Integrand size = 31, antiderivative size = 128 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\frac {\left (2 a^2 C+b^2 (2 A+C)\right ) x}{2 b^3}-\frac {2 a \left (A b^2+a^2 C\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^3 \sqrt {a+b} d}-\frac {a C \sin (c+d x)}{b^2 d}+\frac {C \cos (c+d x) \sin (c+d x)}{2 b d} \]
1/2*(2*a^2*C+b^2*(2*A+C))*x/b^3-a*C*sin(d*x+c)/b^2/d+1/2*C*cos(d*x+c)*sin( d*x+c)/b/d-2*a*(A*b^2+C*a^2)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^( 1/2))/b^3/d/(a-b)^(1/2)/(a+b)^(1/2)
Time = 1.13 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.91 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\frac {2 \left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) (c+d x)+\frac {8 a \left (A b^2+a^2 C\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}-4 a b C \sin (c+d x)+b^2 C \sin (2 (c+d x))}{4 b^3 d} \]
(2*(2*A*b^2 + (2*a^2 + b^2)*C)*(c + d*x) + (8*a*(A*b^2 + a^2*C)*ArcTanh[(( a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] - 4*a*b*C*Sin [c + d*x] + b^2*C*Sin[2*(c + d*x)])/(4*b^3*d)
Time = 0.67 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, Rules used = {3042, 3529, 3042, 3502, 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.
| \(\displaystyle \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 3529 |
| \(\displaystyle \frac {\int \frac {-2 a C \cos ^2(c+d x)+b (2 A+C) \cos (c+d x)+a C}{a+b \cos (c+d x)}dx}{2 b}+\frac {C \sin (c+d x) \cos (c+d x)}{2 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {-2 a C \sin \left (c+d x+\frac {\pi }{2}\right )^2+b (2 A+C) \sin \left (c+d x+\frac {\pi }{2}\right )+a C}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 b}+\frac {C \sin (c+d x) \cos (c+d x)}{2 b d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {\frac {\int \frac {a b C+\left (2 C a^2+b^2 (2 A+C)\right ) \cos (c+d x)}{a+b \cos (c+d x)}dx}{b}-\frac {2 a C \sin (c+d x)}{b d}}{2 b}+\frac {C \sin (c+d x) \cos (c+d x)}{2 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {a b C+\left (2 C a^2+b^2 (2 A+C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {2 a C \sin (c+d x)}{b d}}{2 b}+\frac {C \sin (c+d x) \cos (c+d x)}{2 b d}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {\frac {\frac {x \left (2 a^2 C+b^2 (2 A+C)\right )}{b}-\frac {2 a \left (a^2 C+A b^2\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{b}}{b}-\frac {2 a C \sin (c+d x)}{b d}}{2 b}+\frac {C \sin (c+d x) \cos (c+d x)}{2 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {x \left (2 a^2 C+b^2 (2 A+C)\right )}{b}-\frac {2 a \left (a^2 C+A b^2\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}}{b}-\frac {2 a C \sin (c+d x)}{b d}}{2 b}+\frac {C \sin (c+d x) \cos (c+d x)}{2 b d}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {\frac {\frac {x \left (2 a^2 C+b^2 (2 A+C)\right )}{b}-\frac {4 a \left (a^2 C+A b^2\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{b d}}{b}-\frac {2 a C \sin (c+d x)}{b d}}{2 b}+\frac {C \sin (c+d x) \cos (c+d x)}{2 b d}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {x \left (2 a^2 C+b^2 (2 A+C)\right )}{b}-\frac {4 a \left (a^2 C+A b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b d \sqrt {a-b} \sqrt {a+b}}}{b}-\frac {2 a C \sin (c+d x)}{b d}}{2 b}+\frac {C \sin (c+d x) \cos (c+d x)}{2 b d}\) |
(C*Cos[c + d*x]*Sin[c + d*x])/(2*b*d) + ((((2*a^2*C + b^2*(2*A + C))*x)/b - (4*a*(A*b^2 + a^2*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]]) /(Sqrt[a - b]*b*Sqrt[a + b]*d))/b - (2*a*C*Sin[c + d*x])/(b*d))/(2*b)
3.6.64.3.1 Defintions of rubi rules used
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_.)*((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_.) + (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)) + (A*b*d*(m + n + 2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + C* (a*d*m - b*c*(m + 1))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f , A, 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 = 1.67 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.25
| method | result | size |
| derivativedivides | \(\frac {\frac {\frac {2 \left (\left (-C a b -\frac {1}{2} b^{2} C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-C a b +\frac {1}{2} b^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\left (2 A \,b^{2}+2 a^{2} C +b^{2} C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3}}-\frac {2 a \left (A \,b^{2}+a^{2} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(160\) |
| default | \(\frac {\frac {\frac {2 \left (\left (-C a b -\frac {1}{2} b^{2} C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-C a b +\frac {1}{2} b^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\left (2 A \,b^{2}+2 a^{2} C +b^{2} C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3}}-\frac {2 a \left (A \,b^{2}+a^{2} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(160\) |
| risch | \(\frac {x A}{b}+\frac {x C \,a^{2}}{b^{3}}+\frac {C x}{2 b}+\frac {i a \,{\mathrm e}^{i \left (d x +c \right )} C}{2 b^{2} d}-\frac {i a \,{\mathrm e}^{-i \left (d x +c \right )} C}{2 b^{2} d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+a \sqrt {-a^{2}+b^{2}}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) A}{\sqrt {-a^{2}+b^{2}}\, d b}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+a \sqrt {-a^{2}+b^{2}}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) C}{\sqrt {-a^{2}+b^{2}}\, d \,b^{3}}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, d b}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, d \,b^{3}}+\frac {\sin \left (2 d x +2 c \right ) C}{4 b d}\) | \(376\) |
1/d*(2/b^3*(((-C*a*b-1/2*b^2*C)*tan(1/2*d*x+1/2*c)^3+(-C*a*b+1/2*b^2*C)*ta n(1/2*d*x+1/2*c))/(1+tan(1/2*d*x+1/2*c)^2)^2+1/2*(2*A*b^2+2*C*a^2+C*b^2)*a rctan(tan(1/2*d*x+1/2*c)))-2*a*(A*b^2+C*a^2)/b^3/((a-b)*(a+b))^(1/2)*arcta n((a-b)*tan(1/2*d*x+1/2*c)/((a-b)*(a+b))^(1/2)))
Time = 0.30 (sec) , antiderivative size = 385, normalized size of antiderivative = 3.01 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\left [\frac {{\left (2 \, C a^{4} + {\left (2 \, A - C\right )} a^{2} b^{2} - {\left (2 \, A + C\right )} b^{4}\right )} d x - {\left (C a^{3} + A a b^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - {\left (2 \, C a^{3} b - 2 \, C a b^{3} - {\left (C a^{2} b^{2} - C b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{2} b^{3} - b^{5}\right )} d}, \frac {{\left (2 \, C a^{4} + {\left (2 \, A - C\right )} a^{2} b^{2} - {\left (2 \, A + C\right )} b^{4}\right )} d x - 2 \, {\left (C a^{3} + A a b^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (2 \, C a^{3} b - 2 \, C a b^{3} - {\left (C a^{2} b^{2} - C b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{2} b^{3} - b^{5}\right )} d}\right ] \]
[1/2*((2*C*a^4 + (2*A - C)*a^2*b^2 - (2*A + C)*b^4)*d*x - (C*a^3 + A*a*b^2 )*sqrt(-a^2 + b^2)*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)) - (2*C*a^3*b - 2*C*a*b^3 - (C *a^2*b^2 - C*b^4)*cos(d*x + c))*sin(d*x + c))/((a^2*b^3 - b^5)*d), 1/2*((2 *C*a^4 + (2*A - C)*a^2*b^2 - (2*A + C)*b^4)*d*x - 2*(C*a^3 + A*a*b^2)*sqrt (a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) - (2*C*a^3*b - 2*C*a*b^3 - (C*a^2*b^2 - C*b^4)*cos(d*x + c))*sin(d*x + c))/ ((a^2*b^3 - b^5)*d)]
Leaf count of result is larger than twice the leaf count of optimal. 9489 vs. \(2 (112) = 224\).
Time = 164.05 (sec) , antiderivative size = 9489, normalized size of antiderivative = 74.13 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\text {Too large to display} \]
Piecewise((zoo*x*(A + C*cos(c)**2), Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (2*A* d*x*tan(c/2 + d*x/2)**4/(2*b*d*tan(c/2 + d*x/2)**4 + 4*b*d*tan(c/2 + d*x/2 )**2 + 2*b*d) + 4*A*d*x*tan(c/2 + d*x/2)**2/(2*b*d*tan(c/2 + d*x/2)**4 + 4 *b*d*tan(c/2 + d*x/2)**2 + 2*b*d) + 2*A*d*x/(2*b*d*tan(c/2 + d*x/2)**4 + 4 *b*d*tan(c/2 + d*x/2)**2 + 2*b*d) - 2*A*tan(c/2 + d*x/2)**5/(2*b*d*tan(c/2 + d*x/2)**4 + 4*b*d*tan(c/2 + d*x/2)**2 + 2*b*d) - 4*A*tan(c/2 + d*x/2)** 3/(2*b*d*tan(c/2 + d*x/2)**4 + 4*b*d*tan(c/2 + d*x/2)**2 + 2*b*d) - 2*A*ta n(c/2 + d*x/2)/(2*b*d*tan(c/2 + d*x/2)**4 + 4*b*d*tan(c/2 + d*x/2)**2 + 2* b*d) + 3*C*d*x*tan(c/2 + d*x/2)**4/(2*b*d*tan(c/2 + d*x/2)**4 + 4*b*d*tan( c/2 + d*x/2)**2 + 2*b*d) + 6*C*d*x*tan(c/2 + d*x/2)**2/(2*b*d*tan(c/2 + d* x/2)**4 + 4*b*d*tan(c/2 + d*x/2)**2 + 2*b*d) + 3*C*d*x/(2*b*d*tan(c/2 + d* x/2)**4 + 4*b*d*tan(c/2 + d*x/2)**2 + 2*b*d) - 2*C*tan(c/2 + d*x/2)**5/(2* b*d*tan(c/2 + d*x/2)**4 + 4*b*d*tan(c/2 + d*x/2)**2 + 2*b*d) - 10*C*tan(c/ 2 + d*x/2)**3/(2*b*d*tan(c/2 + d*x/2)**4 + 4*b*d*tan(c/2 + d*x/2)**2 + 2*b *d) - 4*C*tan(c/2 + d*x/2)/(2*b*d*tan(c/2 + d*x/2)**4 + 4*b*d*tan(c/2 + d* x/2)**2 + 2*b*d), Eq(a, b)), (2*A*d*x*tan(c/2 + d*x/2)**5/(2*b*d*tan(c/2 + d*x/2)**5 + 4*b*d*tan(c/2 + d*x/2)**3 + 2*b*d*tan(c/2 + d*x/2)) + 4*A*d*x *tan(c/2 + d*x/2)**3/(2*b*d*tan(c/2 + d*x/2)**5 + 4*b*d*tan(c/2 + d*x/2)** 3 + 2*b*d*tan(c/2 + d*x/2)) + 2*A*d*x*tan(c/2 + d*x/2)/(2*b*d*tan(c/2 + d* x/2)**5 + 4*b*d*tan(c/2 + d*x/2)**3 + 2*b*d*tan(c/2 + d*x/2)) + 2*A*tan...
Exception generated. \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\text {Exception raised: ValueError} \]
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 = 0.31 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.55 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\frac {\frac {{\left (2 \, C a^{2} + 2 \, A b^{2} + C b^{2}\right )} {\left (d x + c\right )}}{b^{3}} + \frac {4 \, {\left (C a^{3} + A a b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} b^{3}} - \frac {2 \, {\left (2 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} b^{2}}}{2 \, d} \]
1/2*((2*C*a^2 + 2*A*b^2 + C*b^2)*(d*x + c)/b^3 + 4*(C*a^3 + A*a*b^2)*(pi*f loor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*d*x + 1/ 2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(a^2 - b^2)))/(sqrt(a^2 - b^2)*b^3) - 2 *(2*C*a*tan(1/2*d*x + 1/2*c)^3 + C*b*tan(1/2*d*x + 1/2*c)^3 + 2*C*a*tan(1/ 2*d*x + 1/2*c) - C*b*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^2 *b^2))/d
Time = 5.17 (sec) , antiderivative size = 2398, normalized size of antiderivative = 18.73 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\text {Too large to display} \]
- ((tan(c/2 + (d*x)/2)*(2*C*a - C*b))/b^2 + (tan(c/2 + (d*x)/2)^3*(2*C*a + C*b))/b^2)/(d*(2*tan(c/2 + (d*x)/2)^2 + tan(c/2 + (d*x)/2)^4 + 1)) - (ata n(((((8*tan(c/2 + (d*x)/2)*(4*A^2*b^7 - 8*C^2*a^7 + C^2*b^7 - 12*A^2*a*b^6 - 3*C^2*a*b^6 + 16*C^2*a^6*b + 16*A^2*a^2*b^5 - 8*A^2*a^3*b^4 + 7*C^2*a^2 *b^5 - 13*C^2*a^3*b^4 + 16*C^2*a^4*b^3 - 16*C^2*a^5*b^2 + 4*A*C*b^7 - 12*A *C*a*b^6 + 20*A*C*a^2*b^5 - 28*A*C*a^3*b^4 + 32*A*C*a^4*b^3 - 16*A*C*a^5*b ^2))/b^4 + ((C*a^2*1i + b^2*(A*1i + (C*1i)/2))*((8*(4*A*b^10 + 2*C*b^10 + 4*A*a^2*b^8 + 2*C*a^2*b^8 - 6*C*a^3*b^7 + 4*C*a^4*b^6 - 8*A*a*b^9 - 2*C*a* b^9))/b^6 - (8*tan(c/2 + (d*x)/2)*(C*a^2*1i + b^2*(A*1i + (C*1i)/2))*(8*a* b^8 - 16*a^2*b^7 + 8*a^3*b^6))/b^7))/b^3)*(C*a^2*1i + b^2*(A*1i + (C*1i)/2 ))*1i)/b^3 + (((8*tan(c/2 + (d*x)/2)*(4*A^2*b^7 - 8*C^2*a^7 + C^2*b^7 - 12 *A^2*a*b^6 - 3*C^2*a*b^6 + 16*C^2*a^6*b + 16*A^2*a^2*b^5 - 8*A^2*a^3*b^4 + 7*C^2*a^2*b^5 - 13*C^2*a^3*b^4 + 16*C^2*a^4*b^3 - 16*C^2*a^5*b^2 + 4*A*C* b^7 - 12*A*C*a*b^6 + 20*A*C*a^2*b^5 - 28*A*C*a^3*b^4 + 32*A*C*a^4*b^3 - 16 *A*C*a^5*b^2))/b^4 - ((C*a^2*1i + b^2*(A*1i + (C*1i)/2))*((8*(4*A*b^10 + 2 *C*b^10 + 4*A*a^2*b^8 + 2*C*a^2*b^8 - 6*C*a^3*b^7 + 4*C*a^4*b^6 - 8*A*a*b^ 9 - 2*C*a*b^9))/b^6 + (8*tan(c/2 + (d*x)/2)*(C*a^2*1i + b^2*(A*1i + (C*1i) /2))*(8*a*b^8 - 16*a^2*b^7 + 8*a^3*b^6))/b^7))/b^3)*(C*a^2*1i + b^2*(A*1i + (C*1i)/2))*1i)/b^3)/((16*(4*C^3*a^8 - 4*A^3*a*b^7 - 6*C^3*a^7*b + 4*A^3* a^2*b^6 - C^3*a^3*b^5 + 2*C^3*a^4*b^4 - 5*C^3*a^5*b^3 + 6*C^3*a^6*b^2 -...