Integrand size = 24, antiderivative size = 123 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^3} \, dx=\frac {\log \left (1+\tan \left (\frac {1}{2} (d+e x)\right )\right )}{4 a^3 e}-\frac {a \cos (d+e x)-a \sin (d+e x)}{16 e \left (a^2+a^2 \cos (d+e x)+a^2 \sin (d+e x)\right )^2}+\frac {3 (\cos (d+e x)-\sin (d+e x))}{16 e \left (a^3+a^3 \cos (d+e x)+a^3 \sin (d+e x)\right )} \] Output:
1/4*ln(1+tan(1/2*e*x+1/2*d))/a^3/e-1/16*(a*cos(e*x+d)-a*sin(e*x+d))/e/(a^2 +a^2*cos(e*x+d)+a^2*sin(e*x+d))^2+3/16*(cos(e*x+d)-sin(e*x+d))/e/(a^3+a^3* cos(e*x+d)+a^3*sin(e*x+d))
Time = 0.46 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.10 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^3} \, dx=\frac {\sec ^2\left (\frac {1}{2} (d+e x)\right )+2 \left (-8 \log \left (\cos \left (\frac {1}{2} (d+e x)\right )\right )+8 \log \left (\cos \left (\frac {1}{2} (d+e x)\right )+\sin \left (\frac {1}{2} (d+e x)\right )\right )-\frac {1}{\left (\cos \left (\frac {1}{2} (d+e x)\right )+\sin \left (\frac {1}{2} (d+e x)\right )\right )^2}-\frac {6 \sin \left (\frac {1}{2} (d+e x)\right )}{\cos \left (\frac {1}{2} (d+e x)\right )+\sin \left (\frac {1}{2} (d+e x)\right )}-3 \tan \left (\frac {1}{2} (d+e x)\right )\right )}{64 a^3 e} \] Input:
Integrate[(2*a + 2*a*Cos[d + e*x] + 2*a*Sin[d + e*x])^(-3),x]
Output:
(Sec[(d + e*x)/2]^2 + 2*(-8*Log[Cos[(d + e*x)/2]] + 8*Log[Cos[(d + e*x)/2] + Sin[(d + e*x)/2]] - (Cos[(d + e*x)/2] + Sin[(d + e*x)/2])^(-2) - (6*Sin [(d + e*x)/2])/(Cos[(d + e*x)/2] + Sin[(d + e*x)/2]) - 3*Tan[(d + e*x)/2]) )/(64*a^3*e)
Time = 0.51 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.09, 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.
| \(\displaystyle \int \frac {1}{(2 a \sin (d+e x)+2 a \cos (d+e x)+2 a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(2 a \sin (d+e x)+2 a \cos (d+e x)+2 a)^3}dx\) |
| \(\Big \downarrow \) 3608 |
| \(\displaystyle \frac {\int -\frac {-\cos (d+e x) a-\sin (d+e x) a+2 a}{2 (\cos (d+e x) a+\sin (d+e x) a+a)^2}dx}{8 a^2}-\frac {a \cos (d+e x)-a \sin (d+e x)}{16 e \left (a^2 \sin (d+e x)+a^2 \cos (d+e x)+a^2\right )^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {-\cos (d+e x) a-\sin (d+e x) a+2 a}{(\cos (d+e x) a+\sin (d+e x) a+a)^2}dx}{16 a^2}-\frac {a \cos (d+e x)-a \sin (d+e x)}{16 e \left (a^2 \sin (d+e x)+a^2 \cos (d+e x)+a^2\right )^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {-\cos (d+e x) a-\sin (d+e x) a+2 a}{(\cos (d+e x) a+\sin (d+e x) a+a)^2}dx}{16 a^2}-\frac {a \cos (d+e x)-a \sin (d+e x)}{16 e \left (a^2 \sin (d+e x)+a^2 \cos (d+e x)+a^2\right )^2}\) |
| \(\Big \downarrow \) 3632 |
| \(\displaystyle -\frac {-4 \int \frac {1}{\cos (d+e x) a+\sin (d+e x) a+a}dx-\frac {3 \left (a^2 \cos (d+e x)-a^2 \sin (d+e x)\right )}{e \left (a^3 \sin (d+e x)+a^3 \cos (d+e x)+a^3\right )}}{16 a^2}-\frac {a \cos (d+e x)-a \sin (d+e x)}{16 e \left (a^2 \sin (d+e x)+a^2 \cos (d+e x)+a^2\right )^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-4 \int \frac {1}{\cos (d+e x) a+\sin (d+e x) a+a}dx-\frac {3 \left (a^2 \cos (d+e x)-a^2 \sin (d+e x)\right )}{e \left (a^3 \sin (d+e x)+a^3 \cos (d+e x)+a^3\right )}}{16 a^2}-\frac {a \cos (d+e x)-a \sin (d+e x)}{16 e \left (a^2 \sin (d+e x)+a^2 \cos (d+e x)+a^2\right )^2}\) |
| \(\Big \downarrow \) 3603 |
| \(\displaystyle -\frac {-\frac {8 \int \frac {1}{2 \tan \left (\frac {1}{2} (d+e x)\right ) a+2 a}d\tan \left (\frac {1}{2} (d+e x)\right )}{e}-\frac {3 \left (a^2 \cos (d+e x)-a^2 \sin (d+e x)\right )}{e \left (a^3 \sin (d+e x)+a^3 \cos (d+e x)+a^3\right )}}{16 a^2}-\frac {a \cos (d+e x)-a \sin (d+e x)}{16 e \left (a^2 \sin (d+e x)+a^2 \cos (d+e x)+a^2\right )^2}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle -\frac {a \cos (d+e x)-a \sin (d+e x)}{16 e \left (a^2 \sin (d+e x)+a^2 \cos (d+e x)+a^2\right )^2}-\frac {-\frac {3 \left (a^2 \cos (d+e x)-a^2 \sin (d+e x)\right )}{e \left (a^3 \sin (d+e x)+a^3 \cos (d+e x)+a^3\right )}-\frac {4 \log \left (\tan \left (\frac {1}{2} (d+e x)\right )+1\right )}{a e}}{16 a^2}\) |
Input:
Int[(2*a + 2*a*Cos[d + e*x] + 2*a*Sin[d + e*x])^(-3),x]
Output:
-1/16*(a*Cos[d + e*x] - a*Sin[d + e*x])/(e*(a^2 + a^2*Cos[d + e*x] + a^2*S in[d + e*x])^2) - ((-4*Log[1 + Tan[(d + e*x)/2]])/(a*e) - (3*(a^2*Cos[d + e*x] - a^2*Sin[d + e*x]))/(e*(a^3 + a^3*Cos[d + e*x] + a^3*Sin[d + e*x]))) /(16*a^2)
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 = 0.62 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.63
| method | result | size |
| derivativedivides | \(\frac {\frac {\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}{2}-3 \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-\frac {2}{\left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )^{2}}+8 \ln \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )+\frac {8}{1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )}}{32 e \,a^{3}}\) | \(78\) |
| default | \(\frac {\frac {\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}{2}-3 \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-\frac {2}{\left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )^{2}}+8 \ln \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )+\frac {8}{1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )}}{32 e \,a^{3}}\) | \(78\) |
| parallelrisch | \(\frac {-12+16 \left (\sin \left (e x +d \right )+1\right ) \ln \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}+\left (-5 \cos \left (e x +d \right )-9\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+12 \cos \left (e x +d \right )}{64 a^{3} e \left (\sin \left (e x +d \right )+1\right )}\) | \(82\) |
| norman | \(\frac {-\frac {\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{3}}{16 a e}+\frac {\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{4}}{64 a e}+\frac {23}{64 a e}+\frac {\tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{2 a e}}{a^{2} \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )^{2}}+\frac {\ln \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{4 a^{3} e}\) | \(103\) |
| risch | \(\frac {\left (\frac {1}{16}-\frac {i}{16}\right ) \left (6 i {\mathrm e}^{2 i \left (e x +d \right )}+4 \,{\mathrm e}^{3 i \left (e x +d \right )}+8 i {\mathrm e}^{i \left (e x +d \right )}+6 \,{\mathrm e}^{2 i \left (e x +d \right )}-3+3 i\right )}{a^{3} e \left (i {\mathrm e}^{i \left (e x +d \right )}+{\mathrm e}^{2 i \left (e x +d \right )}+i+{\mathrm e}^{i \left (e x +d \right )}\right )^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}+1\right )}{4 a^{3} e}+\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}+i\right )}{4 a^{3} e}\) | \(138\) |
Input:
int(1/(2*a+2*a*cos(e*x+d)+2*a*sin(e*x+d))^3,x,method=_RETURNVERBOSE)
Output:
1/32/e/a^3*(1/2*tan(1/2*e*x+1/2*d)^2-3*tan(1/2*e*x+1/2*d)-2/(1+tan(1/2*e*x +1/2*d))^2+8*ln(1+tan(1/2*e*x+1/2*d))+8/(1+tan(1/2*e*x+1/2*d)))
Time = 0.09 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.16 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^3} \, dx=\frac {6 \, \cos \left (e x + d\right )^{2} - 4 \, {\left ({\left (\cos \left (e x + d\right ) + 1\right )} \sin \left (e x + d\right ) + \cos \left (e x + d\right ) + 1\right )} \log \left (\frac {1}{2} \, \cos \left (e x + d\right ) + \frac {1}{2}\right ) + 4 \, {\left ({\left (\cos \left (e x + d\right ) + 1\right )} \sin \left (e x + d\right ) + \cos \left (e x + d\right ) + 1\right )} \log \left (\sin \left (e x + d\right ) + 1\right ) + 2 \, \cos \left (e x + d\right ) - 2 \, \sin \left (e x + d\right ) - 3}{32 \, {\left (a^{3} e \cos \left (e x + d\right ) + a^{3} e + {\left (a^{3} e \cos \left (e x + d\right ) + a^{3} e\right )} \sin \left (e x + d\right )\right )}} \] Input:
integrate(1/(2*a+2*a*cos(e*x+d)+2*a*sin(e*x+d))^3,x, algorithm="fricas")
Output:
1/32*(6*cos(e*x + d)^2 - 4*((cos(e*x + d) + 1)*sin(e*x + d) + cos(e*x + d) + 1)*log(1/2*cos(e*x + d) + 1/2) + 4*((cos(e*x + d) + 1)*sin(e*x + d) + c os(e*x + d) + 1)*log(sin(e*x + d) + 1) + 2*cos(e*x + d) - 2*sin(e*x + d) - 3)/(a^3*e*cos(e*x + d) + a^3*e + (a^3*e*cos(e*x + d) + a^3*e)*sin(e*x + d ))
Leaf count of result is larger than twice the leaf count of optimal. 423 vs. \(2 (112) = 224\).
Time = 3.02 (sec) , antiderivative size = 423, normalized size of antiderivative = 3.44 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^3} \, dx=\begin {cases} \frac {16 \log {\left (\tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 1 \right )} \tan ^{2}{\left (\frac {d}{2} + \frac {e x}{2} \right )}}{64 a^{3} e \tan ^{2}{\left (\frac {d}{2} + \frac {e x}{2} \right )} + 128 a^{3} e \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 64 a^{3} e} + \frac {32 \log {\left (\tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 1 \right )} \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )}}{64 a^{3} e \tan ^{2}{\left (\frac {d}{2} + \frac {e x}{2} \right )} + 128 a^{3} e \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 64 a^{3} e} + \frac {16 \log {\left (\tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 1 \right )}}{64 a^{3} e \tan ^{2}{\left (\frac {d}{2} + \frac {e x}{2} \right )} + 128 a^{3} e \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 64 a^{3} e} + \frac {\tan ^{4}{\left (\frac {d}{2} + \frac {e x}{2} \right )}}{64 a^{3} e \tan ^{2}{\left (\frac {d}{2} + \frac {e x}{2} \right )} + 128 a^{3} e \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 64 a^{3} e} - \frac {4 \tan ^{3}{\left (\frac {d}{2} + \frac {e x}{2} \right )}}{64 a^{3} e \tan ^{2}{\left (\frac {d}{2} + \frac {e x}{2} \right )} + 128 a^{3} e \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 64 a^{3} e} + \frac {32 \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )}}{64 a^{3} e \tan ^{2}{\left (\frac {d}{2} + \frac {e x}{2} \right )} + 128 a^{3} e \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 64 a^{3} e} + \frac {23}{64 a^{3} e \tan ^{2}{\left (\frac {d}{2} + \frac {e x}{2} \right )} + 128 a^{3} e \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 64 a^{3} e} & \text {for}\: e \neq 0 \\\frac {x}{\left (2 a \sin {\left (d \right )} + 2 a \cos {\left (d \right )} + 2 a\right )^{3}} & \text {otherwise} \end {cases} \] Input:
integrate(1/(2*a+2*a*cos(e*x+d)+2*a*sin(e*x+d))**3,x)
Output:
Piecewise((16*log(tan(d/2 + e*x/2) + 1)*tan(d/2 + e*x/2)**2/(64*a**3*e*tan (d/2 + e*x/2)**2 + 128*a**3*e*tan(d/2 + e*x/2) + 64*a**3*e) + 32*log(tan(d /2 + e*x/2) + 1)*tan(d/2 + e*x/2)/(64*a**3*e*tan(d/2 + e*x/2)**2 + 128*a** 3*e*tan(d/2 + e*x/2) + 64*a**3*e) + 16*log(tan(d/2 + e*x/2) + 1)/(64*a**3* e*tan(d/2 + e*x/2)**2 + 128*a**3*e*tan(d/2 + e*x/2) + 64*a**3*e) + tan(d/2 + e*x/2)**4/(64*a**3*e*tan(d/2 + e*x/2)**2 + 128*a**3*e*tan(d/2 + e*x/2) + 64*a**3*e) - 4*tan(d/2 + e*x/2)**3/(64*a**3*e*tan(d/2 + e*x/2)**2 + 128* a**3*e*tan(d/2 + e*x/2) + 64*a**3*e) + 32*tan(d/2 + e*x/2)/(64*a**3*e*tan( d/2 + e*x/2)**2 + 128*a**3*e*tan(d/2 + e*x/2) + 64*a**3*e) + 23/(64*a**3*e *tan(d/2 + e*x/2)**2 + 128*a**3*e*tan(d/2 + e*x/2) + 64*a**3*e), Ne(e, 0)) , (x/(2*a*sin(d) + 2*a*cos(d) + 2*a)**3, True))
Time = 0.04 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.19 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^3} \, dx=\frac {\frac {4 \, {\left (\frac {4 \, \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + 3\right )}}{a^{3} + \frac {2 \, a^{3} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + \frac {a^{3} \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}}} - \frac {\frac {6 \, \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} - \frac {\sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}}}{a^{3}} + \frac {16 \, \log \left (\frac {\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + 1\right )}{a^{3}}}{64 \, e} \] Input:
integrate(1/(2*a+2*a*cos(e*x+d)+2*a*sin(e*x+d))^3,x, algorithm="maxima")
Output:
1/64*(4*(4*sin(e*x + d)/(cos(e*x + d) + 1) + 3)/(a^3 + 2*a^3*sin(e*x + d)/ (cos(e*x + d) + 1) + a^3*sin(e*x + d)^2/(cos(e*x + d) + 1)^2) - (6*sin(e*x + d)/(cos(e*x + d) + 1) - sin(e*x + d)^2/(cos(e*x + d) + 1)^2)/a^3 + 16*l og(sin(e*x + d)/(cos(e*x + d) + 1) + 1)/a^3)/e
Time = 0.13 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.83 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^3} \, dx=\frac {\frac {16 \, \log \left ({\left | \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + 1 \right |}\right )}{a^{3}} - \frac {4 \, {\left (6 \, \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} + 8 \, \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + 3\right )}}{a^{3} {\left (\tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + 1\right )}^{2}} + \frac {a^{3} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} - 6 \, a^{3} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )}{a^{6}}}{64 \, e} \] Input:
integrate(1/(2*a+2*a*cos(e*x+d)+2*a*sin(e*x+d))^3,x, algorithm="giac")
Output:
1/64*(16*log(abs(tan(1/2*e*x + 1/2*d) + 1))/a^3 - 4*(6*tan(1/2*e*x + 1/2*d )^2 + 8*tan(1/2*e*x + 1/2*d) + 3)/(a^3*(tan(1/2*e*x + 1/2*d) + 1)^2) + (a^ 3*tan(1/2*e*x + 1/2*d)^2 - 6*a^3*tan(1/2*e*x + 1/2*d))/a^6)/e
Time = 15.43 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.73 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^3} \, dx=\frac {{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2}{64\,a^3\,e}+\frac {\ln \left (\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )+1\right )}{4\,a^3\,e}-\frac {3\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}{32\,a^3\,e}+\frac {\frac {\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}{4}+\frac {3}{16}}{a^3\,e\,{\left (\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )+1\right )}^2} \] Input:
int(1/(2*a + 2*a*cos(d + e*x) + 2*a*sin(d + e*x))^3,x)
Output:
tan(d/2 + (e*x)/2)^2/(64*a^3*e) + log(tan(d/2 + (e*x)/2) + 1)/(4*a^3*e) - (3*tan(d/2 + (e*x)/2))/(32*a^3*e) + (tan(d/2 + (e*x)/2)/4 + 3/16)/(a^3*e*( tan(d/2 + (e*x)/2) + 1)^2)
Time = 0.18 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.10 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^3} \, dx=\frac {16 \,\mathrm {log}\left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )+1\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}+32 \,\mathrm {log}\left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )+1\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+16 \,\mathrm {log}\left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )+1\right )+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{4}-4 \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{3}-16 \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}+7}{64 a^{3} e \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}+2 \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+1\right )} \] Input:
int(1/(2*a+2*a*cos(e*x+d)+2*a*sin(e*x+d))^3,x)
Output:
(16*log(tan((d + e*x)/2) + 1)*tan((d + e*x)/2)**2 + 32*log(tan((d + e*x)/2 ) + 1)*tan((d + e*x)/2) + 16*log(tan((d + e*x)/2) + 1) + tan((d + e*x)/2)* *4 - 4*tan((d + e*x)/2)**3 - 16*tan((d + e*x)/2)**2 + 7)/(64*a**3*e*(tan(( d + e*x)/2)**2 + 2*tan((d + e*x)/2) + 1))