3.1.11 \(\int \frac {\cos ^4(x)}{a+b \cot (x)} \, dx\) [11]

Optimal. Leaf size=126 \[ \frac {a \left (3 a^4-6 a^2 b^2-b^4\right ) x}{8 \left (a^2+b^2\right )^3}-\frac {a^4 b \log (b \cos (x)+a \sin (x))}{\left (a^2+b^2\right )^3}+\frac {\left (4 b \left (2 a^2+b^2\right )+a \left (5 a^2+b^2\right ) \cot (x)\right ) \sin ^2(x)}{8 \left (a^2+b^2\right )^2}-\frac {(b+a \cot (x)) \sin ^4(x)}{4 \left (a^2+b^2\right )} \]

[Out]

1/8*a*(3*a^4-6*a^2*b^2-b^4)*x/(a^2+b^2)^3-a^4*b*ln(b*cos(x)+a*sin(x))/(a^2+b^2)^3+1/8*(4*b*(2*a^2+b^2)+a*(5*a^
2+b^2)*cot(x))*sin(x)^2/(a^2+b^2)^2-1/4*(b+a*cot(x))*sin(x)^4/(a^2+b^2)

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Rubi [A]
time = 0.24, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {3597, 1661, 815, 649, 209, 266} \begin {gather*} -\frac {\sin ^4(x) (a \cot (x)+b)}{4 \left (a^2+b^2\right )}+\frac {\sin ^2(x) \left (a \left (5 a^2+b^2\right ) \cot (x)+4 b \left (2 a^2+b^2\right )\right )}{8 \left (a^2+b^2\right )^2}-\frac {a^4 b \log (a \sin (x)+b \cos (x))}{\left (a^2+b^2\right )^3}+\frac {a x \left (3 a^4-6 a^2 b^2-b^4\right )}{8 \left (a^2+b^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[x]^4/(a + b*Cot[x]),x]

[Out]

(a*(3*a^4 - 6*a^2*b^2 - b^4)*x)/(8*(a^2 + b^2)^3) - (a^4*b*Log[b*Cos[x] + a*Sin[x]])/(a^2 + b^2)^3 + ((4*b*(2*
a^2 + b^2) + a*(5*a^2 + b^2)*Cot[x])*Sin[x]^2)/(8*(a^2 + b^2)^2) - ((b + a*Cot[x])*Sin[x]^4)/(4*(a^2 + b^2))

Rule 209

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

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 649

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[(-a)*c]

Rule 815

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m*((f + g*x)/(a + c*x^2)), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 1661

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(d +
e*x)^m*Pq, a + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 0], g = Coeff[Polyn
omialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 1]}, Simp[(a*g - c*f*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1)))
, x] + Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSum[(2*a*c*(p + 1)*Q)/(d + e*x)^m +
(c*f*(2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && ILtQ[m, 0]

Rule 3597

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[b/f, Subst[Int
[x^m*((a + x)^n/(b^2 + x^2)^(m/2 + 1)), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && IntegerQ[m/
2]

Rubi steps

\begin {align*} \int \frac {\cos ^4(x)}{a+b \cot (x)} \, dx &=-\left (b \text {Subst}\left (\int \frac {x^4}{(a+x) \left (b^2+x^2\right )^3} \, dx,x,b \cot (x)\right )\right )\\ &=-\frac {(b+a \cot (x)) \sin ^4(x)}{4 \left (a^2+b^2\right )}+\frac {\text {Subst}\left (\int \frac {\frac {a^2 b^4}{a^2+b^2}-\frac {3 a b^4 x}{a^2+b^2}-4 b^2 x^2}{(a+x) \left (b^2+x^2\right )^2} \, dx,x,b \cot (x)\right )}{4 b}\\ &=\frac {\left (4 b \left (2 a^2+b^2\right )+a \left (5 a^2+b^2\right ) \cot (x)\right ) \sin ^2(x)}{8 \left (a^2+b^2\right )^2}-\frac {(b+a \cot (x)) \sin ^4(x)}{4 \left (a^2+b^2\right )}-\frac {\text {Subst}\left (\int \frac {\frac {a^2 b^4 \left (3 a^2-b^2\right )}{\left (a^2+b^2\right )^2}-\frac {a b^4 \left (5 a^2+b^2\right ) x}{\left (a^2+b^2\right )^2}}{(a+x) \left (b^2+x^2\right )} \, dx,x,b \cot (x)\right )}{8 b^3}\\ &=\frac {\left (4 b \left (2 a^2+b^2\right )+a \left (5 a^2+b^2\right ) \cot (x)\right ) \sin ^2(x)}{8 \left (a^2+b^2\right )^2}-\frac {(b+a \cot (x)) \sin ^4(x)}{4 \left (a^2+b^2\right )}-\frac {\text {Subst}\left (\int \left (\frac {8 a^4 b^4}{\left (a^2+b^2\right )^3 (a+x)}+\frac {a b^4 \left (3 a^4-6 a^2 b^2-b^4-8 a^3 x\right )}{\left (a^2+b^2\right )^3 \left (b^2+x^2\right )}\right ) \, dx,x,b \cot (x)\right )}{8 b^3}\\ &=-\frac {a^4 b \log (a+b \cot (x))}{\left (a^2+b^2\right )^3}+\frac {\left (4 b \left (2 a^2+b^2\right )+a \left (5 a^2+b^2\right ) \cot (x)\right ) \sin ^2(x)}{8 \left (a^2+b^2\right )^2}-\frac {(b+a \cot (x)) \sin ^4(x)}{4 \left (a^2+b^2\right )}-\frac {(a b) \text {Subst}\left (\int \frac {3 a^4-6 a^2 b^2-b^4-8 a^3 x}{b^2+x^2} \, dx,x,b \cot (x)\right )}{8 \left (a^2+b^2\right )^3}\\ &=-\frac {a^4 b \log (a+b \cot (x))}{\left (a^2+b^2\right )^3}+\frac {\left (4 b \left (2 a^2+b^2\right )+a \left (5 a^2+b^2\right ) \cot (x)\right ) \sin ^2(x)}{8 \left (a^2+b^2\right )^2}-\frac {(b+a \cot (x)) \sin ^4(x)}{4 \left (a^2+b^2\right )}+\frac {\left (a^4 b\right ) \text {Subst}\left (\int \frac {x}{b^2+x^2} \, dx,x,b \cot (x)\right )}{\left (a^2+b^2\right )^3}-\frac {\left (a b \left (3 a^4-6 a^2 b^2-b^4\right )\right ) \text {Subst}\left (\int \frac {1}{b^2+x^2} \, dx,x,b \cot (x)\right )}{8 \left (a^2+b^2\right )^3}\\ &=\frac {a \left (3 a^4-6 a^2 b^2-b^4\right ) x}{8 \left (a^2+b^2\right )^3}-\frac {a^4 b \log (a+b \cot (x))}{\left (a^2+b^2\right )^3}-\frac {a^4 b \log (\sin (x))}{\left (a^2+b^2\right )^3}+\frac {\left (4 b \left (2 a^2+b^2\right )+a \left (5 a^2+b^2\right ) \cot (x)\right ) \sin ^2(x)}{8 \left (a^2+b^2\right )^2}-\frac {(b+a \cot (x)) \sin ^4(x)}{4 \left (a^2+b^2\right )}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.65, size = 179, normalized size = 1.42 \begin {gather*} \frac {12 a^5 x-32 i a^4 b x-24 a^3 b^2 x-4 a b^4 x+32 i a^4 b \text {ArcTan}(\tan (x))-4 b \left (3 a^4+4 a^2 b^2+b^4\right ) \cos (2 x)-a^4 b \cos (4 x)-2 a^2 b^3 \cos (4 x)-b^5 \cos (4 x)-16 a^4 b \log \left ((b \cos (x)+a \sin (x))^2\right )+8 a^5 \sin (2 x)+8 a^3 b^2 \sin (2 x)+a^5 \sin (4 x)+2 a^3 b^2 \sin (4 x)+a b^4 \sin (4 x)}{32 \left (a^2+b^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[x]^4/(a + b*Cot[x]),x]

[Out]

(12*a^5*x - (32*I)*a^4*b*x - 24*a^3*b^2*x - 4*a*b^4*x + (32*I)*a^4*b*ArcTan[Tan[x]] - 4*b*(3*a^4 + 4*a^2*b^2 +
b^4)*Cos[2*x] - a^4*b*Cos[4*x] - 2*a^2*b^3*Cos[4*x] - b^5*Cos[4*x] - 16*a^4*b*Log[(b*Cos[x] + a*Sin[x])^2] +
8*a^5*Sin[2*x] + 8*a^3*b^2*Sin[2*x] + a^5*Sin[4*x] + 2*a^3*b^2*Sin[4*x] + a*b^4*Sin[4*x])/(32*(a^2 + b^2)^3)

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Maple [A]
time = 0.36, size = 172, normalized size = 1.37

method	result	size
default	\(-\frac {b \,a^{4} \ln \left (a \tan \left (x \right )+b \right )}{\left (a^{2}+b^{2}\right )^{3}}+\frac {\frac {\left (\frac {3}{8} a^{5}+\frac {1}{4} a^{3} b^{2}-\frac {1}{8} a \,b^{4}\right ) \left (\tan ^{3}\left (x \right )\right )+\left (-\frac {1}{2} a^{4} b -\frac {1}{2} a^{2} b^{3}\right ) \left (\tan ^{2}\left (x \right )\right )+\left (\frac {5}{8} a^{5}+\frac {3}{4} a^{3} b^{2}+\frac {1}{8} a \,b^{4}\right ) \tan \left (x \right )-\frac {3 a^{4} b}{4}-a^{2} b^{3}-\frac {b^{5}}{4}}{\left (1+\tan ^{2}\left (x \right )\right )^{2}}+\frac {a \left (4 a^{3} b \ln \left (1+\tan ^{2}\left (x \right )\right )+\left (3 a^{4}-6 a^{2} b^{2}-b^{4}\right ) \arctan \left (\tan \left (x \right )\right )\right )}{8}}{\left (a^{2}+b^{2}\right )^{3}}\)	\(172\)
risch	\(\frac {i a x b}{24 i a^{2} b -8 i b^{3}+8 a^{3}-24 a \,b^{2}}+\frac {3 a^{2} x}{24 i a^{2} b -8 i b^{3}+8 a^{3}-24 a \,b^{2}}+\frac {{\mathrm e}^{2 i x} b}{32 i a b +16 a^{2}-16 b^{2}}-\frac {i {\mathrm e}^{2 i x} a}{8 \left (2 i a b +a^{2}-b^{2}\right )}+\frac {{\mathrm e}^{-2 i x} b}{16 \left (-i b +a \right )^{2}}+\frac {i {\mathrm e}^{-2 i x} a}{8 \left (-i b +a \right )^{2}}+\frac {2 i a^{4} b x}{a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}}-\frac {a^{4} b \ln \left ({\mathrm e}^{2 i x}+\frac {i b -a}{i b +a}\right )}{a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}}+\frac {b \cos \left (4 x \right )}{-32 a^{2}-32 b^{2}}-\frac {a \sin \left (4 x \right )}{32 \left (-a^{2}-b^{2}\right )}\)	\(278\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b*a^4/(a^2+b^2)^3*ln(a*tan(x)+b)+1/(a^2+b^2)^3*(((3/8*a^5+1/4*a^3*b^2-1/8*a*b^4)*tan(x)^3+(-1/2*a^4*b-1/2*a^2
*b^3)*tan(x)^2+(5/8*a^5+3/4*a^3*b^2+1/8*a*b^4)*tan(x)-3/4*a^4*b-a^2*b^3-1/4*b^5)/(1+tan(x)^2)^2+1/8*a*(4*a^3*b
*ln(1+tan(x)^2)+(3*a^4-6*a^2*b^2-b^4)*arctan(tan(x))))

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Maxima [A]
time = 0.50, size = 240, normalized size = 1.90 \begin {gather*} -\frac {a^{4} b \log \left (a \tan \left (x\right ) + b\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {a^{4} b \log \left (\tan \left (x\right )^{2} + 1\right )}{2 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )}} + \frac {{\left (3 \, a^{5} - 6 \, a^{3} b^{2} - a b^{4}\right )} x}{8 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )}} - \frac {4 \, a^{2} b \tan \left (x\right )^{2} - {\left (3 \, a^{3} - a b^{2}\right )} \tan \left (x\right )^{3} + 6 \, a^{2} b + 2 \, b^{3} - {\left (5 \, a^{3} + a b^{2}\right )} \tan \left (x\right )}{8 \, {\left ({\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \tan \left (x\right )^{4} + a^{4} + 2 \, a^{2} b^{2} + b^{4} + 2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \tan \left (x\right )^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^4*b*log(a*tan(x) + b)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + 1/2*a^4*b*log(tan(x)^2 + 1)/(a^6 + 3*a^4*b^2 +
3*a^2*b^4 + b^6) + 1/8*(3*a^5 - 6*a^3*b^2 - a*b^4)*x/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) - 1/8*(4*a^2*b*tan(x)
^2 - (3*a^3 - a*b^2)*tan(x)^3 + 6*a^2*b + 2*b^3 - (5*a^3 + a*b^2)*tan(x))/((a^4 + 2*a^2*b^2 + b^4)*tan(x)^4 +
a^4 + 2*a^2*b^2 + b^4 + 2*(a^4 + 2*a^2*b^2 + b^4)*tan(x)^2)

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Fricas [A]
time = 3.49, size = 178, normalized size = 1.41 \begin {gather*} -\frac {4 \, a^{4} b \log \left (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 )^{4} + 4 \, {\left (a^{4} b + a^{2} b^{3}\right )} \cos \left (x\right )^{2} - {\left (3 \, a^{5} - 6 \, a^{3} b^{2} - a b^{4}\right )} x - {\left (2 \, {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (x\right )^{3} + {\left (3 \, a^{5} + 2 \, a^{3} b^{2} - a b^{4}\right )} \cos \left (x\right )\right )} \sin \left (x\right )}{8 \, {\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)^4/(a+b*cot(x)),x, algorithm="fricas")

[Out]

-1/8*(4*a^4*b*log(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)^4 + 4
*(a^4*b + a^2*b^3)*cos(x)^2 - (3*a^5 - 6*a^3*b^2 - a*b^4)*x - (2*(a^5 + 2*a^3*b^2 + a*b^4)*cos(x)^3 + (3*a^5 +
2*a^3*b^2 - a*b^4)*cos(x))*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 ^{4}{\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)**4/(a+b*cot(x)),x)

[Out]

Integral(cos(x)**4/(a + b*cot(x)), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 270 vs. \(2 (120) = 240\).
time = 0.45, size = 270, normalized size = 2.14 \begin {gather*} -\frac {a^{5} b \log \left ({\left | a \tan \left (x\right ) + b \right |}\right )}{a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}} + \frac {a^{4} b \log \left (\tan \left (x\right )^{2} + 1\right )}{2 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )}} + \frac {{\left (3 \, a^{5} - 6 \, a^{3} b^{2} - a b^{4}\right )} x}{8 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )}} - \frac {6 \, a^{4} b \tan \left (x\right )^{4} - 3 \, a^{5} \tan \left (x\right )^{3} - 2 \, a^{3} b^{2} \tan \left (x\right )^{3} + a b^{4} \tan \left (x\right )^{3} + 16 \, a^{4} b \tan \left (x\right )^{2} + 4 \, a^{2} b^{3} \tan \left (x\right )^{2} - 5 \, a^{5} \tan \left (x\right ) - 6 \, a^{3} b^{2} \tan \left (x\right ) - a b^{4} \tan \left (x\right ) + 12 \, a^{4} b + 8 \, a^{2} b^{3} + 2 \, b^{5}}{8 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} {\left (\tan \left (x\right )^{2} + 1\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^5*b*log(abs(a*tan(x) + b))/(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6) + 1/2*a^4*b*log(tan(x)^2 + 1)/(a^6 + 3*a^4
*b^2 + 3*a^2*b^4 + b^6) + 1/8*(3*a^5 - 6*a^3*b^2 - a*b^4)*x/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) - 1/8*(6*a^4*b
*tan(x)^4 - 3*a^5*tan(x)^3 - 2*a^3*b^2*tan(x)^3 + a*b^4*tan(x)^3 + 16*a^4*b*tan(x)^2 + 4*a^2*b^3*tan(x)^2 - 5*
a^5*tan(x) - 6*a^3*b^2*tan(x) - a*b^4*tan(x) + 12*a^4*b + 8*a^2*b^3 + 2*b^5)/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b
^6)*(tan(x)^2 + 1)^2)

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Mupad [B]
time = 0.50, size = 279, normalized size = 2.21 \begin {gather*} -\frac {\frac {3\,a^2\,b+b^3}{4\,\left (a^4+2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (x\right )}^3\,\left (a\,b^2-3\,a^3\right )}{8\,\left (a^4+2\,a^2\,b^2+b^4\right )}-\frac {a\,\mathrm {tan}\left (x\right )\,\left (5\,a^2+b^2\right )}{8\,\left (a^4+2\,a^2\,b^2+b^4\right )}+\frac {a^2\,b\,{\mathrm {tan}\left (x\right )}^2}{2\,\left (a^4+2\,a^2\,b^2+b^4\right )}}{{\mathrm {tan}\left (x\right )}^4+2\,{\mathrm {tan}\left (x\right )}^2+1}+\frac {\ln \left (\mathrm {tan}\left (x\right )-\mathrm {i}\right )\,\left (-3\,a^2+a\,b\,1{}\mathrm {i}\right )}{16\,\left (-a^3\,1{}\mathrm {i}-3\,a^2\,b+a\,b^2\,3{}\mathrm {i}+b^3\right )}+\frac {\ln \left (\mathrm {tan}\left (x\right )+1{}\mathrm {i}\right )\,\left (a\,b-a^2\,3{}\mathrm {i}\right )}{16\,\left (-a^3-a^2\,b\,3{}\mathrm {i}+3\,a\,b^2+b^3\,1{}\mathrm {i}\right )}-\frac {a^4\,b\,\ln \left (b+a\,\mathrm {tan}\left (x\right )\right )}{a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)^4/(a + b*cot(x)),x)

[Out]

(log(tan(x) - 1i)*(a*b*1i - 3*a^2))/(16*(a*b^2*3i - 3*a^2*b - a^3*1i + b^3)) - ((3*a^2*b + b^3)/(4*(a^4 + b^4
+ 2*a^2*b^2)) + (tan(x)^3*(a*b^2 - 3*a^3))/(8*(a^4 + b^4 + 2*a^2*b^2)) - (a*tan(x)*(5*a^2 + b^2))/(8*(a^4 + b^
4 + 2*a^2*b^2)) + (a^2*b*tan(x)^2)/(2*(a^4 + b^4 + 2*a^2*b^2)))/(2*tan(x)^2 + tan(x)^4 + 1) + (log(tan(x) + 1i
)*(a*b - a^2*3i))/(16*(3*a*b^2 - a^2*b*3i - a^3 + b^3*1i)) - (a^4*b*log(b + a*tan(x)))/(a^6 + b^6 + 3*a^2*b^4
+ 3*a^4*b^2)

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