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

Optimal. Leaf size=159 \[ \frac {x}{a^4}+\frac {b \sin ^3(x)}{2 a \left (a^2-b^2\right ) (a \cos (x)+b)^2}-\frac {b \left (3 a^2-2 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a^4 (a-b)^{3/2} (a+b)^{3/2}}-\frac {\sin (x) \left (2 \left (a^2-b^2\right )-a b \cos (x)\right )}{2 a^3 \left (a^2-b^2\right ) (a \cos (x)+b)}+\frac {\sin ^3(x)}{3 a (a \cos (x)+b)^3} \]

[Out]

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*c
os(x))^2

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Rubi [A]  time = 0.34, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.636, Rules used = {4392, 2693, 2864, 2863, 2735, 2659, 208} \[ \frac {b \sin ^3(x)}{2 a \left (a^2-b^2\right ) (a \cos (x)+b)^2}-\frac {\sin (x) \left (2 \left (a^2-b^2\right )-a b \cos (x)\right )}{2 a^3 \left (a^2-b^2\right ) (a \cos (x)+b)}-\frac {b \left (3 a^2-2 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a^4 (a-b)^{3/2} (a+b)^{3/2}}+\frac {x}{a^4}+\frac {\sin ^3(x)}{3 a (a \cos (x)+b)^3} \]

Antiderivative was successfully verified.

[In]

Int[(a*Cot[x] + b*Csc[x])^(-4),x]

[Out]

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

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(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 2693

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] + Dist[(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] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a
^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && IntegersQ[2*m, 2*p]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(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 2863

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*Cos[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] + Dist[(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)*Si
n[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && N
eQ[m + p + 1, 0] && IntegerQ[2*m]

Rule 2864

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] + Dist[1/((a^2 - b^2)*(m + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1)*Sim
p[(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 4392

Int[(cot[(c_.) + (d_.)*(x_)]^(n_.)*(a_.) + csc[(c_.) + (d_.)*(x_)]^(n_.)*(b_.))^(p_)*(u_.), x_Symbol] :> Int[A
ctivateTrig[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]

Rubi steps

\begin {align*} \int \frac {1}{(a \cot (x)+b \csc (x))^4} \, dx &=\int \frac {\sin ^4(x)}{(b+a \cos (x))^4} \, dx\\ &=\frac {\sin ^3(x)}{3 a (b+a \cos (x))^3}-\frac {\int \frac {\cos (x) \sin ^2(x)}{(b+a \cos (x))^3} \, dx}{a}\\ &=\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}-\frac {\int \frac {(2 a+b \cos (x)) \sin ^2(x)}{(b+a \cos (x))^2} \, dx}{2 a \left (a^2-b^2\right )}\\ &=-\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}+\frac {\int \frac {-a b+2 \left (a^2-b^2\right ) \cos (x)}{b+a \cos (x)} \, dx}{2 a^3 \left (a^2-b^2\right )}\\ &=\frac {x}{a^4}-\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}-\frac {\left (b \left (3 a^2-2 b^2\right )\right ) \int \frac {1}{b+a \cos (x)} \, dx}{2 a^4 \left (a^2-b^2\right )}\\ &=\frac {x}{a^4}-\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}-\frac {\left (b \left (3 a^2-2 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a+b+(-a+b) x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )}{a^4 \left (a^2-b^2\right )}\\ &=\frac {x}{a^4}-\frac {b \left (3 a^2-2 b^2\right ) \tanh ^{-1}\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}\\ \end {align*}

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Mathematica [A]  time = 0.48, size = 150, normalized size = 0.94 \[ \frac {\sin (x) \left (-\frac {a \left (8 a^2-11 b^2\right ) (a \cos (x)+b)^2}{(a-b) (a+b)}-\frac {6 b \left (2 b^2-3 a^2\right ) \csc (x) (a \cos (x)+b)^3 \tanh ^{-1}\left (\frac {(b-a) \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+2 a \left (a^2-b^2\right )+7 a b (a \cos (x)+b)+6 x \csc (x) (a \cos (x)+b)^3\right )}{6 a^4 (a \cos (x)+b)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((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)

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fricas [B]  time = 2.08, size = 878, normalized size = 5.52 \[ \left [\frac {12 \, {\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} x \cos \relax (x)^{3} + 36 \, {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} x \cos \relax (x)^{2} + 36 \, {\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} x \cos \relax (x) + 3 \, {\left (3 \, a^{2} b^{4} - 2 \, b^{6} + {\left (3 \, a^{5} b - 2 \, a^{3} b^{3}\right )} \cos \relax (x)^{3} + 3 \, {\left (3 \, a^{4} b^{2} - 2 \, a^{2} b^{4}\right )} \cos \relax (x)^{2} + 3 \, {\left (3 \, a^{3} b^{3} - 2 \, a b^{5}\right )} \cos \relax (x)\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \relax (x) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \relax (x)^{2} - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \relax (x) + a\right )} \sin \relax (x) + 2 \, a^{2} - b^{2}}{a^{2} \cos \relax (x)^{2} + 2 \, a b \cos \relax (x) + b^{2}}\right ) + 12 \, {\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} x + 2 \, {\left (2 \, a^{7} - 7 \, a^{5} b^{2} + 11 \, a^{3} b^{4} - 6 \, a b^{6} - {\left (8 \, a^{7} - 19 \, a^{5} b^{2} + 11 \, a^{3} b^{4}\right )} \cos \relax (x)^{2} - 3 \, {\left (3 \, a^{6} b - 8 \, a^{4} b^{3} + 5 \, a^{2} b^{5}\right )} \cos \relax (x)\right )} \sin \relax (x)}{12 \, {\left (a^{8} b^{3} - 2 \, a^{6} b^{5} + a^{4} b^{7} + {\left (a^{11} - 2 \, a^{9} b^{2} + a^{7} b^{4}\right )} \cos \relax (x)^{3} + 3 \, {\left (a^{10} b - 2 \, a^{8} b^{3} + a^{6} b^{5}\right )} \cos \relax (x)^{2} + 3 \, {\left (a^{9} b^{2} - 2 \, a^{7} b^{4} + a^{5} b^{6}\right )} \cos \relax (x)\right )}}, \frac {6 \, {\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} x \cos \relax (x)^{3} + 18 \, {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} x \cos \relax (x)^{2} + 18 \, {\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} x \cos \relax (x) - 3 \, {\left (3 \, a^{2} b^{4} - 2 \, b^{6} + {\left (3 \, a^{5} b - 2 \, a^{3} b^{3}\right )} \cos \relax (x)^{3} + 3 \, {\left (3 \, a^{4} b^{2} - 2 \, a^{2} b^{4}\right )} \cos \relax (x)^{2} + 3 \, {\left (3 \, a^{3} b^{3} - 2 \, a b^{5}\right )} \cos \relax (x)\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \relax (x) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \relax (x)}\right ) + 6 \, {\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} x + {\left (2 \, a^{7} - 7 \, a^{5} b^{2} + 11 \, a^{3} b^{4} - 6 \, a b^{6} - {\left (8 \, a^{7} - 19 \, a^{5} b^{2} + 11 \, a^{3} b^{4}\right )} \cos \relax (x)^{2} - 3 \, {\left (3 \, a^{6} b - 8 \, a^{4} b^{3} + 5 \, a^{2} b^{5}\right )} \cos \relax (x)\right )} \sin \relax (x)}{6 \, {\left (a^{8} b^{3} - 2 \, a^{6} b^{5} + a^{4} b^{7} + {\left (a^{11} - 2 \, a^{9} b^{2} + a^{7} b^{4}\right )} \cos \relax (x)^{3} + 3 \, {\left (a^{10} b - 2 \, a^{8} b^{3} + a^{6} b^{5}\right )} \cos \relax (x)^{2} + 3 \, {\left (a^{9} b^{2} - 2 \, a^{7} b^{4} + a^{5} b^{6}\right )} \cos \relax (x)\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[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*s
qrt(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))*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)), 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)*c
os(x))]

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giac [A]  time = 0.20, size = 282, normalized size = 1.77 \[ -\frac {{\left (3 \, a^{2} b - 2 \, b^{3}\right )} {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, x\right ) - b \tan \left (\frac {1}{2} \, x\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{6} - a^{4} b^{2}\right )} \sqrt {-a^{2} + b^{2}}} + \frac {6 \, a^{4} \tan \left (\frac {1}{2} \, x\right )^{5} - 9 \, a^{3} b \tan \left (\frac {1}{2} \, x\right )^{5} - 6 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, x\right )^{5} + 15 \, a b^{3} \tan \left (\frac {1}{2} \, x\right )^{5} - 6 \, b^{4} \tan \left (\frac {1}{2} \, x\right )^{5} - 20 \, a^{4} \tan \left (\frac {1}{2} \, x\right )^{3} + 32 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, x\right )^{3} - 12 \, b^{4} \tan \left (\frac {1}{2} \, x\right )^{3} + 6 \, a^{4} \tan \left (\frac {1}{2} \, x\right ) + 9 \, a^{3} b \tan \left (\frac {1}{2} \, x\right ) - 6 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, x\right ) - 15 \, a b^{3} \tan \left (\frac {1}{2} \, x\right ) - 6 \, b^{4} \tan \left (\frac {1}{2} \, x\right )}{3 \, {\left (a^{5} - a^{3} b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, x\right )^{2} - b \tan \left (\frac {1}{2} \, x\right )^{2} - a - b\right )}^{3}} + \frac {x}{a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(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

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maple [B]  time = 0.14, size = 534, normalized size = 3.36 \[ \frac {2 \left (\tan ^{5}\left (\frac {x}{2}\right )\right )}{\left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-a -b \right )^{3} \left (a +b \right )}-\frac {\left (\tan ^{5}\left (\frac {x}{2}\right )\right ) b}{a \left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-a -b \right )^{3} \left (a +b \right )}-\frac {3 \left (\tan ^{5}\left (\frac {x}{2}\right )\right ) b^{2}}{a^{2} \left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-a -b \right )^{3} \left (a +b \right )}+\frac {2 \left (\tan ^{5}\left (\frac {x}{2}\right )\right ) b^{3}}{a^{3} \left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-a -b \right )^{3} \left (a +b \right )}-\frac {20 \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{3 a \left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-a -b \right )^{3}}+\frac {4 \left (\tan ^{3}\left (\frac {x}{2}\right )\right ) b^{2}}{a^{3} \left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-a -b \right )^{3}}+\frac {2 \tan \left (\frac {x}{2}\right )}{\left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-a -b \right )^{3} \left (a -b \right )}+\frac {\tan \left (\frac {x}{2}\right ) b}{a \left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-a -b \right )^{3} \left (a -b \right )}-\frac {3 \tan \left (\frac {x}{2}\right ) b^{2}}{a^{2} \left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-a -b \right )^{3} \left (a -b \right )}-\frac {2 \tan \left (\frac {x}{2}\right ) b^{3}}{a^{3} \left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-a -b \right )^{3} \left (a -b \right )}-\frac {3 b \arctanh \left (\frac {\tan \left (\frac {x}{2}\right ) \left (a -b \right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{2} \left (a^{2}-b^{2}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {2 b^{3} \arctanh \left (\frac {\tan \left (\frac {x}{2}\right ) \left (a -b \right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{4} \left (a^{2}-b^{2}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {2 \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/(a*tan(1/2*x)^2-b*tan(1/2*x)^2-a-b)^3/(a+b)*tan(1/2*x)^5-1/a/(a*tan(1/2*x)^2-b*tan(1/2*x)^2-a-b)^3/(a+b)*tan
(1/2*x)^5*b-3/a^2/(a*tan(1/2*x)^2-b*tan(1/2*x)^2-a-b)^3/(a+b)*tan(1/2*x)^5*b^2+2/a^3/(a*tan(1/2*x)^2-b*tan(1/2
*x)^2-a-b)^3/(a+b)*tan(1/2*x)^5*b^3-20/3/a/(a*tan(1/2*x)^2-b*tan(1/2*x)^2-a-b)^3*tan(1/2*x)^3+4/a^3/(a*tan(1/2
*x)^2-b*tan(1/2*x)^2-a-b)^3*tan(1/2*x)^3*b^2+2/(a*tan(1/2*x)^2-b*tan(1/2*x)^2-a-b)^3/(a-b)*tan(1/2*x)+1/a/(a*t
an(1/2*x)^2-b*tan(1/2*x)^2-a-b)^3/(a-b)*tan(1/2*x)*b-3/a^2/(a*tan(1/2*x)^2-b*tan(1/2*x)^2-a-b)^3/(a-b)*tan(1/2
*x)*b^2-2/a^3/(a*tan(1/2*x)^2-b*tan(1/2*x)^2-a-b)^3/(a-b)*tan(1/2*x)*b^3-3/a^2*b/(a^2-b^2)/((a+b)*(a-b))^(1/2)
*arctanh(tan(1/2*x)*(a-b)/((a+b)*(a-b))^(1/2))+2/a^4*b^3/(a^2-b^2)/((a+b)*(a-b))^(1/2)*arctanh(tan(1/2*x)*(a-b
)/((a+b)*(a-b))^(1/2))+2/a^4*arctan(tan(1/2*x))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 8.25, size = 3068, normalized size = 19.30 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b/sin(x) + a*cot(x))^4,x)

[Out]

(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 + 1
6*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)/(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)))/a^4 + ((4*tan(x/2)^3*(5*a^2 - 3*b^2))/(3*a^3) - (tan(x/2)*(a + b)*(a*b - 2*a^2 + 2*b^2))/(a^3*b - a
^4) + (tan(x/2)^5*(3*a*b^2 + a^2*b - 2*a^3 - 2*b^3))/(a^3*(a + b)))/(3*a*b^2 - tan(x/2)^6*(3*a*b^2 - 3*a^2*b +
a^3 - b^3) + 3*a^2*b + tan(x/2)^2*(3*a*b^2 - 3*a^2*b - 3*a^3 + 3*b^3) - tan(x/2)^4*(3*a*b^2 + 3*a^2*b - 3*a^3
- 3*b^3) + a^3 + b^3) + (b*atan(((b*(3*a^2 - 2*b^2)*((a + b)^3*(a - b)^3)^(1/2)*((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) + (b*((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) - (4*b*ta
n(x/2)*(3*a^2 - 2*b^2)*((a + b)^3*(a - b)^3)^(1/2)*(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))/((a^8*b + a^9 - a^6*b^3 - a^7*b^2)*(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2)))*(3*a^2 - 2*b^2
)*((a + b)^3*(a - b)^3)^(1/2))/(2*(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2)))*1i)/(2*(a^10 - a^4*b^6 + 3*a^6*b^
4 - 3*a^8*b^2)) + (b*(3*a^2 - 2*b^2)*((a + b)^3*(a - b)^3)^(1/2)*((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) - (b*((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) + (4*b*tan(x/2)*(3*a^2 -
2*b^2)*((a + b)^3*(a - b)^3)^(1/2)*(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)
)/((a^8*b + a^9 - a^6*b^3 - a^7*b^2)*(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2)))*(3*a^2 - 2*b^2)*((a + b)^3*(a
- b)^3)^(1/2))/(2*(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2)))*1i)/(2*(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2)))
/((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) + (b*(3*a^2 -
2*b^2)*((a + b)^3*(a - b)^3)^(1/2)*((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) + (b*((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) - (4*b*tan(x/2)*(3*a^2 - 2*b^2)*((a + b)^3*(a - b)^3)^
(1/2)*(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))/((a^8*b + a^9 - a^6*b^3 - a
^7*b^2)*(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2)))*(3*a^2 - 2*b^2)*((a + b)^3*(a - b)^3)^(1/2))/(2*(a^10 - a^4
*b^6 + 3*a^6*b^4 - 3*a^8*b^2))))/(2*(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2)) - (b*(3*a^2 - 2*b^2)*((a + b)^3*
(a - b)^3)^(1/2)*((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) - (b*((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) + (4*b*tan(x/2)*(3*a^2 - 2*b^2)*((a + b)^3*(a - b)^3)^(1/2)*(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))/((a^8*b + a^9 - a^6*b^3 - a^7*b^2)*(a^10 - a^4
*b^6 + 3*a^6*b^4 - 3*a^8*b^2)))*(3*a^2 - 2*b^2)*((a + b)^3*(a - b)^3)^(1/2))/(2*(a^10 - a^4*b^6 + 3*a^6*b^4 -
3*a^8*b^2))))/(2*(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2))))*(3*a^2 - 2*b^2)*((a + b)^3*(a - b)^3)^(1/2)*1i)/(
a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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