Integrand size = 15, antiderivative size = 75 \[ \int (a+a \cos (x))^3 (A+B \sec (x)) \, dx=\frac {1}{2} a^3 (5 A+7 B) x+a^3 B \text {arctanh}(\sin (x))+\frac {5}{2} a^3 (A+B) \sin (x)+\frac {1}{3} a A (a+a \cos (x))^2 \sin (x)+\frac {1}{6} (5 A+3 B) \left (a^3+a^3 \cos (x)\right ) \sin (x) \]
1/2*a^3*(5*A+7*B)*x+a^3*B*arctanh(sin(x))+5/2*a^3*(A+B)*sin(x)+1/3*a*A*(a+ a*cos(x))^2*sin(x)+1/6*(5*A+3*B)*(a^3+a^3*cos(x))*sin(x)
Time = 0.41 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.07 \[ \int (a+a \cos (x))^3 (A+B \sec (x)) \, dx=\frac {1}{12} a^3 \left (30 A x+42 B x-12 B \log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )+12 B \log \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )+9 (5 A+4 B) \sin (x)+3 (3 A+B) \sin (2 x)+A \sin (3 x)\right ) \]
(a^3*(30*A*x + 42*B*x - 12*B*Log[Cos[x/2] - Sin[x/2]] + 12*B*Log[Cos[x/2] + Sin[x/2]] + 9*(5*A + 4*B)*Sin[x] + 3*(3*A + B)*Sin[2*x] + A*Sin[3*x]))/1 2
Time = 0.82 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.08, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {3042, 3307, 3042, 3455, 3042, 3455, 27, 3042, 3447, 3042, 3502, 3042, 3214, 3042, 4257}
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.
| \(\displaystyle \int (a \cos (x)+a)^3 (A+B \sec (x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (a \sin \left (x+\frac {\pi }{2}\right )+a\right )^3 \left (A+B \csc \left (x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 3307 |
| \(\displaystyle \int \sec (x) (a \cos (x)+a)^3 (A \cos (x)+B)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (x+\frac {\pi }{2}\right )+a\right )^3 \left (A \sin \left (x+\frac {\pi }{2}\right )+B\right )}{\sin \left (x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {1}{3} \int (\cos (x) a+a)^2 (3 a B+a (5 A+3 B) \cos (x)) \sec (x)dx+\frac {1}{3} a A \sin (x) (a \cos (x)+a)^2\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \int \frac {\left (\sin \left (x+\frac {\pi }{2}\right ) a+a\right )^2 \left (3 a B+a (5 A+3 B) \sin \left (x+\frac {\pi }{2}\right )\right )}{\sin \left (x+\frac {\pi }{2}\right )}dx+\frac {1}{3} a A \sin (x) (a \cos (x)+a)^2\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int 3 (\cos (x) a+a) \left (2 B a^2+5 (A+B) \cos (x) a^2\right ) \sec (x)dx+\frac {1}{2} (5 A+3 B) \sin (x) \left (a^3 \cos (x)+a^3\right )\right )+\frac {1}{3} a A \sin (x) (a \cos (x)+a)^2\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (\frac {3}{2} \int (\cos (x) a+a) \left (2 B a^2+5 (A+B) \cos (x) a^2\right ) \sec (x)dx+\frac {1}{2} (5 A+3 B) \sin (x) \left (a^3 \cos (x)+a^3\right )\right )+\frac {1}{3} a A \sin (x) (a \cos (x)+a)^2\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\frac {3}{2} \int \frac {\left (\sin \left (x+\frac {\pi }{2}\right ) a+a\right ) \left (2 B a^2+5 (A+B) \sin \left (x+\frac {\pi }{2}\right ) a^2\right )}{\sin \left (x+\frac {\pi }{2}\right )}dx+\frac {1}{2} (5 A+3 B) \sin (x) \left (a^3 \cos (x)+a^3\right )\right )+\frac {1}{3} a A \sin (x) (a \cos (x)+a)^2\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {1}{3} \left (\frac {3}{2} \int \left (5 (A+B) \cos ^2(x) a^3+2 B a^3+\left (2 B a^3+5 (A+B) a^3\right ) \cos (x)\right ) \sec (x)dx+\frac {1}{2} (5 A+3 B) \sin (x) \left (a^3 \cos (x)+a^3\right )\right )+\frac {1}{3} a A \sin (x) (a \cos (x)+a)^2\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\frac {3}{2} \int \frac {5 (A+B) \sin \left (x+\frac {\pi }{2}\right )^2 a^3+2 B a^3+\left (2 B a^3+5 (A+B) a^3\right ) \sin \left (x+\frac {\pi }{2}\right )}{\sin \left (x+\frac {\pi }{2}\right )}dx+\frac {1}{2} (5 A+3 B) \sin (x) \left (a^3 \cos (x)+a^3\right )\right )+\frac {1}{3} a A \sin (x) (a \cos (x)+a)^2\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {1}{3} \left (\frac {3}{2} \left (\int \left (2 B a^3+(5 A+7 B) \cos (x) a^3\right ) \sec (x)dx+5 a^3 (A+B) \sin (x)\right )+\frac {1}{2} (5 A+3 B) \sin (x) \left (a^3 \cos (x)+a^3\right )\right )+\frac {1}{3} a A \sin (x) (a \cos (x)+a)^2\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\frac {3}{2} \left (\int \frac {2 B a^3+(5 A+7 B) \sin \left (x+\frac {\pi }{2}\right ) a^3}{\sin \left (x+\frac {\pi }{2}\right )}dx+5 a^3 (A+B) \sin (x)\right )+\frac {1}{2} (5 A+3 B) \sin (x) \left (a^3 \cos (x)+a^3\right )\right )+\frac {1}{3} a A \sin (x) (a \cos (x)+a)^2\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {1}{3} \left (\frac {3}{2} \left (2 a^3 B \int \sec (x)dx+a^3 x (5 A+7 B)+5 a^3 (A+B) \sin (x)\right )+\frac {1}{2} (5 A+3 B) \sin (x) \left (a^3 \cos (x)+a^3\right )\right )+\frac {1}{3} a A \sin (x) (a \cos (x)+a)^2\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\frac {3}{2} \left (2 a^3 B \int \csc \left (x+\frac {\pi }{2}\right )dx+a^3 x (5 A+7 B)+5 a^3 (A+B) \sin (x)\right )+\frac {1}{2} (5 A+3 B) \sin (x) \left (a^3 \cos (x)+a^3\right )\right )+\frac {1}{3} a A \sin (x) (a \cos (x)+a)^2\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{3} \left (\frac {3}{2} \left (a^3 x (5 A+7 B)+5 a^3 (A+B) \sin (x)+2 a^3 B \text {arctanh}(\sin (x))\right )+\frac {1}{2} (5 A+3 B) \sin (x) \left (a^3 \cos (x)+a^3\right )\right )+\frac {1}{3} a A \sin (x) (a \cos (x)+a)^2\) |
(a*A*(a + a*Cos[x])^2*Sin[x])/3 + (((5*A + 3*B)*(a^3 + a^3*Cos[x])*Sin[x]) /2 + (3*(a^3*(5*A + 7*B)*x + 2*a^3*B*ArcTanh[Sin[x]] + 5*a^3*(A + B)*Sin[x ]))/2)/3
3.2.87.3.1 Defintions of rubi rules used
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_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d Int[1/(c + d *Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*((d + c*Sin[e + f*x])^n/Sin[e + f*x]^n), x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IntegerQ [n]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(-b)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1 ) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && !LtQ[n, -1 ] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 1.50 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.89
| method | result | size |
| parallelrisch | \(\frac {5 \left (-\frac {2 B \ln \left (-\cot \left (x \right )+\csc \left (x \right )-1\right )}{5}+\frac {2 B \ln \left (\csc \left (x \right )-\cot \left (x \right )+1\right )}{5}+\left (\frac {3 A}{10}+\frac {B}{10}\right ) \sin \left (2 x \right )+\frac {A \sin \left (3 x \right )}{30}+\left (\frac {3 A}{2}+\frac {6 B}{5}\right ) \sin \left (x \right )+x \left (A +\frac {7 B}{5}\right )\right ) a^{3}}{2}\) | \(67\) |
| default | \(\frac {a^{3} A \left (2+\cos \left (x \right )^{2}\right ) \sin \left (x \right )}{3}+a^{3} B \left (\frac {\cos \left (x \right ) \sin \left (x \right )}{2}+\frac {x}{2}\right )+3 a^{3} A \left (\frac {\cos \left (x \right ) \sin \left (x \right )}{2}+\frac {x}{2}\right )+3 a^{3} B \sin \left (x \right )+3 a^{3} A \sin \left (x \right )+3 a^{3} B x +a^{3} A x +a^{3} B \ln \left (\sec \left (x \right )+\tan \left (x \right )\right )\) | \(87\) |
| parts | \(\frac {a^{3} A \left (2+\cos \left (x \right )^{2}\right ) \sin \left (x \right )}{3}+a^{3} B \left (\frac {\cos \left (x \right ) \sin \left (x \right )}{2}+\frac {x}{2}\right )+3 a^{3} A \left (\frac {\cos \left (x \right ) \sin \left (x \right )}{2}+\frac {x}{2}\right )+3 a^{3} B \sin \left (x \right )+3 a^{3} A \sin \left (x \right )+3 a^{3} B x +a^{3} A x +a^{3} B \ln \left (\sec \left (x \right )+\tan \left (x \right )\right )\) | \(87\) |
| risch | \(\frac {5 a^{3} A x}{2}+\frac {7 a^{3} B x}{2}-\frac {15 i A \,{\mathrm e}^{i x} a^{3}}{8}-\frac {3 i B \,{\mathrm e}^{i x} a^{3}}{2}+\frac {15 i A \,{\mathrm e}^{-i x} a^{3}}{8}+\frac {3 i B \,{\mathrm e}^{-i x} a^{3}}{2}+a^{3} B \ln \left (i+{\mathrm e}^{i x}\right )-a^{3} B \ln \left ({\mathrm e}^{i x}-i\right )+\frac {a^{3} A \sin \left (3 x \right )}{12}+\frac {3 A \sin \left (2 x \right ) a^{3}}{4}+\frac {B \sin \left (2 x \right ) a^{3}}{4}\) | \(123\) |
| norman | \(\frac {\left (\frac {5}{2} a^{3} A +\frac {7}{2} a^{3} B \right ) x +\left (\frac {40}{3} a^{3} A +12 a^{3} B \right ) \tan \left (\frac {x}{2}\right )^{3}+\left (5 a^{3} A +5 a^{3} B \right ) \tan \left (\frac {x}{2}\right )^{5}+\left (11 a^{3} A +7 a^{3} B \right ) \tan \left (\frac {x}{2}\right )+\left (\frac {5}{2} a^{3} A +\frac {7}{2} a^{3} B \right ) x \tan \left (\frac {x}{2}\right )^{6}+\left (\frac {15}{2} a^{3} A +\frac {21}{2} a^{3} B \right ) x \tan \left (\frac {x}{2}\right )^{2}+\left (\frac {15}{2} a^{3} A +\frac {21}{2} a^{3} B \right ) x \tan \left (\frac {x}{2}\right )^{4}}{\left (1+\tan \left (\frac {x}{2}\right )^{2}\right )^{3}}+a^{3} B \ln \left (\tan \left (\frac {x}{2}\right )+1\right )-a^{3} B \ln \left (\tan \left (\frac {x}{2}\right )-1\right )\) | \(175\) |
5/2*(-2/5*B*ln(-cot(x)+csc(x)-1)+2/5*B*ln(csc(x)-cot(x)+1)+(3/10*A+1/10*B) *sin(2*x)+1/30*A*sin(3*x)+(3/2*A+6/5*B)*sin(x)+x*(A+7/5*B))*a^3
Time = 0.27 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.03 \[ \int (a+a \cos (x))^3 (A+B \sec (x)) \, dx=\frac {1}{2} \, {\left (5 \, A + 7 \, B\right )} a^{3} x + \frac {1}{2} \, B a^{3} \log \left (\sin \left (x\right ) + 1\right ) - \frac {1}{2} \, B a^{3} \log \left (-\sin \left (x\right ) + 1\right ) + \frac {1}{6} \, {\left (2 \, A a^{3} \cos \left (x\right )^{2} + 3 \, {\left (3 \, A + B\right )} a^{3} \cos \left (x\right ) + 2 \, {\left (11 \, A + 9 \, B\right )} a^{3}\right )} \sin \left (x\right ) \]
1/2*(5*A + 7*B)*a^3*x + 1/2*B*a^3*log(sin(x) + 1) - 1/2*B*a^3*log(-sin(x) + 1) + 1/6*(2*A*a^3*cos(x)^2 + 3*(3*A + B)*a^3*cos(x) + 2*(11*A + 9*B)*a^3 )*sin(x)
Time = 2.88 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.23 \[ \int (a+a \cos (x))^3 (A+B \sec (x)) \, dx=\frac {5 A a^{3} x}{2} - \frac {A a^{3} \sin ^{3}{\left (x \right )}}{3} + 4 A a^{3} \sin {\left (x \right )} + \frac {3 A a^{3} \sin {\left (2 x \right )}}{4} + \frac {7 B a^{3} x}{2} + B a^{3} \log {\left (\tan {\left (x \right )} + \sec {\left (x \right )} \right )} + \frac {B a^{3} \sin {\left (x \right )} \cos {\left (x \right )}}{2} + 3 B a^{3} \sin {\left (x \right )} \]
5*A*a**3*x/2 - A*a**3*sin(x)**3/3 + 4*A*a**3*sin(x) + 3*A*a**3*sin(2*x)/4 + 7*B*a**3*x/2 + B*a**3*log(tan(x) + sec(x)) + B*a**3*sin(x)*cos(x)/2 + 3* B*a**3*sin(x)
Time = 0.20 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.12 \[ \int (a+a \cos (x))^3 (A+B \sec (x)) \, dx=-\frac {1}{3} \, {\left (\sin \left (x\right )^{3} - 3 \, \sin \left (x\right )\right )} A a^{3} + \frac {3}{4} \, A a^{3} {\left (2 \, x + \sin \left (2 \, x\right )\right )} + \frac {1}{4} \, B a^{3} {\left (2 \, x + \sin \left (2 \, x\right )\right )} + A a^{3} x + 3 \, B a^{3} x + B a^{3} \log \left (\sec \left (x\right ) + \tan \left (x\right )\right ) + 3 \, A a^{3} \sin \left (x\right ) + 3 \, B a^{3} \sin \left (x\right ) \]
-1/3*(sin(x)^3 - 3*sin(x))*A*a^3 + 3/4*A*a^3*(2*x + sin(2*x)) + 1/4*B*a^3* (2*x + sin(2*x)) + A*a^3*x + 3*B*a^3*x + B*a^3*log(sec(x) + tan(x)) + 3*A* a^3*sin(x) + 3*B*a^3*sin(x)
Time = 0.29 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.67 \[ \int (a+a \cos (x))^3 (A+B \sec (x)) \, dx=B a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) + 1 \right |}\right ) - B a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) - 1 \right |}\right ) + \frac {1}{2} \, {\left (5 \, A a^{3} + 7 \, B a^{3}\right )} x + \frac {15 \, A a^{3} \tan \left (\frac {1}{2} \, x\right )^{5} + 15 \, B a^{3} \tan \left (\frac {1}{2} \, x\right )^{5} + 40 \, A a^{3} \tan \left (\frac {1}{2} \, x\right )^{3} + 36 \, B a^{3} \tan \left (\frac {1}{2} \, x\right )^{3} + 33 \, A a^{3} \tan \left (\frac {1}{2} \, x\right ) + 21 \, B a^{3} \tan \left (\frac {1}{2} \, x\right )}{3 \, {\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}^{3}} \]
B*a^3*log(abs(tan(1/2*x) + 1)) - B*a^3*log(abs(tan(1/2*x) - 1)) + 1/2*(5*A *a^3 + 7*B*a^3)*x + 1/3*(15*A*a^3*tan(1/2*x)^5 + 15*B*a^3*tan(1/2*x)^5 + 4 0*A*a^3*tan(1/2*x)^3 + 36*B*a^3*tan(1/2*x)^3 + 33*A*a^3*tan(1/2*x) + 21*B* a^3*tan(1/2*x))/(tan(1/2*x)^2 + 1)^3
Time = 28.29 (sec) , antiderivative size = 431, normalized size of antiderivative = 5.75 \[ \int (a+a \cos (x))^3 (A+B \sec (x)) \, dx=\frac {\left (5\,A\,a^3+5\,B\,a^3\right )\,{\mathrm {tan}\left (\frac {x}{2}\right )}^5+\left (\frac {40\,A\,a^3}{3}+12\,B\,a^3\right )\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3+\left (11\,A\,a^3+7\,B\,a^3\right )\,\mathrm {tan}\left (\frac {x}{2}\right )}{{\mathrm {tan}\left (\frac {x}{2}\right )}^6+3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+1}+a^3\,\mathrm {atan}\left (\frac {1000\,A^3\,a^9\,\mathrm {tan}\left (\frac {x}{2}\right )}{1000\,A^3\,a^9+4200\,A^2\,B\,a^9+6040\,A\,B^2\,a^9+2968\,B^3\,a^9}+\frac {2968\,B^3\,a^9\,\mathrm {tan}\left (\frac {x}{2}\right )}{1000\,A^3\,a^9+4200\,A^2\,B\,a^9+6040\,A\,B^2\,a^9+2968\,B^3\,a^9}+\frac {6040\,A\,B^2\,a^9\,\mathrm {tan}\left (\frac {x}{2}\right )}{1000\,A^3\,a^9+4200\,A^2\,B\,a^9+6040\,A\,B^2\,a^9+2968\,B^3\,a^9}+\frac {4200\,A^2\,B\,a^9\,\mathrm {tan}\left (\frac {x}{2}\right )}{1000\,A^3\,a^9+4200\,A^2\,B\,a^9+6040\,A\,B^2\,a^9+2968\,B^3\,a^9}\right )\,\left (5\,A+7\,B\right )+2\,B\,a^3\,\mathrm {atanh}\left (\frac {848\,B^3\,a^9\,\mathrm {tan}\left (\frac {x}{2}\right )}{400\,A^2\,B\,a^9+1120\,A\,B^2\,a^9+848\,B^3\,a^9}+\frac {1120\,A\,B^2\,a^9\,\mathrm {tan}\left (\frac {x}{2}\right )}{400\,A^2\,B\,a^9+1120\,A\,B^2\,a^9+848\,B^3\,a^9}+\frac {400\,A^2\,B\,a^9\,\mathrm {tan}\left (\frac {x}{2}\right )}{400\,A^2\,B\,a^9+1120\,A\,B^2\,a^9+848\,B^3\,a^9}\right ) \]
(tan(x/2)^5*(5*A*a^3 + 5*B*a^3) + tan(x/2)^3*((40*A*a^3)/3 + 12*B*a^3) + t an(x/2)*(11*A*a^3 + 7*B*a^3))/(3*tan(x/2)^2 + 3*tan(x/2)^4 + tan(x/2)^6 + 1) + a^3*atan((1000*A^3*a^9*tan(x/2))/(1000*A^3*a^9 + 2968*B^3*a^9 + 6040* A*B^2*a^9 + 4200*A^2*B*a^9) + (2968*B^3*a^9*tan(x/2))/(1000*A^3*a^9 + 2968 *B^3*a^9 + 6040*A*B^2*a^9 + 4200*A^2*B*a^9) + (6040*A*B^2*a^9*tan(x/2))/(1 000*A^3*a^9 + 2968*B^3*a^9 + 6040*A*B^2*a^9 + 4200*A^2*B*a^9) + (4200*A^2* B*a^9*tan(x/2))/(1000*A^3*a^9 + 2968*B^3*a^9 + 6040*A*B^2*a^9 + 4200*A^2*B *a^9))*(5*A + 7*B) + 2*B*a^3*atanh((848*B^3*a^9*tan(x/2))/(848*B^3*a^9 + 1 120*A*B^2*a^9 + 400*A^2*B*a^9) + (1120*A*B^2*a^9*tan(x/2))/(848*B^3*a^9 + 1120*A*B^2*a^9 + 400*A^2*B*a^9) + (400*A^2*B*a^9*tan(x/2))/(848*B^3*a^9 + 1120*A*B^2*a^9 + 400*A^2*B*a^9))