Integrand size = 21, antiderivative size = 57 \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {x}{a^2}-\frac {5 \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}+\frac {\sin (c+d x)}{3 d (a+a \cos (c+d x))^2} \] Output:
x/a^2-5/3*sin(d*x+c)/a^2/d/(1+cos(d*x+c))+1/3*sin(d*x+c)/d/(a+a*cos(d*x+c) )^2
Time = 0.48 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.53 \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx=-\frac {\sin (c+d x) \left (12 \arcsin (\cos (c+d x)) \cos ^4\left (\frac {1}{2} (c+d x)\right )+(4+5 \cos (c+d x)) \sqrt {\sin ^2(c+d x)}\right )}{3 a^2 d \sqrt {1-\cos (c+d x)} (1+\cos (c+d x))^{5/2}} \] Input:
Integrate[Cos[c + d*x]^2/(a + a*Cos[c + d*x])^2,x]
Output:
-1/3*(Sin[c + d*x]*(12*ArcSin[Cos[c + d*x]]*Cos[(c + d*x)/2]^4 + (4 + 5*Co s[c + d*x])*Sqrt[Sin[c + d*x]^2]))/(a^2*d*Sqrt[1 - Cos[c + d*x]]*(1 + Cos[ c + d*x])^(5/2))
Time = 0.40 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3237, 25, 3042, 3214, 3042, 3127}
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.
\(\displaystyle \int \frac {\cos ^2(c+d x)}{(a \cos (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
\(\Big \downarrow \) 3237 |
\(\displaystyle \frac {\int -\frac {2 a-3 a \cos (c+d x)}{\cos (c+d x) a+a}dx}{3 a^2}+\frac {\sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}-\frac {\int \frac {2 a-3 a \cos (c+d x)}{\cos (c+d x) a+a}dx}{3 a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}-\frac {\int \frac {2 a-3 a \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}\) |
\(\Big \downarrow \) 3214 |
\(\displaystyle \frac {\sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}-\frac {5 a \int \frac {1}{\cos (c+d x) a+a}dx-3 x}{3 a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}-\frac {5 a \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-3 x}{3 a^2}\) |
\(\Big \downarrow \) 3127 |
\(\displaystyle \frac {\sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}-\frac {\frac {5 a \sin (c+d x)}{d (a \cos (c+d x)+a)}-3 x}{3 a^2}\) |
Input:
Int[Cos[c + d*x]^2/(a + a*Cos[c + d*x])^2,x]
Output:
Sin[c + d*x]/(3*d*(a + a*Cos[c + d*x])^2) - (-3*x + (5*a*Sin[c + d*x])/(d* (a + a*Cos[c + d*x])))/(3*a^2)
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b ^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d Int[1/(c + d *Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[sin[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(a*f*(2*m + 1))), x] - Simp[1/(a^2*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(a*m - b*(2* m + 1)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Time = 0.70 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.63
method | result | size |
parallelrisch | \(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+6 d x -9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{6 a^{2} d}\) | \(36\) |
derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\) | \(46\) |
default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\) | \(46\) |
risch | \(\frac {x}{a^{2}}-\frac {2 i \left (6 \,{\mathrm e}^{2 i \left (d x +c \right )}+9 \,{\mathrm e}^{i \left (d x +c \right )}+5\right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}\) | \(53\) |
norman | \(\frac {\frac {x}{a}+\frac {x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{a}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}-\frac {17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 a d}-\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{6 a d}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{6 a d}+\frac {2 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} a}\) | \(133\) |
Input:
int(cos(d*x+c)^2/(a+a*cos(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/6*(tan(1/2*d*x+1/2*c)^3+6*d*x-9*tan(1/2*d*x+1/2*c))/a^2/d
Time = 0.08 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.40 \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {3 \, d x \cos \left (d x + c\right )^{2} + 6 \, d x \cos \left (d x + c\right ) + 3 \, d x - {\left (5 \, \cos \left (d x + c\right ) + 4\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \] Input:
integrate(cos(d*x+c)^2/(a+a*cos(d*x+c))^2,x, algorithm="fricas")
Output:
1/3*(3*d*x*cos(d*x + c)^2 + 6*d*x*cos(d*x + c) + 3*d*x - (5*cos(d*x + c) + 4)*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)
Time = 0.59 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.98 \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\begin {cases} \frac {x}{a^{2}} + \frac {\tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d} - \frac {3 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a^{2} d} & \text {for}\: d \neq 0 \\\frac {x \cos ^{2}{\left (c \right )}}{\left (a \cos {\left (c \right )} + a\right )^{2}} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**2/(a+a*cos(d*x+c))**2,x)
Output:
Piecewise((x/a**2 + tan(c/2 + d*x/2)**3/(6*a**2*d) - 3*tan(c/2 + d*x/2)/(2 *a**2*d), Ne(d, 0)), (x*cos(c)**2/(a*cos(c) + a)**2, True))
Time = 0.14 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.26 \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx=-\frac {\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {12 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{6 \, d} \] Input:
integrate(cos(d*x+c)^2/(a+a*cos(d*x+c))^2,x, algorithm="maxima")
Output:
-1/6*((9*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 12*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d
Time = 0.36 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.88 \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {\frac {6 \, {\left (d x + c\right )}}{a^{2}} + \frac {a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \] Input:
integrate(cos(d*x+c)^2/(a+a*cos(d*x+c))^2,x, algorithm="giac")
Output:
1/6*(6*(d*x + c)/a^2 + (a^4*tan(1/2*d*x + 1/2*c)^3 - 9*a^4*tan(1/2*d*x + 1 /2*c))/a^6)/d
Time = 41.82 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.61 \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-9\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+6\,d\,x}{6\,a^2\,d} \] Input:
int(cos(c + d*x)^2/(a + a*cos(c + d*x))^2,x)
Output:
(tan(c/2 + (d*x)/2)^3 - 9*tan(c/2 + (d*x)/2) + 6*d*x)/(6*a^2*d)
Time = 0.16 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.61 \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+6 d x}{6 a^{2} d} \] Input:
int(cos(d*x+c)^2/(a+a*cos(d*x+c))^2,x)
Output:
(tan((c + d*x)/2)**3 - 9*tan((c + d*x)/2) + 6*d*x)/(6*a**2*d)