Integrand size = 23, antiderivative size = 237 \[ \int \frac {A+B \cos (x)+C \sin (x)}{(a+b \cos (x)+c \sin (x))^3} \, dx=\frac {\left (2 a^2 A+A \left (b^2+c^2\right )-3 a (b B+c C)\right ) \arctan \left (\frac {c+(a-b) \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2-c^2}}\right )}{\left (a^2-b^2-c^2\right )^{5/2}}+\frac {B c-b C+(A c-a C) \cos (x)-(A b-a B) \sin (x)}{2 \left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))^2}+\frac {a (B c-b C)+\left (3 a A c-a^2 C-2 c (b B+c C)\right ) \cos (x)-\left (3 a A b-a^2 B-2 b (b B+c C)\right ) \sin (x)}{2 \left (a^2-b^2-c^2\right )^2 (a+b \cos (x)+c \sin (x))} \]
(2*a^2*A+A*(b^2+c^2)-3*a*(B*b+C*c))*arctan((c+(a-b)*tan(1/2*x))/(a^2-b^2-c ^2)^(1/2))/(a^2-b^2-c^2)^(5/2)+1/2*(B*c-b*C+(A*c-C*a)*cos(x)-(A*b-B*a)*sin (x))/(a^2-b^2-c^2)/(a+b*cos(x)+c*sin(x))^2+1/2*(a*(B*c-C*b)+(3*A*a*c-C*a^2 -2*c*(B*b+C*c))*cos(x)-(3*A*a*b-B*a^2-2*b*(B*b+C*c))*sin(x))/(a^2-b^2-c^2) ^2/(a+b*cos(x)+c*sin(x))
Time = 1.26 (sec) , antiderivative size = 452, normalized size of antiderivative = 1.91 \[ \int \frac {A+B \cos (x)+C \sin (x)}{(a+b \cos (x)+c \sin (x))^3} \, dx=-\frac {\left (2 a^2 A+A \left (b^2+c^2\right )-3 a (b B+c C)\right ) \text {arctanh}\left (\frac {c+(a-b) \tan \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2+c^2}}\right )}{\left (-a^2+b^2+c^2\right )^{5/2}}+\frac {-6 a^3 A c-3 a A b^2 c+9 a^2 b B c-3 a A c^3+2 a^4 C-4 a^2 b^2 C+2 b^4 C+5 a^2 c^2 C+4 b^2 c^2 C+2 c^4 C-2 b c \left (2 a^2 A+A \left (b^2+c^2\right )-3 a (b B+c C)\right ) \cos (x)-c \left (-3 a A \left (b^2+c^2\right )+a^2 (b B+c C)+2 \left (b^2+c^2\right ) (b B+c C)\right ) \cos (2 x)-8 a^2 A b^2 \sin (x)+2 A b^4 \sin (x)+4 a^3 b B \sin (x)+2 a b^3 B \sin (x)-12 a^2 A c^2 \sin (x)+2 A b^2 c^2 \sin (x)+8 a b B c^2 \sin (x)+4 a^3 c C \sin (x)+2 a b^2 c C \sin (x)+8 a c^3 C \sin (x)-3 a A b^3 \sin (2 x)+a^2 b^2 B \sin (2 x)+2 b^4 B \sin (2 x)-3 a A b c^2 \sin (2 x)+2 b^2 B c^2 \sin (2 x)+a^2 b c C \sin (2 x)+2 b^3 c C \sin (2 x)+2 b c^3 C \sin (2 x)}{4 b \left (-a^2+b^2+c^2\right )^2 (a+b \cos (x)+c \sin (x))^2} \]
-(((2*a^2*A + A*(b^2 + c^2) - 3*a*(b*B + c*C))*ArcTanh[(c + (a - b)*Tan[x/ 2])/Sqrt[-a^2 + b^2 + c^2]])/(-a^2 + b^2 + c^2)^(5/2)) + (-6*a^3*A*c - 3*a *A*b^2*c + 9*a^2*b*B*c - 3*a*A*c^3 + 2*a^4*C - 4*a^2*b^2*C + 2*b^4*C + 5*a ^2*c^2*C + 4*b^2*c^2*C + 2*c^4*C - 2*b*c*(2*a^2*A + A*(b^2 + c^2) - 3*a*(b *B + c*C))*Cos[x] - c*(-3*a*A*(b^2 + c^2) + a^2*(b*B + c*C) + 2*(b^2 + c^2 )*(b*B + c*C))*Cos[2*x] - 8*a^2*A*b^2*Sin[x] + 2*A*b^4*Sin[x] + 4*a^3*b*B* Sin[x] + 2*a*b^3*B*Sin[x] - 12*a^2*A*c^2*Sin[x] + 2*A*b^2*c^2*Sin[x] + 8*a *b*B*c^2*Sin[x] + 4*a^3*c*C*Sin[x] + 2*a*b^2*c*C*Sin[x] + 8*a*c^3*C*Sin[x] - 3*a*A*b^3*Sin[2*x] + a^2*b^2*B*Sin[2*x] + 2*b^4*B*Sin[2*x] - 3*a*A*b*c^ 2*Sin[2*x] + 2*b^2*B*c^2*Sin[2*x] + a^2*b*c*C*Sin[2*x] + 2*b^3*c*C*Sin[2*x ] + 2*b*c^3*C*Sin[2*x])/(4*b*(-a^2 + b^2 + c^2)^2*(a + b*Cos[x] + c*Sin[x] )^2)
Time = 0.75 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.11, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 3635, 25, 3042, 3632, 3042, 3603, 1083, 217}
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.
| \(\displaystyle \int \frac {A+B \cos (x)+C \sin (x)}{(a+b \cos (x)+c \sin (x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \cos (x)+C \sin (x)}{(a+b \cos (x)+c \sin (x))^3}dx\) |
| \(\Big \downarrow \) 3635 |
| \(\displaystyle \frac {-\sin (x) (A b-a B)+\cos (x) (A c-a C)-b C+B c}{2 \left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))^2}-\frac {\int -\frac {2 (a A-b B-c C)-(A b-a B) \cos (x)-(A c-a C) \sin (x)}{(a+b \cos (x)+c \sin (x))^2}dx}{2 \left (a^2-b^2-c^2\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {2 (a A-b B-c C)-(A b-a B) \cos (x)-(A c-a C) \sin (x)}{(a+b \cos (x)+c \sin (x))^2}dx}{2 \left (a^2-b^2-c^2\right )}+\frac {-\sin (x) (A b-a B)+\cos (x) (A c-a C)-b C+B c}{2 \left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {2 (a A-b B-c C)-(A b-a B) \cos (x)-(A c-a C) \sin (x)}{(a+b \cos (x)+c \sin (x))^2}dx}{2 \left (a^2-b^2-c^2\right )}+\frac {-\sin (x) (A b-a B)+\cos (x) (A c-a C)-b C+B c}{2 \left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))^2}\) |
| \(\Big \downarrow \) 3632 |
| \(\displaystyle \frac {\frac {\left (2 a^2 A-3 a (b B+c C)+A \left (b^2+c^2\right )\right ) \int \frac {1}{a+b \cos (x)+c \sin (x)}dx}{a^2-b^2-c^2}+\frac {-\sin (x) \left (a^2 (-B)+3 a A b-2 b (b B+c C)\right )+\cos (x) \left (a^2 (-C)+3 a A c-2 c (b B+c C)\right )+a (B c-b C)}{\left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))}}{2 \left (a^2-b^2-c^2\right )}+\frac {-\sin (x) (A b-a B)+\cos (x) (A c-a C)-b C+B c}{2 \left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\left (2 a^2 A-3 a (b B+c C)+A \left (b^2+c^2\right )\right ) \int \frac {1}{a+b \cos (x)+c \sin (x)}dx}{a^2-b^2-c^2}+\frac {-\sin (x) \left (a^2 (-B)+3 a A b-2 b (b B+c C)\right )+\cos (x) \left (a^2 (-C)+3 a A c-2 c (b B+c C)\right )+a (B c-b C)}{\left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))}}{2 \left (a^2-b^2-c^2\right )}+\frac {-\sin (x) (A b-a B)+\cos (x) (A c-a C)-b C+B c}{2 \left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))^2}\) |
| \(\Big \downarrow \) 3603 |
| \(\displaystyle \frac {\frac {2 \left (2 a^2 A-3 a (b B+c C)+A \left (b^2+c^2\right )\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {x}{2}\right )+2 c \tan \left (\frac {x}{2}\right )+a+b}d\tan \left (\frac {x}{2}\right )}{a^2-b^2-c^2}+\frac {-\sin (x) \left (a^2 (-B)+3 a A b-2 b (b B+c C)\right )+\cos (x) \left (a^2 (-C)+3 a A c-2 c (b B+c C)\right )+a (B c-b C)}{\left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))}}{2 \left (a^2-b^2-c^2\right )}+\frac {-\sin (x) (A b-a B)+\cos (x) (A c-a C)-b C+B c}{2 \left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))^2}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {\frac {-\sin (x) \left (a^2 (-B)+3 a A b-2 b (b B+c C)\right )+\cos (x) \left (a^2 (-C)+3 a A c-2 c (b B+c C)\right )+a (B c-b C)}{\left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))}-\frac {4 \left (2 a^2 A-3 a (b B+c C)+A \left (b^2+c^2\right )\right ) \int \frac {1}{-\left (2 c+2 (a-b) \tan \left (\frac {x}{2}\right )\right )^2-4 \left (a^2-b^2-c^2\right )}d\left (2 c+2 (a-b) \tan \left (\frac {x}{2}\right )\right )}{a^2-b^2-c^2}}{2 \left (a^2-b^2-c^2\right )}+\frac {-\sin (x) (A b-a B)+\cos (x) (A c-a C)-b C+B c}{2 \left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))^2}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {\frac {2 \left (2 a^2 A-3 a (b B+c C)+A \left (b^2+c^2\right )\right ) \arctan \left (\frac {2 (a-b) \tan \left (\frac {x}{2}\right )+2 c}{2 \sqrt {a^2-b^2-c^2}}\right )}{\left (a^2-b^2-c^2\right )^{3/2}}+\frac {-\sin (x) \left (a^2 (-B)+3 a A b-2 b (b B+c C)\right )+\cos (x) \left (a^2 (-C)+3 a A c-2 c (b B+c C)\right )+a (B c-b C)}{\left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))}}{2 \left (a^2-b^2-c^2\right )}+\frac {-\sin (x) (A b-a B)+\cos (x) (A c-a C)-b C+B c}{2 \left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))^2}\) |
(B*c - b*C + (A*c - a*C)*Cos[x] - (A*b - a*B)*Sin[x])/(2*(a^2 - b^2 - c^2) *(a + b*Cos[x] + c*Sin[x])^2) + ((2*(2*a^2*A + A*(b^2 + c^2) - 3*a*(b*B + c*C))*ArcTan[(2*c + 2*(a - b)*Tan[x/2])/(2*Sqrt[a^2 - b^2 - c^2])])/(a^2 - b^2 - c^2)^(3/2) + (a*(B*c - b*C) + (3*a*A*c - a^2*C - 2*c*(b*B + c*C))*C os[x] - (3*a*A*b - a^2*B - 2*b*(b*B + c*C))*Sin[x])/((a^2 - b^2 - c^2)*(a + b*Cos[x] + c*Sin[x])))/(2*(a^2 - b^2 - c^2))
3.6.52.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^ (-1), x_Symbol] :> Module[{f = FreeFactors[Tan[(d + e*x)/2], x]}, Simp[2*(f /e) Subst[Int[1/(a + b + 2*c*f*x + (a - b)*f^2*x^2), x], x, Tan[(d + e*x) /2]/f], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0]
Int[((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.) + (C_.)*sin[(d_.) + (e_.)*(x_)]) /((a_.) + cos[(d_.) + (e_.)*(x_)]*(b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)])^2, x_Symbol] :> Simp[(c*B - b*C - (a*C - c*A)*Cos[d + e*x] + (a*B - b*A)*Sin[ d + e*x])/(e*(a^2 - b^2 - c^2)*(a + b*Cos[d + e*x] + c*Sin[d + e*x])), x] + Simp[(a*A - b*B - c*C)/(a^2 - b^2 - c^2) Int[1/(a + b*Cos[d + e*x] + c*S in[d + e*x]), x], x] /; FreeQ[{a, b, c, d, e, A, B, C}, x] && NeQ[a^2 - b^2 - c^2, 0] && NeQ[a*A - b*B - c*C, 0]
Int[((a_.) + cos[(d_.) + (e_.)*(x_)]*(b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)]) ^(n_)*((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.) + (C_.)*sin[(d_.) + (e_.)*(x_) ]), x_Symbol] :> Simp[(-(c*B - b*C - (a*C - c*A)*Cos[d + e*x] + (a*B - b*A) *Sin[d + e*x]))*((a + b*Cos[d + e*x] + c*Sin[d + e*x])^(n + 1)/(e*(n + 1)*( a^2 - b^2 - c^2))), x] + Simp[1/((n + 1)*(a^2 - b^2 - c^2)) Int[(a + b*Co s[d + e*x] + c*Sin[d + e*x])^(n + 1)*Simp[(n + 1)*(a*A - b*B - c*C) + (n + 2)*(a*B - b*A)*Cos[d + e*x] + (n + 2)*(a*C - c*A)*Sin[d + e*x], x], x], x] /; FreeQ[{a, b, c, d, e, A, B, C}, x] && LtQ[n, -1] && NeQ[a^2 - b^2 - c^2, 0] && NeQ[n, -2]
Leaf count of result is larger than twice the leaf count of optimal. \(1079\) vs. \(2(227)=454\).
Time = 1.87 (sec) , antiderivative size = 1080, normalized size of antiderivative = 4.56
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(1080\) |
| risch | \(\text {Expression too large to display}\) | \(2157\) |
2*(-1/2*(4*A*a^3*b-7*A*a^2*b^2-5*A*a^2*c^2+2*A*a*b^3+2*A*a*b*c^2+A*b^4+3*A *b^2*c^2+2*A*c^4-2*B*a^4+3*B*a^3*b-2*B*a^2*b^2+4*B*a^2*c^2+3*B*a*b^3-2*B*b ^4-4*B*b^2*c^2-2*B*c^4+3*C*a^3*c-6*C*a^2*b*c+3*C*a*b^2*c)/(a^4-2*a^2*b^2-2 *a^2*c^2+b^4+2*b^2*c^2+c^4)/(a-b)*tan(1/2*x)^3+1/2*(4*A*a^4*c-12*A*a^3*b*c +13*A*a^2*b^2*c+7*A*a^2*c^3-6*A*a*b^3*c-6*A*a*b*c^3+A*b^4*c-A*b^2*c^3-2*A* c^5+2*B*a^4*c-9*B*a^3*b*c+14*B*a^2*b^2*c-4*B*a^2*c^3-9*B*a*b^3*c+2*B*b^4*c +4*B*b^2*c^3+2*B*c^5-2*C*a^5+2*C*a^4*b+4*C*a^3*b^2-5*C*a^3*c^2-4*C*a^2*b^3 +14*C*a^2*b*c^2-2*C*a*b^4-13*C*a*b^2*c^2-2*C*a*c^4+2*C*b^5+4*C*b^3*c^2+2*C *b*c^4)/(a^4-2*a^2*b^2-2*a^2*c^2+b^4+2*b^2*c^2+c^4)/(a^2-2*a*b+b^2)*tan(1/ 2*x)^2-1/2*(4*A*a^4*b-5*A*a^3*b^2-11*A*a^3*c^2-3*A*a^2*b^3+3*A*a^2*b*c^2+5 *A*a*b^4+7*A*a*b^2*c^2+2*A*a*c^4-A*b^5+A*b^3*c^2+2*A*b*c^4-2*B*a^5+3*B*a^4 *b-B*a^3*b^2+4*B*a^3*c^2-B*a^2*b^3+8*B*a^2*b*c^2+3*B*a*b^4-8*B*a*b^2*c^2-2 *B*a*c^4-2*B*b^5-4*B*b^3*c^2-2*B*b*c^4+5*C*a^4*c-5*C*a^3*b*c-5*C*a^2*b^2*c +4*C*a^2*c^3+5*C*a*b^3*c-4*C*a*b*c^3)/(a^4-2*a^2*b^2-2*a^2*c^2+b^4+2*b^2*c ^2+c^4)/(a^2-2*a*b+b^2)*tan(1/2*x)+1/2*(4*A*a^4*c-3*A*a^2*b^2*c-A*a^2*c^3- A*b^4*c-A*b^2*c^3-5*B*a^3*b*c+5*B*a*b^3*c+2*B*a*b*c^3-2*C*a^5+4*C*a^3*b^2- C*a^3*c^2-2*C*a*b^4+C*a*b^2*c^2)/(a^4-2*a^2*b^2-2*a^2*c^2+b^4+2*b^2*c^2+c^ 4)/(a^2-2*a*b+b^2))/(tan(1/2*x)^2*a-tan(1/2*x)^2*b+2*c*tan(1/2*x)+a+b)^2+( 2*A*a^2+A*b^2+A*c^2-3*B*a*b-3*C*a*c)/(a^4-2*a^2*b^2-2*a^2*c^2+b^4+2*b^2*c^ 2+c^4)/(a^2-b^2-c^2)^(1/2)*arctan(1/2*(2*(a-b)*tan(1/2*x)+2*c)/(a^2-b^2...
Leaf count of result is larger than twice the leaf count of optimal. 2038 vs. \(2 (225) = 450\).
Time = 0.51 (sec) , antiderivative size = 4240, normalized size of antiderivative = 17.89 \[ \int \frac {A+B \cos (x)+C \sin (x)}{(a+b \cos (x)+c \sin (x))^3} \, dx=\text {Too large to display} \]
[1/4*(2*C*a^6*b - 6*C*a^4*b^3 + 6*C*a^2*b^5 - 2*C*b^7 - 6*C*b*c^6 + 2*B*c^ 7 - 2*(3*B*a^2 - 3*A*a*b - B*b^2)*c^5 + 2*(4*C*a^2*b - 7*C*b^3)*c^4 + 2*(3 *B*a^4 - 3*A*a^3*b - 5*B*a^2*b^2 + 6*A*a*b^3 - B*b^4)*c^3 - 2*(2*C*a^4*b - 7*C*a^2*b^3 + 5*C*b^5)*c^2 + 4*(2*C*b*c^6 - (3*A*a*b - 2*B*b^2)*c^5 - (C* a^2*b - 4*C*b^3)*c^4 + (3*A*a^3*b - B*a^2*b^2 - 6*A*a*b^3 + 4*B*b^4)*c^3 - (C*a^4*b + C*a^2*b^3 - 2*C*b^5)*c^2 - (B*a^4*b^2 - 3*A*a^3*b^3 + B*a^2*b^ 4 + 3*A*a*b^5 - 2*B*b^6)*c)*cos(x)^2 - (2*A*a^4*b^2 - 3*B*a^3*b^3 + A*a^2* b^4 - 3*C*a^3*b^2*c - 3*C*a*c^5 + A*c^6 + (3*A*a^2 - 3*B*a*b + 2*A*b^2)*c^ 4 - 3*(C*a^3 + C*a*b^2)*c^3 + (2*A*a^4 - 3*B*a^3*b + 4*A*a^2*b^2 - 3*B*a*b ^3 + A*b^4)*c^2 + (2*A*a^2*b^4 - 3*B*a*b^5 + A*b^6 - 3*C*a*b^4*c + A*b^4*c ^2 + 3*C*a*c^5 - A*c^6 - (2*A*a^2 - 3*B*a*b + A*b^2)*c^4)*cos(x)^2 + 2*(2* A*a^3*b^3 - 3*B*a^2*b^4 + A*a*b^5 - 3*C*a^2*b^3*c - 3*C*a^2*b*c^3 + A*a*b* c^4 + (2*A*a^3*b - 3*B*a^2*b^2 + 2*A*a*b^3)*c^2)*cos(x) - 2*(3*C*a^2*b^2*c ^2 + 3*C*a^2*c^4 - A*a*c^5 - (2*A*a^3 - 3*B*a^2*b + 2*A*a*b^2)*c^3 - (2*A* a^3*b^2 - 3*B*a^2*b^3 + A*a*b^4)*c + (3*C*a*b^3*c^2 + 3*C*a*b*c^4 - A*b*c^ 5 - (2*A*a^2*b - 3*B*a*b^2 + 2*A*b^3)*c^3 - (2*A*a^2*b^3 - 3*B*a*b^4 + A*b ^5)*c)*cos(x))*sin(x))*sqrt(-a^2 + b^2 + c^2)*log(-(a^2*b^2 - 2*b^4 - c^4 - (a^2 + 3*b^2)*c^2 - (2*a^2*b^2 - b^4 - 2*a^2*c^2 + c^4)*cos(x)^2 - 2*(a* b^3 + a*b*c^2)*cos(x) - 2*(a*b^2*c + a*c^3 - (b*c^3 - (2*a^2*b - b^3)*c)*c os(x))*sin(x) + 2*(2*a*b*c*cos(x)^2 - a*b*c + (b^2*c + c^3)*cos(x) - (b...
Timed out. \[ \int \frac {A+B \cos (x)+C \sin (x)}{(a+b \cos (x)+c \sin (x))^3} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {A+B \cos (x)+C \sin (x)}{(a+b \cos (x)+c \sin (x))^3} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(c^2+b^2-a^2>0)', see `assume?` f or more de
Leaf count of result is larger than twice the leaf count of optimal. 1506 vs. \(2 (225) = 450\).
Time = 0.37 (sec) , antiderivative size = 1506, normalized size of antiderivative = 6.35 \[ \int \frac {A+B \cos (x)+C \sin (x)}{(a+b \cos (x)+c \sin (x))^3} \, dx=\text {Too large to display} \]
-(2*A*a^2 - 3*B*a*b + A*b^2 - 3*C*a*c + A*c^2)*(pi*floor(1/2*x/pi + 1/2)*s gn(-2*a + 2*b) + arctan(-(a*tan(1/2*x) - b*tan(1/2*x) + c)/sqrt(a^2 - b^2 - c^2)))/((a^4 - 2*a^2*b^2 + b^4 - 2*a^2*c^2 + 2*b^2*c^2 + c^4)*sqrt(a^2 - b^2 - c^2)) + (2*B*a^5*tan(1/2*x)^3 - 4*A*a^4*b*tan(1/2*x)^3 - 5*B*a^4*b* tan(1/2*x)^3 + 11*A*a^3*b^2*tan(1/2*x)^3 + 5*B*a^3*b^2*tan(1/2*x)^3 - 9*A* a^2*b^3*tan(1/2*x)^3 - 5*B*a^2*b^3*tan(1/2*x)^3 + A*a*b^4*tan(1/2*x)^3 + 5 *B*a*b^4*tan(1/2*x)^3 + A*b^5*tan(1/2*x)^3 - 2*B*b^5*tan(1/2*x)^3 - 3*C*a^ 4*c*tan(1/2*x)^3 + 9*C*a^3*b*c*tan(1/2*x)^3 - 9*C*a^2*b^2*c*tan(1/2*x)^3 + 3*C*a*b^3*c*tan(1/2*x)^3 + 5*A*a^3*c^2*tan(1/2*x)^3 - 4*B*a^3*c^2*tan(1/2 *x)^3 - 7*A*a^2*b*c^2*tan(1/2*x)^3 + 4*B*a^2*b*c^2*tan(1/2*x)^3 - A*a*b^2* c^2*tan(1/2*x)^3 + 4*B*a*b^2*c^2*tan(1/2*x)^3 + 3*A*b^3*c^2*tan(1/2*x)^3 - 4*B*b^3*c^2*tan(1/2*x)^3 - 2*A*a*c^4*tan(1/2*x)^3 + 2*B*a*c^4*tan(1/2*x)^ 3 + 2*A*b*c^4*tan(1/2*x)^3 - 2*B*b*c^4*tan(1/2*x)^3 - 2*C*a^5*tan(1/2*x)^2 + 2*C*a^4*b*tan(1/2*x)^2 + 4*C*a^3*b^2*tan(1/2*x)^2 - 4*C*a^2*b^3*tan(1/2 *x)^2 - 2*C*a*b^4*tan(1/2*x)^2 + 2*C*b^5*tan(1/2*x)^2 + 4*A*a^4*c*tan(1/2* x)^2 + 2*B*a^4*c*tan(1/2*x)^2 - 12*A*a^3*b*c*tan(1/2*x)^2 - 9*B*a^3*b*c*ta n(1/2*x)^2 + 13*A*a^2*b^2*c*tan(1/2*x)^2 + 14*B*a^2*b^2*c*tan(1/2*x)^2 - 6 *A*a*b^3*c*tan(1/2*x)^2 - 9*B*a*b^3*c*tan(1/2*x)^2 + A*b^4*c*tan(1/2*x)^2 + 2*B*b^4*c*tan(1/2*x)^2 - 5*C*a^3*c^2*tan(1/2*x)^2 + 14*C*a^2*b*c^2*tan(1 /2*x)^2 - 13*C*a*b^2*c^2*tan(1/2*x)^2 + 4*C*b^3*c^2*tan(1/2*x)^2 + 7*A*...
Time = 33.52 (sec) , antiderivative size = 1160, normalized size of antiderivative = 4.89 \[ \int \frac {A+B \cos (x)+C \sin (x)}{(a+b \cos (x)+c \sin (x))^3} \, dx=-\frac {\frac {2\,C\,a^5-4\,A\,a^4\,c-4\,C\,a^3\,b^2+5\,B\,a^3\,b\,c+C\,a^3\,c^2+3\,A\,a^2\,b^2\,c+A\,a^2\,c^3+2\,C\,a\,b^4-5\,B\,a\,b^3\,c-C\,a\,b^2\,c^2-2\,B\,a\,b\,c^3+A\,b^4\,c+A\,b^2\,c^3}{{\left (a-b\right )}^2\,\left (a^4-2\,a^2\,b^2-2\,a^2\,c^2+b^4+2\,b^2\,c^2+c^4\right )}+\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^3\,\left (A\,b^4-2\,B\,a^4+2\,A\,c^4-2\,B\,b^4-2\,B\,c^4-7\,A\,a^2\,b^2-5\,A\,a^2\,c^2-2\,B\,a^2\,b^2+3\,A\,b^2\,c^2+4\,B\,a^2\,c^2-4\,B\,b^2\,c^2+2\,A\,a\,b^3+4\,A\,a^3\,b+3\,B\,a\,b^3+3\,B\,a^3\,b+3\,C\,a^3\,c+2\,A\,a\,b\,c^2+3\,C\,a\,b^2\,c-6\,C\,a^2\,b\,c\right )}{\left (a-b\right )\,\left (a^4-2\,a^2\,b^2-2\,a^2\,c^2+b^4+2\,b^2\,c^2+c^4\right )}-\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (2\,B\,c^5-2\,C\,a^5-2\,A\,c^5+2\,C\,b^5+7\,A\,a^2\,c^3-A\,b^2\,c^3-4\,B\,a^2\,c^3-4\,C\,a^2\,b^3+4\,C\,a^3\,b^2+4\,B\,b^2\,c^3-5\,C\,a^3\,c^2+4\,C\,b^3\,c^2+4\,A\,a^4\,c+A\,b^4\,c+2\,B\,a^4\,c-2\,C\,a\,b^4+2\,C\,a^4\,b+2\,B\,b^4\,c-2\,C\,a\,c^4+2\,C\,b\,c^4-6\,A\,a\,b\,c^3-6\,A\,a\,b^3\,c-12\,A\,a^3\,b\,c-9\,B\,a\,b^3\,c-9\,B\,a^3\,b\,c+13\,A\,a^2\,b^2\,c+14\,B\,a^2\,b^2\,c-13\,C\,a\,b^2\,c^2+14\,C\,a^2\,b\,c^2\right )}{{\left (a-b\right )}^2\,\left (a^4-2\,a^2\,b^2-2\,a^2\,c^2+b^4+2\,b^2\,c^2+c^4\right )}-\frac {\mathrm {tan}\left (\frac {x}{2}\right )\,\left (A\,b^5+2\,B\,a^5+2\,B\,b^5+3\,A\,a^2\,b^3+5\,A\,a^3\,b^2+11\,A\,a^3\,c^2+B\,a^2\,b^3+B\,a^3\,b^2-A\,b^3\,c^2-4\,B\,a^3\,c^2+4\,B\,b^3\,c^2-4\,C\,a^2\,c^3-5\,A\,a\,b^4-4\,A\,a^4\,b-2\,A\,a\,c^4-3\,B\,a\,b^4-3\,B\,a^4\,b-2\,A\,b\,c^4+2\,B\,a\,c^4+2\,B\,b\,c^4-5\,C\,a^4\,c+4\,C\,a\,b\,c^3-5\,C\,a\,b^3\,c+5\,C\,a^3\,b\,c-7\,A\,a\,b^2\,c^2-3\,A\,a^2\,b\,c^2+8\,B\,a\,b^2\,c^2-8\,B\,a^2\,b\,c^2+5\,C\,a^2\,b^2\,c\right )}{{\left (a-b\right )}^2\,\left (a^4-2\,a^2\,b^2-2\,a^2\,c^2+b^4+2\,b^2\,c^2+c^4\right )}}{{\mathrm {tan}\left (\frac {x}{2}\right )}^4\,\left (a^2-2\,a\,b+b^2\right )+2\,a\,b+\mathrm {tan}\left (\frac {x}{2}\right )\,\left (4\,a\,c+4\,b\,c\right )+{\mathrm {tan}\left (\frac {x}{2}\right )}^3\,\left (4\,a\,c-4\,b\,c\right )+a^2+b^2+{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (2\,a^2-2\,b^2+4\,c^2\right )}-\frac {\mathrm {atanh}\left (\frac {2\,a^4\,c-4\,a^2\,b^2\,c-4\,a^2\,c^3+2\,b^4\,c+4\,b^2\,c^3+2\,c^5}{2\,{\left (-a^2+b^2+c^2\right )}^{5/2}}+\frac {\mathrm {tan}\left (\frac {x}{2}\right )\,\left (2\,a-2\,b\right )\,\left (a^4-2\,a^2\,b^2-2\,a^2\,c^2+b^4+2\,b^2\,c^2+c^4\right )}{2\,{\left (-a^2+b^2+c^2\right )}^{5/2}}\right )\,\left (2\,A\,a^2-3\,B\,a\,b-3\,C\,a\,c+A\,b^2+A\,c^2\right )}{{\left (-a^2+b^2+c^2\right )}^{5/2}} \]
- ((2*C*a^5 + A*a^2*c^3 + A*b^2*c^3 - 4*C*a^3*b^2 + C*a^3*c^2 - 4*A*a^4*c + A*b^4*c + 2*C*a*b^4 - 2*B*a*b*c^3 - 5*B*a*b^3*c + 5*B*a^3*b*c + 3*A*a^2* b^2*c - C*a*b^2*c^2)/((a - b)^2*(a^4 + b^4 + c^4 - 2*a^2*b^2 - 2*a^2*c^2 + 2*b^2*c^2)) + (tan(x/2)^3*(A*b^4 - 2*B*a^4 + 2*A*c^4 - 2*B*b^4 - 2*B*c^4 - 7*A*a^2*b^2 - 5*A*a^2*c^2 - 2*B*a^2*b^2 + 3*A*b^2*c^2 + 4*B*a^2*c^2 - 4* B*b^2*c^2 + 2*A*a*b^3 + 4*A*a^3*b + 3*B*a*b^3 + 3*B*a^3*b + 3*C*a^3*c + 2* A*a*b*c^2 + 3*C*a*b^2*c - 6*C*a^2*b*c))/((a - b)*(a^4 + b^4 + c^4 - 2*a^2* b^2 - 2*a^2*c^2 + 2*b^2*c^2)) - (tan(x/2)^2*(2*B*c^5 - 2*C*a^5 - 2*A*c^5 + 2*C*b^5 + 7*A*a^2*c^3 - A*b^2*c^3 - 4*B*a^2*c^3 - 4*C*a^2*b^3 + 4*C*a^3*b ^2 + 4*B*b^2*c^3 - 5*C*a^3*c^2 + 4*C*b^3*c^2 + 4*A*a^4*c + A*b^4*c + 2*B*a ^4*c - 2*C*a*b^4 + 2*C*a^4*b + 2*B*b^4*c - 2*C*a*c^4 + 2*C*b*c^4 - 6*A*a*b *c^3 - 6*A*a*b^3*c - 12*A*a^3*b*c - 9*B*a*b^3*c - 9*B*a^3*b*c + 13*A*a^2*b ^2*c + 14*B*a^2*b^2*c - 13*C*a*b^2*c^2 + 14*C*a^2*b*c^2))/((a - b)^2*(a^4 + b^4 + c^4 - 2*a^2*b^2 - 2*a^2*c^2 + 2*b^2*c^2)) - (tan(x/2)*(A*b^5 + 2*B *a^5 + 2*B*b^5 + 3*A*a^2*b^3 + 5*A*a^3*b^2 + 11*A*a^3*c^2 + B*a^2*b^3 + B* a^3*b^2 - A*b^3*c^2 - 4*B*a^3*c^2 + 4*B*b^3*c^2 - 4*C*a^2*c^3 - 5*A*a*b^4 - 4*A*a^4*b - 2*A*a*c^4 - 3*B*a*b^4 - 3*B*a^4*b - 2*A*b*c^4 + 2*B*a*c^4 + 2*B*b*c^4 - 5*C*a^4*c + 4*C*a*b*c^3 - 5*C*a*b^3*c + 5*C*a^3*b*c - 7*A*a*b^ 2*c^2 - 3*A*a^2*b*c^2 + 8*B*a*b^2*c^2 - 8*B*a^2*b*c^2 + 5*C*a^2*b^2*c))/(( a - b)^2*(a^4 + b^4 + c^4 - 2*a^2*b^2 - 2*a^2*c^2 + 2*b^2*c^2)))/(tan(x...