Integrand size = 24, antiderivative size = 134 \[ \int \frac {1}{(2 a-2 a \cos (d+e x)+2 c \sin (d+e x))^3} \, dx=-\frac {\left (3 a^2+c^2\right ) \log \left (a+c \cot \left (\frac {1}{2} (d+e x)\right )\right )}{16 c^5 e}-\frac {c \cos (d+e x)+a \sin (d+e x)}{16 c^2 e (a-a \cos (d+e x)+c \sin (d+e x))^2}+\frac {3 \left (a c \cos (d+e x)+a^2 \sin (d+e x)\right )}{16 c^4 e (a-a \cos (d+e x)+c \sin (d+e x))} \]
-1/16*(3*a^2+c^2)*ln(a+c*cot(1/2*e*x+1/2*d))/c^5/e+1/16*(-c*cos(e*x+d)-a*s in(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* sin(e*x+d))/c^4/e/(a-a*cos(e*x+d)+c*sin(e*x+d))
Result contains complex when optimal does not.
Time = 7.66 (sec) , antiderivative size = 350, normalized size of antiderivative = 2.61 \[ \int \frac {1}{(2 a-2 a \cos (d+e x)+2 c \sin (d+e x))^3} \, dx=\frac {\sin \left (\frac {1}{2} (d+e x)\right ) \left (c \cos \left (\frac {1}{2} (d+e x)\right )+a \sin \left (\frac {1}{2} (d+e x)\right )\right ) \left (c^2 (-i a+c) (i a+c) \sin ^2\left (\frac {1}{2} (d+e x)\right )-6 a \left (a^2+c^2\right ) \sin ^3\left (\frac {1}{2} (d+e x)\right ) \left (c \cos \left (\frac {1}{2} (d+e x)\right )+a \sin \left (\frac {1}{2} (d+e x)\right )\right )-c^2 \left (c \cos \left (\frac {1}{2} (d+e x)\right )+a \sin \left (\frac {1}{2} (d+e x)\right )\right )^2+4 \left (3 a^2+c^2\right ) \log \left (\sin \left (\frac {1}{2} (d+e x)\right )\right ) \sin ^2\left (\frac {1}{2} (d+e x)\right ) \left (c \cos \left (\frac {1}{2} (d+e x)\right )+a \sin \left (\frac {1}{2} (d+e x)\right )\right )^2-4 \left (3 a^2+c^2\right ) \log \left (c \cos \left (\frac {1}{2} (d+e x)\right )+a \sin \left (\frac {1}{2} (d+e x)\right )\right ) \sin ^2\left (\frac {1}{2} (d+e x)\right ) \left (c \cos \left (\frac {1}{2} (d+e x)\right )+a \sin \left (\frac {1}{2} (d+e x)\right )\right )^2+3 a c \left (c \cos \left (\frac {1}{2} (d+e x)\right )+a \sin \left (\frac {1}{2} (d+e x)\right )\right )^2 \sin (d+e x)\right )}{8 c^5 e (a-a \cos (d+e x)+c \sin (d+e x))^3} \]
(Sin[(d + e*x)/2]*(c*Cos[(d + e*x)/2] + a*Sin[(d + e*x)/2])*(c^2*((-I)*a + c)*(I*a + c)*Sin[(d + e*x)/2]^2 - 6*a*(a^2 + c^2)*Sin[(d + e*x)/2]^3*(c*C os[(d + e*x)/2] + a*Sin[(d + e*x)/2]) - c^2*(c*Cos[(d + e*x)/2] + a*Sin[(d + e*x)/2])^2 + 4*(3*a^2 + c^2)*Log[Sin[(d + e*x)/2]]*Sin[(d + e*x)/2]^2*( c*Cos[(d + e*x)/2] + a*Sin[(d + e*x)/2])^2 - 4*(3*a^2 + c^2)*Log[c*Cos[(d + e*x)/2] + a*Sin[(d + e*x)/2]]*Sin[(d + e*x)/2]^2*(c*Cos[(d + e*x)/2] + a *Sin[(d + e*x)/2])^2 + 3*a*c*(c*Cos[(d + e*x)/2] + a*Sin[(d + e*x)/2])^2*S in[d + e*x]))/(8*c^5*e*(a - a*Cos[d + e*x] + c*Sin[d + e*x])^3)
Time = 0.51 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.03, 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, 3600, 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 |
| \(\displaystyle \frac {\int -\frac {\cos (d+e x) a+2 a-c \sin (d+e x)}{2 (-\cos (d+e x) a+a+c \sin (d+e x))^2}dx}{8 c^2}-\frac {a \sin (d+e x)+c \cos (d+e x)}{16 c^2 e (a (-\cos (d+e x))+a+c \sin (d+e x))^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {\cos (d+e x) a+2 a-c \sin (d+e x)}{(-\cos (d+e x) a+a+c \sin (d+e x))^2}dx}{16 c^2}-\frac {a \sin (d+e x)+c \cos (d+e x)}{16 c^2 e (a (-\cos (d+e x))+a+c \sin (d+e x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {\cos (d+e x) a+2 a-c \sin (d+e x)}{(-\cos (d+e x) a+a+c \sin (d+e x))^2}dx}{16 c^2}-\frac {a \sin (d+e x)+c \cos (d+e x)}{16 c^2 e (a (-\cos (d+e x))+a+c \sin (d+e x))^2}\) |
| \(\Big \downarrow \) 3632 |
| \(\displaystyle -\frac {-\left (\frac {3 a^2}{c^2}+1\right ) \int \frac {1}{-\cos (d+e x) a+a+c \sin (d+e x)}dx-\frac {3 \left (a^2 \sin (d+e x)+a c \cos (d+e x)\right )}{c^2 e (a (-\cos (d+e x))+a+c \sin (d+e x))}}{16 c^2}-\frac {a \sin (d+e x)+c \cos (d+e x)}{16 c^2 e (a (-\cos (d+e x))+a+c \sin (d+e x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\left (\frac {3 a^2}{c^2}+1\right ) \int \frac {1}{-\cos (d+e x) a+a+c \sin (d+e x)}dx-\frac {3 \left (a^2 \sin (d+e x)+a c \cos (d+e x)\right )}{c^2 e (a (-\cos (d+e x))+a+c \sin (d+e x))}}{16 c^2}-\frac {a \sin (d+e x)+c \cos (d+e x)}{16 c^2 e (a (-\cos (d+e x))+a+c \sin (d+e x))^2}\) |
| \(\Big \downarrow \) 3600 |
| \(\displaystyle -\frac {\frac {\left (\frac {3 a^2}{c^2}+1\right ) \int \frac {1}{a+c \cot \left (\frac {1}{2} (d+e x)\right )}d\cot \left (\frac {1}{2} (d+e x)\right )}{e}-\frac {3 \left (a^2 \sin (d+e x)+a c \cos (d+e x)\right )}{c^2 e (a (-\cos (d+e x))+a+c \sin (d+e x))}}{16 c^2}-\frac {a \sin (d+e x)+c \cos (d+e x)}{16 c^2 e (a (-\cos (d+e x))+a+c \sin (d+e x))^2}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle -\frac {\frac {\left (\frac {3 a^2}{c^2}+1\right ) \log \left (a+c \cot \left (\frac {1}{2} (d+e x)\right )\right )}{c e}-\frac {3 \left (a^2 \sin (d+e x)+a c \cos (d+e x)\right )}{c^2 e (a (-\cos (d+e x))+a+c \sin (d+e x))}}{16 c^2}-\frac {a \sin (d+e x)+c \cos (d+e x)}{16 c^2 e (a (-\cos (d+e x))+a+c \sin (d+e x))^2}\) |
-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*Cot[(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*Si n[d + e*x])))/(16*c^2)
3.4.79.3.1 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[Cot[(d + e*x)/2], x]}, Simp[-f/e Subst[Int[1/(a + c*f*x), x], x, Cot[(d + e*x)/2]/f], x]] /; FreeQ[{a, b, c, d, e}, x] && EqQ[a + b, 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 = 1.18 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.31
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{8 c^{3} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}+\frac {\left (6 a^{2}+2 c^{2}\right ) \ln \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{4 c^{5}}+\frac {3 a}{4 c^{4} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}-\frac {-3 a^{4}-2 a^{2} c^{2}+c^{4}}{4 a^{2} c^{4} \left (c +a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}-\frac {-a^{4}-2 a^{2} c^{2}-c^{4}}{8 a^{2} c^{3} \left (c +a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )^{2}}-\frac {\left (3 a^{2}+c^{2}\right ) \ln \left (c +a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{2 c^{5}}}{8 e}\) | \(176\) |
| default | \(\frac {-\frac {1}{8 c^{3} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}+\frac {\left (6 a^{2}+2 c^{2}\right ) \ln \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{4 c^{5}}+\frac {3 a}{4 c^{4} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}-\frac {-3 a^{4}-2 a^{2} c^{2}+c^{4}}{4 a^{2} c^{4} \left (c +a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}-\frac {-a^{4}-2 a^{2} c^{2}-c^{4}}{8 a^{2} c^{3} \left (c +a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )^{2}}-\frac {\left (3 a^{2}+c^{2}\right ) \ln \left (c +a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{2 c^{5}}}{8 e}\) | \(176\) |
| norman | \(\frac {-\frac {1}{64 c e}+\frac {a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{16 c^{2} e}-\frac {\left (18 a^{4}+6 a^{2} c^{2}-c^{4}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{4}}{64 c^{5} e}-\frac {\left (3 a^{2}+c^{2}\right ) a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{3}}{8 c^{4} e}}{\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2} \left (c +a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )^{2}}+\frac {\left (3 a^{2}+c^{2}\right ) \ln \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{16 c^{5} e}-\frac {\left (3 a^{2}+c^{2}\right ) \ln \left (c +a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{16 c^{5} e}\) | \(181\) |
| parallelrisch | \(\frac {-12 \left (c +a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )^{2} \left (a^{2}+\frac {c^{2}}{3}\right ) \ln \left (c +a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )+12 \left (c +a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )^{2} \left (a^{2}+\frac {c^{2}}{3}\right ) \ln \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )+\left (-18 a^{4}-6 a^{2} c^{2}+c^{4}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}+\left (-24 a^{3} c -8 a \,c^{3}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-\cot \left (\frac {e x}{2}+\frac {d}{2}\right )^{2} c^{4}+4 \cot \left (\frac {e x}{2}+\frac {d}{2}\right ) a \,c^{3}}{64 c^{5} e \left (c +a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )^{2}}\) | \(184\) |
| risch | \(\frac {-3 i a^{3} {\mathrm e}^{3 i \left (e x +d \right )}-i a \,c^{2} {\mathrm e}^{3 i \left (e x +d \right )}+9 i a^{3} {\mathrm e}^{2 i \left (e x +d \right )}+3 a^{2} c \,{\mathrm e}^{3 i \left (e x +d \right )}+3 i a \,c^{2} {\mathrm e}^{2 i \left (e x +d \right )}+c^{3} {\mathrm e}^{3 i \left (e x +d \right )}-9 i a^{3} {\mathrm e}^{i \left (e x +d \right )}+i a \,c^{2} {\mathrm e}^{i \left (e x +d \right )}+3 i a^{3}-9 a^{2} c \,{\mathrm e}^{i \left (e x +d \right )}-3 i a \,c^{2}+c^{3} {\mathrm e}^{i \left (e x +d \right )}+6 a^{2} c}{8 \left (c \,{\mathrm e}^{2 i \left (e x +d \right )}-i a \,{\mathrm e}^{2 i \left (e x +d \right )}-c +2 i a \,{\mathrm e}^{i \left (e x +d \right )}-i a \right )^{2} c^{4} e}-\frac {3 \ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {i c -a}{i c +a}\right ) a^{2}}{16 c^{5} e}-\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {i c -a}{i c +a}\right )}{16 c^{3} e}+\frac {3 \ln \left ({\mathrm e}^{i \left (e x +d \right )}-1\right ) a^{2}}{16 c^{5} e}+\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}-1\right )}{16 c^{3} e}\) | \(344\) |
1/8/e*(-1/8/c^3/tan(1/2*e*x+1/2*d)^2+1/4*(6*a^2+2*c^2)/c^5*ln(tan(1/2*e*x+ 1/2*d))+3/4/c^4*a/tan(1/2*e*x+1/2*d)-1/4*(-3*a^4-2*a^2*c^2+c^4)/a^2/c^4/(c +a*tan(1/2*e*x+1/2*d))-1/8*(-a^4-2*a^2*c^2-c^4)/a^2/c^3/(c+a*tan(1/2*e*x+1 /2*d))^2-1/2*(3*a^2+c^2)/c^5*ln(c+a*tan(1/2*e*x+1/2*d)))
Leaf count of result is larger than twice the leaf count of optimal. 438 vs. \(2 (131) = 262\).
Time = 0.27 (sec) , antiderivative size = 438, normalized size of antiderivative = 3.27 \[ \int \frac {1}{(2 a-2 a \cos (d+e x)+2 c \sin (d+e x))^3} \, dx=\frac {12 \, a^{2} c^{2} \cos \left (e x + d\right )^{2} - 6 \, a^{2} c^{2} - 2 \, {\left (3 \, a^{2} c^{2} - c^{4}\right )} \cos \left (e x + d\right ) + {\left (3 \, a^{4} + 4 \, a^{2} c^{2} + c^{4} + {\left (3 \, a^{4} - 2 \, a^{2} c^{2} - c^{4}\right )} \cos \left (e x + d\right )^{2} - 2 \, {\left (3 \, a^{4} + a^{2} c^{2}\right )} \cos \left (e x + d\right ) + 2 \, {\left (3 \, a^{3} c + a c^{3} - {\left (3 \, a^{3} c + a c^{3}\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right )\right )} \log \left (a c \sin \left (e x + d\right ) + \frac {1}{2} \, a^{2} + \frac {1}{2} \, c^{2} - \frac {1}{2} \, {\left (a^{2} - c^{2}\right )} \cos \left (e x + d\right )\right ) - {\left (3 \, a^{4} + 4 \, a^{2} c^{2} + c^{4} + {\left (3 \, a^{4} - 2 \, a^{2} c^{2} - c^{4}\right )} \cos \left (e x + d\right )^{2} - 2 \, {\left (3 \, a^{4} + a^{2} c^{2}\right )} \cos \left (e x + d\right ) + 2 \, {\left (3 \, a^{3} c + a c^{3} - {\left (3 \, a^{3} c + a c^{3}\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (e x + d\right ) + \frac {1}{2}\right ) - 2 \, {\left (3 \, a^{3} c - a c^{3} - 3 \, {\left (a^{3} c - a c^{3}\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right )}{32 \, {\left (2 \, a^{2} c^{5} e \cos \left (e x + d\right ) - {\left (a^{2} c^{5} - c^{7}\right )} e \cos \left (e x + d\right )^{2} - {\left (a^{2} c^{5} + c^{7}\right )} e + 2 \, {\left (a c^{6} e \cos \left (e x + d\right ) - a c^{6} e\right )} \sin \left (e x + d\right )\right )}} \]
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. \[ \int \frac {1}{(2 a-2 a \cos (d+e x)+2 c \sin (d+e x))^3} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 265 vs. \(2 (131) = 262\).
Time = 0.24 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.98 \[ \int \frac {1}{(2 a-2 a \cos (d+e x)+2 c \sin (d+e x))^3} \, dx=-\frac {\frac {a^{2} c^{3} - \frac {4 \, a^{3} c^{2} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} - \frac {{\left (18 \, a^{4} c + 6 \, a^{2} c^{3} - c^{5}\right )} \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}} - \frac {2 \, {\left (6 \, a^{5} + 2 \, a^{3} c^{2} - a c^{4}\right )} \sin \left (e x + d\right )^{3}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{3}}}{\frac {a^{2} c^{6} \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}} + \frac {2 \, a^{3} c^{5} \sin \left (e x + d\right )^{3}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{3}} + \frac {a^{4} c^{4} \sin \left (e x + d\right )^{4}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{4}}} + \frac {4 \, {\left (3 \, a^{2} + c^{2}\right )} \log \left (c + \frac {a \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}\right )}{c^{5}} - \frac {4 \, {\left (3 \, a^{2} + c^{2}\right )} \log \left (\frac {\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}\right )}{c^{5}}}{64 \, e} \]
-1/64*((a^2*c^3 - 4*a^3*c^2*sin(e*x + d)/(cos(e*x + d) + 1) - (18*a^4*c + 6*a^2*c^3 - c^5)*sin(e*x + d)^2/(cos(e*x + d) + 1)^2 - 2*(6*a^5 + 2*a^3*c^ 2 - a*c^4)*sin(e*x + d)^3/(cos(e*x + d) + 1)^3)/(a^2*c^6*sin(e*x + d)^2/(c os(e*x + d) + 1)^2 + 2*a^3*c^5*sin(e*x + d)^3/(cos(e*x + d) + 1)^3 + a^4*c ^4*sin(e*x + d)^4/(cos(e*x + d) + 1)^4) + 4*(3*a^2 + c^2)*log(c + a*sin(e* x + d)/(cos(e*x + d) + 1))/c^5 - 4*(3*a^2 + c^2)*log(sin(e*x + d)/(cos(e*x + d) + 1))/c^5)/e
Time = 0.30 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.71 \[ \int \frac {1}{(2 a-2 a \cos (d+e x)+2 c \sin (d+e x))^3} \, dx=\frac {\frac {4 \, {\left (3 \, a^{2} + c^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) \right |}\right )}{c^{5}} - \frac {4 \, {\left (3 \, a^{3} + a c^{2}\right )} \log \left ({\left | a \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + c \right |}\right )}{a c^{5}} + \frac {12 \, a^{5} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{3} + 4 \, a^{3} c^{2} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{3} - 2 \, a c^{4} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{3} + 18 \, a^{4} c \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} + 6 \, a^{2} c^{3} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} - c^{5} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} + 4 \, a^{3} c^{2} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) - a^{2} c^{3}}{{\left (a \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} + c \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )\right )}^{2} a^{2} c^{4}}}{64 \, e} \]
1/64*(4*(3*a^2 + c^2)*log(abs(tan(1/2*e*x + 1/2*d)))/c^5 - 4*(3*a^3 + a*c^ 2)*log(abs(a*tan(1/2*e*x + 1/2*d) + c))/(a*c^5) + (12*a^5*tan(1/2*e*x + 1/ 2*d)^3 + 4*a^3*c^2*tan(1/2*e*x + 1/2*d)^3 - 2*a*c^4*tan(1/2*e*x + 1/2*d)^3 + 18*a^4*c*tan(1/2*e*x + 1/2*d)^2 + 6*a^2*c^3*tan(1/2*e*x + 1/2*d)^2 - c^ 5*tan(1/2*e*x + 1/2*d)^2 + 4*a^3*c^2*tan(1/2*e*x + 1/2*d) - a^2*c^3)/((a*t an(1/2*e*x + 1/2*d)^2 + c*tan(1/2*e*x + 1/2*d))^2*a^2*c^4))/e
Time = 28.77 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.39 \[ \int \frac {1}{(2 a-2 a \cos (d+e x)+2 c \sin (d+e x))^3} \, dx=\frac {\frac {2\,a\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}{c^2}-\frac {1}{2\,c}+\frac {{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^3\,\left (6\,a^4+2\,a^2\,c^2-c^4\right )}{a\,c^4}+\frac {{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2\,\left (18\,a^4+6\,a^2\,c^2-c^4\right )}{2\,a^2\,c^3}}{e\,\left (32\,a^2\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^4+64\,a\,c\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^3+32\,c^2\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2\right )}-\frac {\mathrm {atanh}\left (\frac {2\,a\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}{c}+1\right )\,\left (3\,a^2+c^2\right )}{8\,c^5\,e} \]
((2*a*tan(d/2 + (e*x)/2))/c^2 - 1/(2*c) + (tan(d/2 + (e*x)/2)^3*(6*a^4 - c ^4 + 2*a^2*c^2))/(a*c^4) + (tan(d/2 + (e*x)/2)^2*(18*a^4 - c^4 + 6*a^2*c^2 ))/(2*a^2*c^3))/(e*(32*a^2*tan(d/2 + (e*x)/2)^4 + 32*c^2*tan(d/2 + (e*x)/2 )^2 + 64*a*c*tan(d/2 + (e*x)/2)^3)) - (atanh((2*a*tan(d/2 + (e*x)/2))/c + 1)*(3*a^2 + c^2))/(8*c^5*e)