Integrand size = 23, antiderivative size = 164 \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^3} \, dx=\frac {\left (2 a^2 A+A b^2-3 a b B\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} d}-\frac {(A b-a B) \sin (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac {\left (3 a A b-a^2 B-2 b^2 B\right ) \sin (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))} \]
(2*A*a^2+A*b^2-3*B*a*b)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2)) /(a-b)^(5/2)/(a+b)^(5/2)/d-1/2*(A*b-B*a)*sin(d*x+c)/(a^2-b^2)/d/(a+b*cos(d *x+c))^2-1/2*(3*A*a*b-B*a^2-2*B*b^2)*sin(d*x+c)/(a^2-b^2)^2/d/(a+b*cos(d*x +c))
Time = 0.60 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.96 \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^3} \, dx=\frac {-\frac {2 \left (2 a^2 A+A b^2-3 a b B\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{5/2}}+\frac {(-A b+a B) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))^2}+\frac {\left (-3 a A b+a^2 B+2 b^2 B\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (a+b \cos (c+d x))}}{2 d} \]
((-2*(2*a^2*A + A*b^2 - 3*a*b*B)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[- a^2 + b^2]])/(-a^2 + b^2)^(5/2) + ((-(A*b) + a*B)*Sin[c + d*x])/((a - b)*( a + b)*(a + b*Cos[c + d*x])^2) + ((-3*a*A*b + a^2*B + 2*b^2*B)*Sin[c + d*x ])/((a - b)^2*(a + b)^2*(a + b*Cos[c + d*x])))/(2*d)
Time = 0.57 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.16, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3042, 3233, 25, 3042, 3233, 25, 27, 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.
| \(\displaystyle \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3}dx\) |
| \(\Big \downarrow \) 3233 |
| \(\displaystyle -\frac {\int -\frac {2 (a A-b B)-(A b-a B) \cos (c+d x)}{(a+b \cos (c+d x))^2}dx}{2 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {2 (a A-b B)-(A b-a B) \cos (c+d x)}{(a+b \cos (c+d x))^2}dx}{2 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {2 (a A-b B)+(a B-A b) \sin \left (c+d x+\frac {\pi }{2}\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx}{2 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 3233 |
| \(\displaystyle \frac {-\frac {\int -\frac {2 A a^2-3 b B a+A b^2}{a+b \cos (c+d x)}dx}{a^2-b^2}-\frac {\left (a^2 (-B)+3 a A b-2 b^2 B\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {2 A a^2-3 b B a+A b^2}{a+b \cos (c+d x)}dx}{a^2-b^2}-\frac {\left (a^2 (-B)+3 a A b-2 b^2 B\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\left (2 a^2 A-3 a b B+A b^2\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{a^2-b^2}-\frac {\left (a^2 (-B)+3 a A b-2 b^2 B\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\left (2 a^2 A-3 a b B+A b^2\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2-b^2}-\frac {\left (a^2 (-B)+3 a A b-2 b^2 B\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {\frac {2 \left (2 a^2 A-3 a b B+A b^2\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{d \left (a^2-b^2\right )}-\frac {\left (a^2 (-B)+3 a A b-2 b^2 B\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {2 \left (2 a^2 A-3 a b B+A b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b} \left (a^2-b^2\right )}-\frac {\left (a^2 (-B)+3 a A b-2 b^2 B\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
-1/2*((A*b - a*B)*Sin[c + d*x])/((a^2 - b^2)*d*(a + b*Cos[c + d*x])^2) + ( (2*(2*a^2*A + A*b^2 - 3*a*b*B)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[ a + b]])/(Sqrt[a - b]*Sqrt[a + b]*(a^2 - b^2)*d) - ((3*a*A*b - a^2*B - 2*b ^2*B)*Sin[c + d*x])/((a^2 - b^2)*d*(a + b*Cos[c + d*x])))/(2*(a^2 - b^2))
3.3.69.3.1 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_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(b*c - a*d))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*( m + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]
Time = 1.26 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.41
| method | result | size |
| derivativedivides | \(\frac {\frac {-\frac {\left (4 A a b +A \,b^{2}-2 B \,a^{2}-B a b -2 B \,b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (4 A a b -A \,b^{2}-2 B \,a^{2}+B a b -2 B \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}^{2}}+\frac {\left (2 A \,a^{2}+A \,b^{2}-3 B a b \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(232\) |
| default | \(\frac {\frac {-\frac {\left (4 A a b +A \,b^{2}-2 B \,a^{2}-B a b -2 B \,b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (4 A a b -A \,b^{2}-2 B \,a^{2}+B a b -2 B \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}^{2}}+\frac {\left (2 A \,a^{2}+A \,b^{2}-3 B a b \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(232\) |
| risch | \(\frac {i \left (-2 A \,a^{2} b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-A \,b^{4} {\mathrm e}^{3 i \left (d x +c \right )}+3 B a \,b^{3} {\mathrm e}^{3 i \left (d x +c \right )}-6 A \,a^{3} b \,{\mathrm e}^{2 i \left (d x +c \right )}-3 A a \,b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+2 B \,a^{4} {\mathrm e}^{2 i \left (d x +c \right )}+5 B \,a^{2} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+2 B \,b^{4} {\mathrm e}^{2 i \left (d x +c \right )}-10 A \,a^{2} b^{2} {\mathrm e}^{i \left (d x +c \right )}+A \,b^{4} {\mathrm e}^{i \left (d x +c \right )}+4 B \,a^{3} b \,{\mathrm e}^{i \left (d x +c \right )}+5 B a \,b^{3} {\mathrm e}^{i \left (d x +c \right )}-3 A a \,b^{3}+B \,a^{2} b^{2}+2 B \,b^{4}\right )}{b \left (a^{2}-b^{2}\right )^{2} d \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )^{2}}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}-\frac {b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}+\frac {3 b \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 a}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}+\frac {b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}-\frac {3 b \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 a}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}\) | \(771\) |
1/d*(2*(-1/2*(4*A*a*b+A*b^2-2*B*a^2-B*a*b-2*B*b^2)/(a-b)/(a^2+2*a*b+b^2)*t an(1/2*d*x+1/2*c)^3-1/2*(4*A*a*b-A*b^2-2*B*a^2+B*a*b-2*B*b^2)/(a+b)/(a^2-2 *a*b+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+(2*A*a^2+A*b^2-3*B*a*b)/(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)))
Leaf count of result is larger than twice the leaf count of optimal. 336 vs. \(2 (150) = 300\).
Time = 0.34 (sec) , antiderivative size = 742, normalized size of antiderivative = 4.52 \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^3} \, dx=\left [-\frac {{\left (2 \, A a^{4} - 3 \, B a^{3} b + A a^{2} b^{2} + {\left (2 \, A a^{2} b^{2} - 3 \, B a b^{3} + A b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, A a^{3} b - 3 \, B a^{2} b^{2} + A a b^{3}\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 ) - 2 \, {\left (2 \, B a^{5} - 4 \, A a^{4} b - B a^{3} b^{2} + 5 \, A a^{2} b^{3} - B a b^{4} - A b^{5} + {\left (B a^{4} b - 3 \, A a^{3} b^{2} + B a^{2} b^{3} + 3 \, A a b^{4} - 2 \, B b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} d \cos \left (d x + c\right ) + {\left (a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6}\right )} d\right )}}, \frac {{\left (2 \, A a^{4} - 3 \, B a^{3} b + A a^{2} b^{2} + {\left (2 \, A a^{2} b^{2} - 3 \, B a b^{3} + A b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, A a^{3} b - 3 \, B a^{2} b^{2} + A a b^{3}\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 (2 \, B a^{5} - 4 \, A a^{4} b - B a^{3} b^{2} + 5 \, A a^{2} b^{3} - B a b^{4} - A b^{5} + {\left (B a^{4} b - 3 \, A a^{3} b^{2} + B a^{2} b^{3} + 3 \, A a b^{4} - 2 \, B b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} d \cos \left (d x + c\right ) + {\left (a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6}\right )} d\right )}}\right ] \]
[-1/4*((2*A*a^4 - 3*B*a^3*b + A*a^2*b^2 + (2*A*a^2*b^2 - 3*B*a*b^3 + A*b^4 )*cos(d*x + c)^2 + 2*(2*A*a^3*b - 3*B*a^2*b^2 + A*a*b^3)*cos(d*x + c))*sqr t(-a^2 + b^2)*log((2*a*b*cos(d*x + c) + (2*a^2 - b^2)*cos(d*x + c)^2 + 2*s qrt(-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*(2*B*a^5 - 4*A*a^4*b - B*a^3*b ^2 + 5*A*a^2*b^3 - B*a*b^4 - A*b^5 + (B*a^4*b - 3*A*a^3*b^2 + B*a^2*b^3 + 3*A*a*b^4 - 2*B*b^5)*cos(d*x + c))*sin(d*x + c))/((a^6*b^2 - 3*a^4*b^4 + 3 *a^2*b^6 - b^8)*d*cos(d*x + c)^2 + 2*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^ 7)*d*cos(d*x + c) + (a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6)*d), 1/2*((2*A* a^4 - 3*B*a^3*b + A*a^2*b^2 + (2*A*a^2*b^2 - 3*B*a*b^3 + A*b^4)*cos(d*x + c)^2 + 2*(2*A*a^3*b - 3*B*a^2*b^2 + A*a*b^3)*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))) + (2*B*a^5 - 4*A*a^4*b - B*a^3*b^2 + 5*A*a^2*b^3 - B*a*b^4 - A*b^5 + (B*a^4*b - 3*A*a^ 3*b^2 + B*a^2*b^3 + 3*A*a*b^4 - 2*B*b^5)*cos(d*x + c))*sin(d*x + c))/((a^6 *b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*d*cos(d*x + c)^2 + 2*(a^7*b - 3*a^5*b^ 3 + 3*a^3*b^5 - a*b^7)*d*cos(d*x + c) + (a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2 *b^6)*d)]
Timed out. \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(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. 390 vs. \(2 (150) = 300\).
Time = 0.31 (sec) , antiderivative size = 390, normalized size of antiderivative = 2.38 \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^3} \, dx=\frac {\frac {{\left (2 \, A a^{2} - 3 \, B a b + A b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {2 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}^{2}}}{d} \]
((2*A*a^2 - 3*B*a*b + A*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)))/((a^4 - 2*a^2*b^2 + b^4)*sqrt(a^2 - b^2)) + (2*B*a^3*tan(1/2*d*x + 1/2*c)^3 - 4*A*a^2*b*tan(1/2*d*x + 1/2*c)^3 - B*a^2*b*tan(1/2*d*x + 1/2*c) ^3 + 3*A*a*b^2*tan(1/2*d*x + 1/2*c)^3 + B*a*b^2*tan(1/2*d*x + 1/2*c)^3 + A *b^3*tan(1/2*d*x + 1/2*c)^3 - 2*B*b^3*tan(1/2*d*x + 1/2*c)^3 + 2*B*a^3*tan (1/2*d*x + 1/2*c) - 4*A*a^2*b*tan(1/2*d*x + 1/2*c) + B*a^2*b*tan(1/2*d*x + 1/2*c) - 3*A*a*b^2*tan(1/2*d*x + 1/2*c) + B*a*b^2*tan(1/2*d*x + 1/2*c) + A*b^3*tan(1/2*d*x + 1/2*c) + 2*B*b^3*tan(1/2*d*x + 1/2*c))/((a^4 - 2*a^2*b ^2 + b^4)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)^2) )/d
Time = 3.66 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.51 \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^3} \, dx=\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,B\,a^2-A\,b^2+2\,B\,b^2-4\,A\,a\,b+B\,a\,b\right )}{{\left (a+b\right )}^2\,\left (a-b\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A\,b^2+2\,B\,a^2+2\,B\,b^2-4\,A\,a\,b-B\,a\,b\right )}{\left (a+b\right )\,\left (a^2-2\,a\,b+b^2\right )}}{d\,\left (2\,a\,b+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^2-2\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a^2-2\,a\,b+b^2\right )+a^2+b^2\right )}+\frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a-2\,b\right )\,\left (a^2-2\,a\,b+b^2\right )}{2\,\sqrt {a+b}\,{\left (a-b\right )}^{5/2}}\right )\,\left (2\,A\,a^2-3\,B\,a\,b+A\,b^2\right )}{d\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}} \]
((tan(c/2 + (d*x)/2)^3*(2*B*a^2 - A*b^2 + 2*B*b^2 - 4*A*a*b + B*a*b))/((a + b)^2*(a - b)) + (tan(c/2 + (d*x)/2)*(A*b^2 + 2*B*a^2 + 2*B*b^2 - 4*A*a*b - B*a*b))/((a + b)*(a^2 - 2*a*b + b^2)))/(d*(2*a*b + tan(c/2 + (d*x)/2)^2 *(2*a^2 - 2*b^2) + tan(c/2 + (d*x)/2)^4*(a^2 - 2*a*b + b^2) + a^2 + b^2)) + (atan((tan(c/2 + (d*x)/2)*(2*a - 2*b)*(a^2 - 2*a*b + b^2))/(2*(a + b)^(1 /2)*(a - b)^(5/2)))*(2*A*a^2 + A*b^2 - 3*B*a*b))/(d*(a + b)^(5/2)*(a - b)^ (5/2))