\(\int \frac {\cot ^6(e+f x)}{(a-a \sin ^2(e+f x))^{3/2}} \, dx\) [417]

Optimal result

Integrand size = 26, antiderivative size = 77 \[ \int \frac {\cot ^6(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {\cot (e+f x) \csc ^2(e+f x)}{3 a f \sqrt {a \cos ^2(e+f x)}}-\frac {\cot (e+f x) \csc ^4(e+f x)}{5 a f \sqrt {a \cos ^2(e+f x)}} \] Output:

1/3*cot(f*x+e)*csc(f*x+e)^2/a/f/(a*cos(f*x+e)^2)^(1/2)-1/5*cot(f*x+e)*csc( 
f*x+e)^4/a/f/(a*cos(f*x+e)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.53 \[ \int \frac {\cot ^6(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\cot ^3(e+f x) \left (-5+3 \csc ^2(e+f x)\right )}{15 f \left (a \cos ^2(e+f x)\right )^{3/2}} \] Input:

Integrate[Cot[e + f*x]^6/(a - a*Sin[e + f*x]^2)^(3/2),x]
 

Output:

-1/15*(Cot[e + f*x]^3*(-5 + 3*Csc[e + f*x]^2))/(f*(a*Cos[e + f*x]^2)^(3/2) 
)
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.69, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3042, 3655, 3042, 3686, 3042, 25, 3086, 25, 244, 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.

\(\displaystyle \int \frac {\cot ^6(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\tan (e+f x)^6 \left (a-a \sin (e+f x)^2\right )^{3/2}}dx\)

\(\Big \downarrow \) 3655

\(\displaystyle \int \frac {\cot ^6(e+f x)}{\left (a \cos ^2(e+f x)\right )^{3/2}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan \left (e+f x+\frac {\pi }{2}\right )^6}{\left (a \sin \left (e+f x+\frac {\pi }{2}\right )^2\right )^{3/2}}dx\)

\(\Big \downarrow \) 3686

\(\displaystyle \frac {\cos (e+f x) \int \cot ^3(e+f x) \csc ^3(e+f x)dx}{a \sqrt {a \cos ^2(e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\cos (e+f x) \int -\sec \left (e+f x-\frac {\pi }{2}\right )^3 \tan \left (e+f x-\frac {\pi }{2}\right )^3dx}{a \sqrt {a \cos ^2(e+f x)}}\)

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3086

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 244

\(\displaystyle \frac {\cos (e+f x) \int \left (\csc ^2(e+f x)-\csc ^4(e+f x)\right )d\csc (e+f x)}{a f \sqrt {a \cos ^2(e+f x)}}\)

\(\Big \downarrow \) 2009

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

Input:

Int[Cot[e + f*x]^6/(a - a*Sin[e + f*x]^2)^(3/2),x]
 

Output:

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

Defintions of rubi rules used

rule 25

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

rule 244

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand 
Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p 
, 0]
 

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 3086

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( 
n_.), x_Symbol] :> Simp[a/f   Subst[Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2 
), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2 
] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])
 

rule 3655

Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[A 
ctivateTrig[u*(a*cos[e + f*x]^2)^p], x] /; FreeQ[{a, b, e, f, p}, x] && EqQ 
[a + b, 0]
 

rule 3686

Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff 
= FreeFactors[Sin[e + f*x], x]}, Simp[(b*ff^n)^IntPart[p]*((b*Sin[e + f*x]^ 
n)^FracPart[p]/(Sin[e + f*x]/ff)^(n*FracPart[p]))   Int[ActivateTrig[u]*(Si 
n[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] 
 && IntegerQ[n] && (EqQ[u, 1] || MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) / 
; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig]])
 

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.61

method	result	size
default	\(\frac {\cos \left (f x +e \right ) \left (5 \sin \left (f x +e \right )^{2}-3\right )}{15 a \sin \left (f x +e \right )^{5} \sqrt {a \cos \left (f x +e \right )^{2}}\, f}\)	\(47\)
risch	\(-\frac {8 i \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) \left (5 \,{\mathrm e}^{6 i \left (f x +e \right )}+2 \,{\mathrm e}^{4 i \left (f x +e \right )}+5 \,{\mathrm e}^{2 i \left (f x +e \right )}\right )}{15 \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{5} f \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, a}\)	\(94\)

Input:

int(cot(f*x+e)^6/(a-a*sin(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/15/a*cos(f*x+e)*(5*sin(f*x+e)^2-3)/sin(f*x+e)^5/(a*cos(f*x+e)^2)^(1/2)/f
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.97 \[ \int \frac {\cot ^6(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\sqrt {a \cos \left (f x + e\right )^{2}} {\left (5 \, \cos \left (f x + e\right )^{2} - 2\right )}}{15 \, {\left (a^{2} f \cos \left (f x + e\right )^{5} - 2 \, a^{2} f \cos \left (f x + e\right )^{3} + a^{2} f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )} \] Input:

integrate(cot(f*x+e)^6/(a-a*sin(f*x+e)^2)^(3/2),x, algorithm="fricas")
 

Output:

-1/15*sqrt(a*cos(f*x + e)^2)*(5*cos(f*x + e)^2 - 2)/((a^2*f*cos(f*x + e)^5 
 - 2*a^2*f*cos(f*x + e)^3 + a^2*f*cos(f*x + e))*sin(f*x + e))
 

Sympy [F]

\[ \int \frac {\cot ^6(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\cot ^{6}{\left (e + f x \right )}}{\left (- a \left (\sin {\left (e + f x \right )} - 1\right ) \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(cot(f*x+e)**6/(a-a*sin(f*x+e)**2)**(3/2),x)
 

Output:

Integral(cot(e + f*x)**6/(-a*(sin(e + f*x) - 1)*(sin(e + f*x) + 1))**(3/2) 
, x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1063 vs. \(2 (69) = 138\).

Time = 0.17 (sec) , antiderivative size = 1063, normalized size of antiderivative = 13.81 \[ \int \frac {\cot ^6(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate(cot(f*x+e)^6/(a-a*sin(f*x+e)^2)^(3/2),x, algorithm="maxima")
 

Output:

8/15*((5*sin(7*f*x + 7*e) + 2*sin(5*f*x + 5*e) + 5*sin(3*f*x + 3*e))*cos(1 
0*f*x + 10*e) - 5*(5*sin(7*f*x + 7*e) + 2*sin(5*f*x + 5*e) + 5*sin(3*f*x + 
 3*e))*cos(8*f*x + 8*e) - 25*(2*sin(6*f*x + 6*e) - 2*sin(4*f*x + 4*e) + si 
n(2*f*x + 2*e))*cos(7*f*x + 7*e) + 10*(2*sin(5*f*x + 5*e) + 5*sin(3*f*x + 
3*e))*cos(6*f*x + 6*e) + 10*(2*sin(4*f*x + 4*e) - sin(2*f*x + 2*e))*cos(5* 
f*x + 5*e) - (5*cos(7*f*x + 7*e) + 2*cos(5*f*x + 5*e) + 5*cos(3*f*x + 3*e) 
)*sin(10*f*x + 10*e) + 5*(5*cos(7*f*x + 7*e) + 2*cos(5*f*x + 5*e) + 5*cos( 
3*f*x + 3*e))*sin(8*f*x + 8*e) + 5*(10*cos(6*f*x + 6*e) - 10*cos(4*f*x + 4 
*e) + 5*cos(2*f*x + 2*e) - 1)*sin(7*f*x + 7*e) - 10*(2*cos(5*f*x + 5*e) + 
5*cos(3*f*x + 3*e))*sin(6*f*x + 6*e) - 2*(10*cos(4*f*x + 4*e) - 5*cos(2*f* 
x + 2*e) + 1)*sin(5*f*x + 5*e) + 50*cos(3*f*x + 3*e)*sin(4*f*x + 4*e) + 5* 
(5*cos(2*f*x + 2*e) - 1)*sin(3*f*x + 3*e) - 50*cos(4*f*x + 4*e)*sin(3*f*x 
+ 3*e) - 25*cos(3*f*x + 3*e)*sin(2*f*x + 2*e))*sqrt(a)/((a^2*cos(10*f*x + 
10*e)^2 + 25*a^2*cos(8*f*x + 8*e)^2 + 100*a^2*cos(6*f*x + 6*e)^2 + 100*a^2 
*cos(4*f*x + 4*e)^2 + 25*a^2*cos(2*f*x + 2*e)^2 + a^2*sin(10*f*x + 10*e)^2 
 + 25*a^2*sin(8*f*x + 8*e)^2 + 100*a^2*sin(6*f*x + 6*e)^2 + 100*a^2*sin(4* 
f*x + 4*e)^2 - 100*a^2*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + 25*a^2*sin(2*f* 
x + 2*e)^2 - 10*a^2*cos(2*f*x + 2*e) + a^2 - 2*(5*a^2*cos(8*f*x + 8*e) - 1 
0*a^2*cos(6*f*x + 6*e) + 10*a^2*cos(4*f*x + 4*e) - 5*a^2*cos(2*f*x + 2*e) 
+ a^2)*cos(10*f*x + 10*e) - 10*(10*a^2*cos(6*f*x + 6*e) - 10*a^2*cos(4*...
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\cot ^6(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(cot(f*x+e)^6/(a-a*sin(f*x+e)^2)^(3/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m operator + Error: 
Bad Argument Value
 

Mupad [B] (verification not implemented)

Time = 40.67 (sec) , antiderivative size = 393, normalized size of antiderivative = 5.10 \[ \int \frac {\cot ^6(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\sqrt {a-a\,{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}^2}\,16{}\mathrm {i}}{3\,a^2\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}-1\right )}^2\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\right )}-\frac {{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\sqrt {a-a\,{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}^2}\,272{}\mathrm {i}}{15\,a^2\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}-1\right )}^3\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\right )}-\frac {{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\sqrt {a-a\,{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}^2}\,128{}\mathrm {i}}{5\,a^2\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}-1\right )}^4\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\right )}-\frac {{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\sqrt {a-a\,{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}^2}\,64{}\mathrm {i}}{5\,a^2\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}-1\right )}^5\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\right )} \] Input:

int(cot(e + f*x)^6/(a - a*sin(e + f*x)^2)^(3/2),x)
 

Output:

- (exp(e*3i + f*x*3i)*(a - a*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f* 
x*1i)*1i)/2)^2)^(1/2)*16i)/(3*a^2*f*(exp(e*2i + f*x*2i) - 1)^2*(exp(e*1i + 
 f*x*1i) + exp(e*3i + f*x*3i))) - (exp(e*3i + f*x*3i)*(a - a*((exp(- e*1i 
- f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2)^2)^(1/2)*272i)/(15*a^2*f*(exp 
(e*2i + f*x*2i) - 1)^3*(exp(e*1i + f*x*1i) + exp(e*3i + f*x*3i))) - (exp(e 
*3i + f*x*3i)*(a - a*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i 
)/2)^2)^(1/2)*128i)/(5*a^2*f*(exp(e*2i + f*x*2i) - 1)^4*(exp(e*1i + f*x*1i 
) + exp(e*3i + f*x*3i))) - (exp(e*3i + f*x*3i)*(a - a*((exp(- e*1i - f*x*1 
i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2)^2)^(1/2)*64i)/(5*a^2*f*(exp(e*2i + f 
*x*2i) - 1)^5*(exp(e*1i + f*x*1i) + exp(e*3i + f*x*3i)))
 

Reduce [F]

\[ \int \frac {\cot ^6(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {-\sin \left (f x +e \right )^{2}+1}\, \cot \left (f x +e \right )^{6}}{\sin \left (f x +e \right )^{4}-2 \sin \left (f x +e \right )^{2}+1}d x \right )}{a^{2}} \] Input:

int(cot(f*x+e)^6/(a-a*sin(f*x+e)^2)^(3/2),x)
 

Output:

(sqrt(a)*int((sqrt( - sin(e + f*x)**2 + 1)*cot(e + f*x)**6)/(sin(e + f*x)* 
*4 - 2*sin(e + f*x)**2 + 1),x))/a**2