\(\int \frac {\cot ^3(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx\) [490]

Optimal result

Integrand size = 25, antiderivative size = 70 \[ \int \frac {\cot ^3(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sin ^4(c+d x)}}{\sqrt {a}}\right )}{2 \sqrt {a} d}-\frac {\csc ^2(c+d x) \sqrt {a+b \sin ^4(c+d x)}}{2 a d} \] Output:

1/2*arctanh((a+b*sin(d*x+c)^4)^(1/2)/a^(1/2))/a^(1/2)/d-1/2*csc(d*x+c)^2*( 
a+b*sin(d*x+c)^4)^(1/2)/a/d
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.94 \[ \int \frac {\cot ^3(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\frac {\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b \sin ^4(c+d x)}}{\sqrt {a}}\right )-\csc ^2(c+d x) \sqrt {a+b \sin ^4(c+d x)}}{2 a d} \] Input:

Integrate[Cot[c + d*x]^3/Sqrt[a + b*Sin[c + d*x]^4],x]
 

Output:

(Sqrt[a]*ArcTanh[Sqrt[a + b*Sin[c + d*x]^4]/Sqrt[a]] - Csc[c + d*x]^2*Sqrt 
[a + b*Sin[c + d*x]^4])/(2*a*d)
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.94, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 3708, 534, 243, 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.

\(\displaystyle \int \frac {\cot ^3(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\tan (c+d x)^3 \sqrt {a+b \sin (c+d x)^4}}dx\)

\(\Big \downarrow \) 3708

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

\(\Big \downarrow \) 534

\(\displaystyle \frac {-\int \frac {\csc ^2(c+d x)}{\sqrt {b \sin ^4(c+d x)+a}}d\sin ^2(c+d x)-\frac {\csc ^2(c+d x) \sqrt {a+b \sin ^4(c+d x)}}{a}}{2 d}\)

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sin ^4(c+d x)}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {\csc ^2(c+d x) \sqrt {a+b \sin ^4(c+d x)}}{a}}{2 d}\)

Input:

Int[Cot[c + d*x]^3/Sqrt[a + b*Sin[c + d*x]^4],x]
 

Output:

(ArcTanh[Sqrt[a + b*Sin[c + d*x]^4]/Sqrt[a]]/Sqrt[a] - (Csc[c + d*x]^2*Sqr 
t[a + b*Sin[c + d*x]^4])/a)/(2*d)
 

Defintions of rubi rules used

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 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 534

Int[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
Simp[(-c)*x^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1))), x] + Simp[d   Int[ 
x^(m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x] && ILtQ[m, 
0] && GtQ[p, -1] && EqQ[m + 2*p + 3, 0]
 

rule 3042

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

rule 3708

Int[((a_) + (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)*((a + b*ff^(n/2)*x^(n/2))^p/(1 
- ff*x)^((m + 1)/2)), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{a, b, e, f, 
p}, x] && IntegerQ[(m - 1)/2] && IntegerQ[n/2]
 

Maple [F]

\[\int \frac {\cot \left (d x +c \right )^{3}}{\sqrt {a +b \sin \left (d x +c \right )^{4}}}d x\]

Input:

int(cot(d*x+c)^3/(a+b*sin(d*x+c)^4)^(1/2),x)
 

Output:

int(cot(d*x+c)^3/(a+b*sin(d*x+c)^4)^(1/2),x)
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 244, normalized size of antiderivative = 3.49 \[ \int \frac {\cot ^3(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\left [\frac {{\left (\cos \left (d x + c\right )^{2} - 1\right )} \sqrt {a} \log \left (\frac {8 \, {\left (b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + 2 \, \sqrt {b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b} \sqrt {a} + 2 \, a + b\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1}\right ) + 2 \, \sqrt {b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b}}{4 \, {\left (a d \cos \left (d x + c\right )^{2} - a d\right )}}, -\frac {{\left (\cos \left (d x + c\right )^{2} - 1\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b}}\right ) - \sqrt {b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b}}{2 \, {\left (a d \cos \left (d x + c\right )^{2} - a d\right )}}\right ] \] Input:

integrate(cot(d*x+c)^3/(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="fricas")
 

Output:

[1/4*((cos(d*x + c)^2 - 1)*sqrt(a)*log(8*(b*cos(d*x + c)^4 - 2*b*cos(d*x + 
 c)^2 + 2*sqrt(b*cos(d*x + c)^4 - 2*b*cos(d*x + c)^2 + a + b)*sqrt(a) + 2* 
a + b)/(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)) + 2*sqrt(b*cos(d*x + c)^4 
- 2*b*cos(d*x + c)^2 + a + b))/(a*d*cos(d*x + c)^2 - a*d), -1/2*((cos(d*x 
+ c)^2 - 1)*sqrt(-a)*arctan(sqrt(-a)/sqrt(b*cos(d*x + c)^4 - 2*b*cos(d*x + 
 c)^2 + a + b)) - sqrt(b*cos(d*x + c)^4 - 2*b*cos(d*x + c)^2 + a + b))/(a* 
d*cos(d*x + c)^2 - a*d)]
 

Sympy [F]

\[ \int \frac {\cot ^3(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\int \frac {\cot ^{3}{\left (c + d x \right )}}{\sqrt {a + b \sin ^{4}{\left (c + d x \right )}}}\, dx \] Input:

integrate(cot(d*x+c)**3/(a+b*sin(d*x+c)**4)**(1/2),x)
 

Output:

Integral(cot(c + d*x)**3/sqrt(a + b*sin(c + d*x)**4), x)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.13 \[ \int \frac {\cot ^3(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=-\frac {\frac {\log \left (\frac {\sqrt {b \sin \left (d x + c\right )^{4} + a} - \sqrt {a}}{\sqrt {b \sin \left (d x + c\right )^{4} + a} + \sqrt {a}}\right )}{\sqrt {a}} + \frac {2 \, \sqrt {b \sin \left (d x + c\right )^{4} + a}}{a \sin \left (d x + c\right )^{2}}}{4 \, d} \] Input:

integrate(cot(d*x+c)^3/(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="maxima")
 

Output:

-1/4*(log((sqrt(b*sin(d*x + c)^4 + a) - sqrt(a))/(sqrt(b*sin(d*x + c)^4 + 
a) + sqrt(a)))/sqrt(a) + 2*sqrt(b*sin(d*x + c)^4 + a)/(a*sin(d*x + c)^2))/ 
d
 

Giac [F(-1)]

Timed out. \[ \int \frac {\cot ^3(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\text {Timed out} \] Input:

integrate(cot(d*x+c)^3/(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\cot ^3(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\int \frac {{\mathrm {cot}\left (c+d\,x\right )}^3}{\sqrt {b\,{\sin \left (c+d\,x\right )}^4+a}} \,d x \] Input:

int(cot(c + d*x)^3/(a + b*sin(c + d*x)^4)^(1/2),x)
 

Output:

int(cot(c + d*x)^3/(a + b*sin(c + d*x)^4)^(1/2), x)
 

Reduce [F]

\[ \int \frac {\cot ^3(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\int \frac {\sqrt {\sin \left (d x +c \right )^{4} b +a}\, \cot \left (d x +c \right )^{3}}{\sin \left (d x +c \right )^{4} b +a}d x \] Input:

int(cot(d*x+c)^3/(a+b*sin(d*x+c)^4)^(1/2),x)
                                                                                    
                                                                                    
 

Output:

int((sqrt(sin(c + d*x)**4*b + a)*cot(c + d*x)**3)/(sin(c + d*x)**4*b + a), 
x)