Integrand size = 24, antiderivative size = 75 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^2} \, dx=-\frac {\log \left (1+\tan \left (\frac {1}{2} (d+e x)\right )\right )}{4 a^2 e}-\frac {a \cos (d+e x)-a \sin (d+e x)}{4 e \left (a^3+a^3 \cos (d+e x)+a^3 \sin (d+e x)\right )} \]
-1/4*ln(1+tan(1/2*e*x+1/2*d))/a^2/e+1/4*(-a*cos(e*x+d)+a*sin(e*x+d))/e/(a^ 3+a^3*cos(e*x+d)+a^3*sin(e*x+d))
Time = 0.25 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.24 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^2} \, dx=\frac {2 \log \left (\cos \left (\frac {1}{2} (d+e x)\right )\right )-2 \log \left (\cos \left (\frac {1}{2} (d+e x)\right )+\sin \left (\frac {1}{2} (d+e x)\right )\right )+\frac {2 \sin \left (\frac {1}{2} (d+e x)\right )}{\cos \left (\frac {1}{2} (d+e x)\right )+\sin \left (\frac {1}{2} (d+e x)\right )}+\tan \left (\frac {1}{2} (d+e x)\right )}{8 a^2 e} \]
(2*Log[Cos[(d + e*x)/2]] - 2*Log[Cos[(d + e*x)/2] + Sin[(d + e*x)/2]] + (2 *Sin[(d + e*x)/2])/(Cos[(d + e*x)/2] + Sin[(d + e*x)/2]) + Tan[(d + e*x)/2 ])/(8*a^2*e)
Time = 0.30 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 3608, 25, 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.
\(\displaystyle \int \frac {1}{(2 a \sin (d+e x)+2 a \cos (d+e x)+2 a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(2 a \sin (d+e x)+2 a \cos (d+e x)+2 a)^2}dx\) |
\(\Big \downarrow \) 3608 |
\(\displaystyle \frac {\int -\frac {1}{\cos (d+e x)+\sin (d+e x)+1}dx}{4 a^2}-\frac {a \cos (d+e x)-a \sin (d+e x)}{4 e \left (a^3 \sin (d+e x)+a^3 \cos (d+e x)+a^3\right )}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {1}{\cos (d+e x)+\sin (d+e x)+1}dx}{4 a^2}-\frac {a \cos (d+e x)-a \sin (d+e x)}{4 e \left (a^3 \sin (d+e x)+a^3 \cos (d+e x)+a^3\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int \frac {1}{\cos (d+e x)+\sin (d+e x)+1}dx}{4 a^2}-\frac {a \cos (d+e x)-a \sin (d+e x)}{4 e \left (a^3 \sin (d+e x)+a^3 \cos (d+e x)+a^3\right )}\) |
\(\Big \downarrow \) 3603 |
\(\displaystyle -\frac {\int \frac {1}{2 \tan \left (\frac {1}{2} (d+e x)\right )+2}d\tan \left (\frac {1}{2} (d+e x)\right )}{2 a^2 e}-\frac {a \cos (d+e x)-a \sin (d+e x)}{4 e \left (a^3 \sin (d+e x)+a^3 \cos (d+e x)+a^3\right )}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle -\frac {a \cos (d+e x)-a \sin (d+e x)}{4 e \left (a^3 \sin (d+e x)+a^3 \cos (d+e x)+a^3\right )}-\frac {\log \left (\tan \left (\frac {1}{2} (d+e x)\right )+1\right )}{4 a^2 e}\) |
-1/4*Log[1 + Tan[(d + e*x)/2]]/(a^2*e) - (a*Cos[d + e*x] - a*Sin[d + e*x]) /(4*e*(a^3 + a^3*Cos[d + e*x] + a^3*Sin[d + e*x]))
3.4.71.3.1 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[(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]
Time = 0.92 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.64
method | result | size |
derivativedivides | \(\frac {\tan \left (\frac {e x}{2}+\frac {d}{2}\right )-2 \ln \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )-\frac {2}{1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )}}{8 e \,a^{2}}\) | \(48\) |
default | \(\frac {\tan \left (\frac {e x}{2}+\frac {d}{2}\right )-2 \ln \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )-\frac {2}{1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )}}{8 e \,a^{2}}\) | \(48\) |
norman | \(\frac {\frac {\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}{8 a e}-\frac {3}{8 a e}}{a \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}-\frac {\ln \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{4 a^{2} e}\) | \(67\) |
parallelrisch | \(\frac {\left (-2 \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-2\right ) \ln \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}+3 \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{8 e \,a^{2} \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}\) | \(71\) |
risch | \(\frac {\left (-\frac {1}{4}+\frac {i}{4}\right ) \left ({\mathrm e}^{i \left (e x +d \right )}+1+i\right )}{a^{2} e \left (i {\mathrm e}^{i \left (e x +d \right )}+{\mathrm e}^{2 i \left (e x +d \right )}+i+{\mathrm e}^{i \left (e x +d \right )}\right )}-\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}+i\right )}{4 a^{2} e}+\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}+1\right )}{4 a^{2} e}\) | \(101\) |
Time = 0.25 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.33 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^2} \, dx=\frac {{\left (\cos \left (e x + d\right ) + \sin \left (e x + d\right ) + 1\right )} \log \left (\frac {1}{2} \, \cos \left (e x + d\right ) + \frac {1}{2}\right ) - {\left (\cos \left (e x + d\right ) + \sin \left (e x + d\right ) + 1\right )} \log \left (\sin \left (e x + d\right ) + 1\right ) - 2 \, \cos \left (e x + d\right ) + 2 \, \sin \left (e x + d\right )}{8 \, {\left (a^{2} e \cos \left (e x + d\right ) + a^{2} e \sin \left (e x + d\right ) + a^{2} e\right )}} \]
1/8*((cos(e*x + d) + sin(e*x + d) + 1)*log(1/2*cos(e*x + d) + 1/2) - (cos( e*x + d) + sin(e*x + d) + 1)*log(sin(e*x + d) + 1) - 2*cos(e*x + d) + 2*si n(e*x + d))/(a^2*e*cos(e*x + d) + a^2*e*sin(e*x + d) + a^2*e)
Leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (65) = 130\).
Time = 0.83 (sec) , antiderivative size = 168, normalized size of antiderivative = 2.24 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^2} \, dx=\begin {cases} - \frac {2 \log {\left (\tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 1 \right )} \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )}}{8 a^{2} e \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 8 a^{2} e} - \frac {2 \log {\left (\tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 1 \right )}}{8 a^{2} e \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 8 a^{2} e} + \frac {\tan ^{2}{\left (\frac {d}{2} + \frac {e x}{2} \right )}}{8 a^{2} e \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 8 a^{2} e} - \frac {3}{8 a^{2} e \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 8 a^{2} e} & \text {for}\: e \neq 0 \\\frac {x}{\left (2 a \sin {\left (d \right )} + 2 a \cos {\left (d \right )} + 2 a\right )^{2}} & \text {otherwise} \end {cases} \]
Piecewise((-2*log(tan(d/2 + e*x/2) + 1)*tan(d/2 + e*x/2)/(8*a**2*e*tan(d/2 + e*x/2) + 8*a**2*e) - 2*log(tan(d/2 + e*x/2) + 1)/(8*a**2*e*tan(d/2 + e* x/2) + 8*a**2*e) + tan(d/2 + e*x/2)**2/(8*a**2*e*tan(d/2 + e*x/2) + 8*a**2 *e) - 3/(8*a**2*e*tan(d/2 + e*x/2) + 8*a**2*e), Ne(e, 0)), (x/(2*a*sin(d) + 2*a*cos(d) + 2*a)**2, True))
Time = 0.22 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.07 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^2} \, dx=-\frac {\frac {2}{a^{2} + \frac {a^{2} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}} + \frac {2 \, \log \left (\frac {\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + 1\right )}{a^{2}} - \frac {\sin \left (e x + d\right )}{a^{2} {\left (\cos \left (e x + d\right ) + 1\right )}}}{8 \, e} \]
-1/8*(2/(a^2 + a^2*sin(e*x + d)/(cos(e*x + d) + 1)) + 2*log(sin(e*x + d)/( cos(e*x + d) + 1) + 1)/a^2 - sin(e*x + d)/(a^2*(cos(e*x + d) + 1)))/e
Time = 0.30 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.87 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^2} \, dx=-\frac {\frac {2 \, \log \left ({\left | \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + 1 \right |}\right )}{a^{2}} - \frac {\tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )}{a^{2}} - \frac {2 \, \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )}{a^{2} {\left (\tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + 1\right )}}}{8 \, e} \]
-1/8*(2*log(abs(tan(1/2*e*x + 1/2*d) + 1))/a^2 - tan(1/2*e*x + 1/2*d)/a^2 - 2*tan(1/2*e*x + 1/2*d)/(a^2*(tan(1/2*e*x + 1/2*d) + 1)))/e
Time = 27.88 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.79 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^2} \, dx=\frac {\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}{8\,a^2\,e}-\frac {\ln \left (\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )+1\right )}{4\,a^2\,e}-\frac {1}{4\,a^2\,e\,\left (\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )+1\right )} \]