Integrand size = 24, antiderivative size = 134 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 c \sin (d+e x))^3} \, dx=\frac {\left (3 a^2+c^2\right ) \log \left (a+c \tan \left (\frac {1}{2} (d+e x)\right )\right )}{16 c^5 e}-\frac {c \cos (d+e x)-a \sin (d+e x)}{16 c^2 e (a+a \cos (d+e x)+c \sin (d+e x))^2}+\frac {3 \left (a c \cos (d+e x)-a^2 \sin (d+e x)\right )}{16 c^4 e (a+a \cos (d+e x)+c \sin (d+e x))} \]
1/16*(3*a^2+c^2)*ln(a+c*tan(1/2*e*x+1/2*d))/c^5/e+1/16*(-c*cos(e*x+d)+a*si n(e*x+d))/c^2/e/(a+a*cos(e*x+d)+c*sin(e*x+d))^2+3/16*(a*c*cos(e*x+d)-a^2*s in(e*x+d))/c^4/e/(a+a*cos(e*x+d)+c*sin(e*x+d))
Time = 2.23 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.39 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 c \sin (d+e x))^3} \, dx=-\frac {4 \left (3 a^2+c^2\right ) \log \left (\cos \left (\frac {1}{2} (d+e x)\right )\right )-4 \left (3 a^2+c^2\right ) \log \left (a \cos \left (\frac {1}{2} (d+e x)\right )+c \sin \left (\frac {1}{2} (d+e x)\right )\right )-c^2 \sec ^2\left (\frac {1}{2} (d+e x)\right )+\frac {c^2 \left (a^2+c^2\right )}{\left (a \cos \left (\frac {1}{2} (d+e x)\right )+c \sin \left (\frac {1}{2} (d+e x)\right )\right )^2}+\frac {6 c \left (a^2+c^2\right ) \sin \left (\frac {1}{2} (d+e x)\right )}{a \cos \left (\frac {1}{2} (d+e x)\right )+c \sin \left (\frac {1}{2} (d+e x)\right )}+6 a c \tan \left (\frac {1}{2} (d+e x)\right )}{64 c^5 e} \]
-1/64*(4*(3*a^2 + c^2)*Log[Cos[(d + e*x)/2]] - 4*(3*a^2 + c^2)*Log[a*Cos[( d + e*x)/2] + c*Sin[(d + e*x)/2]] - c^2*Sec[(d + e*x)/2]^2 + (c^2*(a^2 + c ^2))/(a*Cos[(d + e*x)/2] + c*Sin[(d + e*x)/2])^2 + (6*c*(a^2 + c^2)*Sin[(d + e*x)/2])/(a*Cos[(d + e*x)/2] + c*Sin[(d + e*x)/2]) + 6*a*c*Tan[(d + e*x )/2])/(c^5*e)
Time = 0.50 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3608, 27, 3042, 3632, 3042, 3603, 16}
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.
| \(\displaystyle \int \frac {1}{(2 a \cos (d+e x)+2 a+2 c \sin (d+e x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(2 a \cos (d+e x)+2 a+2 c \sin (d+e x))^3}dx\) |
| \(\Big \downarrow \) 3608 |
| \(\displaystyle \frac {\int -\frac {-\cos (d+e x) a+2 a-c \sin (d+e x)}{2 (\cos (d+e x) a+a+c \sin (d+e x))^2}dx}{8 c^2}-\frac {c \cos (d+e x)-a \sin (d+e x)}{16 c^2 e (a \cos (d+e x)+a+c \sin (d+e x))^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {-\cos (d+e x) a+2 a-c \sin (d+e x)}{(\cos (d+e x) a+a+c \sin (d+e x))^2}dx}{16 c^2}-\frac {c \cos (d+e x)-a \sin (d+e x)}{16 c^2 e (a \cos (d+e x)+a+c \sin (d+e x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {-\cos (d+e x) a+2 a-c \sin (d+e x)}{(\cos (d+e x) a+a+c \sin (d+e x))^2}dx}{16 c^2}-\frac {c \cos (d+e x)-a \sin (d+e x)}{16 c^2 e (a \cos (d+e x)+a+c \sin (d+e x))^2}\) |
| \(\Big \downarrow \) 3632 |
| \(\displaystyle -\frac {-\left (\frac {3 a^2}{c^2}+1\right ) \int \frac {1}{\cos (d+e x) a+a+c \sin (d+e x)}dx-\frac {3 \left (a c \cos (d+e x)-a^2 \sin (d+e x)\right )}{c^2 e (a \cos (d+e x)+a+c \sin (d+e x))}}{16 c^2}-\frac {c \cos (d+e x)-a \sin (d+e x)}{16 c^2 e (a \cos (d+e x)+a+c \sin (d+e x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\left (\frac {3 a^2}{c^2}+1\right ) \int \frac {1}{\cos (d+e x) a+a+c \sin (d+e x)}dx-\frac {3 \left (a c \cos (d+e x)-a^2 \sin (d+e x)\right )}{c^2 e (a \cos (d+e x)+a+c \sin (d+e x))}}{16 c^2}-\frac {c \cos (d+e x)-a \sin (d+e x)}{16 c^2 e (a \cos (d+e x)+a+c \sin (d+e x))^2}\) |
| \(\Big \downarrow \) 3603 |
| \(\displaystyle -\frac {-\frac {2 \left (\frac {3 a^2}{c^2}+1\right ) \int \frac {1}{2 a+2 c \tan \left (\frac {1}{2} (d+e x)\right )}d\tan \left (\frac {1}{2} (d+e x)\right )}{e}-\frac {3 \left (a c \cos (d+e x)-a^2 \sin (d+e x)\right )}{c^2 e (a \cos (d+e x)+a+c \sin (d+e x))}}{16 c^2}-\frac {c \cos (d+e x)-a \sin (d+e x)}{16 c^2 e (a \cos (d+e x)+a+c \sin (d+e x))^2}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle -\frac {-\frac {\left (\frac {3 a^2}{c^2}+1\right ) \log \left (a+c \tan \left (\frac {1}{2} (d+e x)\right )\right )}{c e}-\frac {3 \left (a c \cos (d+e x)-a^2 \sin (d+e x)\right )}{c^2 e (a \cos (d+e x)+a+c \sin (d+e x))}}{16 c^2}-\frac {c \cos (d+e x)-a \sin (d+e x)}{16 c^2 e (a \cos (d+e x)+a+c \sin (d+e x))^2}\) |
-1/16*(c*Cos[d + e*x] - a*Sin[d + e*x])/(c^2*e*(a + a*Cos[d + e*x] + c*Sin [d + e*x])^2) - (-(((1 + (3*a^2)/c^2)*Log[a + c*Tan[(d + e*x)/2]])/(c*e)) - (3*(a*c*Cos[d + e*x] - a^2*Sin[d + e*x]))/(c^2*e*(a + a*Cos[d + e*x] + c *Sin[d + e*x])))/(16*c^2)
3.4.68.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^ (-1), x_Symbol] :> Module[{f = FreeFactors[Tan[(d + e*x)/2], x]}, Simp[2*(f /e) Subst[Int[1/(a + b + 2*c*f*x + (a - b)*f^2*x^2), x], x, Tan[(d + e*x) /2]/f], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0]
Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^ (n_), x_Symbol] :> Simp[((-c)*Cos[d + e*x] + b*Sin[d + e*x])*((a + b*Cos[d + e*x] + c*Sin[d + e*x])^(n + 1)/(e*(n + 1)*(a^2 - b^2 - c^2))), x] + Simp[ 1/((n + 1)*(a^2 - b^2 - c^2)) Int[(a*(n + 1) - b*(n + 2)*Cos[d + e*x] - c *(n + 2)*Sin[d + e*x])*(a + b*Cos[d + e*x] + c*Sin[d + e*x])^(n + 1), x], x ] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0] && LtQ[n, -1] && NeQ[n, -3/2]
Int[((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.) + (C_.)*sin[(d_.) + (e_.)*(x_)]) /((a_.) + cos[(d_.) + (e_.)*(x_)]*(b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)])^2, x_Symbol] :> Simp[(c*B - b*C - (a*C - 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] + Simp[(a*A - b*B - c*C)/(a^2 - b^2 - c^2) Int[1/(a + b*Cos[d + e*x] + c*S in[d + e*x]), x], x] /; FreeQ[{a, b, c, d, e, A, B, C}, x] && NeQ[a^2 - b^2 - c^2, 0] && NeQ[a*A - b*B - c*C, 0]
Time = 1.14 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.98
| method | result | size |
| derivativedivides | \(\frac {-\frac {-\frac {c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}{2}+3 a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{4 c^{4}}-\frac {a^{4}+2 a^{2} c^{2}+c^{4}}{8 c^{5} \left (a +c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )^{2}}+\frac {a \left (a^{2}+c^{2}\right )}{c^{5} \left (a +c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}+\frac {\left (6 a^{2}+2 c^{2}\right ) \ln \left (a +c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{4 c^{5}}}{8 e}\) | \(131\) |
| default | \(\frac {-\frac {-\frac {c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}{2}+3 a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{4 c^{4}}-\frac {a^{4}+2 a^{2} c^{2}+c^{4}}{8 c^{5} \left (a +c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )^{2}}+\frac {a \left (a^{2}+c^{2}\right )}{c^{5} \left (a +c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}+\frac {\left (6 a^{2}+2 c^{2}\right ) \ln \left (a +c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{4 c^{5}}}{8 e}\) | \(131\) |
| parallelrisch | \(\frac {12 \left (a^{2}+\frac {c^{2}}{3}\right ) \left (a +c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )^{2} \ln \left (a +c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{4} c^{4}-4 a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{3} c^{3}+8 \left (3 a^{3} c +a \,c^{3}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+18 a^{4}+6 a^{2} c^{2}-c^{4}}{64 c^{5} e \left (a +c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )^{2}}\) | \(138\) |
| norman | \(\frac {\frac {18 a^{4}+6 a^{2} c^{2}-c^{4}}{64 c^{5} e}+\frac {\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{4}}{64 c e}-\frac {a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{3}}{16 c^{2} e}+\frac {\left (3 a^{2}+c^{2}\right ) a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{8 c^{4} e}}{\left (a +c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )^{2}}+\frac {\left (3 a^{2}+c^{2}\right ) \ln \left (a +c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{16 c^{5} e}\) | \(143\) |
| risch | \(\frac {3 i a^{3} {\mathrm e}^{3 i \left (e x +d \right )}+i a \,c^{2} {\mathrm e}^{3 i \left (e x +d \right )}+9 i a^{3} {\mathrm e}^{2 i \left (e x +d \right )}+3 a^{2} c \,{\mathrm e}^{3 i \left (e x +d \right )}+3 i a \,c^{2} {\mathrm e}^{2 i \left (e x +d \right )}+c^{3} {\mathrm e}^{3 i \left (e x +d \right )}+9 i a^{3} {\mathrm e}^{i \left (e x +d \right )}-i a \,c^{2} {\mathrm e}^{i \left (e x +d \right )}+3 i a^{3}-9 a^{2} c \,{\mathrm e}^{i \left (e x +d \right )}-3 i a \,c^{2}+c^{3} {\mathrm e}^{i \left (e x +d \right )}-6 a^{2} c}{8 \left (c \,{\mathrm e}^{2 i \left (e x +d \right )}+i a \,{\mathrm e}^{2 i \left (e x +d \right )}-c +2 i a \,{\mathrm e}^{i \left (e x +d \right )}+i a \right )^{2} c^{4} e}+\frac {3 \ln \left ({\mathrm e}^{i \left (e x +d \right )}-\frac {i c +a}{i c -a}\right ) a^{2}}{16 c^{5} e}+\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}-\frac {i c +a}{i c -a}\right )}{16 c^{3} e}-\frac {3 \ln \left ({\mathrm e}^{i \left (e x +d \right )}+1\right ) a^{2}}{16 c^{5} e}-\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}+1\right )}{16 c^{3} e}\) | \(346\) |
1/8/e*(-1/4/c^4*(-1/2*c*tan(1/2*e*x+1/2*d)^2+3*a*tan(1/2*e*x+1/2*d))-1/8/c ^5*(a^4+2*a^2*c^2+c^4)/(a+c*tan(1/2*e*x+1/2*d))^2+a/c^5*(a^2+c^2)/(a+c*tan (1/2*e*x+1/2*d))+1/4*(6*a^2+2*c^2)/c^5*ln(a+c*tan(1/2*e*x+1/2*d)))
Leaf count of result is larger than twice the leaf count of optimal. 433 vs. \(2 (127) = 254\).
Time = 0.26 (sec) , antiderivative size = 433, normalized size of antiderivative = 3.23 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 c \sin (d+e x))^3} \, dx=\frac {12 \, a^{2} c^{2} \cos \left (e x + d\right )^{2} - 6 \, a^{2} c^{2} + 2 \, {\left (3 \, a^{2} c^{2} - c^{4}\right )} \cos \left (e x + d\right ) + {\left (3 \, a^{4} + 4 \, a^{2} c^{2} + c^{4} + {\left (3 \, a^{4} - 2 \, a^{2} c^{2} - c^{4}\right )} \cos \left (e x + d\right )^{2} + 2 \, {\left (3 \, a^{4} + a^{2} c^{2}\right )} \cos \left (e x + d\right ) + 2 \, {\left (3 \, a^{3} c + a c^{3} + {\left (3 \, a^{3} c + a c^{3}\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right )\right )} \log \left (a c \sin \left (e x + d\right ) + \frac {1}{2} \, a^{2} + \frac {1}{2} \, c^{2} + \frac {1}{2} \, {\left (a^{2} - c^{2}\right )} \cos \left (e x + d\right )\right ) - {\left (3 \, a^{4} + 4 \, a^{2} c^{2} + c^{4} + {\left (3 \, a^{4} - 2 \, a^{2} c^{2} - c^{4}\right )} \cos \left (e x + d\right )^{2} + 2 \, {\left (3 \, a^{4} + a^{2} c^{2}\right )} \cos \left (e x + d\right ) + 2 \, {\left (3 \, a^{3} c + a c^{3} + {\left (3 \, a^{3} c + a c^{3}\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right )\right )} \log \left (\frac {1}{2} \, \cos \left (e x + d\right ) + \frac {1}{2}\right ) - 2 \, {\left (3 \, a^{3} c - a c^{3} + 3 \, {\left (a^{3} c - a c^{3}\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right )}{32 \, {\left (2 \, a^{2} c^{5} e \cos \left (e x + d\right ) + {\left (a^{2} c^{5} - c^{7}\right )} e \cos \left (e x + d\right )^{2} + {\left (a^{2} c^{5} + c^{7}\right )} e + 2 \, {\left (a c^{6} e \cos \left (e x + d\right ) + a c^{6} e\right )} \sin \left (e x + d\right )\right )}} \]
1/32*(12*a^2*c^2*cos(e*x + d)^2 - 6*a^2*c^2 + 2*(3*a^2*c^2 - c^4)*cos(e*x + d) + (3*a^4 + 4*a^2*c^2 + c^4 + (3*a^4 - 2*a^2*c^2 - c^4)*cos(e*x + d)^2 + 2*(3*a^4 + a^2*c^2)*cos(e*x + d) + 2*(3*a^3*c + a*c^3 + (3*a^3*c + a*c^ 3)*cos(e*x + d))*sin(e*x + d))*log(a*c*sin(e*x + d) + 1/2*a^2 + 1/2*c^2 + 1/2*(a^2 - c^2)*cos(e*x + d)) - (3*a^4 + 4*a^2*c^2 + c^4 + (3*a^4 - 2*a^2* c^2 - c^4)*cos(e*x + d)^2 + 2*(3*a^4 + a^2*c^2)*cos(e*x + d) + 2*(3*a^3*c + a*c^3 + (3*a^3*c + a*c^3)*cos(e*x + d))*sin(e*x + d))*log(1/2*cos(e*x + d) + 1/2) - 2*(3*a^3*c - a*c^3 + 3*(a^3*c - a*c^3)*cos(e*x + d))*sin(e*x + d))/(2*a^2*c^5*e*cos(e*x + d) + (a^2*c^5 - c^7)*e*cos(e*x + d)^2 + (a^2*c ^5 + c^7)*e + 2*(a*c^6*e*cos(e*x + d) + a*c^6*e)*sin(e*x + d))
Timed out. \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 c \sin (d+e x))^3} \, dx=\text {Timed out} \]
Time = 0.23 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.42 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 c \sin (d+e x))^3} \, dx=\frac {\frac {7 \, a^{4} + 6 \, a^{2} c^{2} - c^{4} + \frac {8 \, {\left (a^{3} c + a c^{3}\right )} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}}{a^{2} c^{5} + \frac {2 \, a c^{6} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + \frac {c^{7} \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}}} - \frac {\frac {6 \, a \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} - \frac {c \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}}}{c^{4}} + \frac {4 \, {\left (3 \, a^{2} + c^{2}\right )} \log \left (a + \frac {c \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}\right )}{c^{5}}}{64 \, e} \]
1/64*((7*a^4 + 6*a^2*c^2 - c^4 + 8*(a^3*c + a*c^3)*sin(e*x + d)/(cos(e*x + d) + 1))/(a^2*c^5 + 2*a*c^6*sin(e*x + d)/(cos(e*x + d) + 1) + c^7*sin(e*x + d)^2/(cos(e*x + d) + 1)^2) - (6*a*sin(e*x + d)/(cos(e*x + d) + 1) - c*s in(e*x + d)^2/(cos(e*x + d) + 1)^2)/c^4 + 4*(3*a^2 + c^2)*log(a + c*sin(e* x + d)/(cos(e*x + d) + 1))/c^5)/e
Time = 0.29 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.22 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 c \sin (d+e x))^3} \, dx=\frac {\frac {4 \, {\left (3 \, a^{2} + c^{2}\right )} \log \left ({\left | c \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + a \right |}\right )}{c^{5}} + \frac {c^{3} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} - 6 \, a c^{2} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )}{c^{6}} - \frac {18 \, a^{2} c^{2} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} + 6 \, c^{4} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} + 28 \, a^{3} c \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + 4 \, a c^{3} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + 11 \, a^{4} + c^{4}}{{\left (c \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + a\right )}^{2} c^{5}}}{64 \, e} \]
1/64*(4*(3*a^2 + c^2)*log(abs(c*tan(1/2*e*x + 1/2*d) + a))/c^5 + (c^3*tan( 1/2*e*x + 1/2*d)^2 - 6*a*c^2*tan(1/2*e*x + 1/2*d))/c^6 - (18*a^2*c^2*tan(1 /2*e*x + 1/2*d)^2 + 6*c^4*tan(1/2*e*x + 1/2*d)^2 + 28*a^3*c*tan(1/2*e*x + 1/2*d) + 4*a*c^3*tan(1/2*e*x + 1/2*d) + 11*a^4 + c^4)/((c*tan(1/2*e*x + 1/ 2*d) + a)^2*c^5))/e
Time = 26.35 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.21 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 c \sin (d+e x))^3} \, dx=\frac {{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2}{64\,c^3\,e}+\frac {\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (4\,a^3+4\,a\,c^2\right )+\frac {7\,a^4+6\,a^2\,c^2-c^4}{2\,c}}{e\,\left (32\,a^2\,c^4+64\,a\,c^5\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )+32\,c^6\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2\right )}-\frac {3\,a\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}{32\,c^4\,e}+\frac {\ln \left (a+c\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\right )\,\left (3\,a^2+c^2\right )}{16\,c^5\,e} \]