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

Optimal. Leaf size=92 \[ \frac{a \left (\left (a^2+4 b^2\right ) \sin (x)+3 a b \cos (x)\right )}{2 \left (a^2+b^2\right )^2 (a \cos (x)+b \sin (x))^2}-\frac{\left (a^2-2 b^2\right ) \tanh ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )-b}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}} \]

[Out]

-(((a^2 - 2*b^2)*ArcTanh[(-b + a*Tan[x/2])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(5/2)) + (a*(3*a*b*Cos[x] + (a^2 + 4*
b^2)*Sin[x]))/(2*(a^2 + b^2)^2*(a*Cos[x] + b*Sin[x])^2)

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Rubi [B]  time = 0.695363, antiderivative size = 300, normalized size of antiderivative = 3.26, number of steps used = 13, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {4401, 1660, 12, 618, 206, 3155, 3074} \[ -\frac{a b \left (5 a^2+2 b^2\right ) \tan \left (\frac{x}{2}\right )+3 a^2 b^2+4 a^4+2 b^4}{a b \left (a^2+b^2\right )^2 \left (-a \tan ^2\left (\frac{x}{2}\right )+a+2 b \tan \left (\frac{x}{2}\right )\right )}+\frac{2 \left (\left (a^2+2 b^2\right ) \tan \left (\frac{x}{2}\right )+a b\right )}{a \left (a^2+b^2\right ) \left (-a \tan ^2\left (\frac{x}{2}\right )+a+2 b \tan \left (\frac{x}{2}\right )\right )^2}+\frac{2 a}{b \left (a^2+b^2\right ) (a \cos (x)+b \sin (x))}-\frac{a^2 \left (2 a^2-b^2\right ) \tanh ^{-1}\left (\frac{b-a \tan \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{b^2 \left (a^2+b^2\right )^{5/2}}+\frac{2 a^2 \tanh ^{-1}\left (\frac{b \cos (x)-a \sin (x)}{\sqrt{a^2+b^2}}\right )}{b^2 \left (a^2+b^2\right )^{3/2}}-\frac{\tanh ^{-1}\left (\frac{b \cos (x)-a \sin (x)}{\sqrt{a^2+b^2}}\right )}{b^2 \sqrt{a^2+b^2}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]^2/(a*Cos[x] + b*Sin[x])^3,x]

[Out]

(2*a^2*ArcTanh[(b*Cos[x] - a*Sin[x])/Sqrt[a^2 + b^2]])/(b^2*(a^2 + b^2)^(3/2)) - ArcTanh[(b*Cos[x] - a*Sin[x])
/Sqrt[a^2 + b^2]]/(b^2*Sqrt[a^2 + b^2]) - (a^2*(2*a^2 - b^2)*ArcTanh[(b - a*Tan[x/2])/Sqrt[a^2 + b^2]])/(b^2*(
a^2 + b^2)^(5/2)) + (2*a)/(b*(a^2 + b^2)*(a*Cos[x] + b*Sin[x])) + (2*(a*b + (a^2 + 2*b^2)*Tan[x/2]))/(a*(a^2 +
b^2)*(a + 2*b*Tan[x/2] - a*Tan[x/2]^2)^2) - (4*a^4 + 3*a^2*b^2 + 2*b^4 + a*b*(5*a^2 + 2*b^2)*Tan[x/2])/(a*b*(
a^2 + b^2)^2*(a + 2*b*Tan[x/2] - a*Tan[x/2]^2))

Rule 4401

Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /;  !InertTrigFreeQ[u]

Rule 1660

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x + c*
x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b
*x + c*x^2, x], x, 1]}, Simp[((b*f - 2*a*g + (2*c*f - b*g)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c
)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*Q - (
2*p + 3)*(2*c*f - b*g), x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1
]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 206

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

Rule 3155

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

Rule 3074

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]

Rubi steps

\begin{align*} \int \frac{\sin ^2(x)}{(a \cos (x)+b \sin (x))^3} \, dx &=\int \left (\frac{a^2 \cos ^2(x)}{b^2 (a \cos (x)+b \sin (x))^3}-\frac{2 a \cos (x)}{b^2 (a \cos (x)+b \sin (x))^2}+\frac{1}{b^2 (a \cos (x)+b \sin (x))}\right ) \, dx\\ &=\frac{\int \frac{1}{a \cos (x)+b \sin (x)} \, dx}{b^2}-\frac{(2 a) \int \frac{\cos (x)}{(a \cos (x)+b \sin (x))^2} \, dx}{b^2}+\frac{a^2 \int \frac{\cos ^2(x)}{(a \cos (x)+b \sin (x))^3} \, dx}{b^2}\\ &=\frac{2 a}{b \left (a^2+b^2\right ) (a \cos (x)+b \sin (x))}-\frac{\operatorname{Subst}\left (\int \frac{1}{a^2+b^2-x^2} \, dx,x,b \cos (x)-a \sin (x)\right )}{b^2}+\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^2}{\left (a+2 b x-a x^2\right )^3} \, dx,x,\tan \left (\frac{x}{2}\right )\right )}{b^2}-\frac{\left (2 a^2\right ) \int \frac{1}{a \cos (x)+b \sin (x)} \, dx}{b^2 \left (a^2+b^2\right )}\\ &=-\frac{\tanh ^{-1}\left (\frac{b \cos (x)-a \sin (x)}{\sqrt{a^2+b^2}}\right )}{b^2 \sqrt{a^2+b^2}}+\frac{2 a}{b \left (a^2+b^2\right ) (a \cos (x)+b \sin (x))}+\frac{2 \left (a b+\left (a^2+2 b^2\right ) \tan \left (\frac{x}{2}\right )\right )}{a \left (a^2+b^2\right ) \left (a+2 b \tan \left (\frac{x}{2}\right )-a \tan ^2\left (\frac{x}{2}\right )\right )^2}-\frac{a^2 \operatorname{Subst}\left (\int \frac{-\frac{8 \left (a^4+2 b^4\right )}{a^3}+16 b \left (1+\frac{b^2}{a^2}\right ) x+8 \left (a+\frac{b^2}{a}\right ) x^2}{\left (a+2 b x-a x^2\right )^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )}{4 b^2 \left (a^2+b^2\right )}+\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{a^2+b^2-x^2} \, dx,x,b \cos (x)-a \sin (x)\right )}{b^2 \left (a^2+b^2\right )}\\ &=\frac{2 a^2 \tanh ^{-1}\left (\frac{b \cos (x)-a \sin (x)}{\sqrt{a^2+b^2}}\right )}{b^2 \left (a^2+b^2\right )^{3/2}}-\frac{\tanh ^{-1}\left (\frac{b \cos (x)-a \sin (x)}{\sqrt{a^2+b^2}}\right )}{b^2 \sqrt{a^2+b^2}}+\frac{2 a}{b \left (a^2+b^2\right ) (a \cos (x)+b \sin (x))}+\frac{2 \left (a b+\left (a^2+2 b^2\right ) \tan \left (\frac{x}{2}\right )\right )}{a \left (a^2+b^2\right ) \left (a+2 b \tan \left (\frac{x}{2}\right )-a \tan ^2\left (\frac{x}{2}\right )\right )^2}-\frac{4 a^4+3 a^2 b^2+2 b^4+a b \left (5 a^2+2 b^2\right ) \tan \left (\frac{x}{2}\right )}{a b \left (a^2+b^2\right )^2 \left (a+2 b \tan \left (\frac{x}{2}\right )-a \tan ^2\left (\frac{x}{2}\right )\right )}+\frac{a^2 \operatorname{Subst}\left (\int \frac{16 \left (2 a^2-b^2\right )}{a+2 b x-a x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )}{16 b^2 \left (a^2+b^2\right )^2}\\ &=\frac{2 a^2 \tanh ^{-1}\left (\frac{b \cos (x)-a \sin (x)}{\sqrt{a^2+b^2}}\right )}{b^2 \left (a^2+b^2\right )^{3/2}}-\frac{\tanh ^{-1}\left (\frac{b \cos (x)-a \sin (x)}{\sqrt{a^2+b^2}}\right )}{b^2 \sqrt{a^2+b^2}}+\frac{2 a}{b \left (a^2+b^2\right ) (a \cos (x)+b \sin (x))}+\frac{2 \left (a b+\left (a^2+2 b^2\right ) \tan \left (\frac{x}{2}\right )\right )}{a \left (a^2+b^2\right ) \left (a+2 b \tan \left (\frac{x}{2}\right )-a \tan ^2\left (\frac{x}{2}\right )\right )^2}-\frac{4 a^4+3 a^2 b^2+2 b^4+a b \left (5 a^2+2 b^2\right ) \tan \left (\frac{x}{2}\right )}{a b \left (a^2+b^2\right )^2 \left (a+2 b \tan \left (\frac{x}{2}\right )-a \tan ^2\left (\frac{x}{2}\right )\right )}+\frac{\left (a^2 \left (2 a^2-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x-a x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )}{b^2 \left (a^2+b^2\right )^2}\\ &=\frac{2 a^2 \tanh ^{-1}\left (\frac{b \cos (x)-a \sin (x)}{\sqrt{a^2+b^2}}\right )}{b^2 \left (a^2+b^2\right )^{3/2}}-\frac{\tanh ^{-1}\left (\frac{b \cos (x)-a \sin (x)}{\sqrt{a^2+b^2}}\right )}{b^2 \sqrt{a^2+b^2}}+\frac{2 a}{b \left (a^2+b^2\right ) (a \cos (x)+b \sin (x))}+\frac{2 \left (a b+\left (a^2+2 b^2\right ) \tan \left (\frac{x}{2}\right )\right )}{a \left (a^2+b^2\right ) \left (a+2 b \tan \left (\frac{x}{2}\right )-a \tan ^2\left (\frac{x}{2}\right )\right )^2}-\frac{4 a^4+3 a^2 b^2+2 b^4+a b \left (5 a^2+2 b^2\right ) \tan \left (\frac{x}{2}\right )}{a b \left (a^2+b^2\right )^2 \left (a+2 b \tan \left (\frac{x}{2}\right )-a \tan ^2\left (\frac{x}{2}\right )\right )}-\frac{\left (2 a^2 \left (2 a^2-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{4 \left (a^2+b^2\right )-x^2} \, dx,x,2 b-2 a \tan \left (\frac{x}{2}\right )\right )}{b^2 \left (a^2+b^2\right )^2}\\ &=\frac{2 a^2 \tanh ^{-1}\left (\frac{b \cos (x)-a \sin (x)}{\sqrt{a^2+b^2}}\right )}{b^2 \left (a^2+b^2\right )^{3/2}}-\frac{\tanh ^{-1}\left (\frac{b \cos (x)-a \sin (x)}{\sqrt{a^2+b^2}}\right )}{b^2 \sqrt{a^2+b^2}}-\frac{a^2 \left (2 a^2-b^2\right ) \tanh ^{-1}\left (\frac{b-a \tan \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{b^2 \left (a^2+b^2\right )^{5/2}}+\frac{2 a}{b \left (a^2+b^2\right ) (a \cos (x)+b \sin (x))}+\frac{2 \left (a b+\left (a^2+2 b^2\right ) \tan \left (\frac{x}{2}\right )\right )}{a \left (a^2+b^2\right ) \left (a+2 b \tan \left (\frac{x}{2}\right )-a \tan ^2\left (\frac{x}{2}\right )\right )^2}-\frac{4 a^4+3 a^2 b^2+2 b^4+a b \left (5 a^2+2 b^2\right ) \tan \left (\frac{x}{2}\right )}{a b \left (a^2+b^2\right )^2 \left (a+2 b \tan \left (\frac{x}{2}\right )-a \tan ^2\left (\frac{x}{2}\right )\right )}\\ \end{align*}

Mathematica [A]  time = 0.394132, size = 92, normalized size = 1. \[ \frac{a \left (\left (a^2+4 b^2\right ) \sin (x)+3 a b \cos (x)\right )}{2 \left (a^2+b^2\right )^2 (a \cos (x)+b \sin (x))^2}-\frac{\left (a^2-2 b^2\right ) \tanh ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )-b}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]^2/(a*Cos[x] + b*Sin[x])^3,x]

[Out]

-(((a^2 - 2*b^2)*ArcTanh[(-b + a*Tan[x/2])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(5/2)) + (a*(3*a*b*Cos[x] + (a^2 + 4*
b^2)*Sin[x]))/(2*(a^2 + b^2)^2*(a*Cos[x] + b*Sin[x])^2)

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Maple [B]  time = 0.12, size = 212, normalized size = 2.3 \begin{align*} -8\,{\frac{1}{ \left ( \left ( \tan \left ( x/2 \right ) \right ) ^{2}a-2\,b\tan \left ( x/2 \right ) -a \right ) ^{2}} \left ( -1/8\,{\frac{a \left ({a}^{2}-2\,{b}^{2} \right ) \left ( \tan \left ( x/2 \right ) \right ) ^{3}}{{a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4}}}+3/8\,{\frac{b \left ({a}^{2}-2\,{b}^{2} \right ) \left ( \tan \left ( x/2 \right ) \right ) ^{2}}{{a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4}}}-1/8\,{\frac{ \left ({a}^{2}+10\,{b}^{2} \right ) a\tan \left ( x/2 \right ) }{{a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4}}}-3/8\,{\frac{{a}^{2}b}{{a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4}}} \right ) }-{\frac{{a}^{2}-2\,{b}^{2}}{{a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4}}{\it Artanh} \left ({\frac{1}{2} \left ( 2\,a\tan \left ( x/2 \right ) -2\,b \right ){\frac{1}{\sqrt{{a}^{2}+{b}^{2}}}}} \right ){\frac{1}{\sqrt{{a}^{2}+{b}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^2/(a*cos(x)+b*sin(x))^3,x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^2/(a*cos(x)+b*sin(x))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.52851, size = 656, normalized size = 7.13 \begin{align*} -\frac{{\left (a^{2} b^{2} - 2 \, b^{4} +{\left (a^{4} - 3 \, a^{2} b^{2} + 2 \, b^{4}\right )} \cos \left (x\right )^{2} + 2 \,{\left (a^{3} b - 2 \, a b^{3}\right )} \cos \left (x\right ) \sin \left (x\right )\right )} \sqrt{a^{2} + b^{2}} \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 ) - 6 \,{\left (a^{4} b + a^{2} b^{3}\right )} \cos \left (x\right ) - 2 \,{\left (a^{5} + 5 \, a^{3} b^{2} + 4 \, a b^{4}\right )} \sin \left (x\right )}{4 \,{\left (a^{6} b^{2} + 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} + b^{8} +{\left (a^{8} + 2 \, a^{6} b^{2} - 2 \, a^{2} b^{6} - b^{8}\right )} \cos \left (x\right )^{2} + 2 \,{\left (a^{7} b + 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} + a b^{7}\right )} \cos \left (x\right ) \sin \left (x\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^2/(a*cos(x)+b*sin(x))^3,x, algorithm="fricas")

[Out]

-1/4*((a^2*b^2 - 2*b^4 + (a^4 - 3*a^2*b^2 + 2*b^4)*cos(x)^2 + 2*(a^3*b - 2*a*b^3)*cos(x)*sin(x))*sqrt(a^2 + b^
2)*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)) - 6*(a^4*b + a^2*b^3)*cos(x) - 2*(a^5 + 5*a^3*b^2 + 4*a*b^4
)*sin(x))/(a^6*b^2 + 3*a^4*b^4 + 3*a^2*b^6 + b^8 + (a^8 + 2*a^6*b^2 - 2*a^2*b^6 - b^8)*cos(x)^2 + 2*(a^7*b + 3
*a^5*b^3 + 3*a^3*b^5 + a*b^7)*cos(x)*sin(x))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)**2/(a*cos(x)+b*sin(x))**3,x)

[Out]

Timed out

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Giac [B]  time = 1.24283, size = 266, normalized size = 2.89 \begin{align*} \frac{{\left (a^{2} - 2 \, b^{2}\right )} \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 )}{2 \,{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt{a^{2} + b^{2}}} + \frac{a^{3} \tan \left (\frac{1}{2} \, x\right )^{3} - 2 \, a b^{2} \tan \left (\frac{1}{2} \, x\right )^{3} - 3 \, a^{2} b \tan \left (\frac{1}{2} \, x\right )^{2} + 6 \, b^{3} \tan \left (\frac{1}{2} \, x\right )^{2} + a^{3} \tan \left (\frac{1}{2} \, x\right ) + 10 \, a b^{2} \tan \left (\frac{1}{2} \, x\right ) + 3 \, a^{2} b}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}{\left (a \tan \left (\frac{1}{2} \, x\right )^{2} - 2 \, b \tan \left (\frac{1}{2} \, x\right ) - a\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^2/(a*cos(x)+b*sin(x))^3,x, algorithm="giac")

[Out]

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