Integrand size = 21, antiderivative size = 148 \[ \int \frac {\cos ^4(c+d x)}{a+b \cos (c+d x)} \, dx=-\frac {a \left (2 a^2+b^2\right ) x}{2 b^4}+\frac {2 a^4 \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^4 \sqrt {a+b} d}+\frac {\left (3 a^2+2 b^2\right ) \sin (c+d x)}{3 b^3 d}-\frac {a \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac {\cos ^2(c+d x) \sin (c+d x)}{3 b d} \] Output:
-1/2*a*(2*a^2+b^2)*x/b^4+2*a^4*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b) ^(1/2))/(a-b)^(1/2)/b^4/(a+b)^(1/2)/d+1/3*(3*a^2+2*b^2)*sin(d*x+c)/b^3/d-1 /2*a*cos(d*x+c)*sin(d*x+c)/b^2/d+1/3*cos(d*x+c)^2*sin(d*x+c)/b/d
Time = 0.63 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.82 \[ \int \frac {\cos ^4(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {-6 a \left (2 a^2+b^2\right ) (c+d x)-\frac {24 a^4 \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}+3 b \left (4 a^2+3 b^2\right ) \sin (c+d x)-3 a b^2 \sin (2 (c+d x))+b^3 \sin (3 (c+d x))}{12 b^4 d} \] Input:
Integrate[Cos[c + d*x]^4/(a + b*Cos[c + d*x]),x]
Output:
(-6*a*(2*a^2 + b^2)*(c + d*x) - (24*a^4*ArcTanh[((a - b)*Tan[(c + d*x)/2]) /Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] + 3*b*(4*a^2 + 3*b^2)*Sin[c + d*x] - 3*a*b^2*Sin[2*(c + d*x)] + b^3*Sin[3*(c + d*x)])/(12*b^4*d)
Time = 0.86 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.11, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.619, Rules used = {3042, 3272, 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.
| \(\displaystyle \int \frac {\cos ^4(c+d x)}{a+b \cos (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^4}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 3272 |
| \(\displaystyle \frac {\int \frac {\cos (c+d x) \left (-3 a \cos ^2(c+d x)+2 b \cos (c+d x)+2 a\right )}{a+b \cos (c+d x)}dx}{3 b}+\frac {\sin (c+d x) \cos ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \left (-3 a \sin \left (c+d x+\frac {\pi }{2}\right )^2+2 b \sin \left (c+d x+\frac {\pi }{2}\right )+2 a\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{3 b}+\frac {\sin (c+d x) \cos ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 3528 |
| \(\displaystyle \frac {\frac {\int -\frac {3 a^2-b \cos (c+d x) a-2 \left (3 a^2+2 b^2\right ) \cos ^2(c+d x)}{a+b \cos (c+d x)}dx}{2 b}-\frac {3 a \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}+\frac {\sin (c+d x) \cos ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {-\frac {\int \frac {3 a^2-b \cos (c+d x) a-2 \left (3 a^2+2 b^2\right ) \cos ^2(c+d x)}{a+b \cos (c+d x)}dx}{2 b}-\frac {3 a \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}+\frac {\sin (c+d x) \cos ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\int \frac {3 a^2-b \sin \left (c+d x+\frac {\pi }{2}\right ) a-2 \left (3 a^2+2 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 b}-\frac {3 a \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}+\frac {\sin (c+d x) \cos ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {-\frac {\frac {\int \frac {3 \left (b a^2+\left (2 a^2+b^2\right ) \cos (c+d x) a\right )}{a+b \cos (c+d x)}dx}{b}-\frac {2 \left (3 a^2+2 b^2\right ) \sin (c+d x)}{b d}}{2 b}-\frac {3 a \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}+\frac {\sin (c+d x) \cos ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {\frac {3 \int \frac {b a^2+\left (2 a^2+b^2\right ) \cos (c+d x) a}{a+b \cos (c+d x)}dx}{b}-\frac {2 \left (3 a^2+2 b^2\right ) \sin (c+d x)}{b d}}{2 b}-\frac {3 a \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}+\frac {\sin (c+d x) \cos ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\frac {3 \int \frac {b a^2+\left (2 a^2+b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {2 \left (3 a^2+2 b^2\right ) \sin (c+d x)}{b d}}{2 b}-\frac {3 a \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}+\frac {\sin (c+d x) \cos ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {-\frac {\frac {3 \left (\frac {a x \left (2 a^2+b^2\right )}{b}-\frac {2 a^4 \int \frac {1}{a+b \cos (c+d x)}dx}{b}\right )}{b}-\frac {2 \left (3 a^2+2 b^2\right ) \sin (c+d x)}{b d}}{2 b}-\frac {3 a \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}+\frac {\sin (c+d x) \cos ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\frac {3 \left (\frac {a x \left (2 a^2+b^2\right )}{b}-\frac {2 a^4 \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}\right )}{b}-\frac {2 \left (3 a^2+2 b^2\right ) \sin (c+d x)}{b d}}{2 b}-\frac {3 a \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}+\frac {\sin (c+d x) \cos ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {-\frac {\frac {3 \left (\frac {a x \left (2 a^2+b^2\right )}{b}-\frac {4 a^4 \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{b d}\right )}{b}-\frac {2 \left (3 a^2+2 b^2\right ) \sin (c+d x)}{b d}}{2 b}-\frac {3 a \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}+\frac {\sin (c+d x) \cos ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {-\frac {\frac {3 \left (\frac {a x \left (2 a^2+b^2\right )}{b}-\frac {4 a^4 \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b d \sqrt {a-b} \sqrt {a+b}}\right )}{b}-\frac {2 \left (3 a^2+2 b^2\right ) \sin (c+d x)}{b d}}{2 b}-\frac {3 a \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}+\frac {\sin (c+d x) \cos ^2(c+d x)}{3 b d}\) |
Input:
Int[Cos[c + d*x]^4/(a + b*Cos[c + d*x]),x]
Output:
(Cos[c + d*x]^2*Sin[c + d*x])/(3*b*d) + ((-3*a*Cos[c + d*x]*Sin[c + d*x])/ (2*b*d) - ((3*((a*(2*a^2 + b^2)*x)/b - (4*a^4*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*b*Sqrt[a + b]*d)))/b - (2*(3*a^2 + 2* b^2)*Sin[c + d*x])/(b*d))/(2*b))/(3*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 = 1.30 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.21
| method | result | size |
| derivativedivides | \(\frac {\frac {2 a^{4} \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{4} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 \left (\frac {\left (-a^{2} b -\frac {1}{2} b^{2} a -b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-2 a^{2} b -\frac {2}{3} b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-a^{2} b -b^{3}+\frac {1}{2} b^{2} a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+\frac {a \left (2 a^{2}+b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{b^{4}}}{d}\) | \(179\) |
| default | \(\frac {\frac {2 a^{4} \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{4} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 \left (\frac {\left (-a^{2} b -\frac {1}{2} b^{2} a -b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-2 a^{2} b -\frac {2}{3} b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-a^{2} b -b^{3}+\frac {1}{2} b^{2} a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+\frac {a \left (2 a^{2}+b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{b^{4}}}{d}\) | \(179\) |
| risch | \(-\frac {a^{3} x}{b^{4}}-\frac {a x}{2 b^{2}}-\frac {i {\mathrm e}^{i \left (d x +c \right )} a^{2}}{2 d \,b^{3}}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )}}{8 d b}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} a^{2}}{2 d \,b^{3}}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )}}{8 d b}-\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,b^{4}}+\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+a \sqrt {-a^{2}+b^{2}}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,b^{4}}+\frac {\sin \left (3 d x +3 c \right )}{12 b d}-\frac {a \sin \left (2 d x +2 c \right )}{4 b^{2} d}\) | \(278\) |
Input:
int(cos(d*x+c)^4/(a+cos(d*x+c)*b),x,method=_RETURNVERBOSE)
Output:
1/d*(2*a^4/b^4/((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^4*(((-a^2*b-1/2*b^2*a-b^3)*tan(1/2*d*x+1/2*c)^5+(-2*a^2* b-2/3*b^3)*tan(1/2*d*x+1/2*c)^3+(-a^2*b-b^3+1/2*b^2*a)*tan(1/2*d*x+1/2*c)) /(1+tan(1/2*d*x+1/2*c)^2)^3+1/2*a*(2*a^2+b^2)*arctan(tan(1/2*d*x+1/2*c))))
Time = 0.11 (sec) , antiderivative size = 400, normalized size of antiderivative = 2.70 \[ \int \frac {\cos ^4(c+d x)}{a+b \cos (c+d x)} \, dx=\left [-\frac {3 \, \sqrt {-a^{2} + b^{2}} a^{4} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) + 3 \, {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} d x - {\left (6 \, a^{4} b - 2 \, a^{2} b^{3} - 4 \, b^{5} + 2 \, {\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} b^{4} - b^{6}\right )} d}, \frac {6 \, \sqrt {a^{2} - b^{2}} a^{4} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - 3 \, {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} d x + {\left (6 \, a^{4} b - 2 \, a^{2} b^{3} - 4 \, b^{5} + 2 \, {\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} b^{4} - b^{6}\right )} d}\right ] \] Input:
integrate(cos(d*x+c)^4/(a+b*cos(d*x+c)),x, algorithm="fricas")
Output:
[-1/6*(3*sqrt(-a^2 + b^2)*a^4*log((2*a*b*cos(d*x + c) + (2*a^2 - b^2)*cos( 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*(2*a^5 - a^3*b ^2 - a*b^4)*d*x - (6*a^4*b - 2*a^2*b^3 - 4*b^5 + 2*(a^2*b^3 - b^5)*cos(d*x + c)^2 - 3*(a^3*b^2 - a*b^4)*cos(d*x + c))*sin(d*x + c))/((a^2*b^4 - b^6) *d), 1/6*(6*sqrt(a^2 - b^2)*a^4*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b ^2)*sin(d*x + c))) - 3*(2*a^5 - a^3*b^2 - a*b^4)*d*x + (6*a^4*b - 2*a^2*b^ 3 - 4*b^5 + 2*(a^2*b^3 - b^5)*cos(d*x + c)^2 - 3*(a^3*b^2 - a*b^4)*cos(d*x + c))*sin(d*x + c))/((a^2*b^4 - b^6)*d)]
Timed out. \[ \int \frac {\cos ^4(c+d x)}{a+b \cos (c+d x)} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**4/(a+b*cos(d*x+c)),x)
Output:
Timed out
Exception generated. \[ \int \frac {\cos ^4(c+d x)}{a+b \cos (c+d x)} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(cos(d*x+c)^4/(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
Time = 0.53 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.68 \[ \int \frac {\cos ^4(c+d x)}{a+b \cos (c+d x)} \, dx=-\frac {\frac {12 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )} a^{4}}{\sqrt {a^{2} - b^{2}} b^{4}} + \frac {3 \, {\left (2 \, a^{3} + a b^{2}\right )} {\left (d x + c\right )}}{b^{4}} - \frac {2 \, {\left (6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} b^{3}}}{6 \, d} \] Input:
integrate(cos(d*x+c)^4/(a+b*cos(d*x+c)),x, algorithm="giac")
Output:
-1/6*(12*(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^4/(sqrt(a ^2 - b^2)*b^4) + 3*(2*a^3 + a*b^2)*(d*x + c)/b^4 - 2*(6*a^2*tan(1/2*d*x + 1/2*c)^5 + 3*a*b*tan(1/2*d*x + 1/2*c)^5 + 6*b^2*tan(1/2*d*x + 1/2*c)^5 + 1 2*a^2*tan(1/2*d*x + 1/2*c)^3 + 4*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*a^2*tan(1/ 2*d*x + 1/2*c) - 3*a*b*tan(1/2*d*x + 1/2*c) + 6*b^2*tan(1/2*d*x + 1/2*c))/ ((tan(1/2*d*x + 1/2*c)^2 + 1)^3*b^3))/d
Time = 35.52 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.37 \[ \int \frac {\cos ^4(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {3\,\sin \left (c+d\,x\right )}{4\,b\,d}+\frac {\sin \left (3\,c+3\,d\,x\right )}{12\,b\,d}-\frac {a\,\sin \left (2\,c+2\,d\,x\right )}{4\,b^2\,d}+\frac {a^2\,\sin \left (c+d\,x\right )}{b^3\,d}-\frac {2\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b^4\,d}-\frac {a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b^2\,d}+\frac {a^4\,\mathrm {atan}\left (\frac {\left (a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}\right )\,2{}\mathrm {i}}{b^4\,d\,\sqrt {b^2-a^2}} \] Input:
int(cos(c + d*x)^4/(a + b*cos(c + d*x)),x)
Output:
(3*sin(c + d*x))/(4*b*d) + sin(3*c + 3*d*x)/(12*b*d) - (a*sin(2*c + 2*d*x) )/(4*b^2*d) + (a^2*sin(c + d*x))/(b^3*d) - (2*a^3*atan(sin(c/2 + (d*x)/2)/ cos(c/2 + (d*x)/2)))/(b^4*d) - (a*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/ 2)))/(b^2*d) + (a^4*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^4*d*(b^2 - a^2)^(1/2))
Time = 0.18 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.42 \[ \int \frac {\cos ^4(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {12 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) a^{4}-3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{3} b^{2}+3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a \,b^{4}-2 \sin \left (d x +c \right )^{3} a^{2} b^{3}+2 \sin \left (d x +c \right )^{3} b^{5}+6 \sin \left (d x +c \right ) a^{4} b -6 \sin \left (d x +c \right ) b^{5}-6 a^{5} c -6 a^{5} d x +3 a^{3} b^{2} c +3 a^{3} b^{2} d x +3 a \,b^{4} c +3 a \,b^{4} d x}{6 b^{4} d \left (a^{2}-b^{2}\right )} \] Input:
int(cos(d*x+c)^4/(a+b*cos(d*x+c)),x)
Output:
(12*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( a**2 - b**2))*a**4 - 3*cos(c + d*x)*sin(c + d*x)*a**3*b**2 + 3*cos(c + d*x )*sin(c + d*x)*a*b**4 - 2*sin(c + d*x)**3*a**2*b**3 + 2*sin(c + d*x)**3*b* *5 + 6*sin(c + d*x)*a**4*b - 6*sin(c + d*x)*b**5 - 6*a**5*c - 6*a**5*d*x + 3*a**3*b**2*c + 3*a**3*b**2*d*x + 3*a*b**4*c + 3*a*b**4*d*x)/(6*b**4*d*(a **2 - b**2))