\(\int \frac {\cot ^5(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx\) [446]

Optimal result

Integrand size = 25, antiderivative size = 126 \[ \int \frac {\cot ^5(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=-\frac {\left (8 a^2+8 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{8 a^{5/2} f}+\frac {(8 a+3 b) \csc ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{8 a^2 f}-\frac {\csc ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{4 a f} \] Output:

-1/8*(8*a^2+8*a*b+3*b^2)*arctanh((a+b*sin(f*x+e)^2)^(1/2)/a^(1/2))/a^(5/2) 
/f+1/8*(8*a+3*b)*csc(f*x+e)^2*(a+b*sin(f*x+e)^2)^(1/2)/a^2/f-1/4*csc(f*x+e 
)^4*(a+b*sin(f*x+e)^2)^(1/2)/a/f
 

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.80 \[ \int \frac {\cot ^5(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {-\left (\left (8 a^2+8 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )\right )+\sqrt {a} \csc ^2(e+f x) \left (8 a+3 b-2 a \csc ^2(e+f x)\right ) \sqrt {a+b \sin ^2(e+f x)}}{8 a^{5/2} f} \] Input:

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

Output:

(-((8*a^2 + 8*a*b + 3*b^2)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a]]) + 
Sqrt[a]*Csc[e + f*x]^2*(8*a + 3*b - 2*a*Csc[e + f*x]^2)*Sqrt[a + b*Sin[e + 
 f*x]^2])/(8*a^(5/2)*f)
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 3673, 100, 27, 87, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3673

\(\displaystyle \frac {\int \frac {\csc ^6(e+f x) \left (1-\sin ^2(e+f x)\right )^2}{\sqrt {b \sin ^2(e+f x)+a}}d\sin ^2(e+f x)}{2 f}\)

\(\Big \downarrow \) 100

\(\displaystyle \frac {\frac {\int -\frac {\csc ^4(e+f x) \left (-4 a \sin ^2(e+f x)+8 a+3 b\right )}{2 \sqrt {b \sin ^2(e+f x)+a}}d\sin ^2(e+f x)}{2 a}-\frac {\csc ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 a}}{2 f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {\int \frac {\csc ^4(e+f x) \left (-4 a \sin ^2(e+f x)+8 a+3 b\right )}{\sqrt {b \sin ^2(e+f x)+a}}d\sin ^2(e+f x)}{4 a}-\frac {\csc ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 a}}{2 f}\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {-\frac {-\frac {\left (8 a^2+8 a b+3 b^2\right ) \int \frac {\csc ^2(e+f x)}{\sqrt {b \sin ^2(e+f x)+a}}d\sin ^2(e+f x)}{2 a}-\frac {(8 a+3 b) \csc ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{a}}{4 a}-\frac {\csc ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 a}}{2 f}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {-\frac {-\frac {\left (8 a^2+8 a b+3 b^2\right ) \int \frac {1}{\frac {\sin ^4(e+f x)}{b}-\frac {a}{b}}d\sqrt {b \sin ^2(e+f x)+a}}{a b}-\frac {(8 a+3 b) \csc ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{a}}{4 a}-\frac {\csc ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 a}}{2 f}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {-\frac {\frac {\left (8 a^2+8 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {(8 a+3 b) \csc ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{a}}{4 a}-\frac {\csc ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 a}}{2 f}\)

Input:

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

Output:

(-1/2*(Csc[e + f*x]^4*Sqrt[a + b*Sin[e + f*x]^2])/a - (((8*a^2 + 8*a*b + 3 
*b^2)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a]])/a^(3/2) - ((8*a + 3*b)* 
Csc[e + f*x]^2*Sqrt[a + b*Sin[e + f*x]^2])/a)/(4*a))/(2*f)
 

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 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 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 100

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d 
*e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1))   Int[(c + d*x)^ 
(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( 
p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n 
 + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))
 

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 [A] (verified)

Time = 0.54 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.63

method	result	size
default	\(\frac {-\frac {\sqrt {a +b \sin \left (f x +e \right )^{2}}}{4 a \sin \left (f x +e \right )^{4}}+\frac {3 b \sqrt {a +b \sin \left (f x +e \right )^{2}}}{8 a^{2} \sin \left (f x +e \right )^{2}}-\frac {3 b^{2} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \sin \left (f x +e \right )^{2}}}{\sin \left (f x +e \right )}\right )}{8 a^{\frac {5}{2}}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \sin \left (f x +e \right )^{2}}}{\sin \left (f x +e \right )}\right )}{\sqrt {a}}+\frac {\sqrt {a +b \sin \left (f x +e \right )^{2}}}{a \sin \left (f x +e \right )^{2}}-\frac {b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \sin \left (f x +e \right )^{2}}}{\sin \left (f x +e \right )}\right )}{a^{\frac {3}{2}}}}{f}\)	\(205\)

Input:

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

Output:

(-1/4/a/sin(f*x+e)^4*(a+b*sin(f*x+e)^2)^(1/2)+3/8*b/a^2/sin(f*x+e)^2*(a+b* 
sin(f*x+e)^2)^(1/2)-3/8*b^2/a^(5/2)*ln((2*a+2*a^(1/2)*(a+b*sin(f*x+e)^2)^( 
1/2))/sin(f*x+e))-1/a^(1/2)*ln((2*a+2*a^(1/2)*(a+b*sin(f*x+e)^2)^(1/2))/si 
n(f*x+e))+1/a/sin(f*x+e)^2*(a+b*sin(f*x+e)^2)^(1/2)-b/a^(3/2)*ln((2*a+2*a^ 
(1/2)*(a+b*sin(f*x+e)^2)^(1/2))/sin(f*x+e)))/f
 

Fricas [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 403, normalized size of antiderivative = 3.20 \[ \int \frac {\cot ^5(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\left [\frac {{\left ({\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \sqrt {a} \log \left (\frac {2 \, {\left (b \cos \left (f x + e\right )^{2} + 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a} - 2 \, a - b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right ) - 2 \, {\left ({\left (8 \, a^{2} + 3 \, a b\right )} \cos \left (f x + e\right )^{2} - 6 \, a^{2} - 3 \, a b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{16 \, {\left (a^{3} f \cos \left (f x + e\right )^{4} - 2 \, a^{3} f \cos \left (f x + e\right )^{2} + a^{3} f\right )}}, -\frac {{\left ({\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a}}{b \cos \left (f x + e\right )^{2} - a - b}\right ) + {\left ({\left (8 \, a^{2} + 3 \, a b\right )} \cos \left (f x + e\right )^{2} - 6 \, a^{2} - 3 \, a b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{8 \, {\left (a^{3} f \cos \left (f x + e\right )^{4} - 2 \, a^{3} f \cos \left (f x + e\right )^{2} + a^{3} f\right )}}\right ] \] Input:

integrate(cot(f*x+e)^5/(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="fricas")
 

Output:

[1/16*(((8*a^2 + 8*a*b + 3*b^2)*cos(f*x + e)^4 - 2*(8*a^2 + 8*a*b + 3*b^2) 
*cos(f*x + e)^2 + 8*a^2 + 8*a*b + 3*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*((8*a^2 + 3*a*b)*cos(f*x + e)^2 - 6*a^2 - 3*a*b)*sqrt(-b*cos(f*x + e 
)^2 + a + b))/(a^3*f*cos(f*x + e)^4 - 2*a^3*f*cos(f*x + e)^2 + a^3*f), -1/ 
8*(((8*a^2 + 8*a*b + 3*b^2)*cos(f*x + e)^4 - 2*(8*a^2 + 8*a*b + 3*b^2)*cos 
(f*x + e)^2 + 8*a^2 + 8*a*b + 3*b^2)*sqrt(-a)*arctan(sqrt(-b*cos(f*x + e)^ 
2 + a + b)*sqrt(-a)/(b*cos(f*x + e)^2 - a - b)) + ((8*a^2 + 3*a*b)*cos(f*x 
 + e)^2 - 6*a^2 - 3*a*b)*sqrt(-b*cos(f*x + e)^2 + a + b))/(a^3*f*cos(f*x + 
 e)^4 - 2*a^3*f*cos(f*x + e)^2 + a^3*f)]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.25 \[ \int \frac {\cot ^5(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=-\frac {\frac {8 \, \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{\sqrt {a}} + \frac {8 \, b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{a^{\frac {3}{2}}} + \frac {3 \, b^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{a^{\frac {5}{2}}} - \frac {8 \, \sqrt {b \sin \left (f x + e\right )^{2} + a}}{a \sin \left (f x + e\right )^{2}} - \frac {3 \, \sqrt {b \sin \left (f x + e\right )^{2} + a} b}{a^{2} \sin \left (f x + e\right )^{2}} + \frac {2 \, \sqrt {b \sin \left (f x + e\right )^{2} + a}}{a \sin \left (f x + e\right )^{4}}}{8 \, f} \] Input:

integrate(cot(f*x+e)^5/(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="maxima")
 

Output:

-1/8*(8*arcsinh(a/(sqrt(a*b)*abs(sin(f*x + e))))/sqrt(a) + 8*b*arcsinh(a/( 
sqrt(a*b)*abs(sin(f*x + e))))/a^(3/2) + 3*b^2*arcsinh(a/(sqrt(a*b)*abs(sin 
(f*x + e))))/a^(5/2) - 8*sqrt(b*sin(f*x + e)^2 + a)/(a*sin(f*x + e)^2) - 3 
*sqrt(b*sin(f*x + e)^2 + a)*b/(a^2*sin(f*x + e)^2) + 2*sqrt(b*sin(f*x + e) 
^2 + a)/(a*sin(f*x + e)^4))/f
 

Giac [F(-1)]

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

integrate(cot(f*x+e)^5/(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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