Integrand size = 21, antiderivative size = 80 \[ \int \frac {\cos ^3(c+d x)}{(a+a \cos (c+d x))^2} \, dx=-\frac {2 x}{a^2}+\frac {4 \sin (c+d x)}{3 a^2 d}+\frac {2 \sin (c+d x)}{a^2 d (1+\cos (c+d x))}-\frac {\cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2} \]
-2*x/a^2+4/3*sin(d*x+c)/a^2/d+2*sin(d*x+c)/a^2/d/(1+cos(d*x+c))-1/3*cos(d* x+c)^2*sin(d*x+c)/d/(a+a*cos(d*x+c))^2
Time = 0.45 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.21 \[ \int \frac {\cos ^3(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {\sin (c+d x) \left (48 \arcsin (\cos (c+d x)) \cos ^4\left (\frac {1}{2} (c+d x)\right )+(23+28 \cos (c+d x)+3 \cos (2 (c+d x))) \sqrt {\sin ^2(c+d x)}\right )}{6 a^2 d \sqrt {1-\cos (c+d x)} (1+\cos (c+d x))^{5/2}} \]
(Sin[c + d*x]*(48*ArcSin[Cos[c + d*x]]*Cos[(c + d*x)/2]^4 + (23 + 28*Cos[c + d*x] + 3*Cos[2*(c + d*x)])*Sqrt[Sin[c + d*x]^2]))/(6*a^2*d*Sqrt[1 - Cos [c + d*x]]*(1 + Cos[c + d*x])^(5/2))
Time = 0.62 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.06, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 3244, 27, 3042, 3447, 3042, 3502, 27, 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.
| \(\displaystyle \int \frac {\cos ^3(c+d x)}{(a \cos (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^3}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 3244 |
| \(\displaystyle -\frac {\int \frac {2 \cos (c+d x) (a-2 a \cos (c+d x))}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {2 \int \frac {\cos (c+d x) (a-2 a \cos (c+d x))}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {2 \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a-2 a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle -\frac {2 \int \frac {a \cos (c+d x)-2 a \cos ^2(c+d x)}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {2 \int \frac {a \sin \left (c+d x+\frac {\pi }{2}\right )-2 a \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle -\frac {2 \left (\frac {\int \frac {3 a^2 \cos (c+d x)}{\cos (c+d x) a+a}dx}{a}-\frac {2 \sin (c+d x)}{d}\right )}{3 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {2 \left (3 a \int \frac {\cos (c+d x)}{\cos (c+d x) a+a}dx-\frac {2 \sin (c+d x)}{d}\right )}{3 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {2 \left (3 a \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-\frac {2 \sin (c+d x)}{d}\right )}{3 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle -\frac {2 \left (3 a \left (\frac {x}{a}-\int \frac {1}{\cos (c+d x) a+a}dx\right )-\frac {2 \sin (c+d x)}{d}\right )}{3 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {2 \left (3 a \left (\frac {x}{a}-\int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx\right )-\frac {2 \sin (c+d x)}{d}\right )}{3 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3127 |
| \(\displaystyle -\frac {2 \left (3 a \left (\frac {x}{a}-\frac {\sin (c+d x)}{d (a \cos (c+d x)+a)}\right )-\frac {2 \sin (c+d x)}{d}\right )}{3 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
-1/3*(Cos[c + d*x]^2*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^2) - (2*((-2*Si n[c + d*x])/d + 3*a*(x/a - Sin[c + d*x]/(d*(a + a*Cos[c + d*x])))))/(3*a^2 )
3.1.55.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n - 1)/(a*f*(2*m + 1))), x] + Simp[1/(a*b* (2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 2)* Simp[b*(c^2*(m + 1) + d^2*(n - 1)) + a*c*d*(m - n + 1) + d*(a*d*(m - n + 1) + b*c*(m + n))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Time = 0.79 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.69
| method | result | size |
| parallelrisch | \(\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\cos \left (d x +c \right )+\frac {3 \cos \left (2 d x +2 c \right )}{28}+\frac {23}{28}\right ) \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-6 d x}{3 a^{2} d}\) | \(55\) |
| derivativedivides | \(\frac {-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}-8 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\) | \(72\) |
| default | \(\frac {-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}-8 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\) | \(72\) |
| risch | \(-\frac {2 x}{a^{2}}-\frac {i {\mathrm e}^{i \left (d x +c \right )}}{2 a^{2} d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{2 a^{2} d}+\frac {2 i \left (9 \,{\mathrm e}^{2 i \left (d x +c \right )}+15 \,{\mathrm e}^{i \left (d x +c \right )}+8\right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}\) | \(90\) |
| norman | \(\frac {-\frac {2 x}{a}+\frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d a}+\frac {34 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {9 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {2 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )}{6 d a}-\frac {6 x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {6 x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {2 x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} a}\) | \(171\) |
1/3*(7*tan(1/2*d*x+1/2*c)*(cos(d*x+c)+3/28*cos(2*d*x+2*c)+23/28)*sec(1/2*d *x+1/2*c)^2-6*d*x)/a^2/d
Time = 0.25 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.12 \[ \int \frac {\cos ^3(c+d x)}{(a+a \cos (c+d x))^2} \, dx=-\frac {6 \, d x \cos \left (d x + c\right )^{2} + 12 \, d x \cos \left (d x + c\right ) + 6 \, d x - {\left (3 \, \cos \left (d x + c\right )^{2} + 14 \, \cos \left (d x + c\right ) + 10\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]
-1/3*(6*d*x*cos(d*x + c)^2 + 12*d*x*cos(d*x + c) + 6*d*x - (3*cos(d*x + c) ^2 + 14*cos(d*x + c) + 10)*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*c os(d*x + c) + a^2*d)
Leaf count of result is larger than twice the leaf count of optimal. 201 vs. \(2 (73) = 146\).
Time = 1.08 (sec) , antiderivative size = 201, normalized size of antiderivative = 2.51 \[ \int \frac {\cos ^3(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\begin {cases} - \frac {12 d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} - \frac {12 d x}{6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} - \frac {\tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} + \frac {14 \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} + \frac {27 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} & \text {for}\: d \neq 0 \\\frac {x \cos ^{3}{\left (c \right )}}{\left (a \cos {\left (c \right )} + a\right )^{2}} & \text {otherwise} \end {cases} \]
Piecewise((-12*d*x*tan(c/2 + d*x/2)**2/(6*a**2*d*tan(c/2 + d*x/2)**2 + 6*a **2*d) - 12*d*x/(6*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - tan(c/2 + d*x/ 2)**5/(6*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 14*tan(c/2 + d*x/2)**3/( 6*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 27*tan(c/2 + d*x/2)/(6*a**2*d*t an(c/2 + d*x/2)**2 + 6*a**2*d), Ne(d, 0)), (x*cos(c)**3/(a*cos(c) + a)**2, True))
Time = 0.30 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.48 \[ \int \frac {\cos ^3(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {24 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac {12 \, \sin \left (d x + c\right )}{{\left (a^{2} + \frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}}{6 \, d} \]
1/6*((15*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 24*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 + 12*sin(d*x + c)/((a^2 + a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)))/d
Time = 0.32 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.99 \[ \int \frac {\cos ^3(c+d x)}{(a+a \cos (c+d x))^2} \, dx=-\frac {\frac {12 \, {\left (d x + c\right )}}{a^{2}} - \frac {12 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{2}} + \frac {a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]
-1/6*(12*(d*x + c)/a^2 - 12*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 + 1)*a^2) + (a^4*tan(1/2*d*x + 1/2*c)^3 - 15*a^4*tan(1/2*d*x + 1/2*c))/a^6 )/d
Time = 14.33 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.14 \[ \int \frac {\cos ^3(c+d x)}{(a+a \cos (c+d x))^2} \, dx=-\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-16\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-12\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+12\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (c+d\,x\right )}{6\,a^2\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3} \]