Integrand size = 24, antiderivative size = 83 \[ \int \frac {1}{(2 a+2 b \cos (d+e x)-2 a \sin (d+e x))^2} \, dx=-\frac {a \log \left (a+b \tan \left (\frac {d}{2}+\frac {\pi }{4}+\frac {e x}{2}\right )\right )}{4 b^3 e}+\frac {a \cos (d+e x)+b \sin (d+e x)}{4 b^2 e (a+b \cos (d+e x)-a \sin (d+e x))} \]
-1/4*a*ln(a+b*tan(1/2*d+1/4*Pi+1/2*e*x))/b^3/e+1/4*(a*cos(e*x+d)+b*sin(e*x +d))/b^2/e/(a+b*cos(e*x+d)-a*sin(e*x+d))
Time = 0.89 (sec) , antiderivative size = 166, normalized size of antiderivative = 2.00 \[ \int \frac {1}{(2 a+2 b \cos (d+e x)-2 a \sin (d+e x))^2} \, dx=\frac {a \log \left (\cos \left (\frac {1}{2} (d+e x)\right )-\sin \left (\frac {1}{2} (d+e x)\right )\right )-a \log \left ((a+b) \cos \left (\frac {1}{2} (d+e x)\right )+(-a+b) \sin \left (\frac {1}{2} (d+e x)\right )\right )+\frac {b \sin \left (\frac {1}{2} (d+e x)\right )}{\cos \left (\frac {1}{2} (d+e x)\right )-\sin \left (\frac {1}{2} (d+e x)\right )}+\frac {b \left (a^2+b^2\right ) \sin \left (\frac {1}{2} (d+e x)\right )}{(a+b) \left ((a+b) \cos \left (\frac {1}{2} (d+e x)\right )+(-a+b) \sin \left (\frac {1}{2} (d+e x)\right )\right )}}{4 b^3 e} \]
(a*Log[Cos[(d + e*x)/2] - Sin[(d + e*x)/2]] - a*Log[(a + b)*Cos[(d + e*x)/ 2] + (-a + b)*Sin[(d + e*x)/2]] + (b*Sin[(d + e*x)/2])/(Cos[(d + e*x)/2] - Sin[(d + e*x)/2]) + (b*(a^2 + b^2)*Sin[(d + e*x)/2])/((a + b)*((a + b)*Co s[(d + e*x)/2] + (-a + b)*Sin[(d + e*x)/2])))/(4*b^3*e)
Time = 0.31 (sec) , antiderivative size = 83, 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, 3601, 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.
\(\displaystyle \int \frac {1}{(-2 a \sin (d+e x)+2 a+2 b \cos (d+e x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(-2 a \sin (d+e x)+2 a+2 b \cos (d+e x))^2}dx\) |
\(\Big \downarrow \) 3608 |
\(\displaystyle \frac {\int -\frac {a}{-\sin (d+e x) a+a+b \cos (d+e x)}dx}{4 b^2}+\frac {a \cos (d+e x)+b \sin (d+e x)}{4 b^2 e (a (-\sin (d+e x))+a+b \cos (d+e x))}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {a \cos (d+e x)+b \sin (d+e x)}{4 b^2 e (a (-\sin (d+e x))+a+b \cos (d+e x))}-\frac {\int \frac {a}{-\sin (d+e x) a+a+b \cos (d+e x)}dx}{4 b^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a \cos (d+e x)+b \sin (d+e x)}{4 b^2 e (a (-\sin (d+e x))+a+b \cos (d+e x))}-\frac {a \int \frac {1}{-\sin (d+e x) a+a+b \cos (d+e x)}dx}{4 b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {a \cos (d+e x)+b \sin (d+e x)}{4 b^2 e (a (-\sin (d+e x))+a+b \cos (d+e x))}-\frac {a \int \frac {1}{-\sin (d+e x) a+a+b \cos (d+e x)}dx}{4 b^2}\) |
\(\Big \downarrow \) 3601 |
\(\displaystyle \frac {a \cos (d+e x)+b \sin (d+e x)}{4 b^2 e (a (-\sin (d+e x))+a+b \cos (d+e x))}-\frac {a \int \frac {1}{a+b \tan \left (\frac {d}{2}+\frac {e x}{2}+\frac {\pi }{4}\right )}d\tan \left (\frac {d}{2}+\frac {e x}{2}+\frac {\pi }{4}\right )}{4 b^2 e}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {a \cos (d+e x)+b \sin (d+e x)}{4 b^2 e (a (-\sin (d+e x))+a+b \cos (d+e x))}-\frac {a \log \left (a+b \tan \left (\frac {d}{2}+\frac {e x}{2}+\frac {\pi }{4}\right )\right )}{4 b^3 e}\) |
-1/4*(a*Log[a + b*Tan[d/2 + Pi/4 + (e*x)/2]])/(b^3*e) + (a*Cos[d + e*x] + b*Sin[d + e*x])/(4*b^2*e*(a + b*Cos[d + e*x] - a*Sin[d + e*x]))
3.4.92.3.1 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 + Pi/4], x]}, Si mp[f/e Subst[Int[1/(a + b*f*x), x], x, Tan[(d + e*x)/2 + Pi/4]/f], x]] /; FreeQ[{a, b, c, d, e}, x] && EqQ[a + c, 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 = 1.17 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.57
method | result | size |
derivativedivides | \(\frac {-\frac {a^{2}+b^{2}}{b^{2} \left (a -b \right ) \left (a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-b \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-a -b \right )}-\frac {a \ln \left (a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-b \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-a -b \right )}{b^{3}}-\frac {1}{b^{2} \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )-1\right )}+\frac {a \ln \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )-1\right )}{b^{3}}}{4 e}\) | \(130\) |
default | \(\frac {-\frac {a^{2}+b^{2}}{b^{2} \left (a -b \right ) \left (a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-b \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-a -b \right )}-\frac {a \ln \left (a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-b \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-a -b \right )}{b^{3}}-\frac {1}{b^{2} \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )-1\right )}+\frac {a \ln \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )-1\right )}{b^{3}}}{4 e}\) | \(130\) |
risch | \(\frac {i \left (-i a +b +a \,{\mathrm e}^{i \left (e x +d \right )}\right )}{2 b^{2} e \left (i a \,{\mathrm e}^{2 i \left (e x +d \right )}+b \,{\mathrm e}^{2 i \left (e x +d \right )}-i a +2 a \,{\mathrm e}^{i \left (e x +d \right )}+b \right )}+\frac {a \ln \left ({\mathrm e}^{i \left (e x +d \right )}-i\right )}{4 b^{3} e}-\frac {a \ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {i a -b}{i b -a}\right )}{4 b^{3} e}\) | \(133\) |
parallelrisch | \(\frac {\left (\cos \left (e x +d \right ) a^{2} b -a^{3} \left (\sin \left (e x +d \right )-1\right )\right ) \ln \left (-a -b +\left (a -b \right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )+a^{2} \left (a \left (\sin \left (e x +d \right )-1\right )-b \cos \left (e x +d \right )\right ) \ln \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )-1\right )+\left (-a^{2} b -b^{3}\right ) \cos \left (e x +d \right )-a \,b^{2}}{4 b^{3} e a \left (a \left (\sin \left (e x +d \right )-1\right )-b \cos \left (e x +d \right )\right )}\) | \(145\) |
norman | \(\frac {\frac {a^{2}+a b +b^{2}}{4 a \,b^{2} e}-\frac {\left (a^{2}-a b +b^{2}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}{4 a \,b^{2} e}}{\left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )-1\right ) \left (a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-b \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-a -b \right )}+\frac {a \ln \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )-1\right )}{4 b^{3} e}-\frac {a \ln \left (a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-b \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-a -b \right )}{4 b^{3} e}\) | \(164\) |
1/4/e*(-(a^2+b^2)/b^2/(a-b)/(a*tan(1/2*e*x+1/2*d)-b*tan(1/2*e*x+1/2*d)-a-b )-a/b^3*ln(a*tan(1/2*e*x+1/2*d)-b*tan(1/2*e*x+1/2*d)-a-b)-1/b^2/(tan(1/2*e *x+1/2*d)-1)+a/b^3*ln(tan(1/2*e*x+1/2*d)-1))
Leaf count of result is larger than twice the leaf count of optimal. 154 vs. \(2 (73) = 146\).
Time = 0.26 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.86 \[ \int \frac {1}{(2 a+2 b \cos (d+e x)-2 a \sin (d+e x))^2} \, dx=\frac {2 \, a b \cos \left (e x + d\right ) + 2 \, b^{2} \sin \left (e x + d\right ) - {\left (a b \cos \left (e x + d\right ) - a^{2} \sin \left (e x + d\right ) + a^{2}\right )} \log \left (2 \, a b \cos \left (e x + d\right ) + a^{2} + b^{2} - {\left (a^{2} - b^{2}\right )} \sin \left (e x + d\right )\right ) + {\left (a b \cos \left (e x + d\right ) - a^{2} \sin \left (e x + d\right ) + a^{2}\right )} \log \left (-\sin \left (e x + d\right ) + 1\right )}{8 \, {\left (b^{4} e \cos \left (e x + d\right ) - a b^{3} e \sin \left (e x + d\right ) + a b^{3} e\right )}} \]
1/8*(2*a*b*cos(e*x + d) + 2*b^2*sin(e*x + d) - (a*b*cos(e*x + d) - a^2*sin (e*x + d) + a^2)*log(2*a*b*cos(e*x + d) + a^2 + b^2 - (a^2 - b^2)*sin(e*x + d)) + (a*b*cos(e*x + d) - a^2*sin(e*x + d) + a^2)*log(-sin(e*x + d) + 1) )/(b^4*e*cos(e*x + d) - a*b^3*e*sin(e*x + d) + a*b^3*e)
Timed out. \[ \int \frac {1}{(2 a+2 b \cos (d+e x)-2 a \sin (d+e x))^2} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 182 vs. \(2 (73) = 146\).
Time = 0.24 (sec) , antiderivative size = 182, normalized size of antiderivative = 2.19 \[ \int \frac {1}{(2 a+2 b \cos (d+e x)-2 a \sin (d+e x))^2} \, dx=\frac {\frac {2 \, {\left (a^{2} - \frac {{\left (a^{2} - a b + b^{2}\right )} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}\right )}}{a^{2} b^{2} - b^{4} - \frac {2 \, {\left (a^{2} b^{2} - a b^{3}\right )} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + \frac {{\left (a^{2} b^{2} - 2 \, a b^{3} + b^{4}\right )} \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}}} - \frac {a \log \left (a + b - \frac {{\left (a - b\right )} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}\right )}{b^{3}} + \frac {a \log \left (\frac {\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} - 1\right )}{b^{3}}}{4 \, e} \]
1/4*(2*(a^2 - (a^2 - a*b + b^2)*sin(e*x + d)/(cos(e*x + d) + 1))/(a^2*b^2 - b^4 - 2*(a^2*b^2 - a*b^3)*sin(e*x + d)/(cos(e*x + d) + 1) + (a^2*b^2 - 2 *a*b^3 + b^4)*sin(e*x + d)^2/(cos(e*x + d) + 1)^2) - a*log(a + b - (a - b) *sin(e*x + d)/(cos(e*x + d) + 1))/b^3 + a*log(sin(e*x + d)/(cos(e*x + d) + 1) - 1)/b^3)/e
Leaf count of result is larger than twice the leaf count of optimal. 189 vs. \(2 (73) = 146\).
Time = 0.30 (sec) , antiderivative size = 189, normalized size of antiderivative = 2.28 \[ \int \frac {1}{(2 a+2 b \cos (d+e x)-2 a \sin (d+e x))^2} \, dx=-\frac {\frac {2 \, {\left (a^{2} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) - a b \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + b^{2} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) - a^{2}\right )}}{{\left (a b^{2} - b^{3}\right )} {\left (a \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} - b \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} - 2 \, a \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + a + b\right )}} + \frac {a \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) - 2 \, b \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) - 2 \, a - 2 \, {\left | b \right |} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) - 2 \, b \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) - 2 \, a + 2 \, {\left | b \right |} \right |}}\right )}{b^{2} {\left | b \right |}}}{4 \, e} \]
-1/4*(2*(a^2*tan(1/2*e*x + 1/2*d) - a*b*tan(1/2*e*x + 1/2*d) + b^2*tan(1/2 *e*x + 1/2*d) - a^2)/((a*b^2 - b^3)*(a*tan(1/2*e*x + 1/2*d)^2 - b*tan(1/2* e*x + 1/2*d)^2 - 2*a*tan(1/2*e*x + 1/2*d) + a + b)) + a*log(abs(2*a*tan(1/ 2*e*x + 1/2*d) - 2*b*tan(1/2*e*x + 1/2*d) - 2*a - 2*abs(b))/abs(2*a*tan(1/ 2*e*x + 1/2*d) - 2*b*tan(1/2*e*x + 1/2*d) - 2*a + 2*abs(b)))/(b^2*abs(b))) /e
Time = 27.15 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.52 \[ \int \frac {1}{(2 a+2 b \cos (d+e x)-2 a \sin (d+e x))^2} \, dx=\frac {\frac {a^2}{b^2\,\left (a-b\right )}-\frac {\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (a^2-a\,b+b^2\right )}{b^2\,\left (a-b\right )}}{e\,\left (\left (2\,a-2\,b\right )\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2-4\,a\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )+2\,a+2\,b\right )}-\frac {a\,\mathrm {atanh}\left (\frac {a-\frac {\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (2\,a-2\,b\right )}{2}}{b}\right )}{2\,b^3\,e} \]