Integrand size = 24, antiderivative size = 168 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^4} \, dx=-\frac {\log \left (1+\tan \left (\frac {1}{2} (d+e x)\right )\right )}{4 a^4 e}-\frac {\cos (d+e x)-\sin (d+e x)}{48 a e (a+a \cos (d+e x)+a \sin (d+e x))^3}+\frac {5 (\cos (d+e x)-\sin (d+e x))}{96 e \left (a^2+a^2 \cos (d+e x)+a^2 \sin (d+e x)\right )^2}-\frac {19 (a \cos (d+e x)-a \sin (d+e x))}{96 e \left (a^5+a^5 \cos (d+e x)+a^5 \sin (d+e x)\right )} \]
-1/4*ln(1+tan(1/2*e*x+1/2*d))/a^4/e+1/48*(-cos(e*x+d)+sin(e*x+d))/a/e/(a+a *cos(e*x+d)+a*sin(e*x+d))^3+5/96*(cos(e*x+d)-sin(e*x+d))/e/(a^2+a^2*cos(e* x+d)+a^2*sin(e*x+d))^2-19/96*(a*cos(e*x+d)-a*sin(e*x+d))/e/(a^5+a^5*cos(e* x+d)+a^5*sin(e*x+d))
Time = 1.16 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.47 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^4} \, dx=\frac {\log \left (\cos \left (\frac {1}{2} (d+e x)\right )\right )}{4 a^4 e}-\frac {\log \left (\cos \left (\frac {1}{2} (d+e x)\right )+\sin \left (\frac {1}{2} (d+e x)\right )\right )}{4 a^4 e}-\frac {\sec ^2\left (\frac {1}{2} (d+e x)\right )}{64 a^4 e}+\frac {\sin \left (\frac {1}{2} (d+e x)\right )}{96 a^4 e \left (\cos \left (\frac {1}{2} (d+e x)\right )+\sin \left (\frac {1}{2} (d+e x)\right )\right )^3}+\frac {5}{192 a^4 e \left (\cos \left (\frac {1}{2} (d+e x)\right )+\sin \left (\frac {1}{2} (d+e x)\right )\right )^2}+\frac {19 \sin \left (\frac {1}{2} (d+e x)\right )}{96 a^4 e \left (\cos \left (\frac {1}{2} (d+e x)\right )+\sin \left (\frac {1}{2} (d+e x)\right )\right )}+\frac {19 \tan \left (\frac {1}{2} (d+e x)\right )}{192 a^4 e}+\frac {\sec ^2\left (\frac {1}{2} (d+e x)\right ) \tan \left (\frac {1}{2} (d+e x)\right )}{384 a^4 e} \]
Log[Cos[(d + e*x)/2]]/(4*a^4*e) - Log[Cos[(d + e*x)/2] + Sin[(d + e*x)/2]] /(4*a^4*e) - Sec[(d + e*x)/2]^2/(64*a^4*e) + Sin[(d + e*x)/2]/(96*a^4*e*(C os[(d + e*x)/2] + Sin[(d + e*x)/2])^3) + 5/(192*a^4*e*(Cos[(d + e*x)/2] + Sin[(d + e*x)/2])^2) + (19*Sin[(d + e*x)/2])/(96*a^4*e*(Cos[(d + e*x)/2] + Sin[(d + e*x)/2])) + (19*Tan[(d + e*x)/2])/(192*a^4*e) + (Sec[(d + e*x)/2 ]^2*Tan[(d + e*x)/2])/(384*a^4*e)
Time = 0.71 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.12, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.458, Rules used = {3042, 3608, 27, 3042, 3635, 25, 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.
| \(\displaystyle \int \frac {1}{(2 a \sin (d+e x)+2 a \cos (d+e x)+2 a)^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(2 a \sin (d+e x)+2 a \cos (d+e x)+2 a)^4}dx\) |
| \(\Big \downarrow \) 3608 |
| \(\displaystyle \frac {\int -\frac {-2 \cos (d+e x) a-2 \sin (d+e x) a+3 a}{4 (\cos (d+e x) a+\sin (d+e x) a+a)^3}dx}{12 a^2}-\frac {\cos (d+e x)-\sin (d+e x)}{48 a e (a \sin (d+e x)+a \cos (d+e x)+a)^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {-2 \cos (d+e x) a-2 \sin (d+e x) a+3 a}{(\cos (d+e x) a+\sin (d+e x) a+a)^3}dx}{48 a^2}-\frac {\cos (d+e x)-\sin (d+e x)}{48 a e (a \sin (d+e x)+a \cos (d+e x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {-2 \cos (d+e x) a-2 \sin (d+e x) a+3 a}{(\cos (d+e x) a+\sin (d+e x) a+a)^3}dx}{48 a^2}-\frac {\cos (d+e x)-\sin (d+e x)}{48 a e (a \sin (d+e x)+a \cos (d+e x)+a)^3}\) |
| \(\Big \downarrow \) 3635 |
| \(\displaystyle -\frac {\frac {\int -\frac {-5 \cos (d+e x) a^2-5 \sin (d+e x) a^2+14 a^2}{(\cos (d+e x) a+\sin (d+e x) a+a)^2}dx}{2 a^2}-\frac {5 \left (a^2 \cos (d+e x)-a^2 \sin (d+e x)\right )}{2 e \left (a^2 \sin (d+e x)+a^2 \cos (d+e x)+a^2\right )^2}}{48 a^2}-\frac {\cos (d+e x)-\sin (d+e x)}{48 a e (a \sin (d+e x)+a \cos (d+e x)+a)^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {-\frac {\int \frac {-5 \cos (d+e x) a^2-5 \sin (d+e x) a^2+14 a^2}{(\cos (d+e x) a+\sin (d+e x) a+a)^2}dx}{2 a^2}-\frac {5 \left (a^2 \cos (d+e x)-a^2 \sin (d+e x)\right )}{2 e \left (a^2 \sin (d+e x)+a^2 \cos (d+e x)+a^2\right )^2}}{48 a^2}-\frac {\cos (d+e x)-\sin (d+e x)}{48 a e (a \sin (d+e x)+a \cos (d+e x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {\int \frac {-5 \cos (d+e x) a^2-5 \sin (d+e x) a^2+14 a^2}{(\cos (d+e x) a+\sin (d+e x) a+a)^2}dx}{2 a^2}-\frac {5 \left (a^2 \cos (d+e x)-a^2 \sin (d+e x)\right )}{2 e \left (a^2 \sin (d+e x)+a^2 \cos (d+e x)+a^2\right )^2}}{48 a^2}-\frac {\cos (d+e x)-\sin (d+e x)}{48 a e (a \sin (d+e x)+a \cos (d+e x)+a)^3}\) |
| \(\Big \downarrow \) 3632 |
| \(\displaystyle -\frac {-\frac {-24 a \int \frac {1}{\cos (d+e x) a+\sin (d+e x) a+a}dx-\frac {19 \left (a^3 \cos (d+e x)-a^3 \sin (d+e x)\right )}{e \left (a^3 \sin (d+e x)+a^3 \cos (d+e x)+a^3\right )}}{2 a^2}-\frac {5 \left (a^2 \cos (d+e x)-a^2 \sin (d+e x)\right )}{2 e \left (a^2 \sin (d+e x)+a^2 \cos (d+e x)+a^2\right )^2}}{48 a^2}-\frac {\cos (d+e x)-\sin (d+e x)}{48 a e (a \sin (d+e x)+a \cos (d+e x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {-24 a \int \frac {1}{\cos (d+e x) a+\sin (d+e x) a+a}dx-\frac {19 \left (a^3 \cos (d+e x)-a^3 \sin (d+e x)\right )}{e \left (a^3 \sin (d+e x)+a^3 \cos (d+e x)+a^3\right )}}{2 a^2}-\frac {5 \left (a^2 \cos (d+e x)-a^2 \sin (d+e x)\right )}{2 e \left (a^2 \sin (d+e x)+a^2 \cos (d+e x)+a^2\right )^2}}{48 a^2}-\frac {\cos (d+e x)-\sin (d+e x)}{48 a e (a \sin (d+e x)+a \cos (d+e x)+a)^3}\) |
| \(\Big \downarrow \) 3603 |
| \(\displaystyle -\frac {-\frac {-\frac {48 a \int \frac {1}{2 \tan \left (\frac {1}{2} (d+e x)\right ) a+2 a}d\tan \left (\frac {1}{2} (d+e x)\right )}{e}-\frac {19 \left (a^3 \cos (d+e x)-a^3 \sin (d+e x)\right )}{e \left (a^3 \sin (d+e x)+a^3 \cos (d+e x)+a^3\right )}}{2 a^2}-\frac {5 \left (a^2 \cos (d+e x)-a^2 \sin (d+e x)\right )}{2 e \left (a^2 \sin (d+e x)+a^2 \cos (d+e x)+a^2\right )^2}}{48 a^2}-\frac {\cos (d+e x)-\sin (d+e x)}{48 a e (a \sin (d+e x)+a \cos (d+e x)+a)^3}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle -\frac {-\frac {5 \left (a^2 \cos (d+e x)-a^2 \sin (d+e x)\right )}{2 e \left (a^2 \sin (d+e x)+a^2 \cos (d+e x)+a^2\right )^2}-\frac {-\frac {19 \left (a^3 \cos (d+e x)-a^3 \sin (d+e x)\right )}{e \left (a^3 \sin (d+e x)+a^3 \cos (d+e x)+a^3\right )}-\frac {24 \log \left (\tan \left (\frac {1}{2} (d+e x)\right )+1\right )}{e}}{2 a^2}}{48 a^2}-\frac {\cos (d+e x)-\sin (d+e x)}{48 a e (a \sin (d+e x)+a \cos (d+e x)+a)^3}\) |
-1/48*(Cos[d + e*x] - Sin[d + e*x])/(a*e*(a + a*Cos[d + e*x] + a*Sin[d + e *x])^3) - ((-5*(a^2*Cos[d + e*x] - a^2*Sin[d + e*x]))/(2*e*(a^2 + a^2*Cos[ d + e*x] + a^2*Sin[d + e*x])^2) - ((-24*Log[1 + Tan[(d + e*x)/2]])/e - (19 *(a^3*Cos[d + e*x] - a^3*Sin[d + e*x]))/(e*(a^3 + a^3*Cos[d + e*x] + a^3*S in[d + e*x])))/(2*a^2))/(48*a^2)
3.4.73.3.1 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]
Int[((a_.) + cos[(d_.) + (e_.)*(x_)]*(b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)]) ^(n_)*((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.) + (C_.)*sin[(d_.) + (e_.)*(x_) ]), x_Symbol] :> Simp[(-(c*B - b*C - (a*C - c*A)*Cos[d + e*x] + (a*B - b*A) *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 + b*Co s[d + e*x] + c*Sin[d + e*x])^(n + 1)*Simp[(n + 1)*(a*A - b*B - c*C) + (n + 2)*(a*B - b*A)*Cos[d + e*x] + (n + 2)*(a*C - c*A)*Sin[d + e*x], x], x], x] /; FreeQ[{a, b, c, d, e, A, B, C}, x] && LtQ[n, -1] && NeQ[a^2 - b^2 - c^2, 0] && NeQ[n, -2]
Time = 0.95 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.63
| method | result | size |
| derivativedivides | \(\frac {\frac {\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{3}}{3}-2 \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}+13 \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-32 \ln \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )+\frac {12}{\left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )^{2}}-\frac {36}{1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )}-\frac {8}{3 \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )^{3}}}{128 e \,a^{4}}\) | \(106\) |
| default | \(\frac {\frac {\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{3}}{3}-2 \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}+13 \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-32 \ln \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )+\frac {12}{\left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )^{2}}-\frac {36}{1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )}-\frac {8}{3 \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )^{3}}}{128 e \,a^{4}}\) | \(106\) |
| parallelrisch | \(\frac {-96 \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )^{3} \ln \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{6}-3 \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{5}+24 \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{4}-297 \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}-441 \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-180}{384 e \,a^{4} \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )^{3}}\) | \(112\) |
| norman | \(\frac {\frac {\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{4}}{16 a e}-\frac {\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{5}}{128 a e}+\frac {\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{6}}{384 a e}-\frac {15}{32 a e}-\frac {147 \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{128 a e}-\frac {99 \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}{128 a e}}{a^{3} \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )^{3}}-\frac {\ln \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{4 a^{4} e}\) | \(141\) |
| risch | \(\frac {\left (-\frac {1}{96}+\frac {i}{96}\right ) \left (60 i {\mathrm e}^{4 i \left (e x +d \right )}+24 \,{\mathrm e}^{5 i \left (e x +d \right )}+152 i {\mathrm e}^{3 i \left (e x +d \right )}+60 \,{\mathrm e}^{4 i \left (e x +d \right )}+111 i {\mathrm e}^{2 i \left (e x +d \right )}-111 \,{\mathrm e}^{2 i \left (e x +d \right )}-19-19 i-90 \,{\mathrm e}^{i \left (e x +d \right )}\right )}{a^{4} e \left (i {\mathrm e}^{i \left (e x +d \right )}+{\mathrm e}^{2 i \left (e x +d \right )}+i+{\mathrm e}^{i \left (e x +d \right )}\right )^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}+1\right )}{4 a^{4} e}-\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}+i\right )}{4 a^{4} e}\) | \(172\) |
1/128/e/a^4*(1/3*tan(1/2*e*x+1/2*d)^3-2*tan(1/2*e*x+1/2*d)^2+13*tan(1/2*e* x+1/2*d)-32*ln(1+tan(1/2*e*x+1/2*d))+12/(1+tan(1/2*e*x+1/2*d))^2-36/(1+tan (1/2*e*x+1/2*d))-8/3/(1+tan(1/2*e*x+1/2*d))^3)
Time = 0.25 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.41 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^4} \, dx=\frac {38 \, \cos \left (e x + d\right )^{3} + 66 \, \cos \left (e x + d\right )^{2} + 24 \, {\left (\cos \left (e x + d\right )^{3} - {\left (\cos \left (e x + d\right )^{2} + 3 \, \cos \left (e x + d\right ) + 2\right )} \sin \left (e x + d\right ) - 3 \, \cos \left (e x + d\right ) - 2\right )} \log \left (\frac {1}{2} \, \cos \left (e x + d\right ) + \frac {1}{2}\right ) - 24 \, {\left (\cos \left (e x + d\right )^{3} - {\left (\cos \left (e x + d\right )^{2} + 3 \, \cos \left (e x + d\right ) + 2\right )} \sin \left (e x + d\right ) - 3 \, \cos \left (e x + d\right ) - 2\right )} \log \left (\sin \left (e x + d\right ) + 1\right ) + {\left (38 \, \cos \left (e x + d\right )^{2} - 35\right )} \sin \left (e x + d\right ) - 3 \, \cos \left (e x + d\right ) - 33}{192 \, {\left (a^{4} e \cos \left (e x + d\right )^{3} - 3 \, a^{4} e \cos \left (e x + d\right ) - 2 \, a^{4} e - {\left (a^{4} e \cos \left (e x + d\right )^{2} + 3 \, a^{4} e \cos \left (e x + d\right ) + 2 \, a^{4} e\right )} \sin \left (e x + d\right )\right )}} \]
1/192*(38*cos(e*x + d)^3 + 66*cos(e*x + d)^2 + 24*(cos(e*x + d)^3 - (cos(e *x + d)^2 + 3*cos(e*x + d) + 2)*sin(e*x + d) - 3*cos(e*x + d) - 2)*log(1/2 *cos(e*x + d) + 1/2) - 24*(cos(e*x + d)^3 - (cos(e*x + d)^2 + 3*cos(e*x + d) + 2)*sin(e*x + d) - 3*cos(e*x + d) - 2)*log(sin(e*x + d) + 1) + (38*cos (e*x + d)^2 - 35)*sin(e*x + d) - 3*cos(e*x + d) - 33)/(a^4*e*cos(e*x + d)^ 3 - 3*a^4*e*cos(e*x + d) - 2*a^4*e - (a^4*e*cos(e*x + d)^2 + 3*a^4*e*cos(e *x + d) + 2*a^4*e)*sin(e*x + d))
Leaf count of result is larger than twice the leaf count of optimal. 792 vs. \(2 (156) = 312\).
Time = 9.73 (sec) , antiderivative size = 792, normalized size of antiderivative = 4.71 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^4} \, dx=\text {Too large to display} \]
Piecewise((-96*log(tan(d/2 + e*x/2) + 1)*tan(d/2 + e*x/2)**3/(384*a**4*e*t an(d/2 + e*x/2)**3 + 1152*a**4*e*tan(d/2 + e*x/2)**2 + 1152*a**4*e*tan(d/2 + e*x/2) + 384*a**4*e) - 288*log(tan(d/2 + e*x/2) + 1)*tan(d/2 + e*x/2)** 2/(384*a**4*e*tan(d/2 + e*x/2)**3 + 1152*a**4*e*tan(d/2 + e*x/2)**2 + 1152 *a**4*e*tan(d/2 + e*x/2) + 384*a**4*e) - 288*log(tan(d/2 + e*x/2) + 1)*tan (d/2 + e*x/2)/(384*a**4*e*tan(d/2 + e*x/2)**3 + 1152*a**4*e*tan(d/2 + e*x/ 2)**2 + 1152*a**4*e*tan(d/2 + e*x/2) + 384*a**4*e) - 96*log(tan(d/2 + e*x/ 2) + 1)/(384*a**4*e*tan(d/2 + e*x/2)**3 + 1152*a**4*e*tan(d/2 + e*x/2)**2 + 1152*a**4*e*tan(d/2 + e*x/2) + 384*a**4*e) + tan(d/2 + e*x/2)**6/(384*a* *4*e*tan(d/2 + e*x/2)**3 + 1152*a**4*e*tan(d/2 + e*x/2)**2 + 1152*a**4*e*t an(d/2 + e*x/2) + 384*a**4*e) - 3*tan(d/2 + e*x/2)**5/(384*a**4*e*tan(d/2 + e*x/2)**3 + 1152*a**4*e*tan(d/2 + e*x/2)**2 + 1152*a**4*e*tan(d/2 + e*x/ 2) + 384*a**4*e) + 24*tan(d/2 + e*x/2)**4/(384*a**4*e*tan(d/2 + e*x/2)**3 + 1152*a**4*e*tan(d/2 + e*x/2)**2 + 1152*a**4*e*tan(d/2 + e*x/2) + 384*a** 4*e) - 297*tan(d/2 + e*x/2)**2/(384*a**4*e*tan(d/2 + e*x/2)**3 + 1152*a**4 *e*tan(d/2 + e*x/2)**2 + 1152*a**4*e*tan(d/2 + e*x/2) + 384*a**4*e) - 441* tan(d/2 + e*x/2)/(384*a**4*e*tan(d/2 + e*x/2)**3 + 1152*a**4*e*tan(d/2 + e *x/2)**2 + 1152*a**4*e*tan(d/2 + e*x/2) + 384*a**4*e) - 180/(384*a**4*e*ta n(d/2 + e*x/2)**3 + 1152*a**4*e*tan(d/2 + e*x/2)**2 + 1152*a**4*e*tan(d/2 + e*x/2) + 384*a**4*e), Ne(e, 0)), (x/(2*a*sin(d) + 2*a*cos(d) + 2*a)**...
Time = 0.25 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.24 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^4} \, dx=-\frac {\frac {4 \, {\left (\frac {45 \, \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + \frac {27 \, \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}} + 20\right )}}{a^{4} + \frac {3 \, a^{4} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + \frac {3 \, a^{4} \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}} + \frac {a^{4} \sin \left (e x + d\right )^{3}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{3}}} - \frac {\frac {39 \, \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} - \frac {6 \, \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}} + \frac {\sin \left (e x + d\right )^{3}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{3}}}{a^{4}} + \frac {96 \, \log \left (\frac {\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + 1\right )}{a^{4}}}{384 \, e} \]
-1/384*(4*(45*sin(e*x + d)/(cos(e*x + d) + 1) + 27*sin(e*x + d)^2/(cos(e*x + d) + 1)^2 + 20)/(a^4 + 3*a^4*sin(e*x + d)/(cos(e*x + d) + 1) + 3*a^4*si n(e*x + d)^2/(cos(e*x + d) + 1)^2 + a^4*sin(e*x + d)^3/(cos(e*x + d) + 1)^ 3) - (39*sin(e*x + d)/(cos(e*x + d) + 1) - 6*sin(e*x + d)^2/(cos(e*x + d) + 1)^2 + sin(e*x + d)^3/(cos(e*x + d) + 1)^3)/a^4 + 96*log(sin(e*x + d)/(c os(e*x + d) + 1) + 1)/a^4)/e
Time = 0.31 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.79 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^4} \, dx=-\frac {\frac {96 \, \log \left ({\left | \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + 1 \right |}\right )}{a^{4}} - \frac {4 \, {\left (44 \, \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{3} + 105 \, \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} + 87 \, \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + 24\right )}}{a^{4} {\left (\tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + 1\right )}^{3}} - \frac {a^{8} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{3} - 6 \, a^{8} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} + 39 \, a^{8} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )}{a^{12}}}{384 \, e} \]
-1/384*(96*log(abs(tan(1/2*e*x + 1/2*d) + 1))/a^4 - 4*(44*tan(1/2*e*x + 1/ 2*d)^3 + 105*tan(1/2*e*x + 1/2*d)^2 + 87*tan(1/2*e*x + 1/2*d) + 24)/(a^4*( tan(1/2*e*x + 1/2*d) + 1)^3) - (a^8*tan(1/2*e*x + 1/2*d)^3 - 6*a^8*tan(1/2 *e*x + 1/2*d)^2 + 39*a^8*tan(1/2*e*x + 1/2*d))/a^12)/e
Time = 27.12 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.96 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^4} \, dx=\frac {{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^3}{384\,a^4\,e}-\frac {{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2}{64\,a^4\,e}-\frac {\ln \left (\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )+1\right )}{4\,a^4\,e}+\frac {13\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}{128\,a^4\,e}-\frac {9\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2+15\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )+\frac {20}{3}}{e\,\left (32\,a^4\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^3+96\,a^4\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2+96\,a^4\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )+32\,a^4\right )} \]
tan(d/2 + (e*x)/2)^3/(384*a^4*e) - tan(d/2 + (e*x)/2)^2/(64*a^4*e) - log(t an(d/2 + (e*x)/2) + 1)/(4*a^4*e) + (13*tan(d/2 + (e*x)/2))/(128*a^4*e) - ( 15*tan(d/2 + (e*x)/2) + 9*tan(d/2 + (e*x)/2)^2 + 20/3)/(e*(96*a^4*tan(d/2 + (e*x)/2)^2 + 32*a^4*tan(d/2 + (e*x)/2)^3 + 32*a^4 + 96*a^4*tan(d/2 + (e* x)/2)))