Integrand size = 10, antiderivative size = 107 \[ \int \frac {1}{\left (a+b \sin ^2(x)\right )^3} \, dx=\frac {\left (8 a^2+8 a b+3 b^2\right ) \arctan \left (\frac {\sqrt {a+b} \tan (x)}{\sqrt {a}}\right )}{8 a^{5/2} (a+b)^{5/2}}+\frac {b \cos (x) \sin (x)}{4 a (a+b) \left (a+b \sin ^2(x)\right )^2}+\frac {3 b (2 a+b) \cos (x) \sin (x)}{8 a^2 (a+b)^2 \left (a+b \sin ^2(x)\right )} \] Output:
1/8*(8*a^2+8*a*b+3*b^2)*arctan((a+b)^(1/2)*tan(x)/a^(1/2))/a^(5/2)/(a+b)^( 5/2)+1/4*b*cos(x)*sin(x)/a/(a+b)/(a+b*sin(x)^2)^2+3/8*b*(2*a+b)*cos(x)*sin (x)/a^2/(a+b)^2/(a+b*sin(x)^2)
Time = 5.62 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.99 \[ \int \frac {1}{\left (a+b \sin ^2(x)\right )^3} \, dx=\frac {\frac {\left (8 a^2+8 a b+3 b^2\right ) \arctan \left (\frac {\sqrt {a+b} \tan (x)}{\sqrt {a}}\right )}{(a+b)^{5/2}}+\frac {\sqrt {a} b \left (16 a^2+16 a b+3 b^2-3 b (2 a+b) \cos (2 x)\right ) \sin (2 x)}{(a+b)^2 (2 a+b-b \cos (2 x))^2}}{8 a^{5/2}} \] Input:
Integrate[(a + b*Sin[x]^2)^(-3),x]
Output:
(((8*a^2 + 8*a*b + 3*b^2)*ArcTan[(Sqrt[a + b]*Tan[x])/Sqrt[a]])/(a + b)^(5 /2) + (Sqrt[a]*b*(16*a^2 + 16*a*b + 3*b^2 - 3*b*(2*a + b)*Cos[2*x])*Sin[2* x])/((a + b)^2*(2*a + b - b*Cos[2*x])^2))/(8*a^(5/2))
Time = 0.46 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.12, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.900, Rules used = {3042, 3663, 25, 3042, 3652, 27, 3042, 3660, 218}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {1}{\left (a+b \sin ^2(x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\left (a+b \sin (x)^2\right )^3}dx\) |
| \(\Big \downarrow \) 3663 |
| \(\displaystyle \frac {b \sin (x) \cos (x)}{4 a (a+b) \left (a+b \sin ^2(x)\right )^2}-\frac {\int -\frac {-2 b \sin ^2(x)+4 a+3 b}{\left (b \sin ^2(x)+a\right )^2}dx}{4 a (a+b)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {-2 b \sin ^2(x)+4 a+3 b}{\left (b \sin ^2(x)+a\right )^2}dx}{4 a (a+b)}+\frac {b \sin (x) \cos (x)}{4 a (a+b) \left (a+b \sin ^2(x)\right )^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {-2 b \sin (x)^2+4 a+3 b}{\left (b \sin (x)^2+a\right )^2}dx}{4 a (a+b)}+\frac {b \sin (x) \cos (x)}{4 a (a+b) \left (a+b \sin ^2(x)\right )^2}\) |
| \(\Big \downarrow \) 3652 |
| \(\displaystyle \frac {\frac {\int \frac {8 a^2+8 b a+3 b^2}{b \sin ^2(x)+a}dx}{2 a (a+b)}+\frac {3 b (2 a+b) \sin (x) \cos (x)}{2 a (a+b) \left (a+b \sin ^2(x)\right )}}{4 a (a+b)}+\frac {b \sin (x) \cos (x)}{4 a (a+b) \left (a+b \sin ^2(x)\right )^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\left (8 a^2+8 a b+3 b^2\right ) \int \frac {1}{b \sin ^2(x)+a}dx}{2 a (a+b)}+\frac {3 b (2 a+b) \sin (x) \cos (x)}{2 a (a+b) \left (a+b \sin ^2(x)\right )}}{4 a (a+b)}+\frac {b \sin (x) \cos (x)}{4 a (a+b) \left (a+b \sin ^2(x)\right )^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\left (8 a^2+8 a b+3 b^2\right ) \int \frac {1}{b \sin (x)^2+a}dx}{2 a (a+b)}+\frac {3 b (2 a+b) \sin (x) \cos (x)}{2 a (a+b) \left (a+b \sin ^2(x)\right )}}{4 a (a+b)}+\frac {b \sin (x) \cos (x)}{4 a (a+b) \left (a+b \sin ^2(x)\right )^2}\) |
| \(\Big \downarrow \) 3660 |
| \(\displaystyle \frac {\frac {\left (8 a^2+8 a b+3 b^2\right ) \int \frac {1}{(a+b) \tan ^2(x)+a}d\tan (x)}{2 a (a+b)}+\frac {3 b (2 a+b) \sin (x) \cos (x)}{2 a (a+b) \left (a+b \sin ^2(x)\right )}}{4 a (a+b)}+\frac {b \sin (x) \cos (x)}{4 a (a+b) \left (a+b \sin ^2(x)\right )^2}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\left (8 a^2+8 a b+3 b^2\right ) \arctan \left (\frac {\sqrt {a+b} \tan (x)}{\sqrt {a}}\right )}{2 a^{3/2} (a+b)^{3/2}}+\frac {3 b (2 a+b) \sin (x) \cos (x)}{2 a (a+b) \left (a+b \sin ^2(x)\right )}}{4 a (a+b)}+\frac {b \sin (x) \cos (x)}{4 a (a+b) \left (a+b \sin ^2(x)\right )^2}\) |
Input:
Int[(a + b*Sin[x]^2)^(-3),x]
Output:
(b*Cos[x]*Sin[x])/(4*a*(a + b)*(a + b*Sin[x]^2)^2) + (((8*a^2 + 8*a*b + 3* b^2)*ArcTan[(Sqrt[a + b]*Tan[x])/Sqrt[a]])/(2*a^(3/2)*(a + b)^(3/2)) + (3* b*(2*a + b)*Cos[x]*Sin[x])/(2*a*(a + b)*(a + b*Sin[x]^2)))/(4*a*(a + b))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b - a*B))*Cos[e + f*x]*Sin[e + f*x ]*((a + b*Sin[e + f*x]^2)^(p + 1)/(2*a*f*(a + b)*(p + 1))), x] - Simp[1/(2* a*(a + b)*(p + 1)) Int[(a + b*Sin[e + f*x]^2)^(p + 1)*Simp[a*B - A*(2*a*( p + 1) + b*(2*p + 3)) + 2*(A*b - a*B)*(p + 2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B}, x] && LtQ[p, -1] && NeQ[a + b, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(-1), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Subst[Int[1/(a + (a + b)*ff^2*x^ 2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Simp[(-b)*C os[e + f*x]*Sin[e + f*x]*((a + b*Sin[e + f*x]^2)^(p + 1)/(2*a*f*(p + 1)*(a + b))), x] + Simp[1/(2*a*(p + 1)*(a + b)) Int[(a + b*Sin[e + f*x]^2)^(p + 1)*Simp[2*a*(p + 1) + b*(2*p + 3) - 2*b*(p + 2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a + b, 0] && LtQ[p, -1]
Time = 0.82 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.16
| method | result | size |
| default | \(\frac {\frac {\left (8 a +3 b \right ) b \tan \left (x \right )^{3}}{8 a^{2} \left (a +b \right )}+\frac {b \left (8 a +5 b \right ) \tan \left (x \right )}{8 a \left (a^{2}+2 b a +b^{2}\right )}}{\left (\tan \left (x \right )^{2} a +b \tan \left (x \right )^{2}+a \right )^{2}}+\frac {\left (8 a^{2}+8 b a +3 b^{2}\right ) \arctan \left (\frac {\left (a +b \right ) \tan \left (x \right )}{\sqrt {a \left (a +b \right )}}\right )}{8 a^{2} \left (a^{2}+2 b a +b^{2}\right ) \sqrt {a \left (a +b \right )}}\) | \(124\) |
| risch | \(-\frac {i \left (-8 a^{2} b \,{\mathrm e}^{6 i x}-8 a \,b^{2} {\mathrm e}^{6 i x}-3 b^{3} {\mathrm e}^{6 i x}+48 a^{3} {\mathrm e}^{4 i x}+72 a^{2} b \,{\mathrm e}^{4 i x}+42 a \,b^{2} {\mathrm e}^{4 i x}+9 b^{3} {\mathrm e}^{4 i x}-40 a^{2} b \,{\mathrm e}^{2 i x}-40 a \,b^{2} {\mathrm e}^{2 i x}-9 b^{3} {\mathrm e}^{2 i x}+6 b^{2} a +3 b^{3}\right )}{4 a^{2} \left (a +b \right )^{2} \left (-{\mathrm e}^{4 i x} b +4 \,{\mathrm e}^{2 i x} a +2 \,{\mathrm e}^{2 i x} b -b \right )^{2}}-\frac {\ln \left ({\mathrm e}^{2 i x}-\frac {2 i a^{2}+2 i a b +2 a \sqrt {-a^{2}-b a}+b \sqrt {-a^{2}-b a}}{b \sqrt {-a^{2}-b a}}\right )}{2 \sqrt {-a^{2}-b a}\, \left (a +b \right )^{2}}-\frac {\ln \left ({\mathrm e}^{2 i x}-\frac {2 i a^{2}+2 i a b +2 a \sqrt {-a^{2}-b a}+b \sqrt {-a^{2}-b a}}{b \sqrt {-a^{2}-b a}}\right ) b}{2 \sqrt {-a^{2}-b a}\, \left (a +b \right )^{2} a}-\frac {3 \ln \left ({\mathrm e}^{2 i x}-\frac {2 i a^{2}+2 i a b +2 a \sqrt {-a^{2}-b a}+b \sqrt {-a^{2}-b a}}{b \sqrt {-a^{2}-b a}}\right ) b^{2}}{16 \sqrt {-a^{2}-b a}\, \left (a +b \right )^{2} a^{2}}+\frac {\ln \left ({\mathrm e}^{2 i x}-\frac {-2 i a^{2}-2 i a b +2 a \sqrt {-a^{2}-b a}+b \sqrt {-a^{2}-b a}}{\sqrt {-a^{2}-b a}\, b}\right )}{2 \sqrt {-a^{2}-b a}\, \left (a +b \right )^{2}}+\frac {\ln \left ({\mathrm e}^{2 i x}-\frac {-2 i a^{2}-2 i a b +2 a \sqrt {-a^{2}-b a}+b \sqrt {-a^{2}-b a}}{\sqrt {-a^{2}-b a}\, b}\right ) b}{2 \sqrt {-a^{2}-b a}\, \left (a +b \right )^{2} a}+\frac {3 \ln \left ({\mathrm e}^{2 i x}-\frac {-2 i a^{2}-2 i a b +2 a \sqrt {-a^{2}-b a}+b \sqrt {-a^{2}-b a}}{\sqrt {-a^{2}-b a}\, b}\right ) b^{2}}{16 \sqrt {-a^{2}-b a}\, \left (a +b \right )^{2} a^{2}}\) | \(685\) |
Input:
int(1/(a+b*sin(x)^2)^3,x,method=_RETURNVERBOSE)
Output:
(1/8*(8*a+3*b)/a^2*b/(a+b)*tan(x)^3+1/8*b*(8*a+5*b)/a/(a^2+2*a*b+b^2)*tan( x))/(tan(x)^2*a+b*tan(x)^2+a)^2+1/8*(8*a^2+8*a*b+3*b^2)/a^2/(a^2+2*a*b+b^2 )/(a*(a+b))^(1/2)*arctan((a+b)*tan(x)/(a*(a+b))^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 326 vs. \(2 (93) = 186\).
Time = 0.12 (sec) , antiderivative size = 737, normalized size of antiderivative = 6.89 \[ \int \frac {1}{\left (a+b \sin ^2(x)\right )^3} \, dx =\text {Too large to display} \] Input:
integrate(1/(a+b*sin(x)^2)^3,x, algorithm="fricas")
Output:
[-1/32*(((8*a^2*b^2 + 8*a*b^3 + 3*b^4)*cos(x)^4 + 8*a^4 + 24*a^3*b + 27*a^ 2*b^2 + 14*a*b^3 + 3*b^4 - 2*(8*a^3*b + 16*a^2*b^2 + 11*a*b^3 + 3*b^4)*cos (x)^2)*sqrt(-a^2 - a*b)*log(((8*a^2 + 8*a*b + b^2)*cos(x)^4 - 2*(4*a^2 + 5 *a*b + b^2)*cos(x)^2 + 4*((2*a + b)*cos(x)^3 - (a + b)*cos(x))*sqrt(-a^2 - a*b)*sin(x) + a^2 + 2*a*b + b^2)/(b^2*cos(x)^4 - 2*(a*b + b^2)*cos(x)^2 + a^2 + 2*a*b + b^2)) + 4*(3*(2*a^3*b^2 + 3*a^2*b^3 + a*b^4)*cos(x)^3 - (8* a^4*b + 19*a^3*b^2 + 14*a^2*b^3 + 3*a*b^4)*cos(x))*sin(x))/(a^8 + 5*a^7*b + 10*a^6*b^2 + 10*a^5*b^3 + 5*a^4*b^4 + a^3*b^5 + (a^6*b^2 + 3*a^5*b^3 + 3 *a^4*b^4 + a^3*b^5)*cos(x)^4 - 2*(a^7*b + 4*a^6*b^2 + 6*a^5*b^3 + 4*a^4*b^ 4 + a^3*b^5)*cos(x)^2), -1/16*(((8*a^2*b^2 + 8*a*b^3 + 3*b^4)*cos(x)^4 + 8 *a^4 + 24*a^3*b + 27*a^2*b^2 + 14*a*b^3 + 3*b^4 - 2*(8*a^3*b + 16*a^2*b^2 + 11*a*b^3 + 3*b^4)*cos(x)^2)*sqrt(a^2 + a*b)*arctan(1/2*((2*a + b)*cos(x) ^2 - a - b)/(sqrt(a^2 + a*b)*cos(x)*sin(x))) + 2*(3*(2*a^3*b^2 + 3*a^2*b^3 + a*b^4)*cos(x)^3 - (8*a^4*b + 19*a^3*b^2 + 14*a^2*b^3 + 3*a*b^4)*cos(x)) *sin(x))/(a^8 + 5*a^7*b + 10*a^6*b^2 + 10*a^5*b^3 + 5*a^4*b^4 + a^3*b^5 + (a^6*b^2 + 3*a^5*b^3 + 3*a^4*b^4 + a^3*b^5)*cos(x)^4 - 2*(a^7*b + 4*a^6*b^ 2 + 6*a^5*b^3 + 4*a^4*b^4 + a^3*b^5)*cos(x)^2)]
Timed out. \[ \int \frac {1}{\left (a+b \sin ^2(x)\right )^3} \, dx=\text {Timed out} \] Input:
integrate(1/(a+b*sin(x)**2)**3,x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 188 vs. \(2 (93) = 186\).
Time = 0.12 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.76 \[ \int \frac {1}{\left (a+b \sin ^2(x)\right )^3} \, dx=\frac {{\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \arctan \left (\frac {{\left (a + b\right )} \tan \left (x\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{8 \, {\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} \sqrt {{\left (a + b\right )} a}} + \frac {{\left (8 \, a^{2} b + 11 \, a b^{2} + 3 \, b^{3}\right )} \tan \left (x\right )^{3} + {\left (8 \, a^{2} b + 5 \, a b^{2}\right )} \tan \left (x\right )}{8 \, {\left (a^{6} + 2 \, a^{5} b + a^{4} b^{2} + {\left (a^{6} + 4 \, a^{5} b + 6 \, a^{4} b^{2} + 4 \, a^{3} b^{3} + a^{2} b^{4}\right )} \tan \left (x\right )^{4} + 2 \, {\left (a^{6} + 3 \, a^{5} b + 3 \, a^{4} b^{2} + a^{3} b^{3}\right )} \tan \left (x\right )^{2}\right )}} \] Input:
integrate(1/(a+b*sin(x)^2)^3,x, algorithm="maxima")
Output:
1/8*(8*a^2 + 8*a*b + 3*b^2)*arctan((a + b)*tan(x)/sqrt((a + b)*a))/((a^4 + 2*a^3*b + a^2*b^2)*sqrt((a + b)*a)) + 1/8*((8*a^2*b + 11*a*b^2 + 3*b^3)*t an(x)^3 + (8*a^2*b + 5*a*b^2)*tan(x))/(a^6 + 2*a^5*b + a^4*b^2 + (a^6 + 4* a^5*b + 6*a^4*b^2 + 4*a^3*b^3 + a^2*b^4)*tan(x)^4 + 2*(a^6 + 3*a^5*b + 3*a ^4*b^2 + a^3*b^3)*tan(x)^2)
Time = 0.42 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.57 \[ \int \frac {1}{\left (a+b \sin ^2(x)\right )^3} \, dx=\frac {{\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac {a \tan \left (x\right ) + b \tan \left (x\right )}{\sqrt {a^{2} + a b}}\right )\right )} {\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )}}{8 \, {\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} \sqrt {a^{2} + a b}} + \frac {8 \, a^{2} b \tan \left (x\right )^{3} + 11 \, a b^{2} \tan \left (x\right )^{3} + 3 \, b^{3} \tan \left (x\right )^{3} + 8 \, a^{2} b \tan \left (x\right ) + 5 \, a b^{2} \tan \left (x\right )}{8 \, {\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} {\left (a \tan \left (x\right )^{2} + b \tan \left (x\right )^{2} + a\right )}^{2}} \] Input:
integrate(1/(a+b*sin(x)^2)^3,x, algorithm="giac")
Output:
1/8*(pi*floor(x/pi + 1/2)*sgn(2*a + 2*b) + arctan((a*tan(x) + b*tan(x))/sq rt(a^2 + a*b)))*(8*a^2 + 8*a*b + 3*b^2)/((a^4 + 2*a^3*b + a^2*b^2)*sqrt(a^ 2 + a*b)) + 1/8*(8*a^2*b*tan(x)^3 + 11*a*b^2*tan(x)^3 + 3*b^3*tan(x)^3 + 8 *a^2*b*tan(x) + 5*a*b^2*tan(x))/((a^4 + 2*a^3*b + a^2*b^2)*(a*tan(x)^2 + b *tan(x)^2 + a)^2)
Time = 37.07 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.40 \[ \int \frac {1}{\left (a+b \sin ^2(x)\right )^3} \, dx=\frac {\frac {\mathrm {tan}\left (x\right )\,\left (5\,b^2+8\,a\,b\right )}{8\,a\,\left (a^2+2\,a\,b+b^2\right )}+\frac {{\mathrm {tan}\left (x\right )}^3\,\left (3\,b^2+8\,a\,b\right )}{8\,a^2\,\left (a+b\right )}}{{\mathrm {tan}\left (x\right )}^2\,\left (2\,a^2+2\,b\,a\right )+{\mathrm {tan}\left (x\right )}^4\,\left (a^2+2\,a\,b+b^2\right )+a^2}+\frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (x\right )\,\left (2\,a+2\,b\right )\,\left (a^2+2\,a\,b+b^2\right )}{2\,\sqrt {a}\,{\left (a+b\right )}^{5/2}}\right )\,\left (8\,a^2+8\,a\,b+3\,b^2\right )}{8\,a^{5/2}\,{\left (a+b\right )}^{5/2}} \] Input:
int(1/(a + b*sin(x)^2)^3,x)
Output:
((tan(x)*(8*a*b + 5*b^2))/(8*a*(2*a*b + a^2 + b^2)) + (tan(x)^3*(8*a*b + 3 *b^2))/(8*a^2*(a + b)))/(tan(x)^2*(2*a*b + 2*a^2) + tan(x)^4*(2*a*b + a^2 + b^2) + a^2) + (atan((tan(x)*(2*a + 2*b)*(2*a*b + a^2 + b^2))/(2*a^(1/2)* (a + b)^(5/2)))*(8*a*b + 8*a^2 + 3*b^2))/(8*a^(5/2)*(a + b)^(5/2))
Time = 0.59 (sec) , antiderivative size = 3695, normalized size of antiderivative = 34.53 \[ \int \frac {1}{\left (a+b \sin ^2(x)\right )^3} \, dx =\text {Too large to display} \] Input:
int(1/(a+b*sin(x)^2)^3,x)
Output:
( - 16*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*a tan((tan(x/2)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)))*sin(x)** 4*a**2*b**2 - 16*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan((tan(x/2)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)) )*sin(x)**4*a*b**3 - 6*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan((tan(x/2)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)))*sin(x)**4*b**4 - 32*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(b)*sqr t(a + b) + a + 2*b)*atan((tan(x/2)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)))*sin(x)**2*a**3*b - 32*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt (b)*sqrt(a + b) + a + 2*b)*atan((tan(x/2)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt( a + b) + a + 2*b)))*sin(x)**2*a**2*b**2 - 12*sqrt(b)*sqrt(a)*sqrt(a + b)*s qrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan((tan(x/2)*a)/(sqrt(a)*sqrt(2*sqr t(b)*sqrt(a + b) + a + 2*b)))*sin(x)**2*a*b**3 - 16*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan((tan(x/2)*a)/(sqrt(a)*sqr t(2*sqrt(b)*sqrt(a + b) + a + 2*b)))*a**4 - 16*sqrt(b)*sqrt(a)*sqrt(a + b) *sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan((tan(x/2)*a)/(sqrt(a)*sqrt(2*s qrt(b)*sqrt(a + b) + a + 2*b)))*a**3*b - 6*sqrt(b)*sqrt(a)*sqrt(a + b)*sqr t(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan((tan(x/2)*a)/(sqrt(a)*sqrt(2*sqrt( b)*sqrt(a + b) + a + 2*b)))*a**2*b**2 + 16*sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan((tan(x/2)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + ...