\(\int \cot ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx\) [424]

Optimal result

Integrand size = 25, antiderivative size = 165 \[ \int \cot ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=-\frac {\left (8 a^2-8 a b-b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{8 a^{3/2} f}+\frac {\left (8 a^2-8 a b-b^2\right ) \sqrt {a+b \sin ^2(e+f x)}}{8 a^2 f}+\frac {(8 a+b) \csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{8 a^2 f}-\frac {\csc ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 a f} \] Output:

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

Mathematica [A] (verified)

Time = 0.63 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.62 \[ \int \cot ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\frac {\left (-8 a^2+8 a b+b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )+\sqrt {a} \left (8 a+(8 a-b) \csc ^2(e+f x)-2 a \csc ^4(e+f x)\right ) \sqrt {a+b \sin ^2(e+f x)}}{8 a^{3/2} f} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.92, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3673, 100, 27, 87, 60, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3673

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

\(\Big \downarrow \) 100

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

(-1/2*(Csc[e + f*x]^4*(a + b*Sin[e + f*x]^2)^(3/2))/a - (-(((8*a + b)*Csc[ 
e + f*x]^2*(a + b*Sin[e + f*x]^2)^(3/2))/a) - ((8*a^2 - 8*a*b - b^2)*(-2*S 
qrt[a]*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a]] + 2*Sqrt[a + b*Sin[e + 
f*x]^2]))/(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 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[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 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 = 0.52 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.28

method	result	size
default	\(\frac {-\frac {\sqrt {a +b \sin \left (f x +e \right )^{2}}}{4 \sin \left (f x +e \right )^{4}}-\frac {b \sqrt {a +b \sin \left (f x +e \right )^{2}}}{8 a \sin \left (f x +e \right )^{2}}+\frac {b^{2} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \sin \left (f x +e \right )^{2}}}{\sin \left (f x +e \right )}\right )}{8 a^{\frac {3}{2}}}+\sqrt {a +b \sin \left (f x +e \right )^{2}}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \sin \left (f x +e \right )^{2}}}{\sin \left (f x +e \right )}\right )+\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \sin \left (f x +e \right )^{2}}}{\sin \left (f x +e \right )}\right ) b}{\sqrt {a}}+\frac {\sqrt {a +b \sin \left (f x +e \right )^{2}}}{\sin \left (f x +e \right )^{2}}}{f}\)	\(212\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 1.02 (sec) , antiderivative size = 432, normalized size of antiderivative = 2.62 \[ \int \cot ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\left [-\frac {{\left ({\left (8 \, a^{2} - 8 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (8 \, a^{2} - 8 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 8 \, a^{2} - 8 \, a b - b^{2}\right )} \sqrt {a} \log \left (\frac {2 \, {\left (b \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a} - 2 \, a - b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right ) - 2 \, {\left (8 \, a^{2} \cos \left (f x + e\right )^{4} - {\left (24 \, a^{2} - a b\right )} \cos \left (f x + e\right )^{2} + 14 \, a^{2} - a b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{16 \, {\left (a^{2} f \cos \left (f x + e\right )^{4} - 2 \, a^{2} f \cos \left (f x + e\right )^{2} + a^{2} f\right )}}, -\frac {{\left ({\left (8 \, a^{2} - 8 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (8 \, a^{2} - 8 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 8 \, a^{2} - 8 \, a b - b^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a}}{b \cos \left (f x + e\right )^{2} - a - b}\right ) - {\left (8 \, a^{2} \cos \left (f x + e\right )^{4} - {\left (24 \, a^{2} - a b\right )} \cos \left (f x + e\right )^{2} + 14 \, a^{2} - a b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{8 \, {\left (a^{2} f \cos \left (f x + e\right )^{4} - 2 \, a^{2} f \cos \left (f x + e\right )^{2} + a^{2} f\right )}}\right ] \] 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 - b^2)*cos(f*x + e)^4 - 2*(8*a^2 - 8*a*b - b^2)*co 
s(f*x + e)^2 + 8*a^2 - 8*a*b - b^2)*sqrt(a)*log(2*(b*cos(f*x + e)^2 - 2*sq 
rt(-b*cos(f*x + e)^2 + a + b)*sqrt(a) - 2*a - b)/(cos(f*x + e)^2 - 1)) - 2 
*(8*a^2*cos(f*x + e)^4 - (24*a^2 - a*b)*cos(f*x + e)^2 + 14*a^2 - a*b)*sqr 
t(-b*cos(f*x + e)^2 + a + b))/(a^2*f*cos(f*x + e)^4 - 2*a^2*f*cos(f*x + e) 
^2 + a^2*f), -1/8*(((8*a^2 - 8*a*b - b^2)*cos(f*x + e)^4 - 2*(8*a^2 - 8*a* 
b - b^2)*cos(f*x + e)^2 + 8*a^2 - 8*a*b - 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*cos(f*x 
 + e)^4 - (24*a^2 - a*b)*cos(f*x + e)^2 + 14*a^2 - a*b)*sqrt(-b*cos(f*x + 
e)^2 + a + b))/(a^2*f*cos(f*x + e)^4 - 2*a^2*f*cos(f*x + e)^2 + a^2*f)]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.30 \[ \int \cot ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=-\frac {8 \, \sqrt {a} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right ) - \frac {8 \, b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{\sqrt {a}} - \frac {b^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{a^{\frac {3}{2}}} - 8 \, \sqrt {b \sin \left (f x + e\right )^{2} + a} + \frac {8 \, \sqrt {b \sin \left (f x + e\right )^{2} + a} b}{a} + \frac {\sqrt {b \sin \left (f x + e\right )^{2} + a} b^{2}}{a^{2}} - \frac {8 \, {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}}{a \sin \left (f x + e\right )^{2}} - \frac {{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} b}{a^{2} \sin \left (f x + e\right )^{2}} + \frac {2 \, {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}}{a \sin \left (f x + e\right )^{4}}}{8 \, f} \] Input:

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

Output:

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

Giac [F(-1)]

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

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