Integrand size = 40, antiderivative size = 263 \[ \int \frac {\cos ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=-\frac {\left (4 a b B-6 a^2 C-b^2 C\right ) x}{2 b^4}+\frac {2 a^2 \left (2 a^2 b B-3 b^3 B-3 a^3 C+4 a b^2 C\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} b^4 (a+b)^{3/2} d}+\frac {\left (2 a^2 b B-b^3 B-3 a^3 C+2 a b^2 C\right ) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}-\frac {\left (2 a b B-3 a^2 C+b^2 C\right ) \cos (c+d x) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) d}+\frac {a (b B-a C) \cos ^2(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \]
-1/2*(4*B*a*b-6*C*a^2-C*b^2)*x/b^4+2*a^2*(2*B*a^2*b-3*B*b^3-3*C*a^3+4*C*a* b^2)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(3/2)/b^4/(a +b)^(3/2)/d+(2*B*a^2*b-B*b^3-3*C*a^3+2*C*a*b^2)*sin(d*x+c)/b^3/(a^2-b^2)/d -1/2*(2*B*a*b-3*C*a^2+C*b^2)*cos(d*x+c)*sin(d*x+c)/b^2/(a^2-b^2)/d+a*(B*b- C*a)*cos(d*x+c)^2*sin(d*x+c)/b/(a^2-b^2)/d/(a+b*cos(d*x+c))
Time = 2.28 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.70 \[ \int \frac {\cos ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\frac {2 \left (-4 a b B+6 a^2 C+b^2 C\right ) (c+d x)-\frac {8 a^2 \left (-2 a^2 b B+3 b^3 B+3 a^3 C-4 a b^2 C\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{3/2}}+4 b (b B-2 a C) \sin (c+d x)+\frac {4 a^3 b (b B-a C) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))}+b^2 C \sin (2 (c+d x))}{4 b^4 d} \]
(2*(-4*a*b*B + 6*a^2*C + b^2*C)*(c + d*x) - (8*a^2*(-2*a^2*b*B + 3*b^3*B + 3*a^3*C - 4*a*b^2*C)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]] )/(-a^2 + b^2)^(3/2) + 4*b*(b*B - 2*a*C)*Sin[c + d*x] + (4*a^3*b*(b*B - a* C)*Sin[c + d*x])/((a - b)*(a + b)*(a + b*Cos[c + d*x])) + b^2*C*Sin[2*(c + d*x)])/(4*b^4*d)
Time = 1.44 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.05, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3042, 3508, 3042, 3468, 25, 3042, 3528, 25, 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.
| \(\displaystyle \int \frac {\cos ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 3508 |
| \(\displaystyle \int \frac {\cos ^3(c+d x) (B+C \cos (c+d x))}{(a+b \cos (c+d x))^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (B+C \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 3468 |
| \(\displaystyle \frac {a (b B-a C) \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\int -\frac {\cos (c+d x) \left (-\left (\left (-3 C a^2+2 b B a+b^2 C\right ) \cos ^2(c+d x)\right )-b (b B-a C) \cos (c+d x)+2 a (b B-a C)\right )}{a+b \cos (c+d x)}dx}{b \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {\cos (c+d x) \left (-\left (\left (-3 C a^2+2 b B a+b^2 C\right ) \cos ^2(c+d x)\right )-b (b B-a C) \cos (c+d x)+2 a (b B-a C)\right )}{a+b \cos (c+d x)}dx}{b \left (a^2-b^2\right )}+\frac {a (b B-a C) \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \left (\left (3 C a^2-2 b B a-b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-b (b B-a C) \sin \left (c+d x+\frac {\pi }{2}\right )+2 a (b B-a C)\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b \left (a^2-b^2\right )}+\frac {a (b B-a C) \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3528 |
| \(\displaystyle \frac {\frac {\int -\frac {-2 \left (-3 C a^3+2 b B a^2+2 b^2 C a-b^3 B\right ) \cos ^2(c+d x)-b \left (-C a^2+2 b B a-b^2 C\right ) \cos (c+d x)+a \left (-3 C a^2+2 b B a+b^2 C\right )}{a+b \cos (c+d x)}dx}{2 b}-\frac {\left (-3 a^2 C+2 a b B+b^2 C\right ) \sin (c+d x) \cos (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}+\frac {a (b B-a C) \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {-\frac {\int \frac {-2 \left (-3 C a^3+2 b B a^2+2 b^2 C a-b^3 B\right ) \cos ^2(c+d x)-b \left (-C a^2+2 b B a-b^2 C\right ) \cos (c+d x)+a \left (-3 C a^2+2 b B a+b^2 C\right )}{a+b \cos (c+d x)}dx}{2 b}-\frac {\left (-3 a^2 C+2 a b B+b^2 C\right ) \sin (c+d x) \cos (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}+\frac {a (b B-a C) \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\int \frac {-2 \left (-3 C a^3+2 b B a^2+2 b^2 C a-b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-b \left (-C a^2+2 b B a-b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+a \left (-3 C a^2+2 b B a+b^2 C\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 b}-\frac {\left (-3 a^2 C+2 a b B+b^2 C\right ) \sin (c+d x) \cos (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}+\frac {a (b B-a C) \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {-\frac {\frac {\int \frac {a b \left (-3 C a^2+2 b B a+b^2 C\right )+\left (a^2-b^2\right ) \left (-6 C a^2+4 b B a-b^2 C\right ) \cos (c+d x)}{a+b \cos (c+d x)}dx}{b}-\frac {2 \left (-3 a^3 C+2 a^2 b B+2 a b^2 C-b^3 B\right ) \sin (c+d x)}{b d}}{2 b}-\frac {\left (-3 a^2 C+2 a b B+b^2 C\right ) \sin (c+d x) \cos (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}+\frac {a (b B-a C) \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\frac {\int \frac {a b \left (-3 C a^2+2 b B a+b^2 C\right )+\left (a^2-b^2\right ) \left (-6 C a^2+4 b B a-b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {2 \left (-3 a^3 C+2 a^2 b B+2 a b^2 C-b^3 B\right ) \sin (c+d x)}{b d}}{2 b}-\frac {\left (-3 a^2 C+2 a b B+b^2 C\right ) \sin (c+d x) \cos (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}+\frac {a (b B-a C) \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {-\frac {\frac {\frac {x \left (a^2-b^2\right ) \left (-6 a^2 C+4 a b B-b^2 C\right )}{b}-\frac {2 a^2 \left (-3 a^3 C+2 a^2 b B+4 a b^2 C-3 b^3 B\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{b}}{b}-\frac {2 \left (-3 a^3 C+2 a^2 b B+2 a b^2 C-b^3 B\right ) \sin (c+d x)}{b d}}{2 b}-\frac {\left (-3 a^2 C+2 a b B+b^2 C\right ) \sin (c+d x) \cos (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}+\frac {a (b B-a C) \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\frac {\frac {x \left (a^2-b^2\right ) \left (-6 a^2 C+4 a b B-b^2 C\right )}{b}-\frac {2 a^2 \left (-3 a^3 C+2 a^2 b B+4 a b^2 C-3 b^3 B\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}}{b}-\frac {2 \left (-3 a^3 C+2 a^2 b B+2 a b^2 C-b^3 B\right ) \sin (c+d x)}{b d}}{2 b}-\frac {\left (-3 a^2 C+2 a b B+b^2 C\right ) \sin (c+d x) \cos (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}+\frac {a (b B-a C) \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {-\frac {\frac {\frac {x \left (a^2-b^2\right ) \left (-6 a^2 C+4 a b B-b^2 C\right )}{b}-\frac {4 a^2 \left (-3 a^3 C+2 a^2 b B+4 a b^2 C-3 b^3 B\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{b d}}{b}-\frac {2 \left (-3 a^3 C+2 a^2 b B+2 a b^2 C-b^3 B\right ) \sin (c+d x)}{b d}}{2 b}-\frac {\left (-3 a^2 C+2 a b B+b^2 C\right ) \sin (c+d x) \cos (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}+\frac {a (b B-a C) \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {a (b B-a C) \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {-\frac {\left (-3 a^2 C+2 a b B+b^2 C\right ) \sin (c+d x) \cos (c+d x)}{2 b d}-\frac {\frac {\frac {x \left (a^2-b^2\right ) \left (-6 a^2 C+4 a b B-b^2 C\right )}{b}-\frac {4 a^2 \left (-3 a^3 C+2 a^2 b B+4 a b^2 C-3 b^3 B\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b d \sqrt {a-b} \sqrt {a+b}}}{b}-\frac {2 \left (-3 a^3 C+2 a^2 b B+2 a b^2 C-b^3 B\right ) \sin (c+d x)}{b d}}{2 b}}{b \left (a^2-b^2\right )}\) |
(a*(b*B - a*C)*Cos[c + d*x]^2*Sin[c + d*x])/(b*(a^2 - b^2)*d*(a + b*Cos[c + d*x])) + (-1/2*((2*a*b*B - 3*a^2*C + b^2*C)*Cos[c + d*x]*Sin[c + d*x])/( b*d) - ((((a^2 - b^2)*(4*a*b*B - 6*a^2*C - b^2*C)*x)/b - (4*a^2*(2*a^2*b*B - 3*b^3*B - 3*a^3*C + 4*a*b^2*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sq rt[a + b]])/(Sqrt[a - b]*b*Sqrt[a + b]*d))/b - (2*(2*a^2*b*B - b^3*B - 3*a ^3*C + 2*a*b^2*C)*Sin[c + d*x])/(b*d))/(2*b))/(b*(a^2 - b^2))
3.9.1.3.1 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_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[(-(b*c - a*d))*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1)*Simp[b*(b*c - a*d)*(B*c - A*d)*(m - 1) + a*d*(a*A*c + b*B*c - (A*b + a *B)*d)*(n + 1) + (b*(b*d*(B*c - A*d) + a*(A*c*d + B*(c^2 - 2*d^2)))*(n + 1) - a*(b*c - a*d)*(B*c - A*d)*(n + 2))*Sin[e + f*x] + b*(d*(A*b*c + a*B*c - a*A*d)*(m + n + 1) - b*B*(c^2*m + d^2*(n + 1)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2 , 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] && LtQ[n, -1]
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[1/b^2 Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
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 = 1.37 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.02
| method | result | size |
| derivativedivides | \(\frac {\frac {2 a^{2} \left (\frac {a b \left (B b -a C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}+\frac {\left (2 B \,a^{2} b -3 B \,b^{3}-3 C \,a^{3}+4 C a \,b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{4}}-\frac {2 \left (\frac {\left (-B \,b^{2}+2 C a b +\frac {1}{2} b^{2} C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-B \,b^{2}+2 C a b -\frac {1}{2} b^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {\left (4 B a b -6 a^{2} C -b^{2} C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{b^{4}}}{d}\) | \(267\) |
| default | \(\frac {\frac {2 a^{2} \left (\frac {a b \left (B b -a C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}+\frac {\left (2 B \,a^{2} b -3 B \,b^{3}-3 C \,a^{3}+4 C a \,b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{4}}-\frac {2 \left (\frac {\left (-B \,b^{2}+2 C a b +\frac {1}{2} b^{2} C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-B \,b^{2}+2 C a b -\frac {1}{2} b^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {\left (4 B a b -6 a^{2} C -b^{2} C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{b^{4}}}{d}\) | \(267\) |
| risch | \(-\frac {2 x B a}{b^{3}}+\frac {3 x \,a^{2} C}{b^{4}}+\frac {C x}{2 b^{2}}-\frac {i B \,{\mathrm e}^{i \left (d x +c \right )}}{2 b^{2} d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B}{2 b^{2} d}-\frac {i C \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 b^{2} d}+\frac {i a C \,{\mathrm e}^{i \left (d x +c \right )}}{b^{3} d}+\frac {i C \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 b^{2} d}-\frac {2 i a^{3} \left (-B b +a C \right ) \left (a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}{b^{4} \left (a^{2}-b^{2}\right ) d \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} a C}{b^{3} d}-\frac {2 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 ) B}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{3}}+\frac {3 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 ) B}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d b}+\frac {3 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}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{4}}-\frac {4 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}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{2}}+\frac {2 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 ) B}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{3}}-\frac {3 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 ) B}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d b}-\frac {3 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}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{4}}+\frac {4 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}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{2}}\) | \(918\) |
1/d*(2*a^2/b^4*(a*b*(B*b-C*a)/(a^2-b^2)*tan(1/2*d*x+1/2*c)/(tan(1/2*d*x+1/ 2*c)^2*a-b*tan(1/2*d*x+1/2*c)^2+a+b)+(2*B*a^2*b-3*B*b^3-3*C*a^3+4*C*a*b^2) /(a-b)/(a+b)/((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*(((-B*b^2+2*C*a*b+1/2*b^2*C)*tan(1/2*d*x+1/2*c)^3+(-B*b ^2+2*C*a*b-1/2*b^2*C)*tan(1/2*d*x+1/2*c))/(1+tan(1/2*d*x+1/2*c)^2)^2+1/2*( 4*B*a*b-6*C*a^2-C*b^2)*arctan(tan(1/2*d*x+1/2*c))))
Time = 0.38 (sec) , antiderivative size = 965, normalized size of antiderivative = 3.67 \[ \int \frac {\cos ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\left [\frac {{\left (6 \, C a^{6} b - 4 \, B a^{5} b^{2} - 11 \, C a^{4} b^{3} + 8 \, B a^{3} b^{4} + 4 \, C a^{2} b^{5} - 4 \, B a b^{6} + C b^{7}\right )} d x \cos \left (d x + c\right ) + {\left (6 \, C a^{7} - 4 \, B a^{6} b - 11 \, C a^{5} b^{2} + 8 \, B a^{4} b^{3} + 4 \, C a^{3} b^{4} - 4 \, B a^{2} b^{5} + C a b^{6}\right )} d x - {\left (3 \, C a^{6} - 2 \, B a^{5} b - 4 \, C a^{4} b^{2} + 3 \, B a^{3} b^{3} + {\left (3 \, C a^{5} b - 2 \, B a^{4} b^{2} - 4 \, C a^{3} b^{3} + 3 \, B a^{2} b^{4}\right )} \cos \left (d x + c\right )\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^{6} b - 4 \, B a^{5} b^{2} - 10 \, C a^{4} b^{3} + 6 \, B a^{3} b^{4} + 4 \, C a^{2} b^{5} - 2 \, B a b^{6} - {\left (C a^{4} b^{3} - 2 \, C a^{2} b^{5} + C b^{7}\right )} \cos \left (d x + c\right )^{2} + {\left (3 \, C a^{5} b^{2} - 2 \, B a^{4} b^{3} - 6 \, C a^{3} b^{4} + 4 \, B a^{2} b^{5} + 3 \, C a b^{6} - 2 \, B b^{7}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{4} b^{5} - 2 \, a^{2} b^{7} + b^{9}\right )} d \cos \left (d x + c\right ) + {\left (a^{5} b^{4} - 2 \, a^{3} b^{6} + a b^{8}\right )} d\right )}}, \frac {{\left (6 \, C a^{6} b - 4 \, B a^{5} b^{2} - 11 \, C a^{4} b^{3} + 8 \, B a^{3} b^{4} + 4 \, C a^{2} b^{5} - 4 \, B a b^{6} + C b^{7}\right )} d x \cos \left (d x + c\right ) + {\left (6 \, C a^{7} - 4 \, B a^{6} b - 11 \, C a^{5} b^{2} + 8 \, B a^{4} b^{3} + 4 \, C a^{3} b^{4} - 4 \, B a^{2} b^{5} + C a b^{6}\right )} d x - 2 \, {\left (3 \, C a^{6} - 2 \, B a^{5} b - 4 \, C a^{4} b^{2} + 3 \, B a^{3} b^{3} + {\left (3 \, C a^{5} b - 2 \, B a^{4} b^{2} - 4 \, C a^{3} b^{3} + 3 \, B a^{2} b^{4}\right )} \cos \left (d x + c\right )\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^{6} b - 4 \, B a^{5} b^{2} - 10 \, C a^{4} b^{3} + 6 \, B a^{3} b^{4} + 4 \, C a^{2} b^{5} - 2 \, B a b^{6} - {\left (C a^{4} b^{3} - 2 \, C a^{2} b^{5} + C b^{7}\right )} \cos \left (d x + c\right )^{2} + {\left (3 \, C a^{5} b^{2} - 2 \, B a^{4} b^{3} - 6 \, C a^{3} b^{4} + 4 \, B a^{2} b^{5} + 3 \, C a b^{6} - 2 \, B b^{7}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{4} b^{5} - 2 \, a^{2} b^{7} + b^{9}\right )} d \cos \left (d x + c\right ) + {\left (a^{5} b^{4} - 2 \, a^{3} b^{6} + a b^{8}\right )} d\right )}}\right ] \]
integrate(cos(d*x+c)^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2,x, algorithm="fricas")
[1/2*((6*C*a^6*b - 4*B*a^5*b^2 - 11*C*a^4*b^3 + 8*B*a^3*b^4 + 4*C*a^2*b^5 - 4*B*a*b^6 + C*b^7)*d*x*cos(d*x + c) + (6*C*a^7 - 4*B*a^6*b - 11*C*a^5*b^ 2 + 8*B*a^4*b^3 + 4*C*a^3*b^4 - 4*B*a^2*b^5 + C*a*b^6)*d*x - (3*C*a^6 - 2* B*a^5*b - 4*C*a^4*b^2 + 3*B*a^3*b^3 + (3*C*a^5*b - 2*B*a^4*b^2 - 4*C*a^3*b ^3 + 3*B*a^2*b^4)*cos(d*x + c))*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)*si n(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^6*b - 4*B*a^5*b^2 - 10*C*a^4*b^3 + 6*B*a^3*b^4 + 4*C*a^2*b^5 - 2 *B*a*b^6 - (C*a^4*b^3 - 2*C*a^2*b^5 + C*b^7)*cos(d*x + c)^2 + (3*C*a^5*b^2 - 2*B*a^4*b^3 - 6*C*a^3*b^4 + 4*B*a^2*b^5 + 3*C*a*b^6 - 2*B*b^7)*cos(d*x + c))*sin(d*x + c))/((a^4*b^5 - 2*a^2*b^7 + b^9)*d*cos(d*x + c) + (a^5*b^4 - 2*a^3*b^6 + a*b^8)*d), 1/2*((6*C*a^6*b - 4*B*a^5*b^2 - 11*C*a^4*b^3 + 8 *B*a^3*b^4 + 4*C*a^2*b^5 - 4*B*a*b^6 + C*b^7)*d*x*cos(d*x + c) + (6*C*a^7 - 4*B*a^6*b - 11*C*a^5*b^2 + 8*B*a^4*b^3 + 4*C*a^3*b^4 - 4*B*a^2*b^5 + C*a *b^6)*d*x - 2*(3*C*a^6 - 2*B*a^5*b - 4*C*a^4*b^2 + 3*B*a^3*b^3 + (3*C*a^5* b - 2*B*a^4*b^2 - 4*C*a^3*b^3 + 3*B*a^2*b^4)*cos(d*x + c))*sqrt(a^2 - b^2) *arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) - (6*C*a^6*b - 4*B*a^5*b^2 - 10*C*a^4*b^3 + 6*B*a^3*b^4 + 4*C*a^2*b^5 - 2*B*a*b^6 - (C *a^4*b^3 - 2*C*a^2*b^5 + C*b^7)*cos(d*x + c)^2 + (3*C*a^5*b^2 - 2*B*a^4*b^ 3 - 6*C*a^3*b^4 + 4*B*a^2*b^5 + 3*C*a*b^6 - 2*B*b^7)*cos(d*x + c))*sin(...
Timed out. \[ \int \frac {\cos ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\cos ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\text {Exception raised: ValueError} \]
integrate(cos(d*x+c)^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2,x, algorithm="maxima")
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 = 0.33 (sec) , antiderivative size = 338, normalized size of antiderivative = 1.29 \[ \int \frac {\cos ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\frac {\frac {4 \, {\left (3 \, C a^{5} - 2 \, B a^{4} b - 4 \, C a^{3} b^{2} + 3 \, B a^{2} b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{2} b^{4} - b^{6}\right )} \sqrt {a^{2} - b^{2}}} - \frac {4 \, {\left (C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (a^{2} b^{3} - b^{5}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}} + \frac {{\left (6 \, C a^{2} - 4 \, B a b + C b^{2}\right )} {\left (d x + c\right )}}{b^{4}} - \frac {2 \, {\left (4 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} b^{3}}}{2 \, d} \]
1/2*(4*(3*C*a^5 - 2*B*a^4*b - 4*C*a^3*b^2 + 3*B*a^2*b^3)*(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)))/((a^2*b^4 - b^6)*sqrt(a^2 - b^2)) - 4 *(C*a^4*tan(1/2*d*x + 1/2*c) - B*a^3*b*tan(1/2*d*x + 1/2*c))/((a^2*b^3 - b ^5)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)) + (6*C* a^2 - 4*B*a*b + C*b^2)*(d*x + c)/b^4 - 2*(4*C*a*tan(1/2*d*x + 1/2*c)^3 - 2 *B*b*tan(1/2*d*x + 1/2*c)^3 + C*b*tan(1/2*d*x + 1/2*c)^3 + 4*C*a*tan(1/2*d *x + 1/2*c) - 2*B*b*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^3))/d
Time = 11.11 (sec) , antiderivative size = 6743, normalized size of antiderivative = 25.64 \[ \int \frac {\cos ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\text {Too large to display} \]
(atan(((((8*tan(c/2 + (d*x)/2)*(72*C^2*a^10 + C^2*b^10 - 2*C^2*a*b^9 - 72* C^2*a^9*b + 16*B^2*a^2*b^8 - 32*B^2*a^3*b^7 + 20*B^2*a^4*b^6 + 64*B^2*a^5* b^5 - 64*B^2*a^6*b^4 - 32*B^2*a^7*b^3 + 32*B^2*a^8*b^2 + 11*C^2*a^2*b^8 - 20*C^2*a^3*b^7 + 23*C^2*a^4*b^6 - 26*C^2*a^5*b^5 + 17*C^2*a^6*b^4 + 120*C^ 2*a^7*b^3 - 120*C^2*a^8*b^2 - 8*B*C*a*b^9 - 96*B*C*a^9*b + 16*B*C*a^2*b^8 - 40*B*C*a^3*b^7 + 64*B*C*a^4*b^6 - 40*B*C*a^5*b^5 - 176*B*C*a^6*b^4 + 176 *B*C*a^7*b^3 + 96*B*C*a^8*b^2))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) + (((8*( 2*C*b^15 + 12*B*a^2*b^13 + 12*B*a^3*b^12 - 20*B*a^4*b^11 - 4*B*a^5*b^10 + 8*B*a^6*b^9 + 6*C*a^2*b^13 - 16*C*a^3*b^12 - 14*C*a^4*b^11 + 28*C*a^5*b^10 + 6*C*a^6*b^9 - 12*C*a^7*b^8 - 8*B*a*b^14))/(a*b^11 + b^12 - a^2*b^10 - a ^3*b^9) - (4*tan(c/2 + (d*x)/2)*(C*a^2*6i + C*b^2*1i - B*a*b*4i)*(8*a*b^13 - 8*a^2*b^12 - 16*a^3*b^11 + 16*a^4*b^10 + 8*a^5*b^9 - 8*a^6*b^8))/(b^4*( a*b^8 + b^9 - a^2*b^7 - a^3*b^6)))*(C*a^2*6i + C*b^2*1i - B*a*b*4i))/(2*b^ 4))*(C*a^2*6i + C*b^2*1i - B*a*b*4i)*1i)/(2*b^4) + (((8*tan(c/2 + (d*x)/2) *(72*C^2*a^10 + C^2*b^10 - 2*C^2*a*b^9 - 72*C^2*a^9*b + 16*B^2*a^2*b^8 - 3 2*B^2*a^3*b^7 + 20*B^2*a^4*b^6 + 64*B^2*a^5*b^5 - 64*B^2*a^6*b^4 - 32*B^2* a^7*b^3 + 32*B^2*a^8*b^2 + 11*C^2*a^2*b^8 - 20*C^2*a^3*b^7 + 23*C^2*a^4*b^ 6 - 26*C^2*a^5*b^5 + 17*C^2*a^6*b^4 + 120*C^2*a^7*b^3 - 120*C^2*a^8*b^2 - 8*B*C*a*b^9 - 96*B*C*a^9*b + 16*B*C*a^2*b^8 - 40*B*C*a^3*b^7 + 64*B*C*a^4* b^6 - 40*B*C*a^5*b^5 - 176*B*C*a^6*b^4 + 176*B*C*a^7*b^3 + 96*B*C*a^8*b...