Integrand size = 40, antiderivative size = 280 \[ \int \frac {\cos ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^3} \, dx=\frac {(b B-3 a C) x}{b^4}-\frac {a \left (2 a^4 b B-5 a^2 b^3 B+6 b^5 B-6 a^5 C+15 a^3 b^2 C-12 a b^4 C\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} b^4 (a+b)^{5/2} d}-\frac {\left (a b B-3 a^2 C+2 b^2 C\right ) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}+\frac {a (b B-a C) \cos ^2(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac {a^2 \left (a^2 b B-4 b^3 B-3 a^3 C+6 a b^2 C\right ) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))} \]
(B*b-3*C*a)*x/b^4-a*(2*B*a^4*b-5*B*a^2*b^3+6*B*b^5-6*C*a^5+15*C*a^3*b^2-12 *C*a*b^4)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(5/2)/b ^4/(a+b)^(5/2)/d-1/2*(B*a*b-3*C*a^2+2*C*b^2)*sin(d*x+c)/b^3/(a^2-b^2)/d+1/ 2*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))^2-1/2 *a^2*(B*a^2*b-4*B*b^3-3*C*a^3+6*C*a*b^2)*sin(d*x+c)/b^3/(a^2-b^2)^2/d/(a+b *cos(d*x+c))
Time = 3.42 (sec) , antiderivative size = 232, normalized size of antiderivative = 0.83 \[ \int \frac {\cos ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^3} \, dx=\frac {2 (b B-3 a C) (c+d x)-\frac {2 a \left (-2 a^4 b B+5 a^2 b^3 B-6 b^5 B+6 a^5 C-15 a^3 b^2 C+12 a b^4 C\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{5/2}}+2 b C \sin (c+d x)+\frac {a^3 b (b B-a C) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))^2}+\frac {a^2 b \left (-3 a^2 b B+6 b^3 B+5 a^3 C-8 a b^2 C\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (a+b \cos (c+d x))}}{2 b^4 d} \]
(2*(b*B - 3*a*C)*(c + d*x) - (2*a*(-2*a^4*b*B + 5*a^2*b^3*B - 6*b^5*B + 6* a^5*C - 15*a^3*b^2*C + 12*a*b^4*C)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt [-a^2 + b^2]])/(-a^2 + b^2)^(5/2) + 2*b*C*Sin[c + d*x] + (a^3*b*(b*B - a*C )*Sin[c + d*x])/((a - b)*(a + b)*(a + b*Cos[c + d*x])^2) + (a^2*b*(-3*a^2* b*B + 6*b^3*B + 5*a^3*C - 8*a*b^2*C)*Sin[c + d*x])/((a - b)^2*(a + b)^2*(a + b*Cos[c + d*x])))/(2*b^4*d)
Time = 1.54 (sec) , antiderivative size = 317, normalized size of antiderivative = 1.13, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {3042, 3508, 3042, 3468, 25, 3042, 3510, 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.
| \(\displaystyle \int \frac {\cos ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3}dx\) |
| \(\Big \downarrow \) 3508 |
| \(\displaystyle \int \frac {\cos ^3(c+d x) (B+C \cos (c+d x))}{(a+b \cos (c+d x))^3}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (B+C \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3}dx\) |
| \(\Big \downarrow \) 3468 |
| \(\displaystyle \frac {a (b B-a C) \sin (c+d x) \cos ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\int -\frac {\cos (c+d x) \left (-\left (\left (-3 C a^2+b B a+2 b^2 C\right ) \cos ^2(c+d x)\right )-2 b (b B-a C) \cos (c+d x)+2 a (b B-a C)\right )}{(a+b \cos (c+d x))^2}dx}{2 b \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {\cos (c+d x) \left (-\left (\left (-3 C a^2+b B a+2 b^2 C\right ) \cos ^2(c+d x)\right )-2 b (b B-a C) \cos (c+d x)+2 a (b B-a C)\right )}{(a+b \cos (c+d x))^2}dx}{2 b \left (a^2-b^2\right )}+\frac {a (b B-a C) \sin (c+d x) \cos ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \left (\left (3 C a^2-b B a-2 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-2 b (b B-a C) \sin \left (c+d x+\frac {\pi }{2}\right )+2 a (b B-a C)\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx}{2 b \left (a^2-b^2\right )}+\frac {a (b B-a C) \sin (c+d x) \cos ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 3510 |
| \(\displaystyle \frac {\frac {\int \frac {-b \left (a^2-b^2\right ) \left (-3 C a^2+b B a+2 b^2 C\right ) \cos ^2(c+d x)+\left (a^2-b^2\right ) \left (-3 C a^3+b B a^2+4 b^2 C a-2 b^3 B\right ) \cos (c+d x)+a b \left (-3 C a^3+b B a^2+6 b^2 C a-4 b^3 B\right )}{a+b \cos (c+d x)}dx}{b^2 \left (a^2-b^2\right )}-\frac {a^2 \left (-3 a^3 C+a^2 b B+6 a b^2 C-4 b^3 B\right ) \sin (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 b \left (a^2-b^2\right )}+\frac {a (b B-a C) \sin (c+d x) \cos ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {-b \left (a^2-b^2\right ) \left (-3 C a^2+b B a+2 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+\left (a^2-b^2\right ) \left (-3 C a^3+b B a^2+4 b^2 C a-2 b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+a b \left (-3 C a^3+b B a^2+6 b^2 C a-4 b^3 B\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2 \left (a^2-b^2\right )}-\frac {a^2 \left (-3 a^3 C+a^2 b B+6 a b^2 C-4 b^3 B\right ) \sin (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 b \left (a^2-b^2\right )}+\frac {a (b B-a C) \sin (c+d x) \cos ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {a \left (-3 C a^3+b B a^2+6 b^2 C a-4 b^3 B\right ) b^2+2 \left (a^2-b^2\right )^2 (b B-3 a C) \cos (c+d x) b}{a+b \cos (c+d x)}dx}{b}-\frac {\left (a^2-b^2\right ) \left (-3 a^2 C+a b B+2 b^2 C\right ) \sin (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}-\frac {a^2 \left (-3 a^3 C+a^2 b B+6 a b^2 C-4 b^3 B\right ) \sin (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 b \left (a^2-b^2\right )}+\frac {a (b B-a C) \sin (c+d x) \cos ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {a \left (-3 C a^3+b B a^2+6 b^2 C a-4 b^3 B\right ) b^2+2 \left (a^2-b^2\right )^2 (b B-3 a C) \sin \left (c+d x+\frac {\pi }{2}\right ) b}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {\left (a^2-b^2\right ) \left (-3 a^2 C+a b B+2 b^2 C\right ) \sin (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}-\frac {a^2 \left (-3 a^3 C+a^2 b B+6 a b^2 C-4 b^3 B\right ) \sin (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 b \left (a^2-b^2\right )}+\frac {a (b B-a C) \sin (c+d x) \cos ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {\frac {\frac {2 x \left (a^2-b^2\right )^2 (b B-3 a C)-a \left (-6 a^5 C+2 a^4 b B+15 a^3 b^2 C-5 a^2 b^3 B-12 a b^4 C+6 b^5 B\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{b}-\frac {\left (a^2-b^2\right ) \left (-3 a^2 C+a b B+2 b^2 C\right ) \sin (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}-\frac {a^2 \left (-3 a^3 C+a^2 b B+6 a b^2 C-4 b^3 B\right ) \sin (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 b \left (a^2-b^2\right )}+\frac {a (b B-a C) \sin (c+d x) \cos ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {2 x \left (a^2-b^2\right )^2 (b B-3 a C)-a \left (-6 a^5 C+2 a^4 b B+15 a^3 b^2 C-5 a^2 b^3 B-12 a b^4 C+6 b^5 B\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {\left (a^2-b^2\right ) \left (-3 a^2 C+a b B+2 b^2 C\right ) \sin (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}-\frac {a^2 \left (-3 a^3 C+a^2 b B+6 a b^2 C-4 b^3 B\right ) \sin (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 b \left (a^2-b^2\right )}+\frac {a (b B-a C) \sin (c+d x) \cos ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {\frac {\frac {2 x \left (a^2-b^2\right )^2 (b B-3 a C)-\frac {2 a \left (-6 a^5 C+2 a^4 b B+15 a^3 b^2 C-5 a^2 b^3 B-12 a b^4 C+6 b^5 B\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{d}}{b}-\frac {\left (a^2-b^2\right ) \left (-3 a^2 C+a b B+2 b^2 C\right ) \sin (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}-\frac {a^2 \left (-3 a^3 C+a^2 b B+6 a b^2 C-4 b^3 B\right ) \sin (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 b \left (a^2-b^2\right )}+\frac {a (b B-a C) \sin (c+d x) \cos ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {a (b B-a C) \sin (c+d x) \cos ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac {\frac {\frac {2 x \left (a^2-b^2\right )^2 (b B-3 a C)-\frac {2 a \left (-6 a^5 C+2 a^4 b B+15 a^3 b^2 C-5 a^2 b^3 B-12 a b^4 C+6 b^5 B\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b}}}{b}-\frac {\left (a^2-b^2\right ) \left (-3 a^2 C+a b B+2 b^2 C\right ) \sin (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}-\frac {a^2 \left (-3 a^3 C+a^2 b B+6 a b^2 C-4 b^3 B\right ) \sin (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 b \left (a^2-b^2\right )}\) |
(a*(b*B - a*C)*Cos[c + d*x]^2*Sin[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b*Cos[ c + d*x])^2) + (-((a^2*(a^2*b*B - 4*b^3*B - 3*a^3*C + 6*a*b^2*C)*Sin[c + d *x])/(b^2*(a^2 - b^2)*d*(a + b*Cos[c + d*x]))) + ((2*(a^2 - b^2)^2*(b*B - 3*a*C)*x - (2*a*(2*a^4*b*B - 5*a^2*b^3*B + 6*b^5*B - 6*a^5*C + 15*a^3*b^2* C - 12*a*b^4*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[ a - b]*Sqrt[a + b]*d))/b - ((a^2 - b^2)*(a*b*B - 3*a^2*C + 2*b^2*C)*Sin[c + d*x])/d)/(b^2*(a^2 - b^2)))/(2*b*(a^2 - b^2))
3.9.8.3.1 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_)])*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f _.)*(x_)]^2), x_Symbol] :> Simp[(-(b*c - a*d))*(A*b^2 - a*b*B + a^2*C)*Cos[ e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b^2*f*(m + 1)*(a^2 - b^2))), x] - S imp[1/(b^2*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*( m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1)) ))*Sin[e + f*x] - b*C*d*(m + 1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; F reeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]
Time = 1.70 (sec) , antiderivative size = 341, normalized size of antiderivative = 1.22
| method | result | size |
| derivativedivides | \(\frac {-\frac {2 a \left (\frac {\frac {\left (2 B \,a^{2} b -B a \,b^{2}-6 B \,b^{3}-4 C \,a^{3}+C \,a^{2} b +8 C a \,b^{2}\right ) a b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 \left (a^{2}+2 a b +b^{2}\right ) \left (a -b \right )}+\frac {b a \left (2 B \,a^{2} b +B a \,b^{2}-6 B \,b^{3}-4 C \,a^{3}-C \,a^{2} b +8 C a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a -b \right )^{2}}}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}^{2}}+\frac {\left (2 B \,a^{4} b -5 B \,a^{2} b^{3}+6 B \,b^{5}-6 C \,a^{5}+15 C \,a^{3} b^{2}-12 C a \,b^{4}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{4}}+\frac {\frac {2 C b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}+2 \left (B b -3 a C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4}}}{d}\) | \(341\) |
| default | \(\frac {-\frac {2 a \left (\frac {\frac {\left (2 B \,a^{2} b -B a \,b^{2}-6 B \,b^{3}-4 C \,a^{3}+C \,a^{2} b +8 C a \,b^{2}\right ) a b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 \left (a^{2}+2 a b +b^{2}\right ) \left (a -b \right )}+\frac {b a \left (2 B \,a^{2} b +B a \,b^{2}-6 B \,b^{3}-4 C \,a^{3}-C \,a^{2} b +8 C a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a -b \right )^{2}}}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}^{2}}+\frac {\left (2 B \,a^{4} b -5 B \,a^{2} b^{3}+6 B \,b^{5}-6 C \,a^{5}+15 C \,a^{3} b^{2}-12 C a \,b^{4}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{4}}+\frac {\frac {2 C b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}+2 \left (B b -3 a C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4}}}{d}\) | \(341\) |
| risch | \(\text {Expression too large to display}\) | \(1411\) |
1/d*(-2*a/b^4*((1/2*(2*B*a^2*b-B*a*b^2-6*B*b^3-4*C*a^3+C*a^2*b+8*C*a*b^2)* a*b/(a^2+2*a*b+b^2)/(a-b)*tan(1/2*d*x+1/2*c)^3+1/2*b*a*(2*B*a^2*b+B*a*b^2- 6*B*b^3-4*C*a^3-C*a^2*b+8*C*a*b^2)/(a+b)/(a-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+1/2*(2*B*a^4*b-5*B*a^2*b^ 3+6*B*b^5-6*C*a^5+15*C*a^3*b^2-12*C*a*b^4)/(a^4-2*a^2*b^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)))+2/b^4*(C*b* tan(1/2*d*x+1/2*c)/(1+tan(1/2*d*x+1/2*c)^2)+(B*b-3*C*a)*arctan(tan(1/2*d*x +1/2*c))))
Leaf count of result is larger than twice the leaf count of optimal. 746 vs. \(2 (268) = 536\).
Time = 0.41 (sec) , antiderivative size = 1561, normalized size of antiderivative = 5.58 \[ \int \frac {\cos ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^3} \, dx=\text {Too large to display} \]
integrate(cos(d*x+c)^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^3,x, algorithm="fricas")
[-1/4*(4*(3*C*a^7*b^2 - B*a^6*b^3 - 9*C*a^5*b^4 + 3*B*a^4*b^5 + 9*C*a^3*b^ 6 - 3*B*a^2*b^7 - 3*C*a*b^8 + B*b^9)*d*x*cos(d*x + c)^2 + 8*(3*C*a^8*b - B *a^7*b^2 - 9*C*a^6*b^3 + 3*B*a^5*b^4 + 9*C*a^4*b^5 - 3*B*a^3*b^6 - 3*C*a^2 *b^7 + B*a*b^8)*d*x*cos(d*x + c) + 4*(3*C*a^9 - B*a^8*b - 9*C*a^7*b^2 + 3* B*a^6*b^3 + 9*C*a^5*b^4 - 3*B*a^4*b^5 - 3*C*a^3*b^6 + B*a^2*b^7)*d*x - (6* C*a^8 - 2*B*a^7*b - 15*C*a^6*b^2 + 5*B*a^5*b^3 + 12*C*a^4*b^4 - 6*B*a^3*b^ 5 + (6*C*a^6*b^2 - 2*B*a^5*b^3 - 15*C*a^4*b^4 + 5*B*a^3*b^5 + 12*C*a^2*b^6 - 6*B*a*b^7)*cos(d*x + c)^2 + 2*(6*C*a^7*b - 2*B*a^6*b^2 - 15*C*a^5*b^3 + 5*B*a^4*b^4 + 12*C*a^3*b^5 - 6*B*a^2*b^6)*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)*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*(6*C*a^8*b - 2*B*a^7*b^2 - 17*C*a^6*b^3 + 7* B*a^5*b^4 + 13*C*a^4*b^5 - 5*B*a^3*b^6 - 2*C*a^2*b^7 + 2*(C*a^6*b^3 - 3*C* a^4*b^5 + 3*C*a^2*b^7 - C*b^9)*cos(d*x + c)^2 + (9*C*a^7*b^2 - 3*B*a^6*b^3 - 25*C*a^5*b^4 + 9*B*a^4*b^5 + 20*C*a^3*b^6 - 6*B*a^2*b^7 - 4*C*a*b^8)*co s(d*x + c))*sin(d*x + c))/((a^6*b^6 - 3*a^4*b^8 + 3*a^2*b^10 - b^12)*d*cos (d*x + c)^2 + 2*(a^7*b^5 - 3*a^5*b^7 + 3*a^3*b^9 - a*b^11)*d*cos(d*x + c) + (a^8*b^4 - 3*a^6*b^6 + 3*a^4*b^8 - a^2*b^10)*d), -1/2*(2*(3*C*a^7*b^2 - B*a^6*b^3 - 9*C*a^5*b^4 + 3*B*a^4*b^5 + 9*C*a^3*b^6 - 3*B*a^2*b^7 - 3*C*a* b^8 + B*b^9)*d*x*cos(d*x + c)^2 + 4*(3*C*a^8*b - B*a^7*b^2 - 9*C*a^6*b^...
Timed out. \[ \int \frac {\cos ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^3} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\cos ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^3} \, dx=\text {Exception raised: ValueError} \]
integrate(cos(d*x+c)^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^3,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
Leaf count of result is larger than twice the leaf count of optimal. 543 vs. \(2 (268) = 536\).
Time = 0.38 (sec) , antiderivative size = 543, normalized size of antiderivative = 1.94 \[ \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))^3} \, dx=-\frac {\frac {{\left (6 \, C a^{6} - 2 \, B a^{5} b - 15 \, C a^{4} b^{2} + 5 \, B a^{3} b^{3} + 12 \, C a^{2} b^{4} - 6 \, B a b^{5}\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^{4} b^{4} - 2 \, a^{2} b^{6} + b^{8}\right )} \sqrt {a^{2} - b^{2}}} - \frac {4 \, C a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B a^{5} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, C a^{5} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, B a^{4} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 7 \, C a^{4} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5 \, B a^{3} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8 \, C a^{3} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, B a^{2} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, C a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B a^{5} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5 \, C a^{5} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, B a^{4} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 7 \, C a^{4} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5 \, B a^{3} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8 \, C a^{3} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, B a^{2} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\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 )}^{2}} + \frac {{\left (3 \, C a - B b\right )} {\left (d x + c\right )}}{b^{4}} - \frac {2 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} b^{3}}}{d} \]
-((6*C*a^6 - 2*B*a^5*b - 15*C*a^4*b^2 + 5*B*a^3*b^3 + 12*C*a^2*b^4 - 6*B*a *b^5)*(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^4*b^4 - 2* a^2*b^6 + b^8)*sqrt(a^2 - b^2)) - (4*C*a^6*tan(1/2*d*x + 1/2*c)^3 - 2*B*a^ 5*b*tan(1/2*d*x + 1/2*c)^3 - 5*C*a^5*b*tan(1/2*d*x + 1/2*c)^3 + 3*B*a^4*b^ 2*tan(1/2*d*x + 1/2*c)^3 - 7*C*a^4*b^2*tan(1/2*d*x + 1/2*c)^3 + 5*B*a^3*b^ 3*tan(1/2*d*x + 1/2*c)^3 + 8*C*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 - 6*B*a^2*b^ 4*tan(1/2*d*x + 1/2*c)^3 + 4*C*a^6*tan(1/2*d*x + 1/2*c) - 2*B*a^5*b*tan(1/ 2*d*x + 1/2*c) + 5*C*a^5*b*tan(1/2*d*x + 1/2*c) - 3*B*a^4*b^2*tan(1/2*d*x + 1/2*c) - 7*C*a^4*b^2*tan(1/2*d*x + 1/2*c) + 5*B*a^3*b^3*tan(1/2*d*x + 1/ 2*c) - 8*C*a^3*b^3*tan(1/2*d*x + 1/2*c) + 6*B*a^2*b^4*tan(1/2*d*x + 1/2*c) )/((a^4*b^3 - 2*a^2*b^5 + b^7)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)^2) + (3*C*a - B*b)*(d*x + c)/b^4 - 2*C*tan(1/2*d*x + 1/ 2*c)/((tan(1/2*d*x + 1/2*c)^2 + 1)*b^3))/d
Time = 9.48 (sec) , antiderivative size = 5542, normalized size of antiderivative = 19.79 \[ \int \frac {\cos ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^3} \, dx=\text {Too large to display} \]
((tan(c/2 + (d*x)/2)^5*(6*C*a^5 - 2*C*b^5 + 6*B*a^2*b^3 + B*a^3*b^2 + 4*C* a^2*b^3 - 12*C*a^3*b^2 - 2*B*a^4*b + 2*C*a*b^4 - 3*C*a^4*b))/((a*b^3 - b^4 )*(a + b)^2) + (tan(c/2 + (d*x)/2)*(6*C*a^5 + 2*C*b^5 + 6*B*a^2*b^3 - B*a^ 3*b^2 - 4*C*a^2*b^3 - 12*C*a^3*b^2 - 2*B*a^4*b + 2*C*a*b^4 + 3*C*a^4*b))/( (a + b)*(b^5 - 2*a*b^4 + a^2*b^3)) + (2*tan(c/2 + (d*x)/2)^3*(6*C*a^6 - 2* C*b^6 + 5*B*a^3*b^3 + 6*C*a^2*b^4 - 13*C*a^4*b^2 - 2*B*a^5*b))/(b*(a*b^2 - b^3)*(a + b)^2*(a - b)))/(d*(2*a*b + tan(c/2 + (d*x)/2)^2*(2*a*b + 3*a^2 - b^2) + tan(c/2 + (d*x)/2)^6*(a^2 - 2*a*b + b^2) + a^2 + b^2 - tan(c/2 + (d*x)/2)^4*(2*a*b - 3*a^2 + b^2))) + (log(tan(c/2 + (d*x)/2) + 1i)*(B*b - 3*C*a)*1i)/(b^4*d) - (log(tan(c/2 + (d*x)/2) - 1i)*(B*b*1i - C*a*3i))/(b^4 *d) - (a*atan(((a*((8*tan(c/2 + (d*x)/2)*(4*B^2*b^12 + 72*C^2*a^12 - 8*B^2 *a*b^11 - 72*C^2*a^11*b + 24*B^2*a^2*b^10 + 32*B^2*a^3*b^9 - 52*B^2*a^4*b^ 8 - 48*B^2*a^5*b^7 + 57*B^2*a^6*b^6 + 32*B^2*a^7*b^5 - 32*B^2*a^8*b^4 - 8* B^2*a^9*b^3 + 8*B^2*a^10*b^2 + 36*C^2*a^2*b^10 - 72*C^2*a^3*b^9 + 36*C^2*a ^4*b^8 + 288*C^2*a^5*b^7 - 288*C^2*a^6*b^6 - 432*C^2*a^7*b^5 + 441*C^2*a^8 *b^4 + 288*C^2*a^9*b^3 - 288*C^2*a^10*b^2 - 24*B*C*a*b^11 - 48*B*C*a^11*b + 48*B*C*a^2*b^10 - 72*B*C*a^3*b^9 - 192*B*C*a^4*b^8 + 252*B*C*a^5*b^7 + 2 88*B*C*a^6*b^6 - 318*B*C*a^7*b^5 - 192*B*C*a^8*b^4 + 192*B*C*a^9*b^3 + 48* B*C*a^10*b^2))/(a*b^12 + b^13 - 3*a^2*b^11 - 3*a^3*b^10 + 3*a^4*b^9 + 3*a^ 5*b^8 - a^6*b^7 - a^7*b^6) + (a*((8*(4*B*b^18 - 8*B*a^2*b^16 + 34*B*a^3...