Integrand size = 24, antiderivative size = 83 \[ \int \frac {1}{(2 a+2 b \cos (d+e x)+2 a \sin (d+e x))^2} \, dx=\frac {a \log \left (a+b \cot \left (\frac {d}{2}+\frac {\pi }{4}+\frac {e x}{2}\right )\right )}{4 b^3 e}-\frac {a \cos (d+e x)-b \sin (d+e x)}{4 b^2 e (a+b \cos (d+e x)+a \sin (d+e x))} \] Output:
1/4*a*ln(a+b*cot(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 = 0.44 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.95 \[ \int \frac {1}{(2 a+2 b \cos (d+e x)+2 a \sin (d+e x))^2} \, dx=\frac {-a \log \left (\cos \left (\frac {1}{2} (d+e x)\right )+\sin \left (\frac {1}{2} (d+e x)\right )\right )+a \log \left ((a+b) \cos \left (\frac {1}{2} (d+e x)\right )+(a-b) \sin \left (\frac {1}{2} (d+e x)\right )\right )+\frac {b \sin \left (\frac {1}{2} (d+e x)\right )}{\cos \left (\frac {1}{2} (d+e x)\right )+\sin \left (\frac {1}{2} (d+e x)\right )}+\frac {b \left (a^2+b^2\right ) \sin \left (\frac {1}{2} (d+e x)\right )}{(a+b) \left ((a+b) \cos \left (\frac {1}{2} (d+e x)\right )+(a-b) \sin \left (\frac {1}{2} (d+e x)\right )\right )}}{4 b^3 e} \] Input:
Integrate[(2*a + 2*b*Cos[d + e*x] + 2*a*Sin[d + e*x])^(-2),x]
Output:
(-(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)* Cos[(d + e*x)/2] + (a - b)*Sin[(d + e*x)/2])))/(4*b^3*e)
Time = 0.32 (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, 3602, 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 |
| \(\displaystyle -\frac {\int \frac {a}{\sin (d+e x) a+a+b \cos (d+e x)}dx}{4 b^2}-\frac {a \cos (d+e x)-b \sin (d+e x)}{4 b^2 e (a \sin (d+e x)+a+b \cos (d+e x))}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {a \int \frac {1}{\sin (d+e x) a+a+b \cos (d+e x)}dx}{4 b^2}-\frac {a \cos (d+e x)-b \sin (d+e x)}{4 b^2 e (a \sin (d+e x)+a+b \cos (d+e x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {a \int \frac {1}{\sin (d+e x) a+a+b \cos (d+e x)}dx}{4 b^2}-\frac {a \cos (d+e x)-b \sin (d+e x)}{4 b^2 e (a \sin (d+e x)+a+b \cos (d+e x))}\) |
| \(\Big \downarrow \) 3602 |
| \(\displaystyle \frac {a \int \frac {1}{a+b \cot \left (\frac {d}{2}+\frac {e x}{2}+\frac {\pi }{4}\right )}d\cot \left (\frac {d}{2}+\frac {e x}{2}+\frac {\pi }{4}\right )}{4 b^2 e}-\frac {a \cos (d+e x)-b \sin (d+e x)}{4 b^2 e (a \sin (d+e x)+a+b \cos (d+e x))}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {a \log \left (a+b \cot \left (\frac {d}{2}+\frac {e x}{2}+\frac {\pi }{4}\right )\right )}{4 b^3 e}-\frac {a \cos (d+e x)-b \sin (d+e x)}{4 b^2 e (a \sin (d+e x)+a+b \cos (d+e x))}\) |
Input:
Int[(2*a + 2*b*Cos[d + e*x] + 2*a*Sin[d + e*x])^(-2),x]
Output:
(a*Log[a + b*Cot[d/2 + Pi/4 + (e*x)/2]])/(4*b^3*e) - (a*Cos[d + e*x] - b*S in[d + e*x])/(4*b^2*e*(a + b*Cos[d + e*x] + a*Sin[d + e*x]))
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[Cot[(d + e*x)/2 + Pi/4], x]}, Si mp[-f/e Subst[Int[1/(a + b*f*x), x], x, Cot[(d + e*x)/2 + Pi/4]/f], x]] / ; FreeQ[{a, b, c, d, e}, x] && EqQ[a - c, 0] && NeQ[a - b, 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 = 122, normalized size of antiderivative = 1.47
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{b^{2} \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}-\frac {a \ln \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{b^{3}}-\frac {a^{2}+b^{2}}{b^{2} \left (a -b \right ) \left (a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-b \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+a +b \right )}+\frac {a \ln \left (a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-b \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+a +b \right )}{b^{3}}}{4 e}\) | \(122\) |
| default | \(\frac {-\frac {1}{b^{2} \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}-\frac {a \ln \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{b^{3}}-\frac {a^{2}+b^{2}}{b^{2} \left (a -b \right ) \left (a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-b \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+a +b \right )}+\frac {a \ln \left (a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-b \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+a +b \right )}{b^{3}}}{4 e}\) | \(122\) |
| risch | \(\frac {i \left (i a +b +a \,{\mathrm e}^{i \left (e x +d \right )}\right )}{2 b^{2} e \left (-i a \,{\mathrm e}^{2 i \left (e x +d \right )}+b \,{\mathrm e}^{2 i \left (e x +d \right )}+i a +2 a \,{\mathrm e}^{i \left (e x +d \right )}+b \right )}+\frac {a \ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {i a +b}{i b +a}\right )}{4 b^{3} e}-\frac {a \ln \left ({\mathrm e}^{i \left (e x +d \right )}+i\right )}{4 b^{3} e}\) | \(129\) |
| parallelrisch | \(\frac {a^{2} \left (a \left (\sin \left (e x +d \right )+1\right )+b \cos \left (e x +d \right )\right ) \ln \left (a +b +\left (a -b \right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )+\left (-a^{2} b \cos \left (e x +d \right )-a^{3} \left (\sin \left (e x +d \right )+1\right )\right ) \ln \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )+\left (-a^{2} b -b^{3}\right ) \cos \left (e x +d \right )-a \,b^{2}}{4 b^{3} a e \left (a \left (\sin \left (e x +d \right )+1\right )+b \cos \left (e x +d \right )\right )}\) | \(140\) |
| norman | \(\frac {-\frac {a^{2}+b a +b^{2}}{4 a e \,b^{2}}+\frac {\left (a^{2}-b a +b^{2}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}{4 a e \,b^{2}}}{\left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right ) \left (a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-b \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+a +b \right )}-\frac {a \ln \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{4 b^{3} e}+\frac {a \ln \left (a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-b \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+a +b \right )}{4 b^{3} e}\) | \(156\) |
Input:
int(1/(2*a+2*b*cos(e*x+d)+2*a*sin(e*x+d))^2,x,method=_RETURNVERBOSE)
Output:
1/4/e*(-1/b^2/(1+tan(1/2*e*x+1/2*d))-a/b^3*ln(1+tan(1/2*e*x+1/2*d))-(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*t an(1/2*e*x+1/2*d)-b*tan(1/2*e*x+1/2*d)+a+b))
Leaf count of result is larger than twice the leaf count of optimal. 148 vs. \(2 (73) = 146\).
Time = 0.08 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.78 \[ \int \frac {1}{(2 a+2 b \cos (d+e x)+2 a \sin (d+e x))^2} \, dx=-\frac {2 \, a b \cos \left (e x + d\right ) - 2 \, b^{2} \sin \left (e x + d\right ) - {\left (a b \cos \left (e x + d\right ) + a^{2} \sin \left (e x + d\right ) + a^{2}\right )} \log \left (2 \, a b \cos \left (e x + d\right ) + a^{2} + b^{2} + {\left (a^{2} - b^{2}\right )} \sin \left (e x + d\right )\right ) + {\left (a b \cos \left (e x + d\right ) + a^{2} \sin \left (e x + d\right ) + a^{2}\right )} \log \left (\sin \left (e x + d\right ) + 1\right )}{8 \, {\left (b^{4} e \cos \left (e x + d\right ) + a b^{3} e \sin \left (e x + d\right ) + a b^{3} e\right )}} \] Input:
integrate(1/(2*a+2*b*cos(e*x+d)+2*a*sin(e*x+d))^2,x, algorithm="fricas")
Output:
-1/8*(2*a*b*cos(e*x + d) - 2*b^2*sin(e*x + d) - (a*b*cos(e*x + d) + a^2*si n(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. \[ \int \frac {1}{(2 a+2 b \cos (d+e x)+2 a \sin (d+e x))^2} \, dx=\text {Timed out} \] Input:
integrate(1/(2*a+2*b*cos(e*x+d)+2*a*sin(e*x+d))**2,x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 185 vs. \(2 (73) = 146\).
Time = 0.04 (sec) , antiderivative size = 185, normalized size of antiderivative = 2.23 \[ \int \frac {1}{(2 a+2 b \cos (d+e x)+2 a \sin (d+e x))^2} \, dx=-\frac {\frac {2 \, {\left (a^{2} + \frac {{\left (a^{2} - a b + b^{2}\right )} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}\right )}}{a^{2} b^{2} - b^{4} + \frac {2 \, {\left (a^{2} b^{2} - a b^{3}\right )} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + \frac {{\left (a^{2} b^{2} - 2 \, a b^{3} + b^{4}\right )} \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}}} - \frac {a \log \left (-a - b - \frac {{\left (a - b\right )} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}\right )}{b^{3}} + \frac {a \log \left (\frac {\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + 1\right )}{b^{3}}}{4 \, e} \] Input:
integrate(1/(2*a+2*b*cos(e*x+d)+2*a*sin(e*x+d))^2,x, algorithm="maxima")
Output:
-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. 187 vs. \(2 (73) = 146\).
Time = 0.13 (sec) , antiderivative size = 187, normalized size of antiderivative = 2.25 \[ \int \frac {1}{(2 a+2 b \cos (d+e x)+2 a \sin (d+e x))^2} \, dx=-\frac {\frac {2 \, {\left (a^{2} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) - a b \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + b^{2} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + a^{2}\right )}}{{\left (a b^{2} - b^{3}\right )} {\left (a \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} - b \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} + 2 \, a \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + a + b\right )}} + \frac {a \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) - 2 \, b \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + 2 \, a - 2 \, {\left | b \right |} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) - 2 \, b \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + 2 \, a + 2 \, {\left | b \right |} \right |}}\right )}{b^{2} {\left | b \right |}}}{4 \, e} \] Input:
integrate(1/(2*a+2*b*cos(e*x+d)+2*a*sin(e*x+d))^2,x, algorithm="giac")
Output:
-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 = 16.13 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.52 \[ \int \frac {1}{(2 a+2 b \cos (d+e x)+2 a \sin (d+e x))^2} \, dx=\frac {a\,\mathrm {atanh}\left (\frac {a+\frac {\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (2\,a-2\,b\right )}{2}}{b}\right )}{2\,b^3\,e}-\frac {\frac {a^2}{b^2\,\left (a-b\right )}+\frac {\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (a^2-a\,b+b^2\right )}{b^2\,\left (a-b\right )}}{e\,\left (\left (2\,a-2\,b\right )\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2+4\,a\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )+2\,a+2\,b\right )} \] Input:
int(1/(2*a + 2*b*cos(d + e*x) + 2*a*sin(d + e*x))^2,x)
Output:
(a*atanh((a + (tan(d/2 + (e*x)/2)*(2*a - 2*b))/2)/b))/(2*b^3*e) - (a^2/(b^ 2*(a - b)) + (tan(d/2 + (e*x)/2)*(a^2 - a*b + b^2))/(b^2*(a - b)))/(e*(2*a + 2*b + tan(d/2 + (e*x)/2)^2*(2*a - 2*b) + 4*a*tan(d/2 + (e*x)/2)))
Time = 0.18 (sec) , antiderivative size = 231, normalized size of antiderivative = 2.78 \[ \int \frac {1}{(2 a+2 b \cos (d+e x)+2 a \sin (d+e x))^2} \, dx=\frac {-\cos \left (e x +d \right ) \mathrm {log}\left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )+1\right ) a^{2} b +\cos \left (e x +d \right ) \mathrm {log}\left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right ) a -\tan \left (\frac {e x}{2}+\frac {d}{2}\right ) b +a +b \right ) a^{2} b -\cos \left (e x +d \right ) a^{2} b -\cos \left (e x +d \right ) b^{3}-\mathrm {log}\left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )+1\right ) \sin \left (e x +d \right ) a^{3}-\mathrm {log}\left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )+1\right ) a^{3}+\mathrm {log}\left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right ) a -\tan \left (\frac {e x}{2}+\frac {d}{2}\right ) b +a +b \right ) \sin \left (e x +d \right ) a^{3}+\mathrm {log}\left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right ) a -\tan \left (\frac {e x}{2}+\frac {d}{2}\right ) b +a +b \right ) a^{3}-a \,b^{2}}{4 a \,b^{3} e \left (b \cos \left (e x +d \right )+\sin \left (e x +d \right ) a +a \right )} \] Input:
int(1/(2*a+2*b*cos(e*x+d)+2*a*sin(e*x+d))^2,x)
Output:
( - cos(d + e*x)*log(tan((d + e*x)/2) + 1)*a**2*b + cos(d + e*x)*log(tan(( d + e*x)/2)*a - tan((d + e*x)/2)*b + a + b)*a**2*b - cos(d + e*x)*a**2*b - cos(d + e*x)*b**3 - log(tan((d + e*x)/2) + 1)*sin(d + e*x)*a**3 - log(tan ((d + e*x)/2) + 1)*a**3 + log(tan((d + e*x)/2)*a - tan((d + e*x)/2)*b + a + b)*sin(d + e*x)*a**3 + log(tan((d + e*x)/2)*a - tan((d + e*x)/2)*b + a + b)*a**3 - a*b**2)/(4*a*b**3*e*(cos(d + e*x)*b + sin(d + e*x)*a + a))