Integrand size = 31, antiderivative size = 149 \[ \int \frac {\cos (c+d x) \left (1-\cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^3} \, dx=-\frac {x}{b^3}+\frac {a \left (2 a^2-3 b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} b^3 (a+b)^{3/2} d}-\frac {a \sin (c+d x)}{2 b^2 d (a+b \cos (c+d x))^2}+\frac {\left (3 a^2-2 b^2\right ) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \]
-x/b^3+a*(2*a^2-3*b^2)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/ (a-b)^(3/2)/b^3/(a+b)^(3/2)/d-1/2*a*sin(d*x+c)/b^2/d/(a+b*cos(d*x+c))^2+1/ 2*(3*a^2-2*b^2)*sin(d*x+c)/b^2/(a^2-b^2)/d/(a+b*cos(d*x+c))
Time = 1.75 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.95 \[ \int \frac {\cos (c+d x) \left (1-\cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^3} \, dx=\frac {-\frac {8 (c+d x)+\frac {2 a \left (8 a^4-20 a^2 b^2+15 b^4\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{5/2}}+\frac {a b \left (4 a^2-3 b^2\right ) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))^2}-\frac {3 b \left (4 a^4-7 a^2 b^2+2 b^4\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (a+b \cos (c+d x))}}{b^3}+\frac {\frac {6 a b \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}+\frac {\left (a \left (2 a^2+b^2\right )+b \left (a^2+2 b^2\right ) \cos (c+d x)\right ) \sin (c+d x)}{(a+b \cos (c+d x))^2}}{(a-b)^2 (a+b)^2}}{8 d} \]
(-((8*(c + d*x) + (2*a*(8*a^4 - 20*a^2*b^2 + 15*b^4)*ArcTanh[((a - b)*Tan[ (c + d*x)/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(5/2) + (a*b*(4*a^2 - 3*b^2) *Sin[c + d*x])/((a - b)*(a + b)*(a + b*Cos[c + d*x])^2) - (3*b*(4*a^4 - 7* a^2*b^2 + 2*b^4)*Sin[c + d*x])/((a - b)^2*(a + b)^2*(a + b*Cos[c + d*x]))) /b^3) + ((6*a*b*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/Sqrt [-a^2 + b^2] + ((a*(2*a^2 + b^2) + b*(a^2 + 2*b^2)*Cos[c + d*x])*Sin[c + d *x])/(a + b*Cos[c + d*x])^2)/((a - b)^2*(a + b)^2))/(8*d)
Time = 0.78 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.23, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.323, Rules used = {3042, 3511, 25, 3042, 3500, 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.
| \(\displaystyle \int \frac {\cos (c+d x) \left (1-\cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \left (1-\sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3}dx\) |
| \(\Big \downarrow \) 3511 |
| \(\displaystyle -\frac {\int -\frac {-2 b \left (a^2-b^2\right ) \cos ^2(c+d x)+a \left (a^2-b^2\right ) \cos (c+d x)+2 b \left (a^2-b^2\right )}{(a+b \cos (c+d x))^2}dx}{2 b^2 \left (a^2-b^2\right )}-\frac {a \sin (c+d x)}{2 b^2 d (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {-2 b \left (a^2-b^2\right ) \cos ^2(c+d x)+a \left (a^2-b^2\right ) \cos (c+d x)+2 b \left (a^2-b^2\right )}{(a+b \cos (c+d x))^2}dx}{2 b^2 \left (a^2-b^2\right )}-\frac {a \sin (c+d x)}{2 b^2 d (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {-2 b \left (a^2-b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+a \left (a^2-b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+2 b \left (a^2-b^2\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx}{2 b^2 \left (a^2-b^2\right )}-\frac {a \sin (c+d x)}{2 b^2 d (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {\frac {\left (3 a^2-2 b^2\right ) \sin (c+d x)}{d (a+b \cos (c+d x))}-\frac {\int \frac {a \left (a^2-b^2\right ) b^2+2 \left (a^2-b^2\right )^2 \cos (c+d x) b}{a+b \cos (c+d x)}dx}{b \left (a^2-b^2\right )}}{2 b^2 \left (a^2-b^2\right )}-\frac {a \sin (c+d x)}{2 b^2 d (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\left (3 a^2-2 b^2\right ) \sin (c+d x)}{d (a+b \cos (c+d x))}-\frac {\int \frac {a \left (a^2-b^2\right ) b^2+2 \left (a^2-b^2\right )^2 \sin \left (c+d x+\frac {\pi }{2}\right ) b}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b \left (a^2-b^2\right )}}{2 b^2 \left (a^2-b^2\right )}-\frac {a \sin (c+d x)}{2 b^2 d (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {\frac {\left (3 a^2-2 b^2\right ) \sin (c+d x)}{d (a+b \cos (c+d x))}-\frac {2 x \left (a^2-b^2\right )^2-a \left (2 a^2-3 b^2\right ) \left (a^2-b^2\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{b \left (a^2-b^2\right )}}{2 b^2 \left (a^2-b^2\right )}-\frac {a \sin (c+d x)}{2 b^2 d (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\left (3 a^2-2 b^2\right ) \sin (c+d x)}{d (a+b \cos (c+d x))}-\frac {2 x \left (a^2-b^2\right )^2-a \left (2 a^2-3 b^2\right ) \left (a^2-b^2\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b \left (a^2-b^2\right )}}{2 b^2 \left (a^2-b^2\right )}-\frac {a \sin (c+d x)}{2 b^2 d (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {\frac {\left (3 a^2-2 b^2\right ) \sin (c+d x)}{d (a+b \cos (c+d x))}-\frac {2 x \left (a^2-b^2\right )^2-\frac {2 a \left (2 a^2-3 b^2\right ) \left (a^2-b^2\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{d}}{b \left (a^2-b^2\right )}}{2 b^2 \left (a^2-b^2\right )}-\frac {a \sin (c+d x)}{2 b^2 d (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\left (3 a^2-2 b^2\right ) \sin (c+d x)}{d (a+b \cos (c+d x))}-\frac {2 x \left (a^2-b^2\right )^2-\frac {2 a \left (2 a^2-3 b^2\right ) \left (a^2-b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b}}}{b \left (a^2-b^2\right )}}{2 b^2 \left (a^2-b^2\right )}-\frac {a \sin (c+d x)}{2 b^2 d (a+b \cos (c+d x))^2}\) |
-1/2*(a*Sin[c + d*x])/(b^2*d*(a + b*Cos[c + d*x])^2) + (-((2*(a^2 - b^2)^2 *x - (2*a*(2*a^2 - 3*b^2)*(a^2 - b^2)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2] )/Sqrt[a + b]])/(Sqrt[a - b]*Sqrt[a + b]*d))/(b*(a^2 - b^2))) + ((3*a^2 - 2*b^2)*Sin[c + d*x])/(d*(a + b*Cos[c + d*x])))/(2*b^2*(a^2 - b^2))
3.7.13.3.1 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_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* (a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x ])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A *b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[ (-(b*c - a*d))*(A*b^2 + 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] + Simp[1/(b^2*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*(a*C*(b*c - a*d) + A*b*(a*c - b*d )) - ((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] /; FreeQ[{a, b, c, d, e , f, A, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]
Time = 1.47 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.28
| method | result | size |
| derivativedivides | \(\frac {-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3}}+\frac {\frac {2 \left (\frac {\left (2 a^{2}-a b -2 b^{2}\right ) b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a +2 b}+\frac {\left (2 a^{2}+a b -2 b^{2}\right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a -2 b}\right )}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}^{2}}+\frac {a \left (2 a^{2}-3 b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{2}-b^{2}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{b^{3}}}{d}\) | \(191\) |
| default | \(\frac {-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3}}+\frac {\frac {2 \left (\frac {\left (2 a^{2}-a b -2 b^{2}\right ) b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a +2 b}+\frac {\left (2 a^{2}+a b -2 b^{2}\right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a -2 b}\right )}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}^{2}}+\frac {a \left (2 a^{2}-3 b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{2}-b^{2}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{b^{3}}}{d}\) | \(191\) |
| risch | \(-\frac {x}{b^{3}}+\frac {i \left (4 b \,a^{3} {\mathrm e}^{3 i \left (d x +c \right )}-3 b^{3} a \,{\mathrm e}^{3 i \left (d x +c \right )}+6 a^{4} {\mathrm e}^{2 i \left (d x +c \right )}-b^{2} a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-2 b^{4} {\mathrm e}^{2 i \left (d x +c \right )}+8 b \,a^{3} {\mathrm e}^{i \left (d x +c \right )}-5 \,{\mathrm e}^{i \left (d x +c \right )} b^{3} a +3 a^{2} b^{2}-2 b^{4}\right )}{b^{3} \left (a^{2}-b^{2}\right ) d \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )^{2}}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{3}}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d b}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{3}}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d b}\) | \(517\) |
1/d*(-2/b^3*arctan(tan(1/2*d*x+1/2*c))+2/b^3*((1/2*(2*a^2-a*b-2*b^2)*b/(a+ b)*tan(1/2*d*x+1/2*c)^3+1/2*(2*a^2+a*b-2*b^2)*b/(a-b)*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*a*(2*a^2-3*b^2)/ (a^2-b^2)/((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. \(2 (136) = 272\).
Time = 0.35 (sec) , antiderivative size = 740, normalized size of antiderivative = 4.97 \[ \int \frac {\cos (c+d x) \left (1-\cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^3} \, dx=\left [-\frac {4 \, {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} d x \cos \left (d x + c\right )^{2} + 8 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} d x \cos \left (d x + c\right ) + 4 \, {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} d x + {\left (2 \, a^{5} - 3 \, a^{3} b^{2} + {\left (2 \, a^{3} b^{2} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, a^{4} b - 3 \, a^{2} 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 \, a^{5} b - 3 \, a^{3} b^{3} + a b^{5} + {\left (3 \, a^{4} b^{2} - 5 \, a^{2} b^{4} + 2 \, b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{4} b^{5} - 2 \, a^{2} b^{7} + b^{9}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{5} b^{4} - 2 \, a^{3} b^{6} + a b^{8}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} b^{3} - 2 \, a^{4} b^{5} + a^{2} b^{7}\right )} d\right )}}, -\frac {2 \, {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} d x \cos \left (d x + c\right )^{2} + 4 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} d x \cos \left (d x + c\right ) + 2 \, {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} d x - {\left (2 \, a^{5} - 3 \, a^{3} b^{2} + {\left (2 \, a^{3} b^{2} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, a^{4} b - 3 \, a^{2} 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 \, a^{5} b - 3 \, a^{3} b^{3} + a b^{5} + {\left (3 \, a^{4} b^{2} - 5 \, a^{2} b^{4} + 2 \, b^{6}\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 )^{2} + 2 \, {\left (a^{5} b^{4} - 2 \, a^{3} b^{6} + a b^{8}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} b^{3} - 2 \, a^{4} b^{5} + a^{2} b^{7}\right )} d\right )}}\right ] \]
[-1/4*(4*(a^4*b^2 - 2*a^2*b^4 + b^6)*d*x*cos(d*x + c)^2 + 8*(a^5*b - 2*a^3 *b^3 + a*b^5)*d*x*cos(d*x + c) + 4*(a^6 - 2*a^4*b^2 + a^2*b^4)*d*x + (2*a^ 5 - 3*a^3*b^2 + (2*a^3*b^2 - 3*a*b^4)*cos(d*x + c)^2 + 2*(2*a^4*b - 3*a^2* b^3)*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*(2*a^5*b - 3*a^3*b^3 + a*b^5 + (3*a^4*b^2 - 5*a^2*b^4 + 2*b^6)*cos(d*x + c))*sin(d *x + c))/((a^4*b^5 - 2*a^2*b^7 + b^9)*d*cos(d*x + c)^2 + 2*(a^5*b^4 - 2*a^ 3*b^6 + a*b^8)*d*cos(d*x + c) + (a^6*b^3 - 2*a^4*b^5 + a^2*b^7)*d), -1/2*( 2*(a^4*b^2 - 2*a^2*b^4 + b^6)*d*x*cos(d*x + c)^2 + 4*(a^5*b - 2*a^3*b^3 + a*b^5)*d*x*cos(d*x + c) + 2*(a^6 - 2*a^4*b^2 + a^2*b^4)*d*x - (2*a^5 - 3*a ^3*b^2 + (2*a^3*b^2 - 3*a*b^4)*cos(d*x + c)^2 + 2*(2*a^4*b - 3*a^2*b^3)*co s(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*a^5*b - 3*a^3*b^3 + a*b^5 + (3*a^4*b^2 - 5*a^2*b^4 + 2 *b^6)*cos(d*x + c))*sin(d*x + c))/((a^4*b^5 - 2*a^2*b^7 + b^9)*d*cos(d*x + c)^2 + 2*(a^5*b^4 - 2*a^3*b^6 + a*b^8)*d*cos(d*x + c) + (a^6*b^3 - 2*a^4* b^5 + a^2*b^7)*d)]
Timed out. \[ \int \frac {\cos (c+d x) \left (1-\cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^3} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\cos (c+d x) \left (1-\cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^3} \, dx=\text {Exception raised: ValueError} \]
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. 290 vs. \(2 (136) = 272\).
Time = 0.35 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.95 \[ \int \frac {\cos (c+d x) \left (1-\cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^3} \, dx=-\frac {\frac {{\left (2 \, a^{3} - 3 \, a b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{2} b^{3} - b^{5}\right )} \sqrt {a^{2} - b^{2}}} - \frac {2 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{2} b^{2} - b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}^{2}} + \frac {d x + c}{b^{3}}}{d} \]
-((2*a^3 - 3*a*b^2)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + ar ctan(-(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^3 - b^5)*sqrt(a^2 - b^2)) - (2*a^3*tan(1/2*d*x + 1/2*c)^3 - 3*a^2* b*tan(1/2*d*x + 1/2*c)^3 - a*b^2*tan(1/2*d*x + 1/2*c)^3 + 2*b^3*tan(1/2*d* x + 1/2*c)^3 + 2*a^3*tan(1/2*d*x + 1/2*c) + 3*a^2*b*tan(1/2*d*x + 1/2*c) - a*b^2*tan(1/2*d*x + 1/2*c) - 2*b^3*tan(1/2*d*x + 1/2*c))/((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*x + c)/b^3)/d
Time = 7.44 (sec) , antiderivative size = 3095, normalized size of antiderivative = 20.77 \[ \int \frac {\cos (c+d x) \left (1-\cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^3} \, dx=\text {Too large to display} \]
((tan(c/2 + (d*x)/2)*(a*b + 2*a^2 - 2*b^2))/(a*b^2 - b^3) - (tan(c/2 + (d* x)/2)^3*(a*b - 2*a^2 + 2*b^2))/(b^2*(a + b)))/(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)) - (2*atan((((((8*(6*a*b^10 - 4*b^11 + 6*a^2*b^9 - 10*a^3*b^8 - 2*a^4*b ^7 + 4*a^5*b^6))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) - (tan(c/2 + (d*x)/2)*( 8*a*b^11 - 8*a^2*b^10 - 16*a^3*b^9 + 16*a^4*b^8 + 8*a^5*b^7 - 8*a^6*b^6)*8 i)/(b^3*(a*b^6 + b^7 - a^2*b^5 - a^3*b^4)))*1i)/b^3 + (8*tan(c/2 + (d*x)/2 )*(8*a^6 - 8*a^5*b - 8*a*b^5 + 4*b^6 + 5*a^2*b^4 + 16*a^3*b^3 - 16*a^4*b^2 ))/(a*b^6 + b^7 - a^2*b^5 - a^3*b^4))/b^3 - ((((8*(6*a*b^10 - 4*b^11 + 6*a ^2*b^9 - 10*a^3*b^8 - 2*a^4*b^7 + 4*a^5*b^6))/(a*b^8 + b^9 - a^2*b^7 - a^3 *b^6) + (tan(c/2 + (d*x)/2)*(8*a*b^11 - 8*a^2*b^10 - 16*a^3*b^9 + 16*a^4*b ^8 + 8*a^5*b^7 - 8*a^6*b^6)*8i)/(b^3*(a*b^6 + b^7 - a^2*b^5 - a^3*b^4)))*1 i)/b^3 - (8*tan(c/2 + (d*x)/2)*(8*a^6 - 8*a^5*b - 8*a*b^5 + 4*b^6 + 5*a^2* b^4 + 16*a^3*b^3 - 16*a^4*b^2))/(a*b^6 + b^7 - a^2*b^5 - a^3*b^4))/b^3)/(( 16*(6*a*b^4 - 2*a^4*b + 4*a^5 + 3*a^2*b^3 - 10*a^3*b^2))/(a*b^8 + b^9 - a^ 2*b^7 - a^3*b^6) + (((((8*(6*a*b^10 - 4*b^11 + 6*a^2*b^9 - 10*a^3*b^8 - 2* a^4*b^7 + 4*a^5*b^6))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) - (tan(c/2 + (d*x) /2)*(8*a*b^11 - 8*a^2*b^10 - 16*a^3*b^9 + 16*a^4*b^8 + 8*a^5*b^7 - 8*a^6*b ^6)*8i)/(b^3*(a*b^6 + b^7 - a^2*b^5 - a^3*b^4)))*1i)/b^3 + (8*tan(c/2 + (d *x)/2)*(8*a^6 - 8*a^5*b - 8*a*b^5 + 4*b^6 + 5*a^2*b^4 + 16*a^3*b^3 - 16...