Integrand size = 21, antiderivative size = 96 \[ \int \frac {\cos ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {x}{a^3}-\frac {\cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {7 \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {29 \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )} \] Output:
x/a^3-1/5*cos(d*x+c)^2*sin(d*x+c)/d/(a+a*cos(d*x+c))^3+7/15*sin(d*x+c)/a/d /(a+a*cos(d*x+c))^2-29/15*sin(d*x+c)/d/(a^3+a^3*cos(d*x+c))
Time = 1.41 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.12 \[ \int \frac {\cos ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {4 \cos \left (\frac {1}{2} (c+d x)\right ) \csc ^6(c+d x) \sin ^7\left (\frac {1}{2} (c+d x)\right ) \left (480 \arcsin (\cos (c+d x)) \cos ^6\left (\frac {1}{2} (c+d x)\right )+4 (38+51 \cos (c+d x)+16 \cos (2 (c+d x))) \sqrt {\sin ^2(c+d x)}\right )}{15 a^3 d \sqrt {\sin ^2(c+d x)}} \] Input:
Integrate[Cos[c + d*x]^3/(a + a*Cos[c + d*x])^3,x]
Output:
(-4*Cos[(c + d*x)/2]*Csc[c + d*x]^6*Sin[(c + d*x)/2]^7*(480*ArcSin[Cos[c + d*x]]*Cos[(c + d*x)/2]^6 + 4*(38 + 51*Cos[c + d*x] + 16*Cos[2*(c + d*x)]) *Sqrt[Sin[c + d*x]^2]))/(15*a^3*d*Sqrt[Sin[c + d*x]^2])
Time = 0.69 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.10, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {3042, 3244, 3042, 3447, 3042, 3498, 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.
| \(\displaystyle \int \frac {\cos ^3(c+d x)}{(a \cos (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^3}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 3244 |
| \(\displaystyle -\frac {\int \frac {\cos (c+d x) (2 a-5 a \cos (c+d x))}{(\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \left (2 a-5 a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle -\frac {\int \frac {2 a \cos (c+d x)-5 a \cos ^2(c+d x)}{(\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {2 a \sin \left (c+d x+\frac {\pi }{2}\right )-5 a \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3498 |
| \(\displaystyle -\frac {-\frac {\int -\frac {14 a^2-15 a^2 \cos (c+d x)}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {7 a \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\frac {\int \frac {14 a^2-15 a^2 \cos (c+d x)}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {7 a \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {\int \frac {14 a^2-15 a^2 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}-\frac {7 a \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle -\frac {\frac {29 a^2 \int \frac {1}{\cos (c+d x) a+a}dx-15 a x}{3 a^2}-\frac {7 a \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {29 a^2 \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-15 a x}{3 a^2}-\frac {7 a \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3127 |
| \(\displaystyle -\frac {\frac {\frac {29 a^2 \sin (c+d x)}{d (a \cos (c+d x)+a)}-15 a x}{3 a^2}-\frac {7 a \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
Input:
Int[Cos[c + d*x]^3/(a + a*Cos[c + d*x])^3,x]
Output:
-1/5*(Cos[c + d*x]^2*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^3) - ((-7*a*Sin [c + d*x])/(3*d*(a + a*Cos[c + d*x])^2) + (-15*a*x + (29*a^2*Sin[c + d*x]) /(d*(a + a*Cos[c + d*x])))/(3*a^2))/(5*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[((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[(A*b - a* B + b*C)*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)*Simp[a*A*(m + 1) + m*(b *B - a*C) + b*C*(2*m + 1)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && EqQ[a^2 - b^2, 0]
Time = 0.79 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.53
| method | result | size |
| parallelrisch | \(\frac {-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+20 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+60 d x -105 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{60 a^{3} d}\) | \(51\) |
| derivativedivides | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}+\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+8 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\) | \(59\) |
| default | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}+\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+8 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\) | \(59\) |
| risch | \(\frac {x}{a^{3}}-\frac {2 i \left (45 \,{\mathrm e}^{4 i \left (d x +c \right )}+135 \,{\mathrm e}^{3 i \left (d x +c \right )}+185 \,{\mathrm e}^{2 i \left (d x +c \right )}+115 \,{\mathrm e}^{i \left (d x +c \right )}+32\right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}\) | \(75\) |
| norman | \(\frac {\frac {x}{a}+\frac {x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{a}-\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}-\frac {59 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 a d}-\frac {43 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{10 a d}-\frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{10 a d}+\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{60 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{20 a d}+\frac {3 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a}+\frac {3 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} a^{2}}\) | \(188\) |
Input:
int(cos(d*x+c)^3/(a+a*cos(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/60*(-3*tan(1/2*d*x+1/2*c)^5+20*tan(1/2*d*x+1/2*c)^3+60*d*x-105*tan(1/2*d *x+1/2*c))/a^3/d
Time = 0.08 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.21 \[ \int \frac {\cos ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {15 \, d x \cos \left (d x + c\right )^{3} + 45 \, d x \cos \left (d x + c\right )^{2} + 45 \, d x \cos \left (d x + c\right ) + 15 \, d x - {\left (32 \, \cos \left (d x + c\right )^{2} + 51 \, \cos \left (d x + c\right ) + 22\right )} \sin \left (d x + c\right )}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \] Input:
integrate(cos(d*x+c)^3/(a+a*cos(d*x+c))^3,x, algorithm="fricas")
Output:
1/15*(15*d*x*cos(d*x + c)^3 + 45*d*x*cos(d*x + c)^2 + 45*d*x*cos(d*x + c) + 15*d*x - (32*cos(d*x + c)^2 + 51*cos(d*x + c) + 22)*sin(d*x + c))/(a^3*d *cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)
Time = 1.23 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.78 \[ \int \frac {\cos ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\begin {cases} \frac {x}{a^{3}} - \frac {\tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{20 a^{3} d} + \frac {\tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a^{3} d} - \frac {7 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{4 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \cos ^{3}{\left (c \right )}}{\left (a \cos {\left (c \right )} + a\right )^{3}} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**3/(a+a*cos(d*x+c))**3,x)
Output:
Piecewise((x/a**3 - tan(c/2 + d*x/2)**5/(20*a**3*d) + tan(c/2 + d*x/2)**3/ (3*a**3*d) - 7*tan(c/2 + d*x/2)/(4*a**3*d), Ne(d, 0)), (x*cos(c)**3/(a*cos (c) + a)**3, True))
Time = 0.12 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.96 \[ \int \frac {\cos ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {120 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{60 \, d} \] Input:
integrate(cos(d*x+c)^3/(a+a*cos(d*x+c))^3,x, algorithm="maxima")
Output:
-1/60*((105*sin(d*x + c)/(cos(d*x + c) + 1) - 20*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 120*arctan(sin(d *x + c)/(cos(d*x + c) + 1))/a^3)/d
Time = 0.36 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.71 \[ \int \frac {\cos ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\frac {60 \, {\left (d x + c\right )}}{a^{3}} - \frac {3 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 20 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \] Input:
integrate(cos(d*x+c)^3/(a+a*cos(d*x+c))^3,x, algorithm="giac")
Output:
1/60*(60*(d*x + c)/a^3 - (3*a^12*tan(1/2*d*x + 1/2*c)^5 - 20*a^12*tan(1/2* d*x + 1/2*c)^3 + 105*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d
Time = 41.39 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.84 \[ \int \frac {\cos ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {x}{a^3}-\frac {\frac {32\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{15}-\frac {13\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{30}+\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{20}}{a^3\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5} \] Input:
int(cos(c + d*x)^3/(a + a*cos(c + d*x))^3,x)
Output:
x/a^3 - (sin(c/2 + (d*x)/2)/20 - (13*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/ 2))/30 + (32*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2))/15)/(a^3*d*cos(c/2 + (d*x)/2)^5)
Time = 0.15 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.52 \[ \int \frac {\cos ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+20 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-105 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+60 d x}{60 a^{3} d} \] Input:
int(cos(d*x+c)^3/(a+a*cos(d*x+c))^3,x)
Output:
( - 3*tan((c + d*x)/2)**5 + 20*tan((c + d*x)/2)**3 - 105*tan((c + d*x)/2) + 60*d*x)/(60*a**3*d)