3.1.12 \(\int \frac {\cos ^3(x)}{a+b \cot (x)} \, dx\) [12]

Optimal. Leaf size=123 \[ \frac {a^3 b \tanh ^{-1}\left (\frac {a \cos (x)-b \sin (x)}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}-\frac {a^2 b \cos (x)}{\left (a^2+b^2\right )^2}-\frac {b \cos ^3(x)}{3 \left (a^2+b^2\right )}-\frac {a b^2 \sin (x)}{\left (a^2+b^2\right )^2}+\frac {a \sin (x)}{a^2+b^2}-\frac {a \sin ^3(x)}{3 \left (a^2+b^2\right )} \]

[Out]

a^3*b*arctanh((a*cos(x)-b*sin(x))/(a^2+b^2)^(1/2))/(a^2+b^2)^(5/2)-a^2*b*cos(x)/(a^2+b^2)^2-1/3*b*cos(x)^3/(a^
2+b^2)-a*b^2*sin(x)/(a^2+b^2)^2+a*sin(x)/(a^2+b^2)-1/3*a*sin(x)^3/(a^2+b^2)

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Rubi [A]
time = 0.16, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.692, Rules used = {3599, 3188, 2713, 2645, 30, 3179, 2717, 3153, 212} \begin {gather*} -\frac {a \sin ^3(x)}{3 \left (a^2+b^2\right )}+\frac {a \sin (x)}{a^2+b^2}-\frac {a b^2 \sin (x)}{\left (a^2+b^2\right )^2}-\frac {b \cos ^3(x)}{3 \left (a^2+b^2\right )}-\frac {a^2 b \cos (x)}{\left (a^2+b^2\right )^2}+\frac {a^3 b \tanh ^{-1}\left (\frac {a \cos (x)-b \sin (x)}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[x]^3/(a + b*Cot[x]),x]

[Out]

(a^3*b*ArcTanh[(a*Cos[x] - b*Sin[x])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(5/2) - (a^2*b*Cos[x])/(a^2 + b^2)^2 - (b*C
os[x]^3)/(3*(a^2 + b^2)) - (a*b^2*Sin[x])/(a^2 + b^2)^2 + (a*Sin[x])/(a^2 + b^2) - (a*Sin[x]^3)/(3*(a^2 + b^2)
)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2645

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 2713

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3153

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Dist[-d^(-1), Subst[Int
[1/(a^2 + b^2 - x^2), x], x, b*Cos[c + d*x] - a*Sin[c + d*x]], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2,
0]

Rule 3179

Int[cos[(c_.) + (d_.)*(x_)]^(m_)/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)]), x_Symbol] :>
Simp[b*(Cos[c + d*x]^(m - 1)/(d*(a^2 + b^2)*(m - 1))), x] + (Dist[a/(a^2 + b^2), Int[Cos[c + d*x]^(m - 1), x]
, x] + Dist[b^2/(a^2 + b^2), Int[Cos[c + d*x]^(m - 2)/(a*Cos[c + d*x] + b*Sin[c + d*x]), x], x]) /; FreeQ[{a,
b, c, d}, x] && NeQ[a^2 + b^2, 0] && GtQ[m, 1]

Rule 3188

Int[(cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.))/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(
c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[b/(a^2 + b^2), Int[Cos[c + d*x]^m*Sin[c + d*x]^(n - 1), x], x] + (Dist[
a/(a^2 + b^2), Int[Cos[c + d*x]^(m - 1)*Sin[c + d*x]^n, x], x] - Dist[a*(b/(a^2 + b^2)), Int[Cos[c + d*x]^(m -
1)*(Sin[c + d*x]^(n - 1)/(a*Cos[c + d*x] + b*Sin[c + d*x])), x], x]) /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b
^2, 0] && IGtQ[m, 0] && IGtQ[n, 0]

Rule 3599

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[Sin[e + f*x]^
m*((a*Cos[e + f*x] + b*Sin[e + f*x])^n/Cos[e + f*x]^n), x] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] &&
ILtQ[n, 0] && ((LtQ[m, 5] && GtQ[n, -4]) || (EqQ[m, 5] && EqQ[n, -1]))

Rubi steps

\begin {align*} \int \frac {\cos ^3(x)}{a+b \cot (x)} \, dx &=-\int \frac {\cos ^3(x) \sin (x)}{-b \cos (x)-a \sin (x)} \, dx\\ &=\frac {a \int \cos ^3(x) \, dx}{a^2+b^2}+\frac {b \int \cos ^2(x) \sin (x) \, dx}{a^2+b^2}+\frac {(a b) \int \frac {\cos ^2(x)}{-b \cos (x)-a \sin (x)} \, dx}{a^2+b^2}\\ &=-\frac {a^2 b \cos (x)}{\left (a^2+b^2\right )^2}+\frac {\left (a^3 b\right ) \int \frac {1}{-b \cos (x)-a \sin (x)} \, dx}{\left (a^2+b^2\right )^2}-\frac {\left (a b^2\right ) \int \cos (x) \, dx}{\left (a^2+b^2\right )^2}-\frac {a \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (x)\right )}{a^2+b^2}-\frac {b \text {Subst}\left (\int x^2 \, dx,x,\cos (x)\right )}{a^2+b^2}\\ &=-\frac {a^2 b \cos (x)}{\left (a^2+b^2\right )^2}-\frac {b \cos ^3(x)}{3 \left (a^2+b^2\right )}-\frac {a b^2 \sin (x)}{\left (a^2+b^2\right )^2}+\frac {a \sin (x)}{a^2+b^2}-\frac {a \sin ^3(x)}{3 \left (a^2+b^2\right )}-\frac {\left (a^3 b\right ) \text {Subst}\left (\int \frac {1}{a^2+b^2-x^2} \, dx,x,-a \cos (x)+b \sin (x)\right )}{\left (a^2+b^2\right )^2}\\ &=\frac {a^3 b \tanh ^{-1}\left (\frac {a \cos (x)-b \sin (x)}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}-\frac {a^2 b \cos (x)}{\left (a^2+b^2\right )^2}-\frac {b \cos ^3(x)}{3 \left (a^2+b^2\right )}-\frac {a b^2 \sin (x)}{\left (a^2+b^2\right )^2}+\frac {a \sin (x)}{a^2+b^2}-\frac {a \sin ^3(x)}{3 \left (a^2+b^2\right )}\\ \end {align*}

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Mathematica [A]
time = 1.20, size = 112, normalized size = 0.91 \begin {gather*} -\frac {2 a^3 b \tanh ^{-1}\left (\frac {-a+b \tan \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}-\frac {3 b \left (5 a^2+b^2\right ) \cos (x)+b \left (a^2+b^2\right ) \cos (3 x)-2 a \left (5 a^2-b^2+\left (a^2+b^2\right ) \cos (2 x)\right ) \sin (x)}{12 \left (a^2+b^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[x]^3/(a + b*Cot[x]),x]

[Out]

(-2*a^3*b*ArcTanh[(-a + b*Tan[x/2])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(5/2) - (3*b*(5*a^2 + b^2)*Cos[x] + b*(a^2 +
b^2)*Cos[3*x] - 2*a*(5*a^2 - b^2 + (a^2 + b^2)*Cos[2*x])*Sin[x])/(12*(a^2 + b^2)^2)

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Maple [A]
time = 0.39, size = 170, normalized size = 1.38

method	result	size
default	\(-\frac {2 \left (-a^{3} \left (\tan ^{5}\left (\frac {x}{2}\right )\right )+\left (2 a^{2} b +b^{3}\right ) \left (\tan ^{4}\left (\frac {x}{2}\right )\right )+\left (-\frac {2}{3} a^{3}+\frac {4}{3} a \,b^{2}\right ) \left (\tan ^{3}\left (\frac {x}{2}\right )\right )+2 a^{2} b \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-a^{3} \tan \left (\frac {x}{2}\right )+\frac {4 a^{2} b}{3}+\frac {b^{3}}{3}\right )}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (1+\tan ^{2}\left (\frac {x}{2}\right )\right )^{3}}+\frac {4 a^{3} b \arctanh \left (\frac {-2 b \tan \left (\frac {x}{2}\right )+2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (2 a^{4}+4 a^{2} b^{2}+2 b^{4}\right ) \sqrt {a^{2}+b^{2}}}\)	\(170\)
risch	\(\frac {{\mathrm e}^{i x} b}{16 i a b +8 a^{2}-8 b^{2}}-\frac {3 i {\mathrm e}^{i x} a}{8 \left (2 i a b +a^{2}-b^{2}\right )}+\frac {{\mathrm e}^{-i x} b}{8 \left (-i b +a \right )^{2}}+\frac {3 i {\mathrm e}^{-i x} a}{8 \left (-i b +a \right )^{2}}+\frac {b \,a^{3} \ln \left ({\mathrm e}^{i x}-\frac {i a^{4} b +2 i a^{2} b^{3}+i b^{5}-a^{5}-2 a^{3} b^{2}-a \,b^{4}}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}}}-\frac {b \,a^{3} \ln \left ({\mathrm e}^{i x}+\frac {i a^{4} b +2 i a^{2} b^{3}+i b^{5}-a^{5}-2 a^{3} b^{2}-a \,b^{4}}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}}}+\frac {b \cos \left (3 x \right )}{-12 a^{2}-12 b^{2}}-\frac {a \sin \left (3 x \right )}{12 \left (-a^{2}-b^{2}\right )}\)	\(272\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)^3/(a+b*cot(x)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 279 vs. \(2 (115) = 230\).
time = 0.53, size = 279, normalized size = 2.27 \begin {gather*} \frac {a^{3} b \log \left (\frac {a - \frac {b \sin \left (x\right )}{\cos \left (x\right ) + 1} + \sqrt {a^{2} + b^{2}}}{a - \frac {b \sin \left (x\right )}{\cos \left (x\right ) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} - \frac {2 \, {\left (4 \, a^{2} b + b^{3} - \frac {3 \, a^{3} \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac {6 \, a^{2} b \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} - \frac {3 \, a^{3} \sin \left (x\right )^{5}}{{\left (\cos \left (x\right ) + 1\right )}^{5}} - \frac {2 \, {\left (a^{3} - 2 \, a b^{2}\right )} \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} + \frac {3 \, {\left (2 \, a^{2} b + b^{3}\right )} \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}}\right )}}{3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4} + \frac {3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + \frac {3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}} + \frac {{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (x\right )^{6}}{{\left (\cos \left (x\right ) + 1\right )}^{6}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^3/(a+b*cot(x)),x, algorithm="maxima")

[Out]

a^3*b*log((a - b*sin(x)/(cos(x) + 1) + sqrt(a^2 + b^2))/(a - b*sin(x)/(cos(x) + 1) - sqrt(a^2 + b^2)))/((a^4 +
2*a^2*b^2 + b^4)*sqrt(a^2 + b^2)) - 2/3*(4*a^2*b + b^3 - 3*a^3*sin(x)/(cos(x) + 1) + 6*a^2*b*sin(x)^2/(cos(x)
+ 1)^2 - 3*a^3*sin(x)^5/(cos(x) + 1)^5 - 2*(a^3 - 2*a*b^2)*sin(x)^3/(cos(x) + 1)^3 + 3*(2*a^2*b + b^3)*sin(x)
^4/(cos(x) + 1)^4)/(a^4 + 2*a^2*b^2 + b^4 + 3*(a^4 + 2*a^2*b^2 + b^4)*sin(x)^2/(cos(x) + 1)^2 + 3*(a^4 + 2*a^2
*b^2 + b^4)*sin(x)^4/(cos(x) + 1)^4 + (a^4 + 2*a^2*b^2 + b^4)*sin(x)^6/(cos(x) + 1)^6)

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Fricas [A]
time = 2.90, size = 214, normalized size = 1.74 \begin {gather*} \frac {3 \, \sqrt {a^{2} + b^{2}} a^{3} b \log \left (\frac {2 \, a b \cos \left (x\right ) \sin \left (x\right ) - {\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - a^{2} - 2 \, b^{2} - 2 \, \sqrt {a^{2} + b^{2}} {\left (a \cos \left (x\right ) - b \sin \left (x\right )\right )}}{2 \, a b \cos \left (x\right ) \sin \left (x\right ) - {\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + a^{2}}\right ) - 2 \, {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (x\right )^{3} - 6 \, {\left (a^{4} b + a^{2} b^{3}\right )} \cos \left (x\right ) + 2 \, {\left (2 \, a^{5} + a^{3} b^{2} - a b^{4} + {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{6 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^3/(a+b*cot(x)),x, algorithm="fricas")

[Out]

1/6*(3*sqrt(a^2 + b^2)*a^3*b*log((2*a*b*cos(x)*sin(x) - (a^2 - b^2)*cos(x)^2 - a^2 - 2*b^2 - 2*sqrt(a^2 + b^2)
*(a*cos(x) - b*sin(x)))/(2*a*b*cos(x)*sin(x) - (a^2 - b^2)*cos(x)^2 + a^2)) - 2*(a^4*b + 2*a^2*b^3 + b^5)*cos(
x)^3 - 6*(a^4*b + a^2*b^3)*cos(x) + 2*(2*a^5 + a^3*b^2 - a*b^4 + (a^5 + 2*a^3*b^2 + a*b^4)*cos(x)^2)*sin(x))/(
a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cos ^{3}{\left (x \right )}}{a + b \cot {\left (x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)**3/(a+b*cot(x)),x)

[Out]

Integral(cos(x)**3/(a + b*cot(x)), x)

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Giac [A]
time = 0.45, size = 201, normalized size = 1.63 \begin {gather*} \frac {a^{3} b \log \left (\frac {{\left | 2 \, b \tan \left (\frac {1}{2} \, x\right ) - 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b \tan \left (\frac {1}{2} \, x\right ) - 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} + \frac {2 \, {\left (3 \, a^{3} \tan \left (\frac {1}{2} \, x\right )^{5} - 6 \, a^{2} b \tan \left (\frac {1}{2} \, x\right )^{4} - 3 \, b^{3} \tan \left (\frac {1}{2} \, x\right )^{4} + 2 \, a^{3} \tan \left (\frac {1}{2} \, x\right )^{3} - 4 \, a b^{2} \tan \left (\frac {1}{2} \, x\right )^{3} - 6 \, a^{2} b \tan \left (\frac {1}{2} \, x\right )^{2} + 3 \, a^{3} \tan \left (\frac {1}{2} \, x\right ) - 4 \, a^{2} b - b^{3}\right )}}{3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^3/(a+b*cot(x)),x, algorithm="giac")

[Out]

a^3*b*log(abs(2*b*tan(1/2*x) - 2*a - 2*sqrt(a^2 + b^2))/abs(2*b*tan(1/2*x) - 2*a + 2*sqrt(a^2 + b^2)))/((a^4 +
2*a^2*b^2 + b^4)*sqrt(a^2 + b^2)) + 2/3*(3*a^3*tan(1/2*x)^5 - 6*a^2*b*tan(1/2*x)^4 - 3*b^3*tan(1/2*x)^4 + 2*a
^3*tan(1/2*x)^3 - 4*a*b^2*tan(1/2*x)^3 - 6*a^2*b*tan(1/2*x)^2 + 3*a^3*tan(1/2*x) - 4*a^2*b - b^3)/((a^4 + 2*a^
2*b^2 + b^4)*(tan(1/2*x)^2 + 1)^3)

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Mupad [B]
time = 0.67, size = 291, normalized size = 2.37 \begin {gather*} \frac {2\,a^3\,b\,\mathrm {atanh}\left (\frac {2\,a\,b^4+2\,a^5+4\,a^3\,b^2-2\,b\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a^4+2\,a^2\,b^2+b^4\right )}{2\,{\left (a^2+b^2\right )}^{5/2}}\right )}{{\left (a^2+b^2\right )}^{5/2}}-\frac {\frac {2\,\left (4\,a^2\,b+b^3\right )}{3\,\left (a^4+2\,a^2\,b^2+b^4\right )}+\frac {4\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3\,\left (2\,a\,b^2-a^3\right )}{3\,\left (a^4+2\,a^2\,b^2+b^4\right )}-\frac {2\,a^3\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^4+2\,a^2\,b^2+b^4}+\frac {2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4\,\left (2\,a^2\,b+b^3\right )}{a^4+2\,a^2\,b^2+b^4}-\frac {2\,a^3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^5}{a^4+2\,a^2\,b^2+b^4}+\frac {4\,a^2\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{a^4+2\,a^2\,b^2+b^4}}{{\mathrm {tan}\left (\frac {x}{2}\right )}^6+3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)^3/(a + b*cot(x)),x)

[Out]

(2*a^3*b*atanh((2*a*b^4 + 2*a^5 + 4*a^3*b^2 - 2*b*tan(x/2)*(a^4 + b^4 + 2*a^2*b^2))/(2*(a^2 + b^2)^(5/2))))/(a
^2 + b^2)^(5/2) - ((2*(4*a^2*b + b^3))/(3*(a^4 + b^4 + 2*a^2*b^2)) + (4*tan(x/2)^3*(2*a*b^2 - a^3))/(3*(a^4 +
b^4 + 2*a^2*b^2)) - (2*a^3*tan(x/2))/(a^4 + b^4 + 2*a^2*b^2) + (2*tan(x/2)^4*(2*a^2*b + b^3))/(a^4 + b^4 + 2*a
^2*b^2) - (2*a^3*tan(x/2)^5)/(a^4 + b^4 + 2*a^2*b^2) + (4*a^2*b*tan(x/2)^2)/(a^4 + b^4 + 2*a^2*b^2))/(3*tan(x/
2)^2 + 3*tan(x/2)^4 + tan(x/2)^6 + 1)

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