Integrand size = 31, antiderivative size = 112 \[ \int \frac {\cos (c+d x) \left (1-\cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\frac {2 a x}{b^3}-\frac {2 \left (2 a^2-b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^3 \sqrt {a+b} d}-\frac {\sin (c+d x)}{b^2 d}-\frac {a \sin (c+d x)}{b^2 d (a+b \cos (c+d x))} \]
2*a*x/b^3-sin(d*x+c)/b^2/d-a*sin(d*x+c)/b^2/d/(a+b*cos(d*x+c))-2*(2*a^2-b^ 2)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/b^3/d/(a-b)^(1/2)/(a +b)^(1/2)
Time = 0.95 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.18 \[ \int \frac {\cos (c+d x) \left (1-\cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\frac {\frac {4 \left (2 a^2-b^2\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}+\frac {4 a^2 c+4 a^2 d x+4 a b (c+d x) \cos (c+d x)-4 a b \sin (c+d x)-b^2 \sin (2 (c+d x))}{a+b \cos (c+d x)}}{2 b^3 d} \]
((4*(2*a^2 - b^2)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/Sq rt[-a^2 + b^2] + (4*a^2*c + 4*a^2*d*x + 4*a*b*(c + d*x)*Cos[c + d*x] - 4*a *b*Sin[c + d*x] - b^2*Sin[2*(c + d*x)])/(a + b*Cos[c + d*x]))/(2*b^3*d)
Time = 0.72 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.35, 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, 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.
| \(\displaystyle \int \frac {\cos (c+d x) \left (1-\cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \left (1-\sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 3511 |
| \(\displaystyle -\frac {\int -\frac {-b \left (a^2-b^2\right ) \cos ^2(c+d x)+a \left (a^2-b^2\right ) \cos (c+d x)+b \left (a^2-b^2\right )}{a+b \cos (c+d x)}dx}{b^2 \left (a^2-b^2\right )}-\frac {a \sin (c+d x)}{b^2 d (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {-b \left (a^2-b^2\right ) \cos ^2(c+d x)+a \left (a^2-b^2\right ) \cos (c+d x)+b \left (a^2-b^2\right )}{a+b \cos (c+d x)}dx}{b^2 \left (a^2-b^2\right )}-\frac {a \sin (c+d x)}{b^2 d (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {-b \left (a^2-b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+a \left (a^2-b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+b \left (a^2-b^2\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2 \left (a^2-b^2\right )}-\frac {a \sin (c+d x)}{b^2 d (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {\frac {\int \frac {\left (a^2-b^2\right ) b^2+2 a \left (a^2-b^2\right ) \cos (c+d x) b}{a+b \cos (c+d x)}dx}{b}-\frac {\left (a^2-b^2\right ) \sin (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}-\frac {a \sin (c+d x)}{b^2 d (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {\left (a^2-b^2\right ) b^2+2 a \left (a^2-b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) b}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {\left (a^2-b^2\right ) \sin (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}-\frac {a \sin (c+d x)}{b^2 d (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {\frac {2 a x \left (a^2-b^2\right )-\left (a^2-b^2\right ) \left (2 a^2-b^2\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{b}-\frac {\left (a^2-b^2\right ) \sin (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}-\frac {a \sin (c+d x)}{b^2 d (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {2 a x \left (a^2-b^2\right )-\left (a^2-b^2\right ) \left (2 a^2-b^2\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {\left (a^2-b^2\right ) \sin (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}-\frac {a \sin (c+d x)}{b^2 d (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {\frac {2 a x \left (a^2-b^2\right )-\frac {2 \left (a^2-b^2\right ) \left (2 a^2-b^2\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{d}}{b}-\frac {\left (a^2-b^2\right ) \sin (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}-\frac {a \sin (c+d x)}{b^2 d (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {2 a x \left (a^2-b^2\right )-\frac {2 \left (a^2-b^2\right ) \left (2 a^2-b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b}}}{b}-\frac {\left (a^2-b^2\right ) \sin (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}-\frac {a \sin (c+d x)}{b^2 d (a+b \cos (c+d x))}\) |
-((a*Sin[c + d*x])/(b^2*d*(a + b*Cos[c + d*x]))) + ((2*a*(a^2 - b^2)*x - ( 2*(a^2 - b^2)*(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)*Sin[c + d*x])/d)/(b^2* (a^2 - b^2))
3.7.4.3.1 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[(-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_)])*((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 = 1.41 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.36
| method | result | size |
| derivativedivides | \(\frac {\frac {-\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}+4 a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3}}-\frac {2 \left (\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b}+\frac {\left (2 a^{2}-b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{3}}}{d}\) | \(152\) |
| default | \(\frac {\frac {-\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}+4 a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3}}-\frac {2 \left (\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b}+\frac {\left (2 a^{2}-b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{3}}}{d}\) | \(152\) |
| risch | \(\frac {2 a x}{b^{3}}+\frac {i {\mathrm e}^{i \left (d x +c \right )}}{2 b^{2} d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{2 b^{2} d}-\frac {2 i a \left (a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}{b^{3} d \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) a^{2}}{\sqrt {-a^{2}+b^{2}}\, d \,b^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d b}+\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,b^{3}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d b}\) | \(388\) |
1/d*(2/b^3*(-b*tan(1/2*d*x+1/2*c)/(1+tan(1/2*d*x+1/2*c)^2)+2*a*arctan(tan( 1/2*d*x+1/2*c)))-2/b^3*(a*b*tan(1/2*d*x+1/2*c)/(tan(1/2*d*x+1/2*c)^2*a-b*t an(1/2*d*x+1/2*c)^2+a+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))))
Time = 0.31 (sec) , antiderivative size = 463, normalized size of antiderivative = 4.13 \[ \int \frac {\cos (c+d x) \left (1-\cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\left [\frac {4 \, {\left (a^{3} b - a b^{3}\right )} d x \cos \left (d x + c\right ) + 4 \, {\left (a^{4} - a^{2} b^{2}\right )} d x + {\left (2 \, a^{3} - a b^{2} + {\left (2 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - 2 \, {\left (2 \, a^{3} b - 2 \, a b^{3} + {\left (a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right ) + {\left (a^{3} b^{3} - a b^{5}\right )} d\right )}}, \frac {2 \, {\left (a^{3} b - a b^{3}\right )} d x \cos \left (d x + c\right ) + 2 \, {\left (a^{4} - a^{2} b^{2}\right )} d x - {\left (2 \, a^{3} - a b^{2} + {\left (2 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (2 \, a^{3} b - 2 \, a b^{3} + {\left (a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{{\left (a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right ) + {\left (a^{3} b^{3} - a b^{5}\right )} d}\right ] \]
[1/2*(4*(a^3*b - a*b^3)*d*x*cos(d*x + c) + 4*(a^4 - a^2*b^2)*d*x + (2*a^3 - a*b^2 + (2*a^2*b - 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^3*b - 2*a*b^3 + (a^2*b^2 - b^4)*cos(d*x + c))*sin(d*x + c))/((a^2*b^4 - b^6)*d*cos(d*x + c) + (a^3*b^3 - a*b^5)*d), (2*(a^3*b - a* b^3)*d*x*cos(d*x + c) + 2*(a^4 - a^2*b^2)*d*x - (2*a^3 - a*b^2 + (2*a^2*b - 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*a^3*b - 2*a*b^3 + (a^2*b^2 - b^4)*cos(d*x + c ))*sin(d*x + c))/((a^2*b^4 - b^6)*d*cos(d*x + c) + (a^3*b^3 - a*b^5)*d)]
Timed out. \[ \int \frac {\cos (c+d x) \left (1-\cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\cos (c+d x) \left (1-\cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\text {Exception raised: ValueError} \]
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. 422 vs. \(2 (103) = 206\).
Time = 0.32 (sec) , antiderivative size = 422, normalized size of antiderivative = 3.77 \[ \int \frac {\cos (c+d x) \left (1-\cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=-\frac {\frac {{\left (\sqrt {a^{2} - b^{2}} {\left (2 \, a - b\right )} {\left | -a + b \right |} {\left | b \right |} + {\left (4 \, a^{2} - 2 \, a b - b^{2}\right )} \sqrt {a^{2} - b^{2}} {\left | -a + b \right |}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {\frac {a b^{2} + \sqrt {a^{2} b^{4} - {\left (a b^{2} + b^{3}\right )} {\left (a b^{2} - b^{3}\right )}}}{a b^{2} - b^{3}}}}\right )\right )}}{{\left (a^{2} b^{2} - 2 \, a b^{3} + b^{4}\right )} b^{2} + {\left (a^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4}\right )} {\left | b \right |}} + \frac {{\left (4 \, a^{2} - 2 \, a b - b^{2} - 2 \, a {\left | b \right |} + b {\left | b \right |}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {\frac {a b^{2} - \sqrt {a^{2} b^{4} - {\left (a b^{2} + b^{3}\right )} {\left (a b^{2} - b^{3}\right )}}}{a b^{2} - b^{3}}}}\right )\right )}}{b^{4} - a b^{2} {\left | b \right |}} + \frac {2 \, {\left (2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )} b^{2}}}{d} \]
-((sqrt(a^2 - b^2)*(2*a - b)*abs(-a + b)*abs(b) + (4*a^2 - 2*a*b - b^2)*sq rt(a^2 - b^2)*abs(-a + b))*(pi*floor(1/2*(d*x + c)/pi + 1/2) + arctan(tan( 1/2*d*x + 1/2*c)/sqrt((a*b^2 + sqrt(a^2*b^4 - (a*b^2 + b^3)*(a*b^2 - b^3)) )/(a*b^2 - b^3))))/((a^2*b^2 - 2*a*b^3 + b^4)*b^2 + (a^3*b^2 - 2*a^2*b^3 + a*b^4)*abs(b)) + (4*a^2 - 2*a*b - b^2 - 2*a*abs(b) + b*abs(b))*(pi*floor( 1/2*(d*x + c)/pi + 1/2) + arctan(tan(1/2*d*x + 1/2*c)/sqrt((a*b^2 - sqrt(a ^2*b^4 - (a*b^2 + b^3)*(a*b^2 - b^3)))/(a*b^2 - b^3))))/(b^4 - a*b^2*abs(b )) + 2*(2*a*tan(1/2*d*x + 1/2*c)^3 - b*tan(1/2*d*x + 1/2*c)^3 + 2*a*tan(1/ 2*d*x + 1/2*c) + b*tan(1/2*d*x + 1/2*c))/((a*tan(1/2*d*x + 1/2*c)^4 - b*ta n(1/2*d*x + 1/2*c)^4 + 2*a*tan(1/2*d*x + 1/2*c)^2 + a + b)*b^2))/d
Time = 2.00 (sec) , antiderivative size = 314, normalized size of antiderivative = 2.80 \[ \int \frac {\cos (c+d x) \left (1-\cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\frac {4\,a\,\mathrm {atan}\left (\frac {128\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{128\,a-\frac {128\,a^2}{b}}-\frac {128\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{128\,a\,b-128\,a^2}\right )}{b^3\,d}-\frac {\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,a-b\right )}{b^2}+\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a+b\right )}{b^2}}{d\,\left (\left (a-b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a+b\right )}-\frac {\ln \left (b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sqrt {b^2-a^2}\right )\,\sqrt {-\left (a+b\right )\,\left (a-b\right )}\,\left (2\,a^2-b^2\right )}{d\,\left (b^5-a^2\,b^3\right )}-\frac {\ln \left (a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sqrt {b^2-a^2}\right )\,\left (2\,a^2\,\sqrt {b^2-a^2}-b^2\,\sqrt {b^2-a^2}\right )}{b^3\,d\,\left (a^2-b^2\right )} \]
(4*a*atan((128*a*tan(c/2 + (d*x)/2))/(128*a - (128*a^2)/b) - (128*a^2*tan( c/2 + (d*x)/2))/(128*a*b - 128*a^2)))/(b^3*d) - ((2*tan(c/2 + (d*x)/2)^3*( 2*a - b))/b^2 + (2*tan(c/2 + (d*x)/2)*(2*a + b))/b^2)/(d*(a + b + tan(c/2 + (d*x)/2)^4*(a - b) + 2*a*tan(c/2 + (d*x)/2)^2)) - (log(b*tan(c/2 + (d*x) /2) - a*tan(c/2 + (d*x)/2) + (b^2 - a^2)^(1/2))*(-(a + b)*(a - b))^(1/2)*( 2*a^2 - b^2))/(d*(b^5 - a^2*b^3)) - (log(a*tan(c/2 + (d*x)/2) - b*tan(c/2 + (d*x)/2) + (b^2 - a^2)^(1/2))*(2*a^2*(b^2 - a^2)^(1/2) - b^2*(b^2 - a^2) ^(1/2)))/(b^3*d*(a^2 - b^2))