Integrand size = 21, antiderivative size = 118 \[ \int \frac {\cos ^5(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {15 x}{8 a}-\frac {4 \sin (c+d x)}{a d}+\frac {15 \cos (c+d x) \sin (c+d x)}{8 a d}+\frac {5 \cos ^3(c+d x) \sin (c+d x)}{4 a d}-\frac {\cos ^4(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))}+\frac {4 \sin ^3(c+d x)}{3 a d} \]
15/8*x/a-4*sin(d*x+c)/a/d+15/8*cos(d*x+c)*sin(d*x+c)/a/d+5/4*cos(d*x+c)^3* sin(d*x+c)/a/d-cos(d*x+c)^4*sin(d*x+c)/d/(a+a*cos(d*x+c))+4/3*sin(d*x+c)^3 /a/d
Time = 0.84 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.47 \[ \int \frac {\cos ^5(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {\sec \left (\frac {c}{2}\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \left (360 d x \cos \left (\frac {d x}{2}\right )+360 d x \cos \left (c+\frac {d x}{2}\right )-552 \sin \left (\frac {d x}{2}\right )-168 \sin \left (c+\frac {d x}{2}\right )-120 \sin \left (c+\frac {3 d x}{2}\right )-120 \sin \left (2 c+\frac {3 d x}{2}\right )+40 \sin \left (2 c+\frac {5 d x}{2}\right )+40 \sin \left (3 c+\frac {5 d x}{2}\right )-5 \sin \left (3 c+\frac {7 d x}{2}\right )-5 \sin \left (4 c+\frac {7 d x}{2}\right )+3 \sin \left (4 c+\frac {9 d x}{2}\right )+3 \sin \left (5 c+\frac {9 d x}{2}\right )\right )}{384 a d} \]
(Sec[c/2]*Sec[(c + d*x)/2]*(360*d*x*Cos[(d*x)/2] + 360*d*x*Cos[c + (d*x)/2 ] - 552*Sin[(d*x)/2] - 168*Sin[c + (d*x)/2] - 120*Sin[c + (3*d*x)/2] - 120 *Sin[2*c + (3*d*x)/2] + 40*Sin[2*c + (5*d*x)/2] + 40*Sin[3*c + (5*d*x)/2] - 5*Sin[3*c + (7*d*x)/2] - 5*Sin[4*c + (7*d*x)/2] + 3*Sin[4*c + (9*d*x)/2] + 3*Sin[5*c + (9*d*x)/2]))/(384*a*d)
Time = 0.56 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.01, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {3042, 3246, 3042, 3227, 3042, 3113, 2009, 3115, 3042, 3115, 24}
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.
| \(\displaystyle \int \frac {\cos ^5(c+d x)}{a \cos (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^5}{a \sin \left (c+d x+\frac {\pi }{2}\right )+a}dx\) |
| \(\Big \downarrow \) 3246 |
| \(\displaystyle -\frac {\int \cos ^3(c+d x) (4 a-5 a \cos (c+d x))dx}{a^2}-\frac {\sin (c+d x) \cos ^4(c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (4 a-5 a \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}-\frac {\sin (c+d x) \cos ^4(c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle -\frac {4 a \int \cos ^3(c+d x)dx-5 a \int \cos ^4(c+d x)dx}{a^2}-\frac {\sin (c+d x) \cos ^4(c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {4 a \int \sin \left (c+d x+\frac {\pi }{2}\right )^3dx-5 a \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx}{a^2}-\frac {\sin (c+d x) \cos ^4(c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3113 |
| \(\displaystyle -\frac {-\frac {4 a \int \left (1-\sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}-5 a \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx}{a^2}-\frac {\sin (c+d x) \cos ^4(c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {-5 a \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx-\frac {4 a \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}}{a^2}-\frac {\sin (c+d x) \cos ^4(c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle -\frac {-5 a \left (\frac {3}{4} \int \cos ^2(c+d x)dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {4 a \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}}{a^2}-\frac {\sin (c+d x) \cos ^4(c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-5 a \left (\frac {3}{4} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {4 a \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}}{a^2}-\frac {\sin (c+d x) \cos ^4(c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle -\frac {-5 a \left (\frac {3}{4} \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {4 a \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}}{a^2}-\frac {\sin (c+d x) \cos ^4(c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle -\frac {-\frac {4 a \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}-5 a \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )}{a^2}-\frac {\sin (c+d x) \cos ^4(c+d x)}{d (a \cos (c+d x)+a)}\) |
-((Cos[c + d*x]^4*Sin[c + d*x])/(d*(a + a*Cos[c + d*x]))) - ((-4*a*(-Sin[c + d*x] + Sin[c + d*x]^3/3))/d - 5*a*((Cos[c + d*x]^3*Sin[c + d*x])/(4*d) + (3*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)))/4))/a^2
3.1.43.3.1 Defintions of rubi rules used
Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp and[(1 - x^2)^((n - 1)/2), x], x], x, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(b*c - a*d))*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n - 1)/(a*f*(a + b*Sin[e + f*x]))), x] - Simp[d/(a*b) Int[(c + d* Sin[e + f*x])^(n - 2)*Simp[b*d*(n - 1) - a*c*n + (b*c*(n - 1) - a*d*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] && GtQ[n, 1] && (IntegerQ[2*n] || EqQ[c, 0])
Time = 0.80 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.57
| method | result | size |
| parallelrisch | \(\frac {180 d x +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (-221+3 \cos \left (4 d x +4 c \right )-2 \cos \left (3 d x +3 c \right )+38 \cos \left (2 d x +2 c \right )-82 \cos \left (d x +c \right )\right )}{96 a d}\) | \(67\) |
| derivativedivides | \(\frac {-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {-\frac {25 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}-\frac {115 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12}-\frac {109 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12}-\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {15 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{d a}\) | \(100\) |
| default | \(\frac {-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {-\frac {25 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}-\frac {115 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12}-\frac {109 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12}-\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {15 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{d a}\) | \(100\) |
| risch | \(\frac {15 x}{8 a}+\frac {7 i {\mathrm e}^{i \left (d x +c \right )}}{8 a d}-\frac {7 i {\mathrm e}^{-i \left (d x +c \right )}}{8 a d}-\frac {2 i}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}+\frac {\sin \left (4 d x +4 c \right )}{32 a d}-\frac {\sin \left (3 d x +3 c \right )}{12 a d}+\frac {\sin \left (2 d x +2 c \right )}{2 a d}\) | \(117\) |
| norman | \(\frac {\frac {15 x}{8 a}-\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d a}-\frac {95 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d a}-\frac {86 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}-\frac {155 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d a}-\frac {45 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}-\frac {\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}+\frac {75 x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a}+\frac {75 x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a}+\frac {75 x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a}+\frac {75 x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a}+\frac {15 x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}\) | \(221\) |
1/96*(180*d*x+tan(1/2*d*x+1/2*c)*(-221+3*cos(4*d*x+4*c)-2*cos(3*d*x+3*c)+3 8*cos(2*d*x+2*c)-82*cos(d*x+c)))/a/d
Time = 0.26 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.67 \[ \int \frac {\cos ^5(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {45 \, d x \cos \left (d x + c\right ) + 45 \, d x + {\left (6 \, \cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{3} + 13 \, \cos \left (d x + c\right )^{2} - 19 \, \cos \left (d x + c\right ) - 64\right )} \sin \left (d x + c\right )}{24 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \]
1/24*(45*d*x*cos(d*x + c) + 45*d*x + (6*cos(d*x + c)^4 - 2*cos(d*x + c)^3 + 13*cos(d*x + c)^2 - 19*cos(d*x + c) - 64)*sin(d*x + c))/(a*d*cos(d*x + c ) + a*d)
Leaf count of result is larger than twice the leaf count of optimal. 882 vs. \(2 (102) = 204\).
Time = 1.93 (sec) , antiderivative size = 882, normalized size of antiderivative = 7.47 \[ \int \frac {\cos ^5(c+d x)}{a+a \cos (c+d x)} \, dx=\text {Too large to display} \]
Piecewise((45*d*x*tan(c/2 + d*x/2)**8/(24*a*d*tan(c/2 + d*x/2)**8 + 96*a*d *tan(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*x/2)**4 + 96*a*d*tan(c/2 + d*x/ 2)**2 + 24*a*d) + 180*d*x*tan(c/2 + d*x/2)**6/(24*a*d*tan(c/2 + d*x/2)**8 + 96*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*x/2)**4 + 96*a*d*tan(c/ 2 + d*x/2)**2 + 24*a*d) + 270*d*x*tan(c/2 + d*x/2)**4/(24*a*d*tan(c/2 + d* x/2)**8 + 96*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*x/2)**4 + 96*a* d*tan(c/2 + d*x/2)**2 + 24*a*d) + 180*d*x*tan(c/2 + d*x/2)**2/(24*a*d*tan( c/2 + d*x/2)**8 + 96*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*x/2)**4 + 96*a*d*tan(c/2 + d*x/2)**2 + 24*a*d) + 45*d*x/(24*a*d*tan(c/2 + d*x/2)* *8 + 96*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*x/2)**4 + 96*a*d*tan (c/2 + d*x/2)**2 + 24*a*d) - 24*tan(c/2 + d*x/2)**9/(24*a*d*tan(c/2 + d*x/ 2)**8 + 96*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*x/2)**4 + 96*a*d* tan(c/2 + d*x/2)**2 + 24*a*d) - 246*tan(c/2 + d*x/2)**7/(24*a*d*tan(c/2 + d*x/2)**8 + 96*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*x/2)**4 + 96* a*d*tan(c/2 + d*x/2)**2 + 24*a*d) - 374*tan(c/2 + d*x/2)**5/(24*a*d*tan(c/ 2 + d*x/2)**8 + 96*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*x/2)**4 + 96*a*d*tan(c/2 + d*x/2)**2 + 24*a*d) - 314*tan(c/2 + d*x/2)**3/(24*a*d*ta n(c/2 + d*x/2)**8 + 96*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*x/2)* *4 + 96*a*d*tan(c/2 + d*x/2)**2 + 24*a*d) - 66*tan(c/2 + d*x/2)/(24*a*d*ta n(c/2 + d*x/2)**8 + 96*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*x/...
Time = 0.32 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.84 \[ \int \frac {\cos ^5(c+d x)}{a+a \cos (c+d x)} \, dx=-\frac {\frac {\frac {21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {109 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {115 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {75 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a + \frac {4 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {6 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {4 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} - \frac {45 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac {12 \, \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}}{12 \, d} \]
-1/12*((21*sin(d*x + c)/(cos(d*x + c) + 1) + 109*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 115*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 75*sin(d*x + c)^7/( cos(d*x + c) + 1)^7)/(a + 4*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 6*a*si n(d*x + c)^4/(cos(d*x + c) + 1)^4 + 4*a*sin(d*x + c)^6/(cos(d*x + c) + 1)^ 6 + a*sin(d*x + c)^8/(cos(d*x + c) + 1)^8) - 45*arctan(sin(d*x + c)/(cos(d *x + c) + 1))/a + 12*sin(d*x + c)/(a*(cos(d*x + c) + 1)))/d
Time = 0.34 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.86 \[ \int \frac {\cos ^5(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {\frac {45 \, {\left (d x + c\right )}}{a} - \frac {24 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a} - \frac {2 \, {\left (75 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 115 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 109 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 21 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4} a}}{24 \, d} \]
1/24*(45*(d*x + c)/a - 24*tan(1/2*d*x + 1/2*c)/a - 2*(75*tan(1/2*d*x + 1/2 *c)^7 + 115*tan(1/2*d*x + 1/2*c)^5 + 109*tan(1/2*d*x + 1/2*c)^3 + 21*tan(1 /2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^4*a))/d
Time = 16.48 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.83 \[ \int \frac {\cos ^5(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {15\,x}{8\,a}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a\,d}-\frac {\frac {25\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\frac {115\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{12}+\frac {109\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{12}+\frac {7\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{a\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^4} \]