\(\int \cot ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx\) [423]

Optimal result

Integrand size = 25, antiderivative size = 110 \[ \int \cot ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\frac {(2 a-b) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{2 \sqrt {a} f}-\frac {(2 a-b) \sqrt {a+b \sin ^2(e+f x)}}{2 a f}-\frac {\csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{2 a f} \] Output:

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

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.70 \[ \int \cot ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\frac {(2 a-b) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )-\sqrt {a} \left (2+\csc ^2(e+f x)\right ) \sqrt {a+b \sin ^2(e+f x)}}{2 \sqrt {a} f} \] Input:

Integrate[Cot[e + f*x]^3*Sqrt[a + b*Sin[e + f*x]^2],x]
 

Output:

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

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.91, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 3673, 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.

\(\displaystyle \int \cot ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {a+b \sin (e+f x)^2}}{\tan (e+f x)^3}dx\)

\(\Big \downarrow \) 3673

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

Int[Cot[e + f*x]^3*Sqrt[a + b*Sin[e + f*x]^2],x]
 

Output:

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

Defintions of rubi rules used

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 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 = 0.45 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.11

method	result	size
default	\(\frac {-\sqrt {a +b \sin \left (f x +e \right )^{2}}+\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \sin \left (f x +e \right )^{2}}}{\sin \left (f x +e \right )}\right )-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \sin \left (f x +e \right )^{2}}}{\sin \left (f x +e \right )}\right ) b}{2 \sqrt {a}}-\frac {\sqrt {a +b \sin \left (f x +e \right )^{2}}}{2 \sin \left (f x +e \right )^{2}}}{f}\)	\(122\)

Input:

int(cot(f*x+e)^3*(a+b*sin(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

(-(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))-1/2/a^(1/2)*ln((2*a+2*a^(1/2)*(a+b*sin(f*x+e)^2)^(1/2))/s 
in(f*x+e))*b-1/2/sin(f*x+e)^2*(a+b*sin(f*x+e)^2)^(1/2))/f
 

Fricas [A] (verification not implemented)

Time = 0.73 (sec) , antiderivative size = 256, normalized size of antiderivative = 2.33 \[ \int \cot ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\left [-\frac {{\left ({\left (2 \, a - b\right )} \cos \left (f x + e\right )^{2} - 2 \, a + b\right )} \sqrt {a} \log \left (\frac {2 \, {\left (b \cos \left (f x + e\right )^{2} + 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a} - 2 \, a - b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right ) + 2 \, {\left (2 \, a \cos \left (f x + e\right )^{2} - 3 \, a\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{4 \, {\left (a f \cos \left (f x + e\right )^{2} - a f\right )}}, \frac {{\left ({\left (2 \, a - b\right )} \cos \left (f x + e\right )^{2} - 2 \, a + b\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a}}{b \cos \left (f x + e\right )^{2} - a - b}\right ) - {\left (2 \, a \cos \left (f x + e\right )^{2} - 3 \, a\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{2 \, {\left (a f \cos \left (f x + e\right )^{2} - a f\right )}}\right ] \] Input:

integrate(cot(f*x+e)^3*(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="fricas")
                                                                                    
                                                                                    
 

Output:

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

Sympy [F]

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

integrate(cot(f*x+e)**3*(a+b*sin(f*x+e)**2)**(1/2),x)
 

Output:

Integral(sqrt(a + b*sin(e + f*x)**2)*cot(e + f*x)**3, x)
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.03 \[ \int \cot ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\frac {2 \, \sqrt {a} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right ) - \frac {b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{\sqrt {a}} - 2 \, \sqrt {b \sin \left (f x + e\right )^{2} + a} + \frac {\sqrt {b \sin \left (f x + e\right )^{2} + a} b}{a} - \frac {{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}}{a \sin \left (f x + e\right )^{2}}}{2 \, f} \] Input:

integrate(cot(f*x+e)^3*(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [F(-1)]

Timed out. \[ \int \cot ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\text {Timed out} \] Input:

integrate(cot(f*x+e)^3*(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(sqrt(sin(e + f*x)**2*b + a)*cot(e + f*x)**3,x)