\(\int \frac {\csc (e+f x)}{(a+b \sin ^2(e+f x))^{5/2}} \, dx\) [115]

Optimal result

Integrand size = 23, antiderivative size = 129 \[ \int \frac {\csc (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a+b-b \cos ^2(e+f x)}}\right )}{a^{5/2} f}+\frac {b \cos (e+f x)}{3 a (a+b) f \left (a+b-b \cos ^2(e+f x)\right )^{3/2}}+\frac {b (5 a+3 b) \cos (e+f x)}{3 a^2 (a+b)^2 f \sqrt {a+b-b \cos ^2(e+f x)}} \] Output:

-arctanh(a^(1/2)*cos(f*x+e)/(a+b-b*cos(f*x+e)^2)^(1/2))/a^(5/2)/f+1/3*b*co 
s(f*x+e)/a/(a+b)/f/(a+b-b*cos(f*x+e)^2)^(3/2)+1/3*b*(5*a+3*b)*cos(f*x+e)/a 
^2/(a+b)^2/f/(a+b-b*cos(f*x+e)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.98 \[ \int \frac {\csc (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\frac {-\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \cos (e+f x)}{\sqrt {2 a+b-b \cos (2 (e+f x))}}\right )}{a^{5/2}}+\frac {\sqrt {2} b \cos (e+f x) \left (12 a^2+13 a b+3 b^2-b (5 a+3 b) \cos (2 (e+f x))\right )}{3 a^2 (a+b)^2 (2 a+b-b \cos (2 (e+f x)))^{3/2}}}{f} \] Input:

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

Output:

(-(ArcTanh[(Sqrt[2]*Sqrt[a]*Cos[e + f*x])/Sqrt[2*a + b - b*Cos[2*(e + f*x) 
]]]/a^(5/2)) + (Sqrt[2]*b*Cos[e + f*x]*(12*a^2 + 13*a*b + 3*b^2 - b*(5*a + 
 3*b)*Cos[2*(e + f*x)]))/(3*a^2*(a + b)^2*(2*a + b - b*Cos[2*(e + f*x)])^( 
3/2)))/f
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 3665, 316, 25, 402, 27, 291, 219}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3665

\(\displaystyle -\frac {\int \frac {1}{\left (1-\cos ^2(e+f x)\right ) \left (-b \cos ^2(e+f x)+a+b\right )^{5/2}}d\cos (e+f x)}{f}\)

\(\Big \downarrow \) 316

\(\displaystyle -\frac {-\frac {\int -\frac {2 b \cos ^2(e+f x)+3 a+b}{\left (1-\cos ^2(e+f x)\right ) \left (-b \cos ^2(e+f x)+a+b\right )^{3/2}}d\cos (e+f x)}{3 a (a+b)}-\frac {b \cos (e+f x)}{3 a (a+b) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\frac {\int \frac {2 b \cos ^2(e+f x)+3 a+b}{\left (1-\cos ^2(e+f x)\right ) \left (-b \cos ^2(e+f x)+a+b\right )^{3/2}}d\cos (e+f x)}{3 a (a+b)}-\frac {b \cos (e+f x)}{3 a (a+b) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 402

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 219

\(\displaystyle -\frac {\frac {\frac {3 (a+b) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a-b \cos ^2(e+f x)+b}}\right )}{a^{3/2}}-\frac {b (5 a+3 b) \cos (e+f x)}{a (a+b) \sqrt {a-b \cos ^2(e+f x)+b}}}{3 a (a+b)}-\frac {b \cos (e+f x)}{3 a (a+b) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}}}{f}\)

Input:

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

Output:

-((-1/3*(b*Cos[e + f*x])/(a*(a + b)*(a + b - b*Cos[e + f*x]^2)^(3/2)) + (( 
3*(a + b)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/Sqrt[a + b - b*Cos[e + f*x]^2]])/ 
a^(3/2) - (b*(5*a + 3*b)*Cos[e + f*x])/(a*(a + b)*Sqrt[a + b - b*Cos[e + f 
*x]^2]))/(3*a*(a + b)))/f)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 291

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst 
[Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, 
d}, x] && NeQ[b*c - a*d, 0]
 

rule 316

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) 
), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d))   Int[(a + b*x^2)^(p + 1)*(c + d*x 
^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x 
], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  ! 
( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 2, 
 p, q, x]
 

rule 402

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x 
_)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ 
(q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) 
 Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) 
*(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b 
, c, d, e, f, q}, x] && LtQ[p, -1]
 

rule 3042

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

rule 3665

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^ 
(p_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-ff/f 
Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos[e + 
 f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(248\) vs. \(2(115)=230\).

Time = 0.45 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.93

method	result	size
default	\(\frac {\sqrt {-\left (-a -b \sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )^{2}}\, \left (-\frac {\ln \left (\frac {2 a +\left (-a +b \right ) \sin \left (f x +e \right )^{2}+2 \sqrt {a}\, \sqrt {-\left (-a -b \sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )^{2}}}{\sin \left (f x +e \right )^{2}}\right )}{2 a^{\frac {5}{2}}}+\frac {b \left (2 b \sin \left (f x +e \right )^{2}+3 a +b \right ) \cos \left (f x +e \right )^{2}}{3 a \sqrt {-\left (-a -b \sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )^{2}}\, \left (a +b \sin \left (f x +e \right )^{2}\right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {b \cos \left (f x +e \right )^{2}}{a^{2} \left (a +b \right ) \sqrt {-\left (-a -b \sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )^{2}}}\right )}{\cos \left (f x +e \right ) \sqrt {a +b \sin \left (f x +e \right )^{2}}\, f}\)	\(249\)

Input:

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

Output:

(-(-a-b*sin(f*x+e)^2)*cos(f*x+e)^2)^(1/2)*(-1/2/a^(5/2)*ln((2*a+(-a+b)*sin 
(f*x+e)^2+2*a^(1/2)*(-(-a-b*sin(f*x+e)^2)*cos(f*x+e)^2)^(1/2))/sin(f*x+e)^ 
2)+1/3*b/a*(2*b*sin(f*x+e)^2+3*a+b)*cos(f*x+e)^2/(-(-a-b*sin(f*x+e)^2)*cos 
(f*x+e)^2)^(1/2)/(a+b*sin(f*x+e)^2)/(a^2+2*a*b+b^2)+b/a^2*cos(f*x+e)^2/(a+ 
b)/(-(-a-b*sin(f*x+e)^2)*cos(f*x+e)^2)^(1/2))/cos(f*x+e)/(a+b*sin(f*x+e)^2 
)^(1/2)/f
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 348 vs. \(2 (115) = 230\).

Time = 0.39 (sec) , antiderivative size = 752, normalized size of antiderivative = 5.83 \[ \int \frac {\csc (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/12*(3*((a^2*b^2 + 2*a*b^3 + b^4)*cos(f*x + e)^4 + a^4 + 4*a^3*b + 6*a^2 
*b^2 + 4*a*b^3 + b^4 - 2*(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*cos(f*x + e)^ 
2)*sqrt(a)*log(2*((a^2 - 6*a*b + b^2)*cos(f*x + e)^4 + 2*(3*a^2 + 2*a*b - 
b^2)*cos(f*x + e)^2 - 4*((a - b)*cos(f*x + e)^3 + (a + b)*cos(f*x + e))*sq 
rt(-b*cos(f*x + e)^2 + a + b)*sqrt(a) + a^2 + 2*a*b + b^2)/(cos(f*x + e)^4 
 - 2*cos(f*x + e)^2 + 1)) - 4*((5*a^2*b^2 + 3*a*b^3)*cos(f*x + e)^3 - 3*(2 
*a^3*b + 3*a^2*b^2 + a*b^3)*cos(f*x + e))*sqrt(-b*cos(f*x + e)^2 + a + b)) 
/((a^5*b^2 + 2*a^4*b^3 + a^3*b^4)*f*cos(f*x + e)^4 - 2*(a^6*b + 3*a^5*b^2 
+ 3*a^4*b^3 + a^3*b^4)*f*cos(f*x + e)^2 + (a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a 
^4*b^3 + a^3*b^4)*f), 1/6*(3*((a^2*b^2 + 2*a*b^3 + b^4)*cos(f*x + e)^4 + a 
^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4 - 2*(a^3*b + 3*a^2*b^2 + 3*a*b^3 
+ b^4)*cos(f*x + e)^2)*sqrt(-a)*arctan(-1/2*((a - b)*cos(f*x + e)^2 + a + 
b)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a)/(a*b*cos(f*x + e)^3 - (a^2 + a 
*b)*cos(f*x + e))) - 2*((5*a^2*b^2 + 3*a*b^3)*cos(f*x + e)^3 - 3*(2*a^3*b 
+ 3*a^2*b^2 + a*b^3)*cos(f*x + e))*sqrt(-b*cos(f*x + e)^2 + a + b))/((a^5* 
b^2 + 2*a^4*b^3 + a^3*b^4)*f*cos(f*x + e)^4 - 2*(a^6*b + 3*a^5*b^2 + 3*a^4 
*b^3 + a^3*b^4)*f*cos(f*x + e)^2 + (a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3 
+ a^3*b^4)*f)]
 

Sympy [F]

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

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

Output:

Integral(csc(e + f*x)/(a + b*sin(e + f*x)**2)**(5/2), x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 306 vs. \(2 (115) = 230\).

Time = 0.14 (sec) , antiderivative size = 306, normalized size of antiderivative = 2.37 \[ \int \frac {\csc (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\frac {\frac {4 \, b^{3} \cos \left (f x + e\right )}{\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} a^{3} b^{2} + 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} a^{2} b^{3} + \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} a b^{4}} + \frac {2 \, b^{2} \cos \left (f x + e\right )}{{\left (-b \cos \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} a^{2} b + {\left (-b \cos \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} a b^{2}} + \frac {6 \, b^{2} \cos \left (f x + e\right )}{\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} a^{3} b + \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} a^{2} b^{2}} - \frac {3 \, \log \left (b - \frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a}}{\cos \left (f x + e\right ) - 1} - \frac {a}{\cos \left (f x + e\right ) - 1}\right )}{a^{\frac {5}{2}}} + \frac {3 \, \log \left (-b + \frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a}}{\cos \left (f x + e\right ) + 1} + \frac {a}{\cos \left (f x + e\right ) + 1}\right )}{a^{\frac {5}{2}}}}{6 \, f} \] Input:

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

Output:

1/6*(4*b^3*cos(f*x + e)/(sqrt(-b*cos(f*x + e)^2 + a + b)*a^3*b^2 + 2*sqrt( 
-b*cos(f*x + e)^2 + a + b)*a^2*b^3 + sqrt(-b*cos(f*x + e)^2 + a + b)*a*b^4 
) + 2*b^2*cos(f*x + e)/((-b*cos(f*x + e)^2 + a + b)^(3/2)*a^2*b + (-b*cos( 
f*x + e)^2 + a + b)^(3/2)*a*b^2) + 6*b^2*cos(f*x + e)/(sqrt(-b*cos(f*x + e 
)^2 + a + b)*a^3*b + sqrt(-b*cos(f*x + e)^2 + a + b)*a^2*b^2) - 3*log(b - 
sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(a)/(cos(f*x + e) - 1) - a/(cos(f*x + 
e) - 1))/a^(5/2) + 3*log(-b + sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(a)/(cos 
(f*x + e) + 1) + a/(cos(f*x + e) + 1))/a^(5/2))/f
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 277 vs. \(2 (115) = 230\).

Time = 0.68 (sec) , antiderivative size = 277, normalized size of antiderivative = 2.15 \[ \int \frac {\csc (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=-\frac {2 \, {\left ({\left ({\left (\frac {{\left (3 \, a^{6} b^{3} + 2 \, a^{5} b^{4}\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}{a^{8} b^{2} + 2 \, a^{7} b^{3} + a^{6} b^{4}} + \frac {3 \, {\left (a^{6} b^{3} + 4 \, a^{5} b^{4} + 2 \, a^{4} b^{5}\right )}}{a^{8} b^{2} + 2 \, a^{7} b^{3} + a^{6} b^{4}}\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \frac {3 \, {\left (a^{6} b^{3} + 4 \, a^{5} b^{4} + 2 \, a^{4} b^{5}\right )}}{a^{8} b^{2} + 2 \, a^{7} b^{3} + a^{6} b^{4}}\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \frac {3 \, a^{6} b^{3} + 2 \, a^{5} b^{4}}{a^{8} b^{2} + 2 \, a^{7} b^{3} + a^{6} b^{4}}\right )}}{3 \, {\left (a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a\right )}^{\frac {3}{2}} f} \] Input:

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

Output:

-2/3*((((3*a^6*b^3 + 2*a^5*b^4)*tan(1/2*f*x + 1/2*e)^2/(a^8*b^2 + 2*a^7*b^ 
3 + a^6*b^4) + 3*(a^6*b^3 + 4*a^5*b^4 + 2*a^4*b^5)/(a^8*b^2 + 2*a^7*b^3 + 
a^6*b^4))*tan(1/2*f*x + 1/2*e)^2 - 3*(a^6*b^3 + 4*a^5*b^4 + 2*a^4*b^5)/(a^ 
8*b^2 + 2*a^7*b^3 + a^6*b^4))*tan(1/2*f*x + 1/2*e)^2 - (3*a^6*b^3 + 2*a^5* 
b^4)/(a^8*b^2 + 2*a^7*b^3 + a^6*b^4))/((a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan 
(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a)^(3/2)*f)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\csc (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\sqrt {\sin \left (f x +e \right )^{2} b +a}\, \csc \left (f x +e \right )}{\sin \left (f x +e \right )^{6} b^{3}+3 \sin \left (f x +e \right )^{4} a \,b^{2}+3 \sin \left (f x +e \right )^{2} a^{2} b +a^{3}}d x \] Input:

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

Output:

int((sqrt(sin(e + f*x)**2*b + a)*csc(e + f*x))/(sin(e + f*x)**6*b**3 + 3*s 
in(e + f*x)**4*a*b**2 + 3*sin(e + f*x)**2*a**2*b + a**3),x)