Integrand size = 33, antiderivative size = 177 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=-\frac {a \left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) x}{2 b^4}+\frac {2 a^2 \left (A b^2+a^2 C\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^4 \sqrt {a+b} d}+\frac {\left (3 a^2 C+b^2 (3 A+2 C)\right ) \sin (c+d x)}{3 b^3 d}-\frac {a C \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac {C \cos ^2(c+d x) \sin (c+d x)}{3 b d} \]
-1/2*a*(2*A*b^2+(2*a^2+b^2)*C)*x/b^4+1/3*(3*a^2*C+b^2*(3*A+2*C))*sin(d*x+c )/b^3/d-1/2*a*C*cos(d*x+c)*sin(d*x+c)/b^2/d+1/3*C*cos(d*x+c)^2*sin(d*x+c)/ b/d+2*a^2*(A*b^2+C*a^2)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2)) /b^4/d/(a-b)^(1/2)/(a+b)^(1/2)
Time = 1.49 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.86 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\frac {-6 a \left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) (c+d x)-\frac {24 a^2 \left (A b^2+a^2 C\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}+3 b \left (4 A b^2+4 a^2 C+3 b^2 C\right ) \sin (c+d x)-3 a b^2 C \sin (2 (c+d x))+b^3 C \sin (3 (c+d x))}{12 b^4 d} \]
(-6*a*(2*A*b^2 + (2*a^2 + b^2)*C)*(c + d*x) - (24*a^2*(A*b^2 + a^2*C)*ArcT anh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] + 3*b*( 4*A*b^2 + 4*a^2*C + 3*b^2*C)*Sin[c + d*x] - 3*a*b^2*C*Sin[2*(c + d*x)] + b ^3*C*Sin[3*(c + d*x)])/(12*b^4*d)
Time = 1.06 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.10, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.394, Rules used = {3042, 3529, 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.
| \(\displaystyle \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 3529 |
| \(\displaystyle \frac {\int \frac {\cos (c+d x) \left (-3 a C \cos ^2(c+d x)+b (3 A+2 C) \cos (c+d x)+2 a C\right )}{a+b \cos (c+d x)}dx}{3 b}+\frac {C \sin (c+d x) \cos ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \left (-3 a C \sin \left (c+d x+\frac {\pi }{2}\right )^2+b (3 A+2 C) \sin \left (c+d x+\frac {\pi }{2}\right )+2 a C\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{3 b}+\frac {C \sin (c+d x) \cos ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 3528 |
| \(\displaystyle \frac {\frac {\int -\frac {3 C a^2-b C \cos (c+d x) a-2 \left (3 C a^2+b^2 (3 A+2 C)\right ) \cos ^2(c+d x)}{a+b \cos (c+d x)}dx}{2 b}-\frac {3 a C \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}+\frac {C \sin (c+d x) \cos ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {-\frac {\int \frac {3 C a^2-b C \cos (c+d x) a-2 \left (3 C a^2+b^2 (3 A+2 C)\right ) \cos ^2(c+d x)}{a+b \cos (c+d x)}dx}{2 b}-\frac {3 a C \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}+\frac {C \sin (c+d x) \cos ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\int \frac {3 C a^2-b C \sin \left (c+d x+\frac {\pi }{2}\right ) a-2 \left (3 C a^2+b^2 (3 A+2 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 b}-\frac {3 a C \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}+\frac {C \sin (c+d x) \cos ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {-\frac {\frac {\int \frac {3 \left (b C a^2+\left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) \cos (c+d x) a\right )}{a+b \cos (c+d x)}dx}{b}-\frac {2 \left (3 a^2 C+b^2 (3 A+2 C)\right ) \sin (c+d x)}{b d}}{2 b}-\frac {3 a C \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}+\frac {C \sin (c+d x) \cos ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {\frac {3 \int \frac {b C a^2+\left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) \cos (c+d x) a}{a+b \cos (c+d x)}dx}{b}-\frac {2 \left (3 a^2 C+b^2 (3 A+2 C)\right ) \sin (c+d x)}{b d}}{2 b}-\frac {3 a C \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}+\frac {C \sin (c+d x) \cos ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\frac {3 \int \frac {b C a^2+\left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {2 \left (3 a^2 C+b^2 (3 A+2 C)\right ) \sin (c+d x)}{b d}}{2 b}-\frac {3 a C \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}+\frac {C \sin (c+d x) \cos ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {-\frac {\frac {3 \left (\frac {a x \left (C \left (2 a^2+b^2\right )+2 A b^2\right )}{b}-\frac {2 a^2 \left (a^2 C+A b^2\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{b}\right )}{b}-\frac {2 \left (3 a^2 C+b^2 (3 A+2 C)\right ) \sin (c+d x)}{b d}}{2 b}-\frac {3 a C \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}+\frac {C \sin (c+d x) \cos ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\frac {3 \left (\frac {a x \left (C \left (2 a^2+b^2\right )+2 A b^2\right )}{b}-\frac {2 a^2 \left (a^2 C+A b^2\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}\right )}{b}-\frac {2 \left (3 a^2 C+b^2 (3 A+2 C)\right ) \sin (c+d x)}{b d}}{2 b}-\frac {3 a C \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}+\frac {C \sin (c+d x) \cos ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {-\frac {\frac {3 \left (\frac {a x \left (C \left (2 a^2+b^2\right )+2 A b^2\right )}{b}-\frac {4 a^2 \left (a^2 C+A b^2\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{b d}\right )}{b}-\frac {2 \left (3 a^2 C+b^2 (3 A+2 C)\right ) \sin (c+d x)}{b d}}{2 b}-\frac {3 a C \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}+\frac {C \sin (c+d x) \cos ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {-\frac {\frac {3 \left (\frac {a x \left (C \left (2 a^2+b^2\right )+2 A b^2\right )}{b}-\frac {4 a^2 \left (a^2 C+A b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b d \sqrt {a-b} \sqrt {a+b}}\right )}{b}-\frac {2 \left (3 a^2 C+b^2 (3 A+2 C)\right ) \sin (c+d x)}{b d}}{2 b}-\frac {3 a C \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}+\frac {C \sin (c+d x) \cos ^2(c+d x)}{3 b d}\) |
(C*Cos[c + d*x]^2*Sin[c + d*x])/(3*b*d) + ((-3*a*C*Cos[c + d*x]*Sin[c + d* x])/(2*b*d) - ((3*((a*(2*A*b^2 + (2*a^2 + b^2)*C)*x)/b - (4*a^2*(A*b^2 + a ^2*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*b*S qrt[a + b]*d)))/b - (2*(3*a^2*C + b^2*(3*A + 2*C))*Sin[c + d*x])/(b*d))/(2 *b))/(3*b)
3.6.63.3.1 Defintions of rubi rules used
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_.)*((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])))
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 = 2.38 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.27
| method | result | size |
| derivativedivides | \(\frac {-\frac {2 \left (\frac {\left (-A \,b^{3}-C \,a^{2} b -\frac {1}{2} C a \,b^{2}-C \,b^{3}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-2 A \,b^{3}-2 C \,a^{2} b -\frac {2}{3} C \,b^{3}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-A \,b^{3}-C \,a^{2} b -C \,b^{3}+\frac {1}{2} C a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {a \left (2 A \,b^{2}+2 a^{2} C +b^{2} C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{b^{4}}+\frac {2 a^{2} \left (A \,b^{2}+a^{2} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{4} \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(225\) |
| default | \(\frac {-\frac {2 \left (\frac {\left (-A \,b^{3}-C \,a^{2} b -\frac {1}{2} C a \,b^{2}-C \,b^{3}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-2 A \,b^{3}-2 C \,a^{2} b -\frac {2}{3} C \,b^{3}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-A \,b^{3}-C \,a^{2} b -C \,b^{3}+\frac {1}{2} C a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {a \left (2 A \,b^{2}+2 a^{2} C +b^{2} C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{b^{4}}+\frac {2 a^{2} \left (A \,b^{2}+a^{2} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{4} \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(225\) |
| risch | \(-\frac {a x A}{b^{2}}-\frac {a^{3} x C}{b^{4}}-\frac {a C x}{2 b^{2}}-\frac {i {\mathrm e}^{i \left (d x +c \right )} A}{2 b d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} a^{2} C}{2 b^{3} d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} C}{8 b d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} A}{2 b d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} a^{2} C}{2 b^{3} d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} C}{8 b d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, d \,b^{2}}-\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, d \,b^{4}}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, d \,b^{2}}+\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, d \,b^{4}}+\frac {C \sin \left (3 d x +3 c \right )}{12 b d}-\frac {a C \sin \left (2 d x +2 c \right )}{4 b^{2} d}\) | \(487\) |
1/d*(-2/b^4*(((-A*b^3-C*a^2*b-1/2*C*a*b^2-C*b^3)*tan(1/2*d*x+1/2*c)^5+(-2* A*b^3-2*C*a^2*b-2/3*C*b^3)*tan(1/2*d*x+1/2*c)^3+(-A*b^3-C*a^2*b-C*b^3+1/2* C*a*b^2)*tan(1/2*d*x+1/2*c))/(1+tan(1/2*d*x+1/2*c)^2)^3+1/2*a*(2*A*b^2+2*C *a^2+C*b^2)*arctan(tan(1/2*d*x+1/2*c)))+2*a^2*(A*b^2+C*a^2)/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)))
Time = 0.32 (sec) , antiderivative size = 485, normalized size of antiderivative = 2.74 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\left [-\frac {3 \, {\left (2 \, C a^{5} + {\left (2 \, A - C\right )} a^{3} b^{2} - {\left (2 \, A + C\right )} a b^{4}\right )} d x + 3 \, {\left (C a^{4} + A a^{2} b^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - {\left (6 \, C a^{4} b + 2 \, {\left (3 \, A - C\right )} a^{2} b^{3} - 2 \, {\left (3 \, A + 2 \, C\right )} b^{5} + 2 \, {\left (C a^{2} b^{3} - C b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (C a^{3} b^{2} - C a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} b^{4} - b^{6}\right )} d}, -\frac {3 \, {\left (2 \, C a^{5} + {\left (2 \, A - C\right )} a^{3} b^{2} - {\left (2 \, A + C\right )} a b^{4}\right )} d x - 6 \, {\left (C a^{4} + A a^{2} b^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (6 \, C a^{4} b + 2 \, {\left (3 \, A - C\right )} a^{2} b^{3} - 2 \, {\left (3 \, A + 2 \, C\right )} b^{5} + 2 \, {\left (C a^{2} b^{3} - C b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (C a^{3} b^{2} - C a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} b^{4} - b^{6}\right )} d}\right ] \]
[-1/6*(3*(2*C*a^5 + (2*A - C)*a^3*b^2 - (2*A + C)*a*b^4)*d*x + 3*(C*a^4 + A*a^2*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)) - (6*C*a^4*b + 2*(3* A - C)*a^2*b^3 - 2*(3*A + 2*C)*b^5 + 2*(C*a^2*b^3 - C*b^5)*cos(d*x + c)^2 - 3*(C*a^3*b^2 - C*a*b^4)*cos(d*x + c))*sin(d*x + c))/((a^2*b^4 - b^6)*d), -1/6*(3*(2*C*a^5 + (2*A - C)*a^3*b^2 - (2*A + C)*a*b^4)*d*x - 6*(C*a^4 + A*a^2*b^2)*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*s in(d*x + c))) - (6*C*a^4*b + 2*(3*A - C)*a^2*b^3 - 2*(3*A + 2*C)*b^5 + 2*( C*a^2*b^3 - C*b^5)*cos(d*x + c)^2 - 3*(C*a^3*b^2 - C*a*b^4)*cos(d*x + c))* sin(d*x + c))/((a^2*b^4 - b^6)*d)]
Timed out. \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\text {Exception raised: ValueError} \]
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
Leaf count of result is larger than twice the leaf count of optimal. 326 vs. \(2 (160) = 320\).
Time = 0.31 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.84 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=-\frac {\frac {3 \, {\left (2 \, C a^{3} + 2 \, A a b^{2} + C a b^{2}\right )} {\left (d x + c\right )}}{b^{4}} + \frac {12 \, {\left (C a^{4} + A a^{2} b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} b^{4}} - \frac {2 \, {\left (6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} b^{3}}}{6 \, d} \]
-1/6*(3*(2*C*a^3 + 2*A*a*b^2 + C*a*b^2)*(d*x + c)/b^4 + 12*(C*a^4 + A*a^2* b^2)*(pi*floor(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^4) - 2*(6*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 3*C*a*b*tan(1/2*d*x + 1/2*c)^ 5 + 6*A*b^2*tan(1/2*d*x + 1/2*c)^5 + 6*C*b^2*tan(1/2*d*x + 1/2*c)^5 + 12*C *a^2*tan(1/2*d*x + 1/2*c)^3 + 12*A*b^2*tan(1/2*d*x + 1/2*c)^3 + 4*C*b^2*ta n(1/2*d*x + 1/2*c)^3 + 6*C*a^2*tan(1/2*d*x + 1/2*c) - 3*C*a*b*tan(1/2*d*x + 1/2*c) + 6*A*b^2*tan(1/2*d*x + 1/2*c) + 6*C*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 = 5.58 (sec) , antiderivative size = 3953, normalized size of antiderivative = 22.33 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\text {Too large to display} \]
((tan(c/2 + (d*x)/2)^5*(2*A*b^2 + 2*C*a^2 + 2*C*b^2 + C*a*b))/b^3 + (4*tan (c/2 + (d*x)/2)^3*(3*A*b^2 + 3*C*a^2 + C*b^2))/(3*b^3) + (tan(c/2 + (d*x)/ 2)*(2*A*b^2 + 2*C*a^2 + 2*C*b^2 - C*a*b))/b^3)/(d*(3*tan(c/2 + (d*x)/2)^2 + 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 + 1)) - (atan(-((((((8*(4* A*a^3*b^10 - 8*A*a^2*b^11 - 2*C*a^2*b^11 + 2*C*a^3*b^10 - 6*C*a^4*b^9 + 4* C*a^5*b^8 + 4*A*a*b^12 + 2*C*a*b^12))/b^9 - (8*tan(c/2 + (d*x)/2)*(C*a^3*1 i + (a*b^2*(2*A + C)*1i)/2)*(8*a*b^10 - 16*a^2*b^9 + 8*a^3*b^8))/b^10)*(C* a^3*1i + (a*b^2*(2*A + C)*1i)/2))/b^4 - (8*tan(c/2 + (d*x)/2)*(8*C^2*a^9 - 16*C^2*a^8*b - 4*A^2*a^2*b^7 + 12*A^2*a^3*b^6 - 16*A^2*a^4*b^5 + 8*A^2*a^ 5*b^4 - C^2*a^2*b^7 + 3*C^2*a^3*b^6 - 7*C^2*a^4*b^5 + 13*C^2*a^5*b^4 - 16* C^2*a^6*b^3 + 16*C^2*a^7*b^2 - 4*A*C*a^2*b^7 + 12*A*C*a^3*b^6 - 20*A*C*a^4 *b^5 + 28*A*C*a^5*b^4 - 32*A*C*a^6*b^3 + 16*A*C*a^7*b^2))/b^6)*(C*a^3*1i + (a*b^2*(2*A + C)*1i)/2)*1i)/b^4 - (((((8*(4*A*a^3*b^10 - 8*A*a^2*b^11 - 2 *C*a^2*b^11 + 2*C*a^3*b^10 - 6*C*a^4*b^9 + 4*C*a^5*b^8 + 4*A*a*b^12 + 2*C* a*b^12))/b^9 + (8*tan(c/2 + (d*x)/2)*(C*a^3*1i + (a*b^2*(2*A + C)*1i)/2)*( 8*a*b^10 - 16*a^2*b^9 + 8*a^3*b^8))/b^10)*(C*a^3*1i + (a*b^2*(2*A + C)*1i) /2))/b^4 + (8*tan(c/2 + (d*x)/2)*(8*C^2*a^9 - 16*C^2*a^8*b - 4*A^2*a^2*b^7 + 12*A^2*a^3*b^6 - 16*A^2*a^4*b^5 + 8*A^2*a^5*b^4 - C^2*a^2*b^7 + 3*C^2*a ^3*b^6 - 7*C^2*a^4*b^5 + 13*C^2*a^5*b^4 - 16*C^2*a^6*b^3 + 16*C^2*a^7*b^2 - 4*A*C*a^2*b^7 + 12*A*C*a^3*b^6 - 20*A*C*a^4*b^5 + 28*A*C*a^5*b^4 - 32...