\(\int \frac {1}{(a \cot (x)+b \csc (x))^4} \, dx\) [264]

Optimal result

Integrand size = 11, antiderivative size = 159 \[ \int \frac {1}{(a \cot (x)+b \csc (x))^4} \, dx=\frac {x}{a^4}-\frac {b \left (3 a^2-2 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a^4 (a-b)^{3/2} (a+b)^{3/2}}-\frac {\left (2 \left (a^2-b^2\right )-a b \cos (x)\right ) \sin (x)}{2 a^3 \left (a^2-b^2\right ) (b+a \cos (x))}+\frac {\sin ^3(x)}{3 a (b+a \cos (x))^3}+\frac {b \sin ^3(x)}{2 a \left (a^2-b^2\right ) (b+a \cos (x))^2} \] Output:

x/a^4-b*(3*a^2-2*b^2)*arctanh((a-b)^(1/2)*tan(1/2*x)/(a+b)^(1/2))/a^4/(a-b 
)^(3/2)/(a+b)^(3/2)-1/2*(2*a^2-2*b^2-a*b*cos(x))*sin(x)/a^3/(a^2-b^2)/(b+a 
*cos(x))+1/3*sin(x)^3/a/(b+a*cos(x))^3+1/2*b*sin(x)^3/a/(a^2-b^2)/(b+a*cos 
(x))^2
 

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.94 \[ \int \frac {1}{(a \cot (x)+b \csc (x))^4} \, dx=\frac {\left (2 a \left (a^2-b^2\right )+7 a b (b+a \cos (x))-\frac {a \left (8 a^2-11 b^2\right ) (b+a \cos (x))^2}{(a-b) (a+b)}+6 x (b+a \cos (x))^3 \csc (x)-\frac {6 b \left (-3 a^2+2 b^2\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right ) (b+a \cos (x))^3 \csc (x)}{\left (a^2-b^2\right )^{3/2}}\right ) \sin (x)}{6 a^4 (b+a \cos (x))^3} \] Input:

Integrate[(a*Cot[x] + b*Csc[x])^(-4),x]
 

Output:

((2*a*(a^2 - b^2) + 7*a*b*(b + a*Cos[x]) - (a*(8*a^2 - 11*b^2)*(b + a*Cos[ 
x])^2)/((a - b)*(a + b)) + 6*x*(b + a*Cos[x])^3*Csc[x] - (6*b*(-3*a^2 + 2* 
b^2)*ArcTanh[((-a + b)*Tan[x/2])/Sqrt[a^2 - b^2]]*(b + a*Cos[x])^3*Csc[x]) 
/(a^2 - b^2)^(3/2))*Sin[x])/(6*a^4*(b + a*Cos[x])^3)
 

Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.13, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.364, Rules used = {3042, 4892, 3042, 3172, 25, 3042, 3343, 3042, 3342, 25, 3042, 3214, 3042, 3138, 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.

\(\displaystyle \int \frac {1}{(a \cot (x)+b \csc (x))^4} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(a \cot (x)+b \csc (x))^4}dx\)

\(\Big \downarrow \) 4892

\(\displaystyle \int \frac {\sin ^4(x)}{(a \cos (x)+b)^4}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos \left (x-\frac {\pi }{2}\right )^4}{\left (b-a \sin \left (x-\frac {\pi }{2}\right )\right )^4}dx\)

\(\Big \downarrow \) 3172

\(\displaystyle \frac {\int -\frac {\cos (x) \sin ^2(x)}{(b+a \cos (x))^3}dx}{a}+\frac {\sin ^3(x)}{3 a (a \cos (x)+b)^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\sin ^3(x)}{3 a (a \cos (x)+b)^3}-\frac {\int \frac {\cos (x) \sin ^2(x)}{(b+a \cos (x))^3}dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sin ^3(x)}{3 a (a \cos (x)+b)^3}-\frac {\int \frac {\cos \left (x+\frac {\pi }{2}\right )^2 \sin \left (x+\frac {\pi }{2}\right )}{\left (b+a \sin \left (x+\frac {\pi }{2}\right )\right )^3}dx}{a}\)

\(\Big \downarrow \) 3343

\(\displaystyle \frac {\sin ^3(x)}{3 a (a \cos (x)+b)^3}-\frac {\frac {\int \frac {(2 a+b \cos (x)) \sin ^2(x)}{(b+a \cos (x))^2}dx}{2 \left (a^2-b^2\right )}-\frac {b \sin ^3(x)}{2 \left (a^2-b^2\right ) (a \cos (x)+b)^2}}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sin ^3(x)}{3 a (a \cos (x)+b)^3}-\frac {\frac {\int \frac {\cos \left (x+\frac {\pi }{2}\right )^2 \left (2 a+b \sin \left (x+\frac {\pi }{2}\right )\right )}{\left (b+a \sin \left (x+\frac {\pi }{2}\right )\right )^2}dx}{2 \left (a^2-b^2\right )}-\frac {b \sin ^3(x)}{2 \left (a^2-b^2\right ) (a \cos (x)+b)^2}}{a}\)

\(\Big \downarrow \) 3342

\(\displaystyle \frac {\sin ^3(x)}{3 a (a \cos (x)+b)^3}-\frac {\frac {\frac {\sin (x) \left (2 \left (a^2-b^2\right )-a b \cos (x)\right )}{a^2 (a \cos (x)+b)}-\frac {\int -\frac {a b-2 \left (a^2-b^2\right ) \cos (x)}{b+a \cos (x)}dx}{a^2}}{2 \left (a^2-b^2\right )}-\frac {b \sin ^3(x)}{2 \left (a^2-b^2\right ) (a \cos (x)+b)^2}}{a}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\sin ^3(x)}{3 a (a \cos (x)+b)^3}-\frac {\frac {\frac {\int \frac {a b-2 \left (a^2-b^2\right ) \cos (x)}{b+a \cos (x)}dx}{a^2}+\frac {\sin (x) \left (2 \left (a^2-b^2\right )-a b \cos (x)\right )}{a^2 (a \cos (x)+b)}}{2 \left (a^2-b^2\right )}-\frac {b \sin ^3(x)}{2 \left (a^2-b^2\right ) (a \cos (x)+b)^2}}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sin ^3(x)}{3 a (a \cos (x)+b)^3}-\frac {\frac {\frac {\int \frac {a b-2 \left (a^2-b^2\right ) \sin \left (x+\frac {\pi }{2}\right )}{b+a \sin \left (x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {\sin (x) \left (2 \left (a^2-b^2\right )-a b \cos (x)\right )}{a^2 (a \cos (x)+b)}}{2 \left (a^2-b^2\right )}-\frac {b \sin ^3(x)}{2 \left (a^2-b^2\right ) (a \cos (x)+b)^2}}{a}\)

\(\Big \downarrow \) 3214

\(\displaystyle \frac {\sin ^3(x)}{3 a (a \cos (x)+b)^3}-\frac {\frac {\frac {\frac {b \left (3 a^2-2 b^2\right ) \int \frac {1}{b+a \cos (x)}dx}{a}-\frac {2 x \left (a^2-b^2\right )}{a}}{a^2}+\frac {\sin (x) \left (2 \left (a^2-b^2\right )-a b \cos (x)\right )}{a^2 (a \cos (x)+b)}}{2 \left (a^2-b^2\right )}-\frac {b \sin ^3(x)}{2 \left (a^2-b^2\right ) (a \cos (x)+b)^2}}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sin ^3(x)}{3 a (a \cos (x)+b)^3}-\frac {\frac {\frac {\frac {b \left (3 a^2-2 b^2\right ) \int \frac {1}{b+a \sin \left (x+\frac {\pi }{2}\right )}dx}{a}-\frac {2 x \left (a^2-b^2\right )}{a}}{a^2}+\frac {\sin (x) \left (2 \left (a^2-b^2\right )-a b \cos (x)\right )}{a^2 (a \cos (x)+b)}}{2 \left (a^2-b^2\right )}-\frac {b \sin ^3(x)}{2 \left (a^2-b^2\right ) (a \cos (x)+b)^2}}{a}\)

\(\Big \downarrow \) 3138

\(\displaystyle \frac {\sin ^3(x)}{3 a (a \cos (x)+b)^3}-\frac {\frac {\frac {\frac {2 b \left (3 a^2-2 b^2\right ) \int \frac {1}{-\left ((a-b) \tan ^2\left (\frac {x}{2}\right )\right )+a+b}d\tan \left (\frac {x}{2}\right )}{a}-\frac {2 x \left (a^2-b^2\right )}{a}}{a^2}+\frac {\sin (x) \left (2 \left (a^2-b^2\right )-a b \cos (x)\right )}{a^2 (a \cos (x)+b)}}{2 \left (a^2-b^2\right )}-\frac {b \sin ^3(x)}{2 \left (a^2-b^2\right ) (a \cos (x)+b)^2}}{a}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\sin ^3(x)}{3 a (a \cos (x)+b)^3}-\frac {\frac {\frac {\frac {2 b \left (3 a^2-2 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a \sqrt {a-b} \sqrt {a+b}}-\frac {2 x \left (a^2-b^2\right )}{a}}{a^2}+\frac {\sin (x) \left (2 \left (a^2-b^2\right )-a b \cos (x)\right )}{a^2 (a \cos (x)+b)}}{2 \left (a^2-b^2\right )}-\frac {b \sin ^3(x)}{2 \left (a^2-b^2\right ) (a \cos (x)+b)^2}}{a}\)

Input:

Int[(a*Cot[x] + b*Csc[x])^(-4),x]
 

Output:

Sin[x]^3/(3*a*(b + a*Cos[x])^3) - (-1/2*(b*Sin[x]^3)/((a^2 - b^2)*(b + a*C 
os[x])^2) + (((-2*(a^2 - b^2)*x)/a + (2*b*(3*a^2 - 2*b^2)*ArcTanh[(Sqrt[a 
- b]*Tan[x/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]))/a^2 + ((2*(a^2 - 
 b^2) - a*b*Cos[x])*Sin[x])/(a^2*(b + a*Cos[x])))/(2*(a^2 - b^2)))/a
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 3042

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

rule 3138

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ 
e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d)   Subst[Int[1/(a + b + 
(a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] 
 && NeQ[a^2 - b^2, 0]
 

rule 3172

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x 
])^(m + 1)/(b*f*(m + 1))), x] + Simp[g^2*((p - 1)/(b*(m + 1)))   Int[(g*Cos 
[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1)*Sin[e + f*x], x], x] /; Fre 
eQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && I 
ntegersQ[2*m, 2*p]
 

rule 3214

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. 
)*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d   Int[1/(c + d 
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
 

rule 3342

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[g*(g*C 
os[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*((b*c*(m + p + 1) - a*d*p 
 + b*d*(m + 1)*Sin[e + f*x])/(b^2*f*(m + 1)*(m + p + 1))), x] + Simp[g^2*(( 
p - 1)/(b^2*(m + 1)*(m + p + 1)))   Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin 
[e + f*x])^(m + 1)*Simp[b*d*(m + 1) + (b*c*(m + p + 1) - a*d*p)*Sin[e + f*x 
], x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2, 0] && Lt 
Q[m, -1] && GtQ[p, 1] && NeQ[m + p + 1, 0] && IntegerQ[2*m]
 

rule 3343

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(b*c 
 - a*d))*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m + 1)/(f*g*(a^2 - 
 b^2)*(m + 1))), x] + Simp[1/((a^2 - b^2)*(m + 1))   Int[(g*Cos[e + f*x])^p 
*(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + p 
 + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ 
[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]
 

rule 4892

Int[(cot[(c_.) + (d_.)*(x_)]^(n_.)*(a_.) + csc[(c_.) + (d_.)*(x_)]^(n_.)*(b 
_.))^(p_)*(u_.), x_Symbol] :> Int[ActivateTrig[u]*Csc[c + d*x]^(n*p)*(b + a 
*Cos[c + d*x]^n)^p, x] /; FreeQ[{a, b, c, d}, x] && IntegersQ[n, p]
 

Maple [A] (verified)

Time = 28.22 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.24

method	result	size
default	\(\frac {2 \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{a^{4}}-\frac {2 \left (\frac {-\frac {\left (2 a^{3}-a^{2} b -3 a \,b^{2}+2 b^{3}\right ) a \tan \left (\frac {x}{2}\right )^{5}}{2 \left (a +b \right )}+\frac {2 a \left (5 a^{2}-3 b^{2}\right ) \tan \left (\frac {x}{2}\right )^{3}}{3}-\frac {\left (2 a^{3}+a^{2} b -3 a \,b^{2}-2 b^{3}\right ) a \tan \left (\frac {x}{2}\right )}{2 \left (a -b \right )}}{\left (\tan \left (\frac {x}{2}\right )^{2} a -b \tan \left (\frac {x}{2}\right )^{2}-a -b \right )^{3}}+\frac {b \left (3 a^{2}-2 b^{2}\right ) \operatorname {arctanh}\left (\frac {\tan \left (\frac {x}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{2 \left (a^{2}-b^{2}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{4}}\)	\(197\)
risch	\(\frac {x}{a^{4}}-\frac {i \left (-15 a^{4} b \,{\mathrm e}^{5 i x}+18 a^{2} b^{3} {\mathrm e}^{5 i x}-12 a^{5} {\mathrm e}^{4 i x}-27 a^{3} b^{2} {\mathrm e}^{4 i x}+54 a \,b^{4} {\mathrm e}^{4 i x}-48 a^{4} b \,{\mathrm e}^{3 i x}+34 a^{2} b^{3} {\mathrm e}^{3 i x}+44 b^{5} {\mathrm e}^{3 i x}-12 a^{5} {\mathrm e}^{2 i x}-36 a^{3} b^{2} {\mathrm e}^{2 i x}+78 a \,b^{4} {\mathrm e}^{2 i x}-33 a^{4} b \,{\mathrm e}^{i x}+48 a^{2} b^{3} {\mathrm e}^{i x}-8 a^{5}+11 b^{2} a^{3}\right )}{3 a^{4} \left (a \,{\mathrm e}^{2 i x}+2 b \,{\mathrm e}^{i x}+a \right )^{3} \left (-a^{2}+b^{2}\right )}+\frac {3 b \ln \left ({\mathrm e}^{i x}-\frac {i a^{2}-i b^{2}-b \sqrt {a^{2}-b^{2}}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) a^{2}}-\frac {b^{3} \ln \left ({\mathrm e}^{i x}-\frac {i a^{2}-i b^{2}-b \sqrt {a^{2}-b^{2}}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) a^{4}}-\frac {3 b \ln \left ({\mathrm e}^{i x}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) a^{2}}+\frac {b^{3} \ln \left ({\mathrm e}^{i x}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) a^{4}}\)	\(522\)

Input:

int(1/(a*cot(x)+b*csc(x))^4,x,method=_RETURNVERBOSE)
 

Output:

2/a^4*arctan(tan(1/2*x))-2/a^4*((-1/2*(2*a^3-a^2*b-3*a*b^2+2*b^3)*a/(a+b)* 
tan(1/2*x)^5+2/3*a*(5*a^2-3*b^2)*tan(1/2*x)^3-1/2*(2*a^3+a^2*b-3*a*b^2-2*b 
^3)*a/(a-b)*tan(1/2*x))/(tan(1/2*x)^2*a-b*tan(1/2*x)^2-a-b)^3+1/2*b*(3*a^2 
-2*b^2)/(a^2-b^2)/((a-b)*(a+b))^(1/2)*arctanh(tan(1/2*x)*(a-b)/((a-b)*(a+b 
))^(1/2)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 418 vs. \(2 (141) = 282\).

Time = 0.13 (sec) , antiderivative size = 878, normalized size of antiderivative = 5.52 \[ \int \frac {1}{(a \cot (x)+b \csc (x))^4} \, dx =\text {Too large to display} \] Input:

integrate(1/(a*cot(x)+b*csc(x))^4,x, algorithm="fricas")
 

Output:

[1/12*(12*(a^7 - 2*a^5*b^2 + a^3*b^4)*x*cos(x)^3 + 36*(a^6*b - 2*a^4*b^3 + 
 a^2*b^5)*x*cos(x)^2 + 36*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*x*cos(x) + 3*(3*a^ 
2*b^4 - 2*b^6 + (3*a^5*b - 2*a^3*b^3)*cos(x)^3 + 3*(3*a^4*b^2 - 2*a^2*b^4) 
*cos(x)^2 + 3*(3*a^3*b^3 - 2*a*b^5)*cos(x))*sqrt(a^2 - b^2)*log((2*a*b*cos 
(x) - (a^2 - 2*b^2)*cos(x)^2 - 2*sqrt(a^2 - b^2)*(b*cos(x) + a)*sin(x) + 2 
*a^2 - b^2)/(a^2*cos(x)^2 + 2*a*b*cos(x) + b^2)) + 12*(a^4*b^3 - 2*a^2*b^5 
 + b^7)*x + 2*(2*a^7 - 7*a^5*b^2 + 11*a^3*b^4 - 6*a*b^6 - (8*a^7 - 19*a^5* 
b^2 + 11*a^3*b^4)*cos(x)^2 - 3*(3*a^6*b - 8*a^4*b^3 + 5*a^2*b^5)*cos(x))*s 
in(x))/(a^8*b^3 - 2*a^6*b^5 + a^4*b^7 + (a^11 - 2*a^9*b^2 + a^7*b^4)*cos(x 
)^3 + 3*(a^10*b - 2*a^8*b^3 + a^6*b^5)*cos(x)^2 + 3*(a^9*b^2 - 2*a^7*b^4 + 
 a^5*b^6)*cos(x)), 1/6*(6*(a^7 - 2*a^5*b^2 + a^3*b^4)*x*cos(x)^3 + 18*(a^6 
*b - 2*a^4*b^3 + a^2*b^5)*x*cos(x)^2 + 18*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*x* 
cos(x) - 3*(3*a^2*b^4 - 2*b^6 + (3*a^5*b - 2*a^3*b^3)*cos(x)^3 + 3*(3*a^4* 
b^2 - 2*a^2*b^4)*cos(x)^2 + 3*(3*a^3*b^3 - 2*a*b^5)*cos(x))*sqrt(-a^2 + b^ 
2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(x) + a)/((a^2 - b^2)*sin(x))) + 6*(a^4* 
b^3 - 2*a^2*b^5 + b^7)*x + (2*a^7 - 7*a^5*b^2 + 11*a^3*b^4 - 6*a*b^6 - (8* 
a^7 - 19*a^5*b^2 + 11*a^3*b^4)*cos(x)^2 - 3*(3*a^6*b - 8*a^4*b^3 + 5*a^2*b 
^5)*cos(x))*sin(x))/(a^8*b^3 - 2*a^6*b^5 + a^4*b^7 + (a^11 - 2*a^9*b^2 + a 
^7*b^4)*cos(x)^3 + 3*(a^10*b - 2*a^8*b^3 + a^6*b^5)*cos(x)^2 + 3*(a^9*b^2 
- 2*a^7*b^4 + a^5*b^6)*cos(x))]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(a \cot (x)+b \csc (x))^4} \, dx=\text {Timed out} \] Input:

integrate(1/(a*cot(x)+b*csc(x))**4,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(a \cot (x)+b \csc (x))^4} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(1/(a*cot(x)+b*csc(x))^4,x, algorithm="maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?` f 
or more de
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.77 \[ \int \frac {1}{(a \cot (x)+b \csc (x))^4} \, dx=-\frac {{\left (3 \, a^{2} b - 2 \, b^{3}\right )} {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, x\right ) - b \tan \left (\frac {1}{2} \, x\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{6} - a^{4} b^{2}\right )} \sqrt {-a^{2} + b^{2}}} + \frac {6 \, a^{4} \tan \left (\frac {1}{2} \, x\right )^{5} - 9 \, a^{3} b \tan \left (\frac {1}{2} \, x\right )^{5} - 6 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, x\right )^{5} + 15 \, a b^{3} \tan \left (\frac {1}{2} \, x\right )^{5} - 6 \, b^{4} \tan \left (\frac {1}{2} \, x\right )^{5} - 20 \, a^{4} \tan \left (\frac {1}{2} \, x\right )^{3} + 32 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, x\right )^{3} - 12 \, b^{4} \tan \left (\frac {1}{2} \, x\right )^{3} + 6 \, a^{4} \tan \left (\frac {1}{2} \, x\right ) + 9 \, a^{3} b \tan \left (\frac {1}{2} \, x\right ) - 6 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, x\right ) - 15 \, a b^{3} \tan \left (\frac {1}{2} \, x\right ) - 6 \, b^{4} \tan \left (\frac {1}{2} \, x\right )}{3 \, {\left (a^{5} - a^{3} b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, x\right )^{2} - b \tan \left (\frac {1}{2} \, x\right )^{2} - a - b\right )}^{3}} + \frac {x}{a^{4}} \] Input:

integrate(1/(a*cot(x)+b*csc(x))^4,x, algorithm="giac")
 

Output:

-(3*a^2*b - 2*b^3)*(pi*floor(1/2*x/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a* 
tan(1/2*x) - b*tan(1/2*x))/sqrt(-a^2 + b^2)))/((a^6 - a^4*b^2)*sqrt(-a^2 + 
 b^2)) + 1/3*(6*a^4*tan(1/2*x)^5 - 9*a^3*b*tan(1/2*x)^5 - 6*a^2*b^2*tan(1/ 
2*x)^5 + 15*a*b^3*tan(1/2*x)^5 - 6*b^4*tan(1/2*x)^5 - 20*a^4*tan(1/2*x)^3 
+ 32*a^2*b^2*tan(1/2*x)^3 - 12*b^4*tan(1/2*x)^3 + 6*a^4*tan(1/2*x) + 9*a^3 
*b*tan(1/2*x) - 6*a^2*b^2*tan(1/2*x) - 15*a*b^3*tan(1/2*x) - 6*b^4*tan(1/2 
*x))/((a^5 - a^3*b^2)*(a*tan(1/2*x)^2 - b*tan(1/2*x)^2 - a - b)^3) + x/a^4
 

Mupad [B] (verification not implemented)

Time = 21.79 (sec) , antiderivative size = 3068, normalized size of antiderivative = 19.30 \[ \int \frac {1}{(a \cot (x)+b \csc (x))^4} \, dx=\text {Too large to display} \] Input:

int(1/(b/sin(x) + a*cot(x))^4,x)
 

Output:

(2*atan((((((8*(6*a^12*b - 4*a^13 + 4*a^8*b^5 - 2*a^9*b^4 - 10*a^10*b^3 + 
6*a^11*b^2))/(a^11*b + a^12 - a^9*b^3 - a^10*b^2) - (tan(x/2)*(8*a^13*b - 
8*a^8*b^6 + 8*a^9*b^5 + 16*a^10*b^4 - 16*a^11*b^3 - 8*a^12*b^2)*8i)/(a^4*( 
a^8*b + a^9 - a^6*b^3 - a^7*b^2)))*1i)/a^4 + (8*tan(x/2)*(4*a^6 - 8*a^5*b 
- 8*a*b^5 + 8*b^6 - 16*a^2*b^4 + 16*a^3*b^3 + 5*a^4*b^2))/(a^8*b + a^9 - a 
^6*b^3 - a^7*b^2))/a^4 - ((((8*(6*a^12*b - 4*a^13 + 4*a^8*b^5 - 2*a^9*b^4 
- 10*a^10*b^3 + 6*a^11*b^2))/(a^11*b + a^12 - a^9*b^3 - a^10*b^2) + (tan(x 
/2)*(8*a^13*b - 8*a^8*b^6 + 8*a^9*b^5 + 16*a^10*b^4 - 16*a^11*b^3 - 8*a^12 
*b^2)*8i)/(a^4*(a^8*b + a^9 - a^6*b^3 - a^7*b^2)))*1i)/a^4 - (8*tan(x/2)*( 
4*a^6 - 8*a^5*b - 8*a*b^5 + 8*b^6 - 16*a^2*b^4 + 16*a^3*b^3 + 5*a^4*b^2))/ 
(a^8*b + a^9 - a^6*b^3 - a^7*b^2))/a^4)/((16*(6*a^4*b - 2*a*b^4 + 4*b^5 - 
10*a^2*b^3 + 3*a^3*b^2))/(a^11*b + a^12 - a^9*b^3 - a^10*b^2) + (((((8*(6* 
a^12*b - 4*a^13 + 4*a^8*b^5 - 2*a^9*b^4 - 10*a^10*b^3 + 6*a^11*b^2))/(a^11 
*b + a^12 - a^9*b^3 - a^10*b^2) - (tan(x/2)*(8*a^13*b - 8*a^8*b^6 + 8*a^9* 
b^5 + 16*a^10*b^4 - 16*a^11*b^3 - 8*a^12*b^2)*8i)/(a^4*(a^8*b + a^9 - a^6* 
b^3 - a^7*b^2)))*1i)/a^4 + (8*tan(x/2)*(4*a^6 - 8*a^5*b - 8*a*b^5 + 8*b^6 
- 16*a^2*b^4 + 16*a^3*b^3 + 5*a^4*b^2))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2)) 
*1i)/a^4 + (((((8*(6*a^12*b - 4*a^13 + 4*a^8*b^5 - 2*a^9*b^4 - 10*a^10*b^3 
 + 6*a^11*b^2))/(a^11*b + a^12 - a^9*b^3 - a^10*b^2) + (tan(x/2)*(8*a^13*b 
 - 8*a^8*b^6 + 8*a^9*b^5 + 16*a^10*b^4 - 16*a^11*b^3 - 8*a^12*b^2)*8i)/...
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 860, normalized size of antiderivative = 5.41 \[ \int \frac {1}{(a \cot (x)+b \csc (x))^4} \, dx =\text {Too large to display} \] Input:

int(1/(a*cot(x)+b*csc(x))^4,x)
 

Output:

( - 18*sqrt( - a**2 + b**2)*atan((tan(x/2)*a - tan(x/2)*b)/sqrt( - a**2 + 
b**2))*cos(x)*sin(x)**2*a**5*b + 12*sqrt( - a**2 + b**2)*atan((tan(x/2)*a 
- tan(x/2)*b)/sqrt( - a**2 + b**2))*cos(x)*sin(x)**2*a**3*b**3 + 18*sqrt( 
- a**2 + b**2)*atan((tan(x/2)*a - tan(x/2)*b)/sqrt( - a**2 + b**2))*cos(x) 
*a**5*b + 42*sqrt( - a**2 + b**2)*atan((tan(x/2)*a - tan(x/2)*b)/sqrt( - a 
**2 + b**2))*cos(x)*a**3*b**3 - 36*sqrt( - a**2 + b**2)*atan((tan(x/2)*a - 
 tan(x/2)*b)/sqrt( - a**2 + b**2))*cos(x)*a*b**5 - 54*sqrt( - a**2 + b**2) 
*atan((tan(x/2)*a - tan(x/2)*b)/sqrt( - a**2 + b**2))*sin(x)**2*a**4*b**2 
+ 36*sqrt( - a**2 + b**2)*atan((tan(x/2)*a - tan(x/2)*b)/sqrt( - a**2 + b* 
*2))*sin(x)**2*a**2*b**4 + 54*sqrt( - a**2 + b**2)*atan((tan(x/2)*a - tan( 
x/2)*b)/sqrt( - a**2 + b**2))*a**4*b**2 - 18*sqrt( - a**2 + b**2)*atan((ta 
n(x/2)*a - tan(x/2)*b)/sqrt( - a**2 + b**2))*a**2*b**4 - 12*sqrt( - a**2 + 
 b**2)*atan((tan(x/2)*a - tan(x/2)*b)/sqrt( - a**2 + b**2))*b**6 + 6*cos(x 
)*sin(x)**2*a**7*x - 12*cos(x)*sin(x)**2*a**5*b**2*x + 6*cos(x)*sin(x)**2* 
a**3*b**4*x + 9*cos(x)*sin(x)*a**6*b - 24*cos(x)*sin(x)*a**4*b**3 + 15*cos 
(x)*sin(x)*a**2*b**5 - 6*cos(x)*a**7*x - 6*cos(x)*a**5*b**2*x + 30*cos(x)* 
a**3*b**4*x - 18*cos(x)*a*b**6*x - 8*sin(x)**3*a**7 + 19*sin(x)**3*a**5*b* 
*2 - 11*sin(x)**3*a**3*b**4 + 18*sin(x)**2*a**6*b*x - 36*sin(x)**2*a**4*b* 
*3*x + 18*sin(x)**2*a**2*b**5*x + 6*sin(x)*a**7 - 12*sin(x)*a**5*b**2 + 6* 
sin(x)*a*b**6 - 18*a**6*b*x + 30*a**4*b**3*x - 6*a**2*b**5*x - 6*b**7*x...