\(\int \frac {\cot ^2(e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx\) [113]

Optimal result

Integrand size = 23, antiderivative size = 141 \[ \int \frac {\cot ^2(e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {5 \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a+a \sin (e+f x)}}\right )}{a^{5/2} f}-\frac {7 \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {2} a^{5/2} f}-\frac {2 \cos (e+f x)}{a f (a+a \sin (e+f x))^{3/2}}-\frac {\cot (e+f x)}{a f (a+a \sin (e+f x))^{3/2}} \] Output:

5*arctanh(a^(1/2)*cos(f*x+e)/(a+a*sin(f*x+e))^(1/2))/a^(5/2)/f-7/2*arctanh 
(1/2*a^(1/2)*cos(f*x+e)*2^(1/2)/(a+a*sin(f*x+e))^(1/2))*2^(1/2)/a^(5/2)/f- 
2*cos(f*x+e)/a/f/(a+a*sin(f*x+e))^(3/2)-cot(f*x+e)/a/f/(a+a*sin(f*x+e))^(3 
/2)
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 1.45 (sec) , antiderivative size = 451, normalized size of antiderivative = 3.20 \[ \int \frac {\cot ^2(e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \left (8 \sin \left (\frac {1}{2} (e+f x)\right )-4 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+(28+28 i) (-1)^{3/4} \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2-\cot \left (\frac {1}{4} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+10 \log \left (1+\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2-10 \log \left (1-\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+\frac {2 \sin \left (\frac {1}{4} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2}{\cos \left (\frac {1}{4} (e+f x)\right )-\sin \left (\frac {1}{4} (e+f x)\right )}-\frac {2 \sin \left (\frac {1}{4} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2}{\cos \left (\frac {1}{4} (e+f x)\right )+\sin \left (\frac {1}{4} (e+f x)\right )}-\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \tan \left (\frac {1}{4} (e+f x)\right )\right )}{4 f (a (1+\sin (e+f x)))^{5/2}} \] Input:

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

Output:

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*(8*Sin[(e + f*x)/2] - 4*(Cos[(e + 
 f*x)/2] + Sin[(e + f*x)/2]) + 2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 + 
 (28 + 28*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(e + f*x) 
/4])]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 - Cot[(e + f*x)/4]*(Cos[(e + 
 f*x)/2] + Sin[(e + f*x)/2])^2 + 10*Log[1 + Cos[(e + f*x)/2] - Sin[(e + f* 
x)/2]]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 - 10*Log[1 - Cos[(e + f*x)/ 
2] + Sin[(e + f*x)/2]]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 + (2*Sin[(e 
 + f*x)/4]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2)/(Cos[(e + f*x)/4] - Si 
n[(e + f*x)/4]) - (2*Sin[(e + f*x)/4]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2] 
)^2)/(Cos[(e + f*x)/4] + Sin[(e + f*x)/4]) - (Cos[(e + f*x)/2] + Sin[(e + 
f*x)/2])^2*Tan[(e + f*x)/4]))/(4*f*(a*(1 + Sin[e + f*x]))^(5/2))
 

Rubi [A] (verified)

Time = 0.88 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.08, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {3042, 3194, 27, 3042, 3457, 27, 3042, 3464, 3042, 3128, 219, 3252, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3194

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\int \frac {5 a-3 a \sin (e+f x)}{\sin (e+f x) (\sin (e+f x) a+a)^{3/2}}dx}{2 a^2}-\frac {\cot (e+f x)}{a f (a \sin (e+f x)+a)^{3/2}}\)

\(\Big \downarrow \) 3457

\(\displaystyle -\frac {\frac {\int \frac {2 \csc (e+f x) \left (5 a^2-2 a^2 \sin (e+f x)\right )}{\sqrt {\sin (e+f x) a+a}}dx}{2 a^2}+\frac {4 a \cos (e+f x)}{f (a \sin (e+f x)+a)^{3/2}}}{2 a^2}-\frac {\cot (e+f x)}{a f (a \sin (e+f x)+a)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {\int \frac {\csc (e+f x) \left (5 a^2-2 a^2 \sin (e+f x)\right )}{\sqrt {\sin (e+f x) a+a}}dx}{a^2}+\frac {4 a \cos (e+f x)}{f (a \sin (e+f x)+a)^{3/2}}}{2 a^2}-\frac {\cot (e+f x)}{a f (a \sin (e+f x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {\int \frac {5 a^2-2 a^2 \sin (e+f x)}{\sin (e+f x) \sqrt {\sin (e+f x) a+a}}dx}{a^2}+\frac {4 a \cos (e+f x)}{f (a \sin (e+f x)+a)^{3/2}}}{2 a^2}-\frac {\cot (e+f x)}{a f (a \sin (e+f x)+a)^{3/2}}\)

\(\Big \downarrow \) 3464

\(\displaystyle -\frac {\frac {5 a \int \csc (e+f x) \sqrt {\sin (e+f x) a+a}dx-7 a^2 \int \frac {1}{\sqrt {\sin (e+f x) a+a}}dx}{a^2}+\frac {4 a \cos (e+f x)}{f (a \sin (e+f x)+a)^{3/2}}}{2 a^2}-\frac {\cot (e+f x)}{a f (a \sin (e+f x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {5 a \int \frac {\sqrt {\sin (e+f x) a+a}}{\sin (e+f x)}dx-7 a^2 \int \frac {1}{\sqrt {\sin (e+f x) a+a}}dx}{a^2}+\frac {4 a \cos (e+f x)}{f (a \sin (e+f x)+a)^{3/2}}}{2 a^2}-\frac {\cot (e+f x)}{a f (a \sin (e+f x)+a)^{3/2}}\)

\(\Big \downarrow \) 3128

\(\displaystyle -\frac {\frac {\frac {14 a^2 \int \frac {1}{2 a-\frac {a^2 \cos ^2(e+f x)}{\sin (e+f x) a+a}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a}}}{f}+5 a \int \frac {\sqrt {\sin (e+f x) a+a}}{\sin (e+f x)}dx}{a^2}+\frac {4 a \cos (e+f x)}{f (a \sin (e+f x)+a)^{3/2}}}{2 a^2}-\frac {\cot (e+f x)}{a f (a \sin (e+f x)+a)^{3/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {\frac {5 a \int \frac {\sqrt {\sin (e+f x) a+a}}{\sin (e+f x)}dx+\frac {7 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{f}}{a^2}+\frac {4 a \cos (e+f x)}{f (a \sin (e+f x)+a)^{3/2}}}{2 a^2}-\frac {\cot (e+f x)}{a f (a \sin (e+f x)+a)^{3/2}}\)

\(\Big \downarrow \) 3252

\(\displaystyle -\frac {\frac {\frac {7 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{f}-\frac {10 a^2 \int \frac {1}{a-\frac {a^2 \cos ^2(e+f x)}{\sin (e+f x) a+a}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a}}}{f}}{a^2}+\frac {4 a \cos (e+f x)}{f (a \sin (e+f x)+a)^{3/2}}}{2 a^2}-\frac {\cot (e+f x)}{a f (a \sin (e+f x)+a)^{3/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {\frac {\frac {7 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{f}-\frac {10 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a}}\right )}{f}}{a^2}+\frac {4 a \cos (e+f x)}{f (a \sin (e+f x)+a)^{3/2}}}{2 a^2}-\frac {\cot (e+f x)}{a f (a \sin (e+f x)+a)^{3/2}}\)

Input:

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

Output:

-(Cot[e + f*x]/(a*f*(a + a*Sin[e + f*x])^(3/2))) - (((-10*a^(3/2)*ArcTanh[ 
(Sqrt[a]*Cos[e + f*x])/Sqrt[a + a*Sin[e + f*x]]])/f + (7*Sqrt[2]*a^(3/2)*A 
rcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/f)/a^2 
+ (4*a*Cos[e + f*x])/(f*(a + a*Sin[e + f*x])^(3/2)))/(2*a^2)
 

Defintions of rubi rules used

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 3042

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

rule 3128

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

rule 3194

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)/tan[(e_.) + (f_.)*(x_)]^2, 
x_Symbol] :> Simp[-(a + b*Sin[e + f*x])^(m + 1)/(a*f*Tan[e + f*x]), x] + Si 
mp[1/b^2   Int[(a + b*Sin[e + f*x])^(m + 1)*((b*m - a*(m + 1)*Sin[e + f*x]) 
/Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && In 
tegerQ[m - 1/2] && LtQ[m, -1]
 

rule 3252

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

rule 3457

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^( 
n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/(a*(2*m + 1)*(b*c - a*d)) 
  Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b 
*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 
2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ 
[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] 
 &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
 

rule 3464

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + 
(f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(A 
*b - a*B)/(b*c - a*d)   Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Simp[(B*c 
- A*d)/(b*c - a*d)   Int[Sqrt[a + b*Sin[e + f*x]]/(c + d*Sin[e + f*x]), x], 
 x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - 
 b^2, 0] && NeQ[c^2 - d^2, 0]
 

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.55

method	result	size
default	\(-\frac {\left (7 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sin \left (f x +e \right )^{2} a +7 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a \sin \left (f x +e \right )-10 \,\operatorname {arctanh}\left (\frac {\sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}}{\sqrt {a}}\right ) \sin \left (f x +e \right )^{2} a +4 \sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, \sqrt {a}\, \sin \left (f x +e \right )-10 \,\operatorname {arctanh}\left (\frac {\sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}}{\sqrt {a}}\right ) \sin \left (f x +e \right ) a +2 \sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, \sqrt {a}\right ) \sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}}{2 a^{\frac {7}{2}} \sin \left (f x +e \right ) \cos \left (f x +e \right ) \sqrt {a +\sin \left (f x +e \right ) a}\, f}\)	\(219\)

Input:

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

Output:

-1/2/a^(7/2)*(7*2^(1/2)*arctanh(1/2*(-a*(-1+sin(f*x+e)))^(1/2)*2^(1/2)/a^( 
1/2))*sin(f*x+e)^2*a+7*2^(1/2)*arctanh(1/2*(-a*(-1+sin(f*x+e)))^(1/2)*2^(1 
/2)/a^(1/2))*a*sin(f*x+e)-10*arctanh((-a*(-1+sin(f*x+e)))^(1/2)/a^(1/2))*s 
in(f*x+e)^2*a+4*(-a*(-1+sin(f*x+e)))^(1/2)*a^(1/2)*sin(f*x+e)-10*arctanh(( 
-a*(-1+sin(f*x+e)))^(1/2)/a^(1/2))*sin(f*x+e)*a+2*(-a*(-1+sin(f*x+e)))^(1/ 
2)*a^(1/2))*(-a*(-1+sin(f*x+e)))^(1/2)/sin(f*x+e)/cos(f*x+e)/(a+sin(f*x+e) 
*a)^(1/2)/f
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 539 vs. \(2 (122) = 244\).

Time = 0.13 (sec) , antiderivative size = 539, normalized size of antiderivative = 3.82 \[ \int \frac {\cot ^2(e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

1/4*(5*(cos(f*x + e)^3 + 2*cos(f*x + e)^2 + (cos(f*x + e)^2 - cos(f*x + e) 
 - 2)*sin(f*x + e) - cos(f*x + e) - 2)*sqrt(a)*log((a*cos(f*x + e)^3 - 7*a 
*cos(f*x + e)^2 + 4*(cos(f*x + e)^2 + (cos(f*x + e) + 3)*sin(f*x + e) - 2* 
cos(f*x + e) - 3)*sqrt(a*sin(f*x + e) + a)*sqrt(a) - 9*a*cos(f*x + e) + (a 
*cos(f*x + e)^2 + 8*a*cos(f*x + e) - a)*sin(f*x + e) - a)/(cos(f*x + e)^3 
+ cos(f*x + e)^2 + (cos(f*x + e)^2 - 1)*sin(f*x + e) - cos(f*x + e) - 1)) 
+ 7*sqrt(2)*(a*cos(f*x + e)^3 + 2*a*cos(f*x + e)^2 - a*cos(f*x + e) + (a*c 
os(f*x + e)^2 - a*cos(f*x + e) - 2*a)*sin(f*x + e) - 2*a)*log(-(cos(f*x + 
e)^2 - (cos(f*x + e) - 2)*sin(f*x + e) - 2*sqrt(2)*sqrt(a*sin(f*x + e) + a 
)*(cos(f*x + e) - sin(f*x + e) + 1)/sqrt(a) + 3*cos(f*x + e) + 2)/(cos(f*x 
 + e)^2 - (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2))/sqrt(a) + 4 
*(2*cos(f*x + e)^2 + (2*cos(f*x + e) + 1)*sin(f*x + e) + cos(f*x + e) - 1) 
*sqrt(a*sin(f*x + e) + a))/(a^3*f*cos(f*x + e)^3 + 2*a^3*f*cos(f*x + e)^2 
- a^3*f*cos(f*x + e) - 2*a^3*f + (a^3*f*cos(f*x + e)^2 - a^3*f*cos(f*x + e 
) - 2*a^3*f)*sin(f*x + e))
 

Sympy [F]

\[ \int \frac {\cot ^2(e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx=\int \frac {\cot ^{2}{\left (e + f x \right )}}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \] Input:

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

Output:

Integral(cot(e + f*x)**2/(a*(sin(e + f*x) + 1))**(5/2), x)
 

Maxima [F]

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

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

Output:

integrate(cot(f*x + e)^2/(a*sin(f*x + e) + a)^(5/2), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 262 vs. \(2 (122) = 244\).

Time = 0.17 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.86 \[ \int \frac {\cot ^2(e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {\sqrt {a} {\left (\frac {7 \, \sqrt {2} \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {7 \, \sqrt {2} \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {10 \, \log \left ({\left | \sqrt {2} + 2 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \right |}\right )}{a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {10 \, \log \left ({\left | -\sqrt {2} + 2 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \right |}\right )}{a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {2 \, {\left (4 \, \sqrt {2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 3 \, \sqrt {2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (2 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 3 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )} a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}\right )}}{4 \, f} \] Input:

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

Output:

1/4*sqrt(a)*(7*sqrt(2)*log(sin(-1/4*pi + 1/2*f*x + 1/2*e) + 1)/(a^3*sgn(co 
s(-1/4*pi + 1/2*f*x + 1/2*e))) - 7*sqrt(2)*log(-sin(-1/4*pi + 1/2*f*x + 1/ 
2*e) + 1)/(a^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))) - 10*log(abs(sqrt(2) + 
 2*sin(-1/4*pi + 1/2*f*x + 1/2*e)))/(a^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e 
))) + 10*log(abs(-sqrt(2) + 2*sin(-1/4*pi + 1/2*f*x + 1/2*e)))/(a^3*sgn(co 
s(-1/4*pi + 1/2*f*x + 1/2*e))) - 2*(4*sqrt(2)*sin(-1/4*pi + 1/2*f*x + 1/2* 
e)^3 - 3*sqrt(2)*sin(-1/4*pi + 1/2*f*x + 1/2*e))/((2*sin(-1/4*pi + 1/2*f*x 
 + 1/2*e)^4 - 3*sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 + 1)*a^3*sgn(cos(-1/4*pi 
+ 1/2*f*x + 1/2*e))))/f
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\cot ^2(e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \cot \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{3}+3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )+1}d x \right )}{a^{3}} \] Input:

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

Output:

(sqrt(a)*int((sqrt(sin(e + f*x) + 1)*cot(e + f*x)**2)/(sin(e + f*x)**3 + 3 
*sin(e + f*x)**2 + 3*sin(e + f*x) + 1),x))/a**3