\(\int \frac {\cot ^5(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx\) [491]

Optimal result

Integrand size = 25, antiderivative size = 108 \[ \int \frac {\cot ^5(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=-\frac {(2 a-b) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^4(c+d x)}}{\sqrt {a}}\right )}{4 a^{3/2} d}+\frac {\csc ^2(c+d x) \sqrt {a+b \sin ^4(c+d x)}}{a d}-\frac {\csc ^4(c+d x) \sqrt {a+b \sin ^4(c+d x)}}{4 a d} \] Output:

-1/4*(2*a-b)*arctanh((a+b*sin(d*x+c)^4)^(1/2)/a^(1/2))/a^(3/2)/d+csc(d*x+c 
)^2*(a+b*sin(d*x+c)^4)^(1/2)/a/d-1/4*csc(d*x+c)^4*(a+b*sin(d*x+c)^4)^(1/2) 
/a/d
 

Mathematica [A] (verified)

Time = 3.29 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.31 \[ \int \frac {\cot ^5(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=-\frac {2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \sin ^4(c+d x)}}{\sqrt {a}}\right )-4 a \csc ^2(c+d x) \sqrt {a+b \sin ^4(c+d x)}+b \sqrt {a+b \sin ^4(c+d x)} \left (\frac {a \csc ^4(c+d x)}{b}-\frac {\text {arctanh}\left (\sqrt {1+\frac {b \sin ^4(c+d x)}{a}}\right )}{\sqrt {1+\frac {b \sin ^4(c+d x)}{a}}}\right )}{4 a^2 d} \] Input:

Integrate[Cot[c + d*x]^5/Sqrt[a + b*Sin[c + d*x]^4],x]
 

Output:

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

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 109, 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, 3708, 540, 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.

\(\displaystyle \int \frac {\cot ^5(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\tan (c+d x)^5 \sqrt {a+b \sin (c+d x)^4}}dx\)

\(\Big \downarrow \) 3708

\(\displaystyle \frac {\int \frac {\csc ^6(c+d x) \left (1-\sin ^2(c+d x)\right )^2}{\sqrt {b \sin ^4(c+d x)+a}}d\sin ^2(c+d x)}{2 d}\)

\(\Big \downarrow \) 540

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

\(\Big \downarrow \) 534

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

Int[Cot[c + d*x]^5/Sqrt[a + b*Sin[c + d*x]^4],x]
 

Output:

(-1/2*(Csc[c + d*x]^4*Sqrt[a + b*Sin[c + d*x]^4])/a - (((2*a - b)*ArcTanh[ 
Sqrt[a + b*Sin[c + d*x]^4]/Sqrt[a]])/Sqrt[a] - 4*Csc[c + d*x]^2*Sqrt[a + b 
*Sin[c + d*x]^4])/(2*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 540

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol 
] :> With[{Qx = PolynomialQuotient[(c + d*x)^n, x, x], R = PolynomialRemain 
der[(c + d*x)^n, x, x]}, Simp[R*x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))) 
, x] + Simp[1/(a*(m + 1))   Int[x^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*(m + 
1)*Qx - b*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d, p}, x] && IG 
tQ[n, 1] && ILtQ[m, -1] && GtQ[p, -1] && IntegerQ[2*p]
 

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]

\[\int \frac {\cot \left (d x +c \right )^{5}}{\sqrt {a +b \sin \left (d x +c \right )^{4}}}d x\]

Input:

int(cot(d*x+c)^5/(a+b*sin(d*x+c)^4)^(1/2),x)
 

Output:

int(cot(d*x+c)^5/(a+b*sin(d*x+c)^4)^(1/2),x)
 

Fricas [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 368, normalized size of antiderivative = 3.41 \[ \int \frac {\cot ^5(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\left [-\frac {{\left ({\left (2 \, a - b\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (2 \, a - b\right )} \cos \left (d x + c\right )^{2} + 2 \, a - b\right )} \sqrt {a} \log \left (\frac {8 \, {\left (b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + 2 \, \sqrt {b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b} \sqrt {a} + 2 \, a + b\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1}\right ) + 2 \, \sqrt {b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b} {\left (4 \, a \cos \left (d x + c\right )^{2} - 3 \, a\right )}}{8 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} - 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d\right )}}, \frac {{\left ({\left (2 \, a - b\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (2 \, a - b\right )} \cos \left (d x + c\right )^{2} + 2 \, a - b\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b}}\right ) - \sqrt {b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b} {\left (4 \, a \cos \left (d x + c\right )^{2} - 3 \, a\right )}}{4 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} - 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d\right )}}\right ] \] Input:

integrate(cot(d*x+c)^5/(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="fricas")
 

Output:

[-1/8*(((2*a - b)*cos(d*x + c)^4 - 2*(2*a - b)*cos(d*x + c)^2 + 2*a - b)*s 
qrt(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*c 
os(d*x + c)^2 + 1)) + 2*sqrt(b*cos(d*x + c)^4 - 2*b*cos(d*x + c)^2 + a + b 
)*(4*a*cos(d*x + c)^2 - 3*a))/(a^2*d*cos(d*x + c)^4 - 2*a^2*d*cos(d*x + c) 
^2 + a^2*d), 1/4*(((2*a - b)*cos(d*x + c)^4 - 2*(2*a - b)*cos(d*x + c)^2 + 
 2*a - b)*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)*(4*a*c 
os(d*x + c)^2 - 3*a))/(a^2*d*cos(d*x + c)^4 - 2*a^2*d*cos(d*x + c)^2 + a^2 
*d)]
 

Sympy [F]

\[ \int \frac {\cot ^5(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\int \frac {\cot ^{5}{\left (c + d x \right )}}{\sqrt {a + b \sin ^{4}{\left (c + d x \right )}}}\, dx \] Input:

integrate(cot(d*x+c)**5/(a+b*sin(d*x+c)**4)**(1/2),x)
 

Output:

Integral(cot(c + d*x)**5/sqrt(a + b*sin(c + d*x)**4), x)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.54 \[ \int \frac {\cot ^5(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=-\frac {\frac {2 \, \sqrt {b \sin \left (d x + c\right )^{4} + a} b}{{\left (b \sin \left (d x + c\right )^{4} + a\right )} a - a^{2}} - \frac {2 \, \log \left (\frac {\sqrt {b \sin \left (d x + c\right )^{4} + a} - \sqrt {a}}{\sqrt {b \sin \left (d x + c\right )^{4} + a} + \sqrt {a}}\right )}{\sqrt {a}} + \frac {b \log \left (\frac {\sqrt {b \sin \left (d x + c\right )^{4} + a} - \sqrt {a}}{\sqrt {b \sin \left (d x + c\right )^{4} + a} + \sqrt {a}}\right )}{a^{\frac {3}{2}}} - \frac {8 \, \sqrt {b \sin \left (d x + c\right )^{4} + a}}{a \sin \left (d x + c\right )^{2}}}{8 \, d} \] Input:

integrate(cot(d*x+c)^5/(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="maxima")
 

Output:

-1/8*(2*sqrt(b*sin(d*x + c)^4 + a)*b/((b*sin(d*x + c)^4 + a)*a - a^2) - 2* 
log((sqrt(b*sin(d*x + c)^4 + a) - sqrt(a))/(sqrt(b*sin(d*x + c)^4 + a) + s 
qrt(a)))/sqrt(a) + b*log((sqrt(b*sin(d*x + c)^4 + a) - sqrt(a))/(sqrt(b*si 
n(d*x + c)^4 + a) + sqrt(a)))/a^(3/2) - 8*sqrt(b*sin(d*x + c)^4 + a)/(a*si 
n(d*x + c)^2))/d
 

Giac [F(-1)]

Timed out. \[ \int \frac {\cot ^5(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\text {Timed out} \] Input:

integrate(cot(d*x+c)^5/(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\cot ^5(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\int \frac {{\mathrm {cot}\left (c+d\,x\right )}^5}{\sqrt {b\,{\sin \left (c+d\,x\right )}^4+a}} \,d x \] Input:

int(cot(c + d*x)^5/(a + b*sin(c + d*x)^4)^(1/2),x)
 

Output:

int(cot(c + d*x)^5/(a + b*sin(c + d*x)^4)^(1/2), x)
 

Reduce [F]

\[ \int \frac {\cot ^5(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\int \frac {\sqrt {\sin \left (d x +c \right )^{4} b +a}\, \cot \left (d x +c \right )^{5}}{\sin \left (d x +c \right )^{4} b +a}d x \] Input:

int(cot(d*x+c)^5/(a+b*sin(d*x+c)^4)^(1/2),x)
 

Output:

int((sqrt(sin(c + d*x)**4*b + a)*cot(c + d*x)**5)/(sin(c + d*x)**4*b + a), 
x)