Integrand size = 33, antiderivative size = 233 \[ \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\frac {\left (8 a^4 C+4 a^2 b^2 (2 A+C)+b^4 (4 A+3 C)\right ) x}{8 b^5}-\frac {2 a^3 \left (A b^2+a^2 C\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^5 \sqrt {a+b} d}-\frac {a \left (3 A b^2+3 a^2 C+2 b^2 C\right ) \sin (c+d x)}{3 b^4 d}+\frac {\left (4 a^2 C+b^2 (4 A+3 C)\right ) \cos (c+d x) \sin (c+d x)}{8 b^3 d}-\frac {a C \cos ^2(c+d x) \sin (c+d x)}{3 b^2 d}+\frac {C \cos ^3(c+d x) \sin (c+d x)}{4 b d} \]
1/8*(8*a^4*C+4*a^2*b^2*(2*A+C)+b^4*(4*A+3*C))*x/b^5-1/3*a*(3*A*b^2+3*C*a^2 +2*C*b^2)*sin(d*x+c)/b^4/d+1/8*(4*a^2*C+b^2*(4*A+3*C))*cos(d*x+c)*sin(d*x+ c)/b^3/d-1/3*a*C*cos(d*x+c)^2*sin(d*x+c)/b^2/d+1/4*C*cos(d*x+c)^3*sin(d*x+ c)/b/d-2*a^3*(A*b^2+C*a^2)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/ 2))/b^5/d/(a-b)^(1/2)/(a+b)^(1/2)
Time = 1.92 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.83 \[ \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\frac {12 \left (8 a^4 C+4 a^2 b^2 (2 A+C)+b^4 (4 A+3 C)\right ) (c+d x)+\frac {192 a^3 \left (A b^2+a^2 C\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}-24 a b \left (4 A b^2+4 a^2 C+3 b^2 C\right ) \sin (c+d x)+24 b^2 \left (A b^2+\left (a^2+b^2\right ) C\right ) \sin (2 (c+d x))-8 a b^3 C \sin (3 (c+d x))+3 b^4 C \sin (4 (c+d x))}{96 b^5 d} \]
(12*(8*a^4*C + 4*a^2*b^2*(2*A + C) + b^4*(4*A + 3*C))*(c + d*x) + (192*a^3 *(A*b^2 + a^2*C)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/Sqr t[-a^2 + b^2] - 24*a*b*(4*A*b^2 + 4*a^2*C + 3*b^2*C)*Sin[c + d*x] + 24*b^2 *(A*b^2 + (a^2 + b^2)*C)*Sin[2*(c + d*x)] - 8*a*b^3*C*Sin[3*(c + d*x)] + 3 *b^4*C*Sin[4*(c + d*x)])/(96*b^5*d)
Time = 1.52 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.11, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.485, Rules used = {3042, 3529, 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.
| \(\displaystyle \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 3529 |
| \(\displaystyle \frac {\int \frac {\cos ^2(c+d x) \left (-4 a C \cos ^2(c+d x)+b (4 A+3 C) \cos (c+d x)+3 a C\right )}{a+b \cos (c+d x)}dx}{4 b}+\frac {C \sin (c+d x) \cos ^3(c+d x)}{4 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (-4 a C \sin \left (c+d x+\frac {\pi }{2}\right )^2+b (4 A+3 C) \sin \left (c+d x+\frac {\pi }{2}\right )+3 a C\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{4 b}+\frac {C \sin (c+d x) \cos ^3(c+d x)}{4 b d}\) |
| \(\Big \downarrow \) 3528 |
| \(\displaystyle \frac {\frac {\int -\frac {\cos (c+d x) \left (8 C a^2-b C \cos (c+d x) a-3 \left (4 C a^2+b^2 (4 A+3 C)\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)}dx}{3 b}-\frac {4 a C \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{4 b}+\frac {C \sin (c+d x) \cos ^3(c+d x)}{4 b d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {-\frac {\int \frac {\cos (c+d x) \left (8 C a^2-b C \cos (c+d x) a-3 \left (4 C a^2+b^2 (4 A+3 C)\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)}dx}{3 b}-\frac {4 a C \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{4 b}+\frac {C \sin (c+d x) \cos ^3(c+d x)}{4 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \left (8 C a^2-b C \sin \left (c+d x+\frac {\pi }{2}\right ) a-3 \left (4 C a^2+b^2 (4 A+3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{3 b}-\frac {4 a C \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{4 b}+\frac {C \sin (c+d x) \cos ^3(c+d x)}{4 b d}\) |
| \(\Big \downarrow \) 3528 |
| \(\displaystyle \frac {-\frac {\frac {\int -\frac {-8 a \left (3 C a^2+3 A b^2+2 b^2 C\right ) \cos ^2(c+d x)+b \left (-4 C a^2+12 A b^2+9 b^2 C\right ) \cos (c+d x)+3 a \left (4 C a^2+b^2 (4 A+3 C)\right )}{a+b \cos (c+d x)}dx}{2 b}-\frac {3 \left (4 a^2 C+b^2 (4 A+3 C)\right ) \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}-\frac {4 a C \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{4 b}+\frac {C \sin (c+d x) \cos ^3(c+d x)}{4 b d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {-\frac {-\frac {\int \frac {-8 a \left (3 C a^2+3 A b^2+2 b^2 C\right ) \cos ^2(c+d x)+b \left (-4 C a^2+12 A b^2+9 b^2 C\right ) \cos (c+d x)+3 a \left (4 C a^2+b^2 (4 A+3 C)\right )}{a+b \cos (c+d x)}dx}{2 b}-\frac {3 \left (4 a^2 C+b^2 (4 A+3 C)\right ) \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}-\frac {4 a C \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{4 b}+\frac {C \sin (c+d x) \cos ^3(c+d x)}{4 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {-\frac {\int \frac {-8 a \left (3 C a^2+3 A b^2+2 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+b \left (-4 C a^2+12 A b^2+9 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+3 a \left (4 C a^2+b^2 (4 A+3 C)\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 b}-\frac {3 \left (4 a^2 C+b^2 (4 A+3 C)\right ) \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}-\frac {4 a C \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{4 b}+\frac {C \sin (c+d x) \cos ^3(c+d x)}{4 b d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {-\frac {-\frac {\frac {\int \frac {3 \left (a b \left (4 C a^2+b^2 (4 A+3 C)\right )+\left (8 C a^4+4 b^2 (2 A+C) a^2+b^4 (4 A+3 C)\right ) \cos (c+d x)\right )}{a+b \cos (c+d x)}dx}{b}-\frac {8 a \left (3 a^2 C+3 A b^2+2 b^2 C\right ) \sin (c+d x)}{b d}}{2 b}-\frac {3 \left (4 a^2 C+b^2 (4 A+3 C)\right ) \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}-\frac {4 a C \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{4 b}+\frac {C \sin (c+d x) \cos ^3(c+d x)}{4 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {-\frac {\frac {3 \int \frac {a b \left (4 C a^2+b^2 (4 A+3 C)\right )+\left (8 C a^4+4 b^2 (2 A+C) a^2+b^4 (4 A+3 C)\right ) \cos (c+d x)}{a+b \cos (c+d x)}dx}{b}-\frac {8 a \left (3 a^2 C+3 A b^2+2 b^2 C\right ) \sin (c+d x)}{b d}}{2 b}-\frac {3 \left (4 a^2 C+b^2 (4 A+3 C)\right ) \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}-\frac {4 a C \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{4 b}+\frac {C \sin (c+d x) \cos ^3(c+d x)}{4 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {-\frac {\frac {3 \int \frac {a b \left (4 C a^2+b^2 (4 A+3 C)\right )+\left (8 C a^4+4 b^2 (2 A+C) a^2+b^4 (4 A+3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {8 a \left (3 a^2 C+3 A b^2+2 b^2 C\right ) \sin (c+d x)}{b d}}{2 b}-\frac {3 \left (4 a^2 C+b^2 (4 A+3 C)\right ) \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}-\frac {4 a C \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{4 b}+\frac {C \sin (c+d x) \cos ^3(c+d x)}{4 b d}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {-\frac {-\frac {\frac {3 \left (\frac {x \left (8 a^4 C+4 a^2 b^2 (2 A+C)+b^4 (4 A+3 C)\right )}{b}-\frac {8 a^3 \left (a^2 C+A b^2\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{b}\right )}{b}-\frac {8 a \left (3 a^2 C+3 A b^2+2 b^2 C\right ) \sin (c+d x)}{b d}}{2 b}-\frac {3 \left (4 a^2 C+b^2 (4 A+3 C)\right ) \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}-\frac {4 a C \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{4 b}+\frac {C \sin (c+d x) \cos ^3(c+d x)}{4 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {-\frac {\frac {3 \left (\frac {x \left (8 a^4 C+4 a^2 b^2 (2 A+C)+b^4 (4 A+3 C)\right )}{b}-\frac {8 a^3 \left (a^2 C+A b^2\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}\right )}{b}-\frac {8 a \left (3 a^2 C+3 A b^2+2 b^2 C\right ) \sin (c+d x)}{b d}}{2 b}-\frac {3 \left (4 a^2 C+b^2 (4 A+3 C)\right ) \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}-\frac {4 a C \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{4 b}+\frac {C \sin (c+d x) \cos ^3(c+d x)}{4 b d}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {-\frac {-\frac {\frac {3 \left (\frac {x \left (8 a^4 C+4 a^2 b^2 (2 A+C)+b^4 (4 A+3 C)\right )}{b}-\frac {16 a^3 \left (a^2 C+A b^2\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{b d}\right )}{b}-\frac {8 a \left (3 a^2 C+3 A b^2+2 b^2 C\right ) \sin (c+d x)}{b d}}{2 b}-\frac {3 \left (4 a^2 C+b^2 (4 A+3 C)\right ) \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}-\frac {4 a C \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{4 b}+\frac {C \sin (c+d x) \cos ^3(c+d x)}{4 b d}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {-\frac {-\frac {3 \left (4 a^2 C+b^2 (4 A+3 C)\right ) \sin (c+d x) \cos (c+d x)}{2 b d}-\frac {\frac {3 \left (\frac {x \left (8 a^4 C+4 a^2 b^2 (2 A+C)+b^4 (4 A+3 C)\right )}{b}-\frac {16 a^3 \left (a^2 C+A b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b d \sqrt {a-b} \sqrt {a+b}}\right )}{b}-\frac {8 a \left (3 a^2 C+3 A b^2+2 b^2 C\right ) \sin (c+d x)}{b d}}{2 b}}{3 b}-\frac {4 a C \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{4 b}+\frac {C \sin (c+d x) \cos ^3(c+d x)}{4 b d}\) |
(C*Cos[c + d*x]^3*Sin[c + d*x])/(4*b*d) + ((-4*a*C*Cos[c + d*x]^2*Sin[c + d*x])/(3*b*d) - ((-3*(4*a^2*C + b^2*(4*A + 3*C))*Cos[c + d*x]*Sin[c + d*x] )/(2*b*d) - ((3*(((8*a^4*C + 4*a^2*b^2*(2*A + C) + b^4*(4*A + 3*C))*x)/b - (16*a^3*(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 - (8*a*(3*A*b^2 + 3*a^2*C + 2*b^2*C)* Sin[c + d*x])/(b*d))/(2*b))/(3*b))/(4*b)
3.6.62.3.1 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 = 2.51 (sec) , antiderivative size = 353, normalized size of antiderivative = 1.52
| method | result | size |
| derivativedivides | \(\frac {\frac {\frac {2 \left (\left (-A a \,b^{3}-\frac {1}{2} A \,b^{4}-C \,a^{3} b -\frac {1}{2} C \,a^{2} b^{2}-C a \,b^{3}-\frac {5}{8} C \,b^{4}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-3 A a \,b^{3}-3 C \,a^{3} b -\frac {5}{3} C a \,b^{3}+\frac {3}{8} C \,b^{4}-\frac {1}{2} A \,b^{4}-\frac {1}{2} C \,a^{2} b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {1}{2} A \,b^{4}+\frac {1}{2} C \,a^{2} b^{2}-\frac {3}{8} C \,b^{4}-3 A a \,b^{3}-3 C \,a^{3} b -\frac {5}{3} C a \,b^{3}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {1}{2} A \,b^{4}+\frac {1}{2} C \,a^{2} b^{2}+\frac {5}{8} C \,b^{4}-A a \,b^{3}-C \,a^{3} b -C a \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {\left (8 A \,a^{2} b^{2}+4 A \,b^{4}+8 C \,a^{4}+4 C \,a^{2} b^{2}+3 C \,b^{4}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{b^{5}}-\frac {2 a^{3} \left (A \,b^{2}+a^{2} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{5} \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(353\) |
| default | \(\frac {\frac {\frac {2 \left (\left (-A a \,b^{3}-\frac {1}{2} A \,b^{4}-C \,a^{3} b -\frac {1}{2} C \,a^{2} b^{2}-C a \,b^{3}-\frac {5}{8} C \,b^{4}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-3 A a \,b^{3}-3 C \,a^{3} b -\frac {5}{3} C a \,b^{3}+\frac {3}{8} C \,b^{4}-\frac {1}{2} A \,b^{4}-\frac {1}{2} C \,a^{2} b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {1}{2} A \,b^{4}+\frac {1}{2} C \,a^{2} b^{2}-\frac {3}{8} C \,b^{4}-3 A a \,b^{3}-3 C \,a^{3} b -\frac {5}{3} C a \,b^{3}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {1}{2} A \,b^{4}+\frac {1}{2} C \,a^{2} b^{2}+\frac {5}{8} C \,b^{4}-A a \,b^{3}-C \,a^{3} b -C a \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {\left (8 A \,a^{2} b^{2}+4 A \,b^{4}+8 C \,a^{4}+4 C \,a^{2} b^{2}+3 C \,b^{4}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{b^{5}}-\frac {2 a^{3} \left (A \,b^{2}+a^{2} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{5} \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(353\) |
| risch | \(\frac {x A \,a^{2}}{b^{3}}+\frac {x A}{2 b}+\frac {x C \,a^{4}}{b^{5}}+\frac {x C \,a^{2}}{2 b^{3}}+\frac {3 C x}{8 b}+\frac {i a^{3} {\mathrm e}^{i \left (d x +c \right )} C}{2 b^{4} d}+\frac {3 i a \,{\mathrm e}^{i \left (d x +c \right )} C}{8 b^{2} d}-\frac {i a \,{\mathrm e}^{-i \left (d x +c \right )} A}{2 b^{2} d}+\frac {i a \,{\mathrm e}^{i \left (d x +c \right )} A}{2 b^{2} d}-\frac {3 i a \,{\mathrm e}^{-i \left (d x +c \right )} C}{8 b^{2} d}-\frac {i a^{3} {\mathrm e}^{-i \left (d x +c \right )} C}{2 b^{4} d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, d \,b^{3}}-\frac {a^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, d \,b^{5}}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, d \,b^{3}}+\frac {a^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, d \,b^{5}}+\frac {C \sin \left (4 d x +4 c \right )}{32 b d}-\frac {a C \sin \left (3 d x +3 c \right )}{12 b^{2} d}+\frac {\sin \left (2 d x +2 c \right ) A}{4 b d}+\frac {\sin \left (2 d x +2 c \right ) a^{2} C}{4 b^{3} d}+\frac {\sin \left (2 d x +2 c \right ) C}{4 b d}\) | \(564\) |
1/d*(2/b^5*(((-A*a*b^3-1/2*A*b^4-C*a^3*b-1/2*C*a^2*b^2-C*a*b^3-5/8*C*b^4)* tan(1/2*d*x+1/2*c)^7+(-3*A*a*b^3-3*C*a^3*b-5/3*C*a*b^3+3/8*C*b^4-1/2*A*b^4 -1/2*C*a^2*b^2)*tan(1/2*d*x+1/2*c)^5+(1/2*A*b^4+1/2*C*a^2*b^2-3/8*C*b^4-3* A*a*b^3-3*C*a^3*b-5/3*C*a*b^3)*tan(1/2*d*x+1/2*c)^3+(1/2*A*b^4+1/2*C*a^2*b ^2+5/8*C*b^4-A*a*b^3-C*a^3*b-C*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*a^2*b^2+4*A*b^4+8*C*a^4+4*C*a^2*b^2+3*C*b^4)*arctan(ta n(1/2*d*x+1/2*c)))-2*a^3*(A*b^2+C*a^2)/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)))
Time = 0.34 (sec) , antiderivative size = 609, normalized size of antiderivative = 2.61 \[ \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\left [\frac {3 \, {\left (8 \, C a^{6} + 4 \, {\left (2 \, A - C\right )} a^{4} b^{2} - {\left (4 \, A + C\right )} a^{2} b^{4} - {\left (4 \, A + 3 \, C\right )} b^{6}\right )} d x - 12 \, {\left (C a^{5} + A a^{3} 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 (24 \, C a^{5} b + 8 \, {\left (3 \, A - C\right )} a^{3} b^{3} - 8 \, {\left (3 \, A + 2 \, C\right )} a b^{5} - 6 \, {\left (C a^{2} b^{4} - C b^{6}\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left (C a^{3} b^{3} - C a b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (4 \, C a^{4} b^{2} + {\left (4 \, A - C\right )} a^{2} b^{4} - {\left (4 \, A + 3 \, C\right )} 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 {3 \, {\left (8 \, C a^{6} + 4 \, {\left (2 \, A - C\right )} a^{4} b^{2} - {\left (4 \, A + C\right )} a^{2} b^{4} - {\left (4 \, A + 3 \, C\right )} b^{6}\right )} d x - 24 \, {\left (C a^{5} + A a^{3} 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 (24 \, C a^{5} b + 8 \, {\left (3 \, A - C\right )} a^{3} b^{3} - 8 \, {\left (3 \, A + 2 \, C\right )} a b^{5} - 6 \, {\left (C a^{2} b^{4} - C b^{6}\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left (C a^{3} b^{3} - C a b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (4 \, C a^{4} b^{2} + {\left (4 \, A - C\right )} a^{2} b^{4} - {\left (4 \, A + 3 \, C\right )} 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 ] \]
[1/24*(3*(8*C*a^6 + 4*(2*A - C)*a^4*b^2 - (4*A + C)*a^2*b^4 - (4*A + 3*C)* b^6)*d*x - 12*(C*a^5 + A*a^3*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 )) - (24*C*a^5*b + 8*(3*A - C)*a^3*b^3 - 8*(3*A + 2*C)*a*b^5 - 6*(C*a^2*b^ 4 - C*b^6)*cos(d*x + c)^3 + 8*(C*a^3*b^3 - C*a*b^5)*cos(d*x + c)^2 - 3*(4* C*a^4*b^2 + (4*A - C)*a^2*b^4 - (4*A + 3*C)*b^6)*cos(d*x + c))*sin(d*x + c ))/((a^2*b^5 - b^7)*d), 1/24*(3*(8*C*a^6 + 4*(2*A - C)*a^4*b^2 - (4*A + C) *a^2*b^4 - (4*A + 3*C)*b^6)*d*x - 24*(C*a^5 + A*a^3*b^2)*sqrt(a^2 - b^2)*a rctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) - (24*C*a^5*b + 8*(3*A - C)*a^3*b^3 - 8*(3*A + 2*C)*a*b^5 - 6*(C*a^2*b^4 - C*b^6)*cos(d* x + c)^3 + 8*(C*a^3*b^3 - C*a*b^5)*cos(d*x + c)^2 - 3*(4*C*a^4*b^2 + (4*A - C)*a^2*b^4 - (4*A + 3*C)*b^6)*cos(d*x + c))*sin(d*x + c))/((a^2*b^5 - b^ 7)*d)]
Timed out. \[ \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\text {Exception raised: ValueError} \]
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. 574 vs. \(2 (214) = 428\).
Time = 0.32 (sec) , antiderivative size = 574, normalized size of antiderivative = 2.46 \[ \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\frac {\frac {3 \, {\left (8 \, C a^{4} + 8 \, A a^{2} b^{2} + 4 \, C a^{2} b^{2} + 4 \, A b^{4} + 3 \, C b^{4}\right )} {\left (d x + c\right )}}{b^{5}} + \frac {48 \, {\left (C a^{5} + A a^{3} 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^{5}} - \frac {2 \, {\left (24 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 72 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 72 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 72 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 72 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, C 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} \]
1/24*(3*(8*C*a^4 + 8*A*a^2*b^2 + 4*C*a^2*b^2 + 4*A*b^4 + 3*C*b^4)*(d*x + c )/b^5 + 48*(C*a^5 + A*a^3*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^5) - 2*(24*C*a^3*tan(1/2*d*x + 1/2*c)^7 + 12 *C*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 24*A*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 24*C *a*b^2*tan(1/2*d*x + 1/2*c)^7 + 12*A*b^3*tan(1/2*d*x + 1/2*c)^7 + 15*C*b^3 *tan(1/2*d*x + 1/2*c)^7 + 72*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 12*C*a^2*b*tan (1/2*d*x + 1/2*c)^5 + 72*A*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 40*C*a*b^2*tan(1 /2*d*x + 1/2*c)^5 + 12*A*b^3*tan(1/2*d*x + 1/2*c)^5 - 9*C*b^3*tan(1/2*d*x + 1/2*c)^5 + 72*C*a^3*tan(1/2*d*x + 1/2*c)^3 - 12*C*a^2*b*tan(1/2*d*x + 1/ 2*c)^3 + 72*A*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 40*C*a*b^2*tan(1/2*d*x + 1/2* c)^3 - 12*A*b^3*tan(1/2*d*x + 1/2*c)^3 + 9*C*b^3*tan(1/2*d*x + 1/2*c)^3 + 24*C*a^3*tan(1/2*d*x + 1/2*c) - 12*C*a^2*b*tan(1/2*d*x + 1/2*c) + 24*A*a*b ^2*tan(1/2*d*x + 1/2*c) + 24*C*a*b^2*tan(1/2*d*x + 1/2*c) - 12*A*b^3*tan(1 /2*d*x + 1/2*c) - 15*C*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 = 8.39 (sec) , antiderivative size = 5844, normalized size of antiderivative = 25.08 \[ \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\text {Too large to display} \]
- ((tan(c/2 + (d*x)/2)^7*(4*A*b^3 + 8*C*a^3 + 5*C*b^3 + 8*A*a*b^2 + 8*C*a* b^2 + 4*C*a^2*b))/(4*b^4) + (tan(c/2 + (d*x)/2)^3*(72*C*a^3 - 12*A*b^3 + 9 *C*b^3 + 72*A*a*b^2 + 40*C*a*b^2 - 12*C*a^2*b))/(12*b^4) + (tan(c/2 + (d*x )/2)^5*(12*A*b^3 + 72*C*a^3 - 9*C*b^3 + 72*A*a*b^2 + 40*C*a*b^2 + 12*C*a^2 *b))/(12*b^4) - (tan(c/2 + (d*x)/2)*(4*A*b^3 - 8*C*a^3 + 5*C*b^3 - 8*A*a*b ^2 - 8*C*a*b^2 + 4*C*a^2*b))/(4*b^4))/(d*(4*tan(c/2 + (d*x)/2)^2 + 6*tan(c /2 + (d*x)/2)^4 + 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1)) - (a tan(((((tan(c/2 + (d*x)/2)*(16*A^2*b^11 - 128*C^2*a^11 + 9*C^2*b^11 - 48*A ^2*a*b^10 - 27*C^2*a*b^10 + 256*C^2*a^10*b + 112*A^2*a^2*b^9 - 208*A^2*a^3 *b^8 + 256*A^2*a^4*b^7 - 256*A^2*a^5*b^6 + 256*A^2*a^6*b^5 - 128*A^2*a^7*b ^4 + 51*C^2*a^2*b^9 - 81*C^2*a^3*b^8 + 136*C^2*a^4*b^7 - 216*C^2*a^5*b^6 + 256*C^2*a^6*b^5 - 256*C^2*a^7*b^4 + 256*C^2*a^8*b^3 - 256*C^2*a^9*b^2 + 2 4*A*C*b^11 - 72*A*C*a*b^10 + 152*A*C*a^2*b^9 - 264*A*C*a^3*b^8 + 368*A*C*a ^4*b^7 - 464*A*C*a^5*b^6 + 512*A*C*a^6*b^5 - 512*A*C*a^7*b^4 + 512*A*C*a^8 *b^3 - 256*A*C*a^9*b^2))/(2*b^8) + (((16*A*b^16 + 12*C*b^16 + 16*A*a^2*b^1 4 - 48*A*a^3*b^13 + 32*A*a^4*b^12 + 4*C*a^2*b^14 - 4*C*a^3*b^13 + 16*C*a^4 *b^12 - 48*C*a^5*b^11 + 32*C*a^6*b^10 - 16*A*a*b^15 - 12*C*a*b^15)/b^12 - (tan(c/2 + (d*x)/2)*(128*a*b^12 - 256*a^2*b^11 + 128*a^3*b^10)*(b^2*(A*a^2 *1i + (C*a^2*1i)/2) + C*a^4*1i + b^4*((A*1i)/2 + (C*3i)/8)))/(2*b^13))*(b^ 2*(A*a^2*1i + (C*a^2*1i)/2) + C*a^4*1i + b^4*((A*1i)/2 + (C*3i)/8)))/b^...