\(\int \frac {\cos (x) \sin ^3(x)}{a \cos (x)+b \sin (x)} \, dx\) [277]

Optimal result

Integrand size = 18, antiderivative size = 122 \[ \int \frac {\cos (x) \sin ^3(x)}{a \cos (x)+b \sin (x)} \, dx=\frac {a^3 b \text {arctanh}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}+\frac {a b^2 \cos (x)}{\left (a^2+b^2\right )^2}-\frac {a \cos (x)}{a^2+b^2}+\frac {a \cos ^3(x)}{3 \left (a^2+b^2\right )}+\frac {a^2 b \sin (x)}{\left (a^2+b^2\right )^2}+\frac {b \sin ^3(x)}{3 \left (a^2+b^2\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3188, 2644, 30, 2713, 3178, 3153, 212, 2718} \[ \int \frac {\cos (x) \sin ^3(x)}{a \cos (x)+b \sin (x)} \, dx=\frac {b \sin ^3(x)}{3 \left (a^2+b^2\right )}+\frac {a^2 b \sin (x)}{\left (a^2+b^2\right )^2}+\frac {a \cos ^3(x)}{3 \left (a^2+b^2\right )}-\frac {a \cos (x)}{a^2+b^2}+\frac {a b^2 \cos (x)}{\left (a^2+b^2\right )^2}+\frac {a^3 b \text {arctanh}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}} \]

[In]

Int[(Cos[x]*Sin[x]^3)/(a*Cos[x] + b*Sin[x]),x]

[Out]

(a^3*b*ArcTanh[(b*Cos[x] - a*Sin[x])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(5/2) + (a*b^2*Cos[x])/(a^2 + b^2)^2 - (a*C
os[x])/(a^2 + b^2) + (a*Cos[x]^3)/(3*(a^2 + b^2)) + (a^2*b*Sin[x])/(a^2 + b^2)^2 + (b*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 2644

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, 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 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[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 3178

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

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

Mathematica [A] (verified)

Time = 1.16 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.93 \[ \int \frac {\cos (x) \sin ^3(x)}{a \cos (x)+b \sin (x)} \, dx=-\frac {2 a^3 b \text {arctanh}\left (\frac {-b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}+\frac {\left (-9 a^3+3 a b^2\right ) \cos (x)+a \left (a^2+b^2\right ) \cos (3 x)-2 b \left (-7 a^2-b^2+\left (a^2+b^2\right ) \cos (2 x)\right ) \sin (x)}{12 \left (a^2+b^2\right )^2} \]

[In]

Integrate[(Cos[x]*Sin[x]^3)/(a*Cos[x] + b*Sin[x]),x]

[Out]

(-2*a^3*b*ArcTanh[(-b + a*Tan[x/2])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(5/2) + ((-9*a^3 + 3*a*b^2)*Cos[x] + a*(a^2
+ b^2)*Cos[3*x] - 2*b*(-7*a^2 - b^2 + (a^2 + b^2)*Cos[2*x])*Sin[x])/(12*(a^2 + b^2)^2)

Maple [A] (verified)

Time = 0.63 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.34

method	result	size
default	\(-\frac {16 a^{3} b \,\operatorname {arctanh}\left (\frac {2 a \tan \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (8 a^{4}+16 a^{2} b^{2}+8 b^{4}\right ) \sqrt {a^{2}+b^{2}}}+\frac {2 a^{2} b \tan \left (\frac {x}{2}\right )^{5}+2 a \,b^{2} \tan \left (\frac {x}{2}\right )^{4}+2 \left (\frac {10}{3} a^{2} b +\frac {4}{3} b^{3}\right ) \tan \left (\frac {x}{2}\right )^{3}-4 \tan \left (\frac {x}{2}\right )^{2} a^{3}+2 a^{2} b \tan \left (\frac {x}{2}\right )-\frac {4 a^{3}}{3}+\frac {2 a \,b^{2}}{3}}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (1+\tan \left (\frac {x}{2}\right )^{2}\right )^{3}}\)	\(163\)
risch	\(\frac {i {\mathrm e}^{i x} b}{-16 i b a +8 a^{2}-8 b^{2}}-\frac {3 \,{\mathrm e}^{i x} a}{8 \left (-2 i b a +a^{2}-b^{2}\right )}-\frac {i {\mathrm e}^{-i x} b}{8 \left (i b +a \right )^{2}}-\frac {3 \,{\mathrm e}^{-i x} a}{8 \left (i b +a \right )^{2}}-\frac {i b \,a^{3} \ln \left ({\mathrm e}^{i x}-\frac {i b +a}{\sqrt {-a^{2}-b^{2}}}\right )}{\sqrt {-a^{2}-b^{2}}\, \left (a^{2}+b^{2}\right )^{2}}+\frac {i b \,a^{3} \ln \left ({\mathrm e}^{i x}+\frac {i b +a}{\sqrt {-a^{2}-b^{2}}}\right )}{\sqrt {-a^{2}-b^{2}}\, \left (a^{2}+b^{2}\right )^{2}}-\frac {a \cos \left (3 x \right )}{12 \left (-a^{2}-b^{2}\right )}+\frac {b \sin \left (3 x \right )}{-12 a^{2}-12 b^{2}}\)	\(237\)

[In]

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

[Out]

-16*a^3*b/(8*a^4+16*a^2*b^2+8*b^4)/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tan(1/2*x)-2*b)/(a^2+b^2)^(1/2))+2/(a^4+2*
a^2*b^2+b^4)*(a^2*b*tan(1/2*x)^5+a*b^2*tan(1/2*x)^4+(10/3*a^2*b+4/3*b^3)*tan(1/2*x)^3-2*tan(1/2*x)^2*a^3+a^2*b
*tan(1/2*x)-2/3*a^3+1/3*a*b^2)/(1+tan(1/2*x)^2)^3

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.72 \[ \int \frac {\cos (x) \sin ^3(x)}{a \cos (x)+b \sin (x)} \, dx=\frac {3 \, \sqrt {a^{2} + b^{2}} a^{3} b \log \left (\frac {2 \, a b \cos \left (x\right ) \sin \left (x\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, a^{2} - b^{2} - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \left (x\right ) - a \sin \left (x\right )\right )}}{2 \, a b \cos \left (x\right ) \sin \left (x\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + b^{2}}\right ) + 2 \, {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (x\right )^{3} - 6 \, {\left (a^{5} + a^{3} b^{2}\right )} \cos \left (x\right ) + 2 \, {\left (4 \, a^{4} b + 5 \, a^{2} b^{3} + b^{5} - {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{6 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )}} \]

[In]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos (x) \sin ^3(x)}{a \cos (x)+b \sin (x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 278 vs. \(2 (114) = 228\).

Time = 0.32 (sec) , antiderivative size = 278, normalized size of antiderivative = 2.28 \[ \int \frac {\cos (x) \sin ^3(x)}{a \cos (x)+b \sin (x)} \, dx=\frac {a^{3} b \log \left (\frac {b - \frac {a \sin \left (x\right )}{\cos \left (x\right ) + 1} + \sqrt {a^{2} + b^{2}}}{b - \frac {a \sin \left (x\right )}{\cos \left (x\right ) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} - \frac {2 \, {\left (2 \, a^{3} - a b^{2} - \frac {3 \, a^{2} b \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac {6 \, a^{3} \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} - \frac {3 \, a b^{2} \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}} - \frac {3 \, a^{2} b \sin \left (x\right )^{5}}{{\left (\cos \left (x\right ) + 1\right )}^{5}} - \frac {2 \, {\left (5 \, a^{2} b + 2 \, b^{3}\right )} \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}}\right )}}{3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4} + \frac {3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + \frac {3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}} + \frac {{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (x\right )^{6}}{{\left (\cos \left (x\right ) + 1\right )}^{6}}\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.56 \[ \int \frac {\cos (x) \sin ^3(x)}{a \cos (x)+b \sin (x)} \, dx=\frac {a^{3} b \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, x\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, x\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} + \frac {2 \, {\left (3 \, a^{2} b \tan \left (\frac {1}{2} \, x\right )^{5} + 3 \, a b^{2} \tan \left (\frac {1}{2} \, x\right )^{4} + 10 \, a^{2} b \tan \left (\frac {1}{2} \, x\right )^{3} + 4 \, b^{3} \tan \left (\frac {1}{2} \, x\right )^{3} - 6 \, a^{3} \tan \left (\frac {1}{2} \, x\right )^{2} + 3 \, a^{2} b \tan \left (\frac {1}{2} \, x\right ) - 2 \, a^{3} + a b^{2}\right )}}{3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}^{3}} \]

[In]

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

[Out]

a^3*b*log(abs(2*a*tan(1/2*x) - 2*b - 2*sqrt(a^2 + b^2))/abs(2*a*tan(1/2*x) - 2*b + 2*sqrt(a^2 + b^2)))/((a^4 +
2*a^2*b^2 + b^4)*sqrt(a^2 + b^2)) + 2/3*(3*a^2*b*tan(1/2*x)^5 + 3*a*b^2*tan(1/2*x)^4 + 10*a^2*b*tan(1/2*x)^3
+ 4*b^3*tan(1/2*x)^3 - 6*a^3*tan(1/2*x)^2 + 3*a^2*b*tan(1/2*x) - 2*a^3 + a*b^2)/((a^4 + 2*a^2*b^2 + b^4)*(tan(
1/2*x)^2 + 1)^3)

Mupad [B] (verification not implemented)

Time = 23.40 (sec) , antiderivative size = 286, normalized size of antiderivative = 2.34 \[ \int \frac {\cos (x) \sin ^3(x)}{a \cos (x)+b \sin (x)} \, dx=\frac {\frac {2\,\left (a\,b^2-2\,a^3\right )}{3\,\left (a^4+2\,a^2\,b^2+b^4\right )}+\frac {4\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3\,\left (5\,a^2\,b+2\,b^3\right )}{3\,\left (a^4+2\,a^2\,b^2+b^4\right )}-\frac {4\,a^3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{a^4+2\,a^2\,b^2+b^4}+\frac {2\,a\,b^2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4}{a^4+2\,a^2\,b^2+b^4}+\frac {2\,a^2\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^5}{a^4+2\,a^2\,b^2+b^4}+\frac {2\,a^2\,b\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^4+2\,a^2\,b^2+b^4}}{{\mathrm {tan}\left (\frac {x}{2}\right )}^6+3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+1}+\frac {2\,a^3\,b\,\mathrm {atanh}\left (\frac {2\,a^4\,b+2\,b^5+4\,a^2\,b^3-2\,a\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a^4+2\,a^2\,b^2+b^4\right )}{2\,{\left (a^2+b^2\right )}^{5/2}}\right )}{{\left (a^2+b^2\right )}^{5/2}} \]

[In]

int((cos(x)*sin(x)^3)/(a*cos(x) + b*sin(x)),x)

[Out]

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