\(\int \frac {\cot (e+f x)}{(a+b \sin ^2(e+f x))^{5/2}} \, dx\) [466]

Optimal result

Integrand size = 23, antiderivative size = 83 \[ \int \frac {\cot (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{a^{5/2} f}+\frac {1}{3 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {1}{a^2 f \sqrt {a+b \sin ^2(e+f x)}} \] Output:

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.07 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.59 \[ \int \frac {\cot (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\frac {\operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},1+\frac {b \sin ^2(e+f x)}{a}\right )}{3 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}} \] Input:

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

Output:

Hypergeometric2F1[-3/2, 1, -1/2, 1 + (b*Sin[e + f*x]^2)/a]/(3*a*f*(a + b*S 
in[e + f*x]^2)^(3/2))
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 3673, 61, 61, 73, 221}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3673

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 61

\(\displaystyle \frac {\frac {\frac {\int \frac {\csc ^2(e+f x)}{\sqrt {b \sin ^2(e+f x)+a}}d\sin ^2(e+f x)}{a}+\frac {2}{a \sqrt {a+b \sin ^2(e+f x)}}}{a}+\frac {2}{3 a \left (a+b \sin ^2(e+f x)\right )^{3/2}}}{2 f}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {\frac {2 \int \frac {1}{\frac {\sin ^4(e+f x)}{b}-\frac {a}{b}}d\sqrt {b \sin ^2(e+f x)+a}}{a b}+\frac {2}{a \sqrt {a+b \sin ^2(e+f x)}}}{a}+\frac {2}{3 a \left (a+b \sin ^2(e+f x)\right )^{3/2}}}{2 f}\)

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 61

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] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d 
, m, n, x]
 

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 221

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

rule 3042

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

rule 3673

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(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)*((a + b*ff*x)^p/(1 - ff*x)^((m + 1 
)/2)), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && Integ 
erQ[(m - 1)/2]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(259\) vs. \(2(71)=142\).

Time = 0.57 (sec) , antiderivative size = 260, normalized size of antiderivative = 3.13

method	result	size
default	\(\frac {-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \sin \left (f x +e \right )^{2}}}{\sin \left (f x +e \right )}\right )}{a^{\frac {5}{2}}}+\frac {7 \sqrt {-b \cos \left (f x +e \right )^{2}+\frac {a b +b^{2}}{b}}}{12 a^{2} \sqrt {-a b}\, \left (\sin \left (f x +e \right )-\frac {\sqrt {-a b}}{b}\right )}-\frac {7 \sqrt {-b \cos \left (f x +e \right )^{2}+\frac {a b +b^{2}}{b}}}{12 a^{2} \sqrt {-a b}\, \left (\sin \left (f x +e \right )+\frac {\sqrt {-a b}}{b}\right )}-\frac {\sqrt {-b \cos \left (f x +e \right )^{2}+\frac {a b +b^{2}}{b}}}{12 a^{2} b \left (\sin \left (f x +e \right )-\frac {\sqrt {-a b}}{b}\right )^{2}}-\frac {\sqrt {-b \cos \left (f x +e \right )^{2}+\frac {a b +b^{2}}{b}}}{12 a^{2} b \left (\sin \left (f x +e \right )+\frac {\sqrt {-a b}}{b}\right )^{2}}}{f}\)	\(260\)

Input:

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

Output:

(-1/a^(5/2)*ln((2*a+2*a^(1/2)*(a+b*sin(f*x+e)^2)^(1/2))/sin(f*x+e))+7/12/a 
^2/(-a*b)^(1/2)/(sin(f*x+e)-(-a*b)^(1/2)/b)*(-b*cos(f*x+e)^2+(a*b+b^2)/b)^ 
(1/2)-7/12/a^2/(-a*b)^(1/2)/(sin(f*x+e)+(-a*b)^(1/2)/b)*(-b*cos(f*x+e)^2+( 
a*b+b^2)/b)^(1/2)-1/12/a^2/b/(sin(f*x+e)-(-a*b)^(1/2)/b)^2*(-b*cos(f*x+e)^ 
2+(a*b+b^2)/b)^(1/2)-1/12/a^2/b/(sin(f*x+e)+(-a*b)^(1/2)/b)^2*(-b*cos(f*x+ 
e)^2+(a*b+b^2)/b)^(1/2))/f
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 193 vs. \(2 (71) = 142\).

Time = 0.17 (sec) , antiderivative size = 397, normalized size of antiderivative = 4.78 \[ \int \frac {\cot (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\left [\frac {3 \, {\left (b^{2} \cos \left (f x + e\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a} \log \left (\frac {2 \, {\left (b \cos \left (f x + e\right )^{2} + 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a} - 2 \, a - b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right ) - 2 \, {\left (3 \, a b \cos \left (f x + e\right )^{2} - 4 \, a^{2} - 3 \, a b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{6 \, {\left (a^{3} b^{2} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{4} b + a^{3} b^{2}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} f\right )}}, -\frac {3 \, {\left (b^{2} \cos \left (f x + e\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a}}{b \cos \left (f x + e\right )^{2} - a - b}\right ) + {\left (3 \, a b \cos \left (f x + e\right )^{2} - 4 \, a^{2} - 3 \, a b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{3 \, {\left (a^{3} b^{2} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{4} b + a^{3} b^{2}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} f\right )}}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.80 \[ \int \frac {\cot (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\frac {3 \, \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{a^{\frac {5}{2}}} - \frac {3}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a^{2}} - \frac {1}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a}}{3 \, f} \] Input:

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

Output:

-1/3*(3*arcsinh(a/(sqrt(a*b)*abs(sin(f*x + e))))/a^(5/2) - 3/(sqrt(b*sin(f 
*x + e)^2 + a)*a^2) - 1/((b*sin(f*x + e)^2 + a)^(3/2)*a))/f
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int((sqrt(sin(e + f*x)**2*b + a)*cot(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)