/4)))/(Sqrt[1 - Cos[e + f*x]^2]*(b + a*Csc[e + f*x 
])) - (b^2*(-1 + Cos[e + f*x]^2)*(a + b*Sqrt[1 - Cos[e + f*x]^2])*Cos[2*(e 
 + f*x)]*Csc[e + f*x]*((-10*Sqrt[2]*(2*a^2 - b^2)*ArcTan[1 - (Sqrt[2]*Sqrt 
[b]*Sqrt[Cos[e + f*x]])/(a^2 - b^2)^(1/4)])/(a*Sqrt[b]*(a^2 - b^2)^(3/4)) 
+ (10*Sqrt[2]*(2*a^2 - b^2)*ArcTan[1 + (Sqrt[2]*Sqrt[b]*Sqrt[Cos[e + f*x]] 
)/(a^2 - b^2)^(1/4)])/(a*Sqrt[b]*(a^2 - b^2)^(3/4)) - (20*ArcTan[Sqrt[Cos[ 
e + f*x]]])/a - (16*b*AppellF1[5/4, 1/2, 1, 9/4, Cos[e + f*x]^2, (b^2*Cos[ 
e + f*x]^2)/(-a^2 + b^2)]*Cos[e + f*x]^(5/2))/(-a^2 + b^2) - (200*b*(a^...
 

Rubi [A] (verified)

Time = 1.75 (sec) , antiderivative size = 527, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {3042, 3377, 2009}

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.

\(\displaystyle \int \frac {\csc (e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sin (e+f x) (g \cos (e+f x))^{5/2} (a+b \sin (e+f x))}dx\)

\(\Big \downarrow \) 3377

\(\displaystyle \int \left (\frac {\csc (e+f x)}{a (g \cos (e+f x))^{5/2}}-\frac {b}{a (g \cos (e+f x))^{5/2} (a+b \sin (e+f x))}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {b^{7/2} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{a f g^{5/2} \left (b^2-a^2\right )^{7/4}}+\frac {b^{7/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{a f g^{5/2} \left (b^2-a^2\right )^{7/4}}-\frac {2 b \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f g^2 \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}+\frac {2 b (b-a \sin (e+f x))}{3 a f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}+\frac {b^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{f g^2 \left (a^2-b^2\right ) \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {g \cos (e+f x)}}+\frac {b^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{f g^2 \left (a^2-b^2\right ) \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {g \cos (e+f x)}}-\frac {\arctan \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f g^{5/2}}-\frac {\text {arctanh}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f g^{5/2}}+\frac {2}{3 a f g (g \cos (e+f x))^{3/2}}\)

Input:

Int[Csc[e + f*x]/((g*Cos[e + f*x])^(5/2)*(a + b*Sin[e + f*x])),x]
 

Output:

-(ArcTan[Sqrt[g*Cos[e + f*x]]/Sqrt[g]]/(a*f*g^(5/2))) + (b^(7/2)*ArcTan[(S 
qrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(a*(-a^2 + b^2 
)^(7/4)*f*g^(5/2)) - ArcTanh[Sqrt[g*Cos[e + f*x]]/Sqrt[g]]/(a*f*g^(5/2)) + 
 (b^(7/2)*ArcTanh[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[ 
g])])/(a*(-a^2 + b^2)^(7/4)*f*g^(5/2)) + 2/(3*a*f*g*(g*Cos[e + f*x])^(3/2) 
) - (2*b*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2])/(3*(a^2 - b^2)*f*g^ 
2*Sqrt[g*Cos[e + f*x]]) + (b^3*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b - Sq 
rt[-a^2 + b^2]), (e + f*x)/2, 2])/((a^2 - b^2)*(a^2 - b*(b - Sqrt[-a^2 + b 
^2]))*f*g^2*Sqrt[g*Cos[e + f*x]]) + (b^3*Sqrt[Cos[e + f*x]]*EllipticPi[(2* 
b)/(b + Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/((a^2 - b^2)*(a^2 - b*(b + Sqr 
t[-a^2 + b^2]))*f*g^2*Sqrt[g*Cos[e + f*x]]) + (2*b*(b - a*Sin[e + f*x]))/( 
3*a*(a^2 - b^2)*f*g*(g*Cos[e + f*x])^(3/2))
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 3377

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^(n_))/((a 
_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[ExpandTrig[(g*cos[e + 
 f*x])^p, sin[e + f*x]^n/(a + b*sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, 
 g, p}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[n] && (LtQ[n, 0] || IGtQ[p + 1/ 
2, 0])
 

Maple [A] (verified)

Time = 1.06 (sec) , antiderivative size = 652, normalized size of antiderivative = 1.24

method	result	size
default	\(\frac {24 \ln \left (\frac {2 \sqrt {-g}\, \sqrt {-2 g \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+g}-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )}\right ) g^{\frac {3}{2}} \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-12 \ln \left (\frac {4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right ) g +2 \sqrt {g}\, \sqrt {-2 g \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+g}-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1}\right ) \sqrt {-g}\, \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{4} g -12 \ln \left (-\frac {2 \left (2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right ) g -\sqrt {g}\, \sqrt {-2 g \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+g}+g \right )}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right ) \sqrt {-g}\, \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{4} g -24 g^{\frac {3}{2}} \ln \left (\frac {2 \sqrt {-g}\, \sqrt {-2 g \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+g}-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )}\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+12 \sqrt {-g}\, \ln \left (\frac {4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right ) g +2 \sqrt {g}\, \sqrt {-2 g \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+g}-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1}\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} g +12 \sqrt {-g}\, \ln \left (-\frac {2 \left (2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right ) g -\sqrt {g}\, \sqrt {-2 g \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+g}+g \right )}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} g +6 \ln \left (\frac {2 \sqrt {-g}\, \sqrt {-2 g \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+g}-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )}\right ) g^{\frac {3}{2}}-3 \ln \left (\frac {4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right ) g +2 \sqrt {g}\, \sqrt {-2 g \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+g}-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1}\right ) \sqrt {-g}\, g -3 \ln \left (-\frac {2 \left (2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right ) g -\sqrt {g}\, \sqrt {-2 g \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+g}+g \right )}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right ) \sqrt {-g}\, g +4 \sqrt {-2 g \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+g}\, \sqrt {-g}\, \sqrt {g}}{6 g^{\frac {7}{2}} a \sqrt {-g}\, \left (4 \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-4 \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right ) f}\)	\(652\)

Input:

int(csc(f*x+e)/(g*cos(f*x+e))^(5/2)/(a+b*sin(f*x+e)),x,method=_RETURNVERBO 
SE)
 

Output:

1/6/g^(7/2)/a/(-g)^(1/2)/(4*sin(1/2*f*x+1/2*e)^4-4*sin(1/2*f*x+1/2*e)^2+1) 
*(24*ln(2/cos(1/2*f*x+1/2*e)*((-g)^(1/2)*(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/ 
2)-g))*g^(3/2)*sin(1/2*f*x+1/2*e)^4-12*ln(2/(cos(1/2*f*x+1/2*e)-1)*(2*cos( 
1/2*f*x+1/2*e)*g+g^(1/2)*(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2)-g))*(-g)^(1/2 
)*sin(1/2*f*x+1/2*e)^4*g-12*ln(-2/(cos(1/2*f*x+1/2*e)+1)*(2*cos(1/2*f*x+1/ 
2*e)*g-g^(1/2)*(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2)+g))*(-g)^(1/2)*sin(1/2* 
f*x+1/2*e)^4*g-24*g^(3/2)*ln(2/cos(1/2*f*x+1/2*e)*((-g)^(1/2)*(-2*g*sin(1/ 
2*f*x+1/2*e)^2+g)^(1/2)-g))*sin(1/2*f*x+1/2*e)^2+12*(-g)^(1/2)*ln(2/(cos(1 
/2*f*x+1/2*e)-1)*(2*cos(1/2*f*x+1/2*e)*g+g^(1/2)*(-2*g*sin(1/2*f*x+1/2*e)^ 
2+g)^(1/2)-g))*sin(1/2*f*x+1/2*e)^2*g+12*(-g)^(1/2)*ln(-2/(cos(1/2*f*x+1/2 
*e)+1)*(2*cos(1/2*f*x+1/2*e)*g-g^(1/2)*(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2) 
+g))*sin(1/2*f*x+1/2*e)^2*g+6*ln(2/cos(1/2*f*x+1/2*e)*((-g)^(1/2)*(-2*g*si 
n(1/2*f*x+1/2*e)^2+g)^(1/2)-g))*g^(3/2)-3*ln(2/(cos(1/2*f*x+1/2*e)-1)*(2*c 
os(1/2*f*x+1/2*e)*g+g^(1/2)*(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2)-g))*(-g)^( 
1/2)*g-3*ln(-2/(cos(1/2*f*x+1/2*e)+1)*(2*cos(1/2*f*x+1/2*e)*g-g^(1/2)*(-2* 
g*sin(1/2*f*x+1/2*e)^2+g)^(1/2)+g))*(-g)^(1/2)*g+4*(-2*g*sin(1/2*f*x+1/2*e 
)^2+g)^(1/2)*(-g)^(1/2)*g^(1/2))/f
 

Fricas [F(-1)]

Timed out. \[ \int \frac {\csc (e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx=\text {Timed out} \] Input:

integrate(csc(f*x+e)/(g*cos(f*x+e))^(5/2)/(a+b*sin(f*x+e)),x, algorithm="f 
ricas")
 

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\csc (e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx=\text {Timed out} \] Input:

integrate(csc(f*x+e)/(g*cos(f*x+e))**(5/2)/(a+b*sin(f*x+e)),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate(csc(f*x+e)/(g*cos(f*x+e))^(5/2)/(a+b*sin(f*x+e)),x, algorithm="m 
axima")
 

Output:

integrate(csc(f*x + e)/((g*cos(f*x + e))^(5/2)*(b*sin(f*x + e) + a)), x)
 

Giac [F]

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

integrate(csc(f*x+e)/(g*cos(f*x+e))^(5/2)/(a+b*sin(f*x+e)),x, algorithm="g 
iac")
 

Output:

integrate(csc(f*x + e)/((g*cos(f*x + e))^(5/2)*(b*sin(f*x + e) + a)), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\csc (e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx=\frac {\sqrt {g}\, \left (\int \frac {\sqrt {\cos \left (f x +e \right )}\, \csc \left (f x +e \right )}{\cos \left (f x +e \right )^{3} \sin \left (f x +e \right ) b +\cos \left (f x +e \right )^{3} a}d x \right )}{g^{3}} \] Input:

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

Output:

(sqrt(g)*int((sqrt(cos(e + f*x))*csc(e + f*x))/(cos(e + f*x)**3*sin(e + f* 
x)*b + cos(e + f*x)**3*a),x))/g**3