Integrand size = 21, antiderivative size = 193 \[ \int \frac {\cos ^5(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {\left (8 a^4+4 a^2 b^2+3 b^4\right ) x}{8 b^5}-\frac {2 a^5 \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^5 \sqrt {a+b} d}-\frac {a \left (3 a^2+2 b^2\right ) \sin (c+d x)}{3 b^4 d}+\frac {\left (4 a^2+3 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^3 d}-\frac {a \cos ^2(c+d x) \sin (c+d x)}{3 b^2 d}+\frac {\cos ^3(c+d x) \sin (c+d x)}{4 b d} \] Output:
1/8*(8*a^4+4*a^2*b^2+3*b^4)*x/b^5-2*a^5*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2 *c)/(a+b)^(1/2))/(a-b)^(1/2)/b^5/(a+b)^(1/2)/d-1/3*a*(3*a^2+2*b^2)*sin(d*x +c)/b^4/d+1/8*(4*a^2+3*b^2)*cos(d*x+c)*sin(d*x+c)/b^3/d-1/3*a*cos(d*x+c)^2 *sin(d*x+c)/b^2/d+1/4*cos(d*x+c)^3*sin(d*x+c)/b/d
Time = 1.02 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.79 \[ \int \frac {\cos ^5(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {12 \left (8 a^4+4 a^2 b^2+3 b^4\right ) (c+d x)+\frac {192 a^5 \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}-24 a b \left (4 a^2+3 b^2\right ) \sin (c+d x)+24 b^2 \left (a^2+b^2\right ) \sin (2 (c+d x))-8 a b^3 \sin (3 (c+d x))+3 b^4 \sin (4 (c+d x))}{96 b^5 d} \] Input:
Integrate[Cos[c + d*x]^5/(a + b*Cos[c + d*x]),x]
Output:
(12*(8*a^4 + 4*a^2*b^2 + 3*b^4)*(c + d*x) + (192*a^5*ArcTanh[((a - b)*Tan[ (c + d*x)/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] - 24*a*b*(4*a^2 + 3*b^2) *Sin[c + d*x] + 24*b^2*(a^2 + b^2)*Sin[2*(c + d*x)] - 8*a*b^3*Sin[3*(c + d *x)] + 3*b^4*Sin[4*(c + d*x)])/(96*b^5*d)
Time = 1.26 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.13, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.762, Rules used = {3042, 3272, 3042, 3528, 25, 3042, 3528, 25, 3042, 3502, 27, 3042, 3214, 3042, 3138, 218}
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.
| \(\displaystyle \int \frac {\cos ^5(c+d x)}{a+b \cos (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^5}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 3272 |
| \(\displaystyle \frac {\int \frac {\cos ^2(c+d x) \left (-4 a \cos ^2(c+d x)+3 b \cos (c+d x)+3 a\right )}{a+b \cos (c+d x)}dx}{4 b}+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (-4 a \sin \left (c+d x+\frac {\pi }{2}\right )^2+3 b \sin \left (c+d x+\frac {\pi }{2}\right )+3 a\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{4 b}+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 b d}\) |
| \(\Big \downarrow \) 3528 |
| \(\displaystyle \frac {\frac {\int -\frac {\cos (c+d x) \left (8 a^2-b \cos (c+d x) a-3 \left (4 a^2+3 b^2\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)}dx}{3 b}-\frac {4 a \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{4 b}+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 b d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {-\frac {\int \frac {\cos (c+d x) \left (8 a^2-b \cos (c+d x) a-3 \left (4 a^2+3 b^2\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)}dx}{3 b}-\frac {4 a \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{4 b}+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \left (8 a^2-b \sin \left (c+d x+\frac {\pi }{2}\right ) a-3 \left (4 a^2+3 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{3 b}-\frac {4 a \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{4 b}+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 b d}\) |
| \(\Big \downarrow \) 3528 |
| \(\displaystyle \frac {-\frac {\frac {\int -\frac {-8 a \left (3 a^2+2 b^2\right ) \cos ^2(c+d x)-b \left (4 a^2-9 b^2\right ) \cos (c+d x)+3 a \left (4 a^2+3 b^2\right )}{a+b \cos (c+d x)}dx}{2 b}-\frac {3 \left (4 a^2+3 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}-\frac {4 a \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{4 b}+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 b d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {-\frac {-\frac {\int \frac {-8 a \left (3 a^2+2 b^2\right ) \cos ^2(c+d x)-b \left (4 a^2-9 b^2\right ) \cos (c+d x)+3 a \left (4 a^2+3 b^2\right )}{a+b \cos (c+d x)}dx}{2 b}-\frac {3 \left (4 a^2+3 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}-\frac {4 a \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{4 b}+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {-\frac {\int \frac {-8 a \left (3 a^2+2 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-b \left (4 a^2-9 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+3 a \left (4 a^2+3 b^2\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 b}-\frac {3 \left (4 a^2+3 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}-\frac {4 a \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{4 b}+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 b d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {-\frac {-\frac {\frac {\int \frac {3 \left (a b \left (4 a^2+3 b^2\right )+\left (8 a^4+4 b^2 a^2+3 b^4\right ) \cos (c+d x)\right )}{a+b \cos (c+d x)}dx}{b}-\frac {8 a \left (3 a^2+2 b^2\right ) \sin (c+d x)}{b d}}{2 b}-\frac {3 \left (4 a^2+3 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}-\frac {4 a \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{4 b}+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {-\frac {\frac {3 \int \frac {a b \left (4 a^2+3 b^2\right )+\left (8 a^4+4 b^2 a^2+3 b^4\right ) \cos (c+d x)}{a+b \cos (c+d x)}dx}{b}-\frac {8 a \left (3 a^2+2 b^2\right ) \sin (c+d x)}{b d}}{2 b}-\frac {3 \left (4 a^2+3 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}-\frac {4 a \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{4 b}+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {-\frac {\frac {3 \int \frac {a b \left (4 a^2+3 b^2\right )+\left (8 a^4+4 b^2 a^2+3 b^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {8 a \left (3 a^2+2 b^2\right ) \sin (c+d x)}{b d}}{2 b}-\frac {3 \left (4 a^2+3 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}-\frac {4 a \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{4 b}+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 b d}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {-\frac {-\frac {\frac {3 \left (\frac {x \left (8 a^4+4 a^2 b^2+3 b^4\right )}{b}-\frac {8 a^5 \int \frac {1}{a+b \cos (c+d x)}dx}{b}\right )}{b}-\frac {8 a \left (3 a^2+2 b^2\right ) \sin (c+d x)}{b d}}{2 b}-\frac {3 \left (4 a^2+3 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}-\frac {4 a \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{4 b}+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {-\frac {\frac {3 \left (\frac {x \left (8 a^4+4 a^2 b^2+3 b^4\right )}{b}-\frac {8 a^5 \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}\right )}{b}-\frac {8 a \left (3 a^2+2 b^2\right ) \sin (c+d x)}{b d}}{2 b}-\frac {3 \left (4 a^2+3 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}-\frac {4 a \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{4 b}+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 b d}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {-\frac {-\frac {\frac {3 \left (\frac {x \left (8 a^4+4 a^2 b^2+3 b^4\right )}{b}-\frac {16 a^5 \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{b d}\right )}{b}-\frac {8 a \left (3 a^2+2 b^2\right ) \sin (c+d x)}{b d}}{2 b}-\frac {3 \left (4 a^2+3 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}-\frac {4 a \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{4 b}+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 b d}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {-\frac {-\frac {3 \left (4 a^2+3 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 b d}-\frac {\frac {3 \left (\frac {x \left (8 a^4+4 a^2 b^2+3 b^4\right )}{b}-\frac {16 a^5 \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b d \sqrt {a-b} \sqrt {a+b}}\right )}{b}-\frac {8 a \left (3 a^2+2 b^2\right ) \sin (c+d x)}{b d}}{2 b}}{3 b}-\frac {4 a \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{4 b}+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 b d}\) |
Input:
Int[Cos[c + d*x]^5/(a + b*Cos[c + d*x]),x]
Output:
(Cos[c + d*x]^3*Sin[c + d*x])/(4*b*d) + ((-4*a*Cos[c + d*x]^2*Sin[c + d*x] )/(3*b*d) - ((-3*(4*a^2 + 3*b^2)*Cos[c + d*x]*Sin[c + d*x])/(2*b*d) - ((3* (((8*a^4 + 4*a^2*b^2 + 3*b^4)*x)/b - (16*a^5*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*b*Sqrt[a + b]*d)))/b - (8*a*(3*a^2 + 2 *b^2)*Sin[c + d*x])/(b*d))/(2*b))/(3*b))/(4*b)
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_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[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^2)*Cos[e + f*x]*(a + b*Sin[e + f* x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^n*Simp[a^3*d *(m + n) + b^2*(b*c*(m - 2) + a*d*(n + 1)) - b*(a*b*c - b^2*d*(m + n - 1) - 3*a^2*d*(m + n))*Sin[e + f*x] - b^2*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Si n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a* d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && (IntegerQ[m ] || IntegersQ[2*m, 2*n]) && !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 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]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_ .) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x ])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Simp[1/(d*(m + n + 2)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A* d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2) - C*(a *c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n} , x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[ m, 0] && !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
Time = 1.73 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.33
| method | result | size |
| derivativedivides | \(\frac {-\frac {2 a^{5} \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{5} \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {\frac {2 \left (\left (-a^{3} b -\frac {1}{2} a^{2} b^{2}-a \,b^{3}-\frac {5}{8} b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}+\left (-3 a^{3} b -\frac {5}{3} a \,b^{3}+\frac {3}{8} b^{4}-\frac {1}{2} a^{2} b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (\frac {1}{2} a^{2} b^{2}-\frac {3}{8} b^{4}-3 a^{3} b -\frac {5}{3} a \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (\frac {1}{2} a^{2} b^{2}+\frac {5}{8} b^{4}-a^{3} b -a \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}+\frac {\left (8 a^{4}+4 a^{2} b^{2}+3 b^{4}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{b^{5}}}{d}\) | \(256\) |
| default | \(\frac {-\frac {2 a^{5} \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{5} \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {\frac {2 \left (\left (-a^{3} b -\frac {1}{2} a^{2} b^{2}-a \,b^{3}-\frac {5}{8} b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}+\left (-3 a^{3} b -\frac {5}{3} a \,b^{3}+\frac {3}{8} b^{4}-\frac {1}{2} a^{2} b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (\frac {1}{2} a^{2} b^{2}-\frac {3}{8} b^{4}-3 a^{3} b -\frac {5}{3} a \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (\frac {1}{2} a^{2} b^{2}+\frac {5}{8} b^{4}-a^{3} b -a \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}+\frac {\left (8 a^{4}+4 a^{2} b^{2}+3 b^{4}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{b^{5}}}{d}\) | \(256\) |
| risch | \(\frac {x \,a^{4}}{b^{5}}+\frac {x \,a^{2}}{2 b^{3}}+\frac {3 x}{8 b}+\frac {i a^{3} {\mathrm e}^{i \left (d x +c \right )}}{2 d \,b^{4}}+\frac {3 i a \,{\mathrm e}^{i \left (d x +c \right )}}{8 d \,b^{2}}-\frac {i a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{2 d \,b^{4}}-\frac {3 i a \,{\mathrm e}^{-i \left (d x +c \right )}}{8 d \,b^{2}}-\frac {a^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,b^{5}}+\frac {a^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,b^{5}}+\frac {\sin \left (4 d x +4 c \right )}{32 b d}-\frac {a \sin \left (3 d x +3 c \right )}{12 b^{2} d}+\frac {\sin \left (2 d x +2 c \right ) a^{2}}{4 d \,b^{3}}+\frac {\sin \left (2 d x +2 c \right )}{4 d b}\) | \(326\) |
Input:
int(cos(d*x+c)^5/(a+cos(d*x+c)*b),x,method=_RETURNVERBOSE)
Output:
1/d*(-2*a^5/b^5/((a-b)*(a+b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a-b) *(a+b))^(1/2))+2/b^5*(((-a^3*b-1/2*a^2*b^2-a*b^3-5/8*b^4)*tan(1/2*d*x+1/2* c)^7+(-3*a^3*b-5/3*a*b^3+3/8*b^4-1/2*a^2*b^2)*tan(1/2*d*x+1/2*c)^5+(1/2*a^ 2*b^2-3/8*b^4-3*a^3*b-5/3*a*b^3)*tan(1/2*d*x+1/2*c)^3+(1/2*a^2*b^2+5/8*b^4 -a^3*b-a*b^3)*tan(1/2*d*x+1/2*c))/(1+tan(1/2*d*x+1/2*c)^2)^4+1/8*(8*a^4+4* a^2*b^2+3*b^4)*arctan(tan(1/2*d*x+1/2*c))))
Time = 0.11 (sec) , antiderivative size = 479, normalized size of antiderivative = 2.48 \[ \int \frac {\cos ^5(c+d x)}{a+b \cos (c+d x)} \, dx=\left [-\frac {12 \, \sqrt {-a^{2} + b^{2}} a^{5} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - 3 \, {\left (8 \, a^{6} - 4 \, a^{4} b^{2} - a^{2} b^{4} - 3 \, b^{6}\right )} d x + {\left (24 \, a^{5} b - 8 \, a^{3} b^{3} - 16 \, a b^{5} - 6 \, {\left (a^{2} b^{4} - b^{6}\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left (a^{3} b^{3} - a b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (4 \, a^{4} b^{2} - a^{2} b^{4} - 3 \, b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, {\left (a^{2} b^{5} - b^{7}\right )} d}, -\frac {24 \, \sqrt {a^{2} - b^{2}} a^{5} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - 3 \, {\left (8 \, a^{6} - 4 \, a^{4} b^{2} - a^{2} b^{4} - 3 \, b^{6}\right )} d x + {\left (24 \, a^{5} b - 8 \, a^{3} b^{3} - 16 \, a b^{5} - 6 \, {\left (a^{2} b^{4} - b^{6}\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left (a^{3} b^{3} - a b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (4 \, a^{4} b^{2} - a^{2} b^{4} - 3 \, b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, {\left (a^{2} b^{5} - b^{7}\right )} d}\right ] \] Input:
integrate(cos(d*x+c)^5/(a+b*cos(d*x+c)),x, algorithm="fricas")
Output:
[-1/24*(12*sqrt(-a^2 + b^2)*a^5*log((2*a*b*cos(d*x + c) + (2*a^2 - b^2)*co s(d*x + c)^2 - 2*sqrt(-a^2 + b^2)*(a*cos(d*x + c) + b)*sin(d*x + c) - a^2 + 2*b^2)/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)) - 3*(8*a^6 - 4*a ^4*b^2 - a^2*b^4 - 3*b^6)*d*x + (24*a^5*b - 8*a^3*b^3 - 16*a*b^5 - 6*(a^2* b^4 - b^6)*cos(d*x + c)^3 + 8*(a^3*b^3 - a*b^5)*cos(d*x + c)^2 - 3*(4*a^4* b^2 - a^2*b^4 - 3*b^6)*cos(d*x + c))*sin(d*x + c))/((a^2*b^5 - b^7)*d), -1 /24*(24*sqrt(a^2 - b^2)*a^5*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)* sin(d*x + c))) - 3*(8*a^6 - 4*a^4*b^2 - a^2*b^4 - 3*b^6)*d*x + (24*a^5*b - 8*a^3*b^3 - 16*a*b^5 - 6*(a^2*b^4 - b^6)*cos(d*x + c)^3 + 8*(a^3*b^3 - a* b^5)*cos(d*x + c)^2 - 3*(4*a^4*b^2 - a^2*b^4 - 3*b^6)*cos(d*x + c))*sin(d* x + c))/((a^2*b^5 - b^7)*d)]
Timed out. \[ \int \frac {\cos ^5(c+d x)}{a+b \cos (c+d x)} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**5/(a+b*cos(d*x+c)),x)
Output:
Timed out
Exception generated. \[ \int \frac {\cos ^5(c+d x)}{a+b \cos (c+d x)} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(cos(d*x+c)^5/(a+b*cos(d*x+c)),x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f or more de
Leaf count of result is larger than twice the leaf count of optimal. 393 vs. \(2 (174) = 348\).
Time = 0.52 (sec) , antiderivative size = 393, normalized size of antiderivative = 2.04 \[ \int \frac {\cos ^5(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {\frac {48 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )} a^{5}}{\sqrt {a^{2} - b^{2}} b^{5}} + \frac {3 \, {\left (8 \, a^{4} + 4 \, a^{2} b^{2} + 3 \, b^{4}\right )} {\left (d x + c\right )}}{b^{5}} - \frac {2 \, {\left (24 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 72 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 72 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4} b^{4}}}{24 \, d} \] Input:
integrate(cos(d*x+c)^5/(a+b*cos(d*x+c)),x, algorithm="giac")
Output:
1/24*(48*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*ta n(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(a^2 - b^2)))*a^5/(sqrt(a ^2 - b^2)*b^5) + 3*(8*a^4 + 4*a^2*b^2 + 3*b^4)*(d*x + c)/b^5 - 2*(24*a^3*t an(1/2*d*x + 1/2*c)^7 + 12*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 24*a*b^2*tan(1/2 *d*x + 1/2*c)^7 + 15*b^3*tan(1/2*d*x + 1/2*c)^7 + 72*a^3*tan(1/2*d*x + 1/2 *c)^5 + 12*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 40*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 9*b^3*tan(1/2*d*x + 1/2*c)^5 + 72*a^3*tan(1/2*d*x + 1/2*c)^3 - 12*a^2*b* tan(1/2*d*x + 1/2*c)^3 + 40*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 9*b^3*tan(1/2*d *x + 1/2*c)^3 + 24*a^3*tan(1/2*d*x + 1/2*c) - 12*a^2*b*tan(1/2*d*x + 1/2*c ) + 24*a*b^2*tan(1/2*d*x + 1/2*c) - 15*b^3*tan(1/2*d*x + 1/2*c))/((tan(1/2 *d*x + 1/2*c)^2 + 1)^4*b^4))/d
Time = 42.77 (sec) , antiderivative size = 474, normalized size of antiderivative = 2.46 \[ \int \frac {\cos ^5(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {\sin \left (2\,c+2\,d\,x\right )}{4\,b\,d}+\frac {\sin \left (4\,c+4\,d\,x\right )}{32\,b\,d}+\frac {3\,\mathrm {atan}\left (\frac {40\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^4\,b^6+15\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^8+9\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^{10}}{b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (40\,a^4\,b^5+15\,a^2\,b^7+9\,b^9\right )}\right )}{4\,b\,d}-\frac {a\,\sin \left (3\,c+3\,d\,x\right )}{12\,b^2\,d}-\frac {a^3\,\sin \left (c+d\,x\right )}{b^4\,d}+\frac {a^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,b^3\,d}+\frac {a^2\,\mathrm {atan}\left (\frac {40\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^4\,b^6+15\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^8+9\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^{10}}{b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (40\,a^4\,b^5+15\,a^2\,b^7+9\,b^9\right )}\right )}{b^3\,d}+\frac {2\,a^4\,\mathrm {atan}\left (\frac {40\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^4\,b^6+15\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^8+9\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^{10}}{b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (40\,a^4\,b^5+15\,a^2\,b^7+9\,b^9\right )}\right )}{b^5\,d}-\frac {3\,a\,\sin \left (c+d\,x\right )}{4\,b^2\,d}-\frac {a^5\,\mathrm {atan}\left (\frac {\left (a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}\right )\,2{}\mathrm {i}}{b^5\,d\,\sqrt {b^2-a^2}} \] Input:
int(cos(c + d*x)^5/(a + b*cos(c + d*x)),x)
Output:
sin(2*c + 2*d*x)/(4*b*d) + sin(4*c + 4*d*x)/(32*b*d) + (3*atan((9*b^10*sin (c/2 + (d*x)/2) + 15*a^2*b^8*sin(c/2 + (d*x)/2) + 40*a^4*b^6*sin(c/2 + (d* x)/2))/(b*cos(c/2 + (d*x)/2)*(9*b^9 + 15*a^2*b^7 + 40*a^4*b^5))))/(4*b*d) - (a*sin(3*c + 3*d*x))/(12*b^2*d) - (a^3*sin(c + d*x))/(b^4*d) + (a^2*sin( 2*c + 2*d*x))/(4*b^3*d) + (a^2*atan((9*b^10*sin(c/2 + (d*x)/2) + 15*a^2*b^ 8*sin(c/2 + (d*x)/2) + 40*a^4*b^6*sin(c/2 + (d*x)/2))/(b*cos(c/2 + (d*x)/2 )*(9*b^9 + 15*a^2*b^7 + 40*a^4*b^5))))/(b^3*d) + (2*a^4*atan((9*b^10*sin(c /2 + (d*x)/2) + 15*a^2*b^8*sin(c/2 + (d*x)/2) + 40*a^4*b^6*sin(c/2 + (d*x) /2))/(b*cos(c/2 + (d*x)/2)*(9*b^9 + 15*a^2*b^7 + 40*a^4*b^5))))/(b^5*d) - (3*a*sin(c + d*x))/(4*b^2*d) - (a^5*atan(((a*sin(c/2 + (d*x)/2) - b*sin(c/ 2 + (d*x)/2))*1i)/(cos(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)))*2i)/(b^5*d*(b^2 - a^2)^(1/2))
Time = 0.20 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.50 \[ \int \frac {\cos ^5(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {-48 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) a^{5}-6 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a^{2} b^{4}+6 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b^{6}+12 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{4} b^{2}+3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b^{4}-15 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{6}+8 \sin \left (d x +c \right )^{3} a^{3} b^{3}-8 \sin \left (d x +c \right )^{3} a \,b^{5}-24 \sin \left (d x +c \right ) a^{5} b +24 \sin \left (d x +c \right ) a \,b^{5}+24 a^{6} c +24 a^{6} d x -12 a^{4} b^{2} c -12 a^{4} b^{2} d x -3 a^{2} b^{4} c -3 a^{2} b^{4} d x -9 b^{6} c -9 b^{6} d x}{24 b^{5} d \left (a^{2}-b^{2}\right )} \] Input:
int(cos(d*x+c)^5/(a+b*cos(d*x+c)),x)
Output:
( - 48*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sq rt(a**2 - b**2))*a**5 - 6*cos(c + d*x)*sin(c + d*x)**3*a**2*b**4 + 6*cos(c + d*x)*sin(c + d*x)**3*b**6 + 12*cos(c + d*x)*sin(c + d*x)*a**4*b**2 + 3* cos(c + d*x)*sin(c + d*x)*a**2*b**4 - 15*cos(c + d*x)*sin(c + d*x)*b**6 + 8*sin(c + d*x)**3*a**3*b**3 - 8*sin(c + d*x)**3*a*b**5 - 24*sin(c + d*x)*a **5*b + 24*sin(c + d*x)*a*b**5 + 24*a**6*c + 24*a**6*d*x - 12*a**4*b**2*c - 12*a**4*b**2*d*x - 3*a**2*b**4*c - 3*a**2*b**4*d*x - 9*b**6*c - 9*b**6*d *x)/(24*b**5*d*(a**2 - b**2))