Integrand size = 24, antiderivative size = 75 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 c \sin (d+e x))^2} \, dx=-\frac {a \log \left (a+c \tan \left (\frac {1}{2} (d+e x)\right )\right )}{4 c^3 e}-\frac {c \cos (d+e x)-a \sin (d+e x)}{4 c^2 e (a+a \cos (d+e x)+c \sin (d+e x))} \]
-1/4*a*ln(a+c*tan(1/2*e*x+1/2*d))/c^3/e+1/4*(-c*cos(e*x+d)+a*sin(e*x+d))/c ^2/e/(a+a*cos(e*x+d)+c*sin(e*x+d))
Time = 0.48 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.53 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 c \sin (d+e x))^2} \, dx=\frac {2 a \left (\log \left (\cos \left (\frac {1}{2} (d+e x)\right )\right )-\log \left (a \cos \left (\frac {1}{2} (d+e x)\right )+c \sin \left (\frac {1}{2} (d+e x)\right )\right )\right )+\frac {c \left (a^2+c^2\right ) \sin \left (\frac {1}{2} (d+e x)\right )}{a \left (a \cos \left (\frac {1}{2} (d+e x)\right )+c \sin \left (\frac {1}{2} (d+e x)\right )\right )}+c \tan \left (\frac {1}{2} (d+e x)\right )}{8 c^3 e} \]
(2*a*(Log[Cos[(d + e*x)/2]] - Log[a*Cos[(d + e*x)/2] + c*Sin[(d + e*x)/2]] ) + (c*(a^2 + c^2)*Sin[(d + e*x)/2])/(a*(a*Cos[(d + e*x)/2] + c*Sin[(d + e *x)/2])) + c*Tan[(d + e*x)/2])/(8*c^3*e)
Time = 0.31 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {3042, 3608, 25, 27, 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.
\(\displaystyle \int \frac {1}{(2 a \cos (d+e x)+2 a+2 c \sin (d+e x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(2 a \cos (d+e x)+2 a+2 c \sin (d+e x))^2}dx\) |
\(\Big \downarrow \) 3608 |
\(\displaystyle \frac {\int -\frac {a}{\cos (d+e x) a+a+c \sin (d+e x)}dx}{4 c^2}-\frac {c \cos (d+e x)-a \sin (d+e x)}{4 c^2 e (a \cos (d+e x)+a+c \sin (d+e x))}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {a}{\cos (d+e x) a+a+c \sin (d+e x)}dx}{4 c^2}-\frac {c \cos (d+e x)-a \sin (d+e x)}{4 c^2 e (a \cos (d+e x)+a+c \sin (d+e x))}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {a \int \frac {1}{\cos (d+e x) a+a+c \sin (d+e x)}dx}{4 c^2}-\frac {c \cos (d+e x)-a \sin (d+e x)}{4 c^2 e (a \cos (d+e x)+a+c \sin (d+e x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {a \int \frac {1}{\cos (d+e x) a+a+c \sin (d+e x)}dx}{4 c^2}-\frac {c \cos (d+e x)-a \sin (d+e x)}{4 c^2 e (a \cos (d+e x)+a+c \sin (d+e x))}\) |
\(\Big \downarrow \) 3603 |
\(\displaystyle -\frac {a \int \frac {1}{2 a+2 c \tan \left (\frac {1}{2} (d+e x)\right )}d\tan \left (\frac {1}{2} (d+e x)\right )}{2 c^2 e}-\frac {c \cos (d+e x)-a \sin (d+e x)}{4 c^2 e (a \cos (d+e x)+a+c \sin (d+e x))}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle -\frac {a \log \left (a+c \tan \left (\frac {1}{2} (d+e x)\right )\right )}{4 c^3 e}-\frac {c \cos (d+e x)-a \sin (d+e x)}{4 c^2 e (a \cos (d+e x)+a+c \sin (d+e x))}\) |
-1/4*(a*Log[a + c*Tan[(d + e*x)/2]])/(c^3*e) - (c*Cos[d + e*x] - a*Sin[d + e*x])/(4*c^2*e*(a + a*Cos[d + e*x] + c*Sin[d + e*x]))
3.4.67.3.1 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]
Time = 0.98 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.91
method | result | size |
derivativedivides | \(\frac {\frac {\tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{2 c^{2}}-\frac {a \ln \left (a +c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{c^{3}}-\frac {a^{2}+c^{2}}{2 c^{3} \left (a +c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}}{4 e}\) | \(68\) |
default | \(\frac {\frac {\tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{2 c^{2}}-\frac {a \ln \left (a +c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{c^{3}}-\frac {a^{2}+c^{2}}{2 c^{3} \left (a +c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}}{4 e}\) | \(68\) |
norman | \(\frac {-\frac {2 a^{2}+c^{2}}{8 c^{3} e}+\frac {\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}{8 c e}}{a +c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}-\frac {a \ln \left (a +c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{4 c^{3} e}\) | \(78\) |
parallelrisch | \(\frac {-2 \ln \left (a +c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right ) a^{2} c +\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2} a \,c^{2}-2 \ln \left (a +c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right ) a^{3}+2 \tan \left (\frac {e x}{2}+\frac {d}{2}\right ) a^{2} c +\tan \left (\frac {e x}{2}+\frac {d}{2}\right ) c^{3}}{8 a \,c^{3} e \left (a +c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}\) | \(120\) |
risch | \(-\frac {i \left (-i a \,{\mathrm e}^{i \left (e x +d \right )}-i a +c \right )}{2 c^{2} e \left (c \,{\mathrm e}^{2 i \left (e x +d \right )}+i a \,{\mathrm e}^{2 i \left (e x +d \right )}-c +2 i a \,{\mathrm e}^{i \left (e x +d \right )}+i a \right )}+\frac {a \ln \left ({\mathrm e}^{i \left (e x +d \right )}+1\right )}{4 c^{3} e}-\frac {a \ln \left ({\mathrm e}^{i \left (e x +d \right )}-\frac {i c +a}{i c -a}\right )}{4 c^{3} e}\) | \(136\) |
1/4/e*(1/2*tan(1/2*e*x+1/2*d)/c^2-1/c^3*a*ln(a+c*tan(1/2*e*x+1/2*d))-1/2/c ^3*(a^2+c^2)/(a+c*tan(1/2*e*x+1/2*d)))
Leaf count of result is larger than twice the leaf count of optimal. 154 vs. \(2 (70) = 140\).
Time = 0.26 (sec) , antiderivative size = 154, normalized size of antiderivative = 2.05 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 c \sin (d+e x))^2} \, dx=-\frac {2 \, c^{2} \cos \left (e x + d\right ) - 2 \, a c \sin \left (e x + d\right ) + {\left (a^{2} \cos \left (e x + d\right ) + a c \sin \left (e x + d\right ) + a^{2}\right )} \log \left (a c \sin \left (e x + d\right ) + \frac {1}{2} \, a^{2} + \frac {1}{2} \, c^{2} + \frac {1}{2} \, {\left (a^{2} - c^{2}\right )} \cos \left (e x + d\right )\right ) - {\left (a^{2} \cos \left (e x + d\right ) + a c \sin \left (e x + d\right ) + a^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (e x + d\right ) + \frac {1}{2}\right )}{8 \, {\left (a c^{3} e \cos \left (e x + d\right ) + c^{4} e \sin \left (e x + d\right ) + a c^{3} e\right )}} \]
-1/8*(2*c^2*cos(e*x + d) - 2*a*c*sin(e*x + d) + (a^2*cos(e*x + d) + a*c*si n(e*x + d) + a^2)*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)) - (a^2*cos(e*x + d) + a*c*sin(e*x + d) + a^2)*log(1/2*cos (e*x + d) + 1/2))/(a*c^3*e*cos(e*x + d) + c^4*e*sin(e*x + d) + a*c^3*e)
Timed out. \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 c \sin (d+e x))^2} \, dx=\text {Timed out} \]
Time = 0.23 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.20 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 c \sin (d+e x))^2} \, dx=-\frac {\frac {a^{2} + c^{2}}{a c^{3} + \frac {c^{4} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}} + \frac {2 \, a \log \left (a + \frac {c \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}\right )}{c^{3}} - \frac {\sin \left (e x + d\right )}{c^{2} {\left (\cos \left (e x + d\right ) + 1\right )}}}{8 \, e} \]
-1/8*((a^2 + c^2)/(a*c^3 + c^4*sin(e*x + d)/(cos(e*x + d) + 1)) + 2*a*log( a + c*sin(e*x + d)/(cos(e*x + d) + 1))/c^3 - sin(e*x + d)/(c^2*(cos(e*x + d) + 1)))/e
Time = 0.28 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.11 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 c \sin (d+e x))^2} \, dx=-\frac {\frac {2 \, a \log \left ({\left | c \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + a \right |}\right )}{c^{3}} - \frac {\tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )}{c^{2}} - \frac {2 \, a c \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + a^{2} - c^{2}}{{\left (c \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + a\right )} c^{3}}}{8 \, e} \]
-1/8*(2*a*log(abs(c*tan(1/2*e*x + 1/2*d) + a))/c^3 - tan(1/2*e*x + 1/2*d)/ c^2 - (2*a*c*tan(1/2*e*x + 1/2*d) + a^2 - c^2)/((c*tan(1/2*e*x + 1/2*d) + a)*c^3))/e
Time = 26.77 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.05 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 c \sin (d+e x))^2} \, dx=\frac {\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}{8\,c^2\,e}-\frac {a\,\ln \left (a+c\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\right )}{4\,c^3\,e}-\frac {a^2+c^2}{c\,e\,\left (8\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,c^3+8\,a\,c^2\right )} \]