\(\int \frac {\cot ^3(e+f x)}{(a-a \sin ^2(e+f x))^{3/2}} \, dx\) [413]

Optimal result

Integrand size = 26, antiderivative size = 66 \[ \int \frac {\cot ^3(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a \cos ^2(e+f x)}}{\sqrt {a}}\right )}{2 a^{3/2} f}-\frac {\sqrt {a \cos ^2(e+f x)} \csc ^2(e+f x)}{2 a^2 f} \] Output:

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.89 \[ \int \frac {\cot ^3(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\sqrt {a \cos ^2(e+f x)} \left (\frac {\text {arctanh}\left (\sqrt {\cos ^2(e+f x)}\right )}{\sqrt {\cos ^2(e+f x)}}+\csc ^2(e+f x)\right )}{2 a^2 f} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.06, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {3042, 3655, 3042, 25, 3684, 8, 52, 73, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3655

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3684

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

\(\Big \downarrow \) 8

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 73

\(\displaystyle -\frac {\frac {\int \frac {1}{1-\frac {\cos ^4(e+f x)}{a}}d\sqrt {a \cos ^2(e+f x)}}{a}+\frac {\sqrt {a \cos ^2(e+f x)}}{a \left (1-\cos ^2(e+f x)\right )}}{2 a f}\)

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 8

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_))^(p_), x_Symbol] :> Simp[1/a^m   Int[u*(a* 
x)^(m + p), x], x] /; FreeQ[{a, m, p}, x] && IntegerQ[m]
 

rule 25

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

rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, 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 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 3684

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

Maple [A] (verified)

Time = 0.60 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.02

method	result	size
default	\(\frac {-\frac {\sqrt {a \cos \left (f x +e \right )^{2}}}{2 a^{2} \sin \left (f x +e \right )^{2}}-\frac {\ln \left (\frac {2 \sqrt {a}\, \sqrt {a \cos \left (f x +e \right )^{2}}+2 a}{\sin \left (f x +e \right )}\right )}{2 a^{\frac {3}{2}}}}{f}\)	\(67\)
risch	\(\frac {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2}}{a \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, f \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2}}+\frac {\ln \left ({\mathrm e}^{i f x}-{\mathrm e}^{-i e}\right ) \cos \left (f x +e \right )}{f \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, a}-\frac {\ln \left ({\mathrm e}^{i f x}+{\mathrm e}^{-i e}\right ) \cos \left (f x +e \right )}{f \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, a}\)	\(168\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.26 \[ \int \frac {\cot ^3(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\sqrt {a \cos \left (f x + e\right )^{2}} {\left ({\left (\cos \left (f x + e\right )^{2} - 1\right )} \log \left (-\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1}\right ) - 2 \, \cos \left (f x + e\right )\right )}}{4 \, {\left (a^{2} f \cos \left (f x + e\right )^{3} - a^{2} f \cos \left (f x + e\right )\right )}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.52 \[ \int \frac {\cot ^3(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\frac {\log \left (\frac {2 \, \sqrt {-a \sin \left (f x + e\right )^{2} + a} \sqrt {a}}{{\left | \sin \left (f x + e\right ) \right |}} + \frac {2 \, a}{{\left | \sin \left (f x + e\right ) \right |}}\right )}{a^{\frac {3}{2}}} - \frac {1}{\sqrt {-a \sin \left (f x + e\right )^{2} + a} a} + \frac {1}{\sqrt {-a \sin \left (f x + e\right )^{2} + a} a \sin \left (f x + e\right )^{2}}}{2 \, f} \] Input:

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

Output:

-1/2*(log(2*sqrt(-a*sin(f*x + e)^2 + a)*sqrt(a)/abs(sin(f*x + e)) + 2*a/ab 
s(sin(f*x + e)))/a^(3/2) - 1/(sqrt(-a*sin(f*x + e)^2 + a)*a) + 1/(sqrt(-a* 
sin(f*x + e)^2 + a)*a*sin(f*x + e)^2))/f
 

Giac [F(-2)]

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

integrate(cot(f*x+e)^3/(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 i_lex_is_greater E 
rror: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int(cot(f*x+e)^3/(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)**3)/(sin(e + f*x)* 
*4 - 2*sin(e + f*x)**2 + 1),x))/a**2