Integrand size = 33, antiderivative size = 150 \[ \int \frac {\cos ^2(c+d x) \left (1-\cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\frac {a \left (2 a^2-b^2\right ) x}{2 b^4}-\frac {2 a^2 \sqrt {a-b} \sqrt {a+b} \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^4 d}-\frac {\left (3 a^2-b^2\right ) \sin (c+d x)}{3 b^3 d}+\frac {a \cos (c+d x) \sin (c+d x)}{2 b^2 d}-\frac {\cos ^2(c+d x) \sin (c+d x)}{3 b d} \]
1/2*a*(2*a^2-b^2)*x/b^4-1/3*(3*a^2-b^2)*sin(d*x+c)/b^3/d+1/2*a*cos(d*x+c)* sin(d*x+c)/b^2/d-1/3*cos(d*x+c)^2*sin(d*x+c)/b/d-2*a^2*arctan((a-b)^(1/2)* tan(1/2*d*x+1/2*c)/(a+b)^(1/2))*(a-b)^(1/2)*(a+b)^(1/2)/b^4/d
Time = 0.74 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.83 \[ \int \frac {\cos ^2(c+d x) \left (1-\cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=-\frac {-6 a \left (2 a^2-b^2\right ) (c+d x)+24 a^2 \sqrt {-a^2+b^2} \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )+3 (2 a-b) b (2 a+b) \sin (c+d x)-3 a b^2 \sin (2 (c+d x))+b^3 \sin (3 (c+d x))}{12 b^4 d} \]
-1/12*(-6*a*(2*a^2 - b^2)*(c + d*x) + 24*a^2*Sqrt[-a^2 + b^2]*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]] + 3*(2*a - b)*b*(2*a + b)*Sin[c + d*x] - 3*a*b^2*Sin[2*(c + d*x)] + b^3*Sin[3*(c + d*x)])/(b^4*d)
Time = 0.90 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.17, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.424, Rules used = {3042, 3529, 25, 3042, 3528, 25, 3042, 3502, 27, 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.
| \(\displaystyle \int \frac {\cos ^2(c+d x) \left (1-\cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (1-\sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 3529 |
| \(\displaystyle \frac {\int -\frac {\cos (c+d x) \left (-3 a \cos ^2(c+d x)-b \cos (c+d x)+2 a\right )}{a+b \cos (c+d x)}dx}{3 b}-\frac {\sin (c+d x) \cos ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\int \frac {\cos (c+d x) \left (-3 a \cos ^2(c+d x)-b \cos (c+d x)+2 a\right )}{a+b \cos (c+d x)}dx}{3 b}-\frac {\sin (c+d x) \cos ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \left (-3 a \sin \left (c+d x+\frac {\pi }{2}\right )^2-b \sin \left (c+d x+\frac {\pi }{2}\right )+2 a\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{3 b}-\frac {\sin (c+d x) \cos ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 3528 |
| \(\displaystyle -\frac {\frac {\int -\frac {3 a^2-b \cos (c+d x) a-2 \left (3 a^2-b^2\right ) \cos ^2(c+d x)}{a+b \cos (c+d x)}dx}{2 b}-\frac {3 a \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}-\frac {\sin (c+d x) \cos ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {-\frac {\int \frac {3 a^2-b \cos (c+d x) a-2 \left (3 a^2-b^2\right ) \cos ^2(c+d x)}{a+b \cos (c+d x)}dx}{2 b}-\frac {3 a \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}-\frac {\sin (c+d x) \cos ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {\int \frac {3 a^2-b \sin \left (c+d x+\frac {\pi }{2}\right ) a-2 \left (3 a^2-b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 b}-\frac {3 a \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}-\frac {\sin (c+d x) \cos ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle -\frac {-\frac {\frac {\int \frac {3 \left (b a^2+\left (2 a^2-b^2\right ) \cos (c+d x) a\right )}{a+b \cos (c+d x)}dx}{b}-\frac {2 \left (3 a^2-b^2\right ) \sin (c+d x)}{b d}}{2 b}-\frac {3 a \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}-\frac {\sin (c+d x) \cos ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {-\frac {\frac {3 \int \frac {b a^2+\left (2 a^2-b^2\right ) \cos (c+d x) a}{a+b \cos (c+d x)}dx}{b}-\frac {2 \left (3 a^2-b^2\right ) \sin (c+d x)}{b d}}{2 b}-\frac {3 a \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}-\frac {\sin (c+d x) \cos ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {\frac {3 \int \frac {b a^2+\left (2 a^2-b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {2 \left (3 a^2-b^2\right ) \sin (c+d x)}{b d}}{2 b}-\frac {3 a \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}-\frac {\sin (c+d x) \cos ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle -\frac {-\frac {\frac {3 \left (\frac {a x \left (2 a^2-b^2\right )}{b}-\frac {2 a^2 \left (a^2-b^2\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{b}\right )}{b}-\frac {2 \left (3 a^2-b^2\right ) \sin (c+d x)}{b d}}{2 b}-\frac {3 a \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}-\frac {\sin (c+d x) \cos ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {\frac {3 \left (\frac {a x \left (2 a^2-b^2\right )}{b}-\frac {2 a^2 \left (a^2-b^2\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}\right )}{b}-\frac {2 \left (3 a^2-b^2\right ) \sin (c+d x)}{b d}}{2 b}-\frac {3 a \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}-\frac {\sin (c+d x) \cos ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle -\frac {-\frac {\frac {3 \left (\frac {a x \left (2 a^2-b^2\right )}{b}-\frac {4 a^2 \left (a^2-b^2\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{b d}\right )}{b}-\frac {2 \left (3 a^2-b^2\right ) \sin (c+d x)}{b d}}{2 b}-\frac {3 a \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}-\frac {\sin (c+d x) \cos ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle -\frac {-\frac {\frac {3 \left (\frac {a x \left (2 a^2-b^2\right )}{b}-\frac {4 a^2 \left (a^2-b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b d \sqrt {a-b} \sqrt {a+b}}\right )}{b}-\frac {2 \left (3 a^2-b^2\right ) \sin (c+d x)}{b d}}{2 b}-\frac {3 a \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}-\frac {\sin (c+d x) \cos ^2(c+d x)}{3 b d}\) |
-1/3*(Cos[c + d*x]^2*Sin[c + d*x])/(b*d) - ((-3*a*Cos[c + d*x]*Sin[c + d*x ])/(2*b*d) - ((3*((a*(2*a^2 - b^2)*x)/b - (4*a^2*(a^2 - b^2)*ArcTan[(Sqrt[ a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*b*Sqrt[a + b]*d)))/b - (2*(3*a^2 - b^2)*Sin[c + d*x])/(b*d))/(2*b))/(3*b)
3.6.94.3.1 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_)])/((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_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_ .) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x ])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Simp[1/(d*(m + n + 2)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A* d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2) - C*(a *c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n} , x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[ m, 0] && !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] : > Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Simp[1/(d*(m + n + 2)) Int[(a + b*Sin[e + f*x ])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*( n + 1)) + (A*b*d*(m + n + 2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + C* (a*d*m - b*c*(m + 1))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f , A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
Time = 1.63 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.19
| method | result | size |
| derivativedivides | \(\frac {-\frac {2 a^{2} \left (a -b \right ) \left (a +b \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{4} \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {\frac {2 \left (\left (-a^{2} b -\frac {1}{2} a \,b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-2 a^{2} b +\frac {4}{3} b^{3}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-a^{2} b +\frac {1}{2} a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+a \left (2 a^{2}-b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4}}}{d}\) | \(179\) |
| default | \(\frac {-\frac {2 a^{2} \left (a -b \right ) \left (a +b \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{4} \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {\frac {2 \left (\left (-a^{2} b -\frac {1}{2} a \,b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-2 a^{2} b +\frac {4}{3} b^{3}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-a^{2} b +\frac {1}{2} a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+a \left (2 a^{2}-b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4}}}{d}\) | \(179\) |
| risch | \(\frac {a^{3} x}{b^{4}}-\frac {a x}{2 b^{2}}+\frac {i {\mathrm e}^{i \left (d x +c \right )} a^{2}}{2 b^{3} d}-\frac {i {\mathrm e}^{i \left (d x +c \right )}}{8 b d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} a^{2}}{2 b^{3} d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{8 b d}+\frac {\sqrt {-a^{2}+b^{2}}\, a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}+a}{b}\right )}{d \,b^{4}}-\frac {\sqrt {-a^{2}+b^{2}}\, a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i \sqrt {-a^{2}+b^{2}}-a}{b}\right )}{d \,b^{4}}-\frac {\sin \left (3 d x +3 c \right )}{12 d b}+\frac {a \sin \left (2 d x +2 c \right )}{4 b^{2} d}\) | \(238\) |
1/d*(-2*a^2*(a-b)*(a+b)/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*(((-a^2*b-1/2*a*b^2)*tan(1/2*d*x+1/2*c)^5 +(-2*a^2*b+4/3*b^3)*tan(1/2*d*x+1/2*c)^3+(-a^2*b+1/2*a*b^2)*tan(1/2*d*x+1/ 2*c))/(1+tan(1/2*d*x+1/2*c)^2)^3+1/2*a*(2*a^2-b^2)*arctan(tan(1/2*d*x+1/2* c))))
Time = 0.31 (sec) , antiderivative size = 304, normalized size of antiderivative = 2.03 \[ \int \frac {\cos ^2(c+d x) \left (1-\cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\left [\frac {3 \, \sqrt {-a^{2} + b^{2}} a^{2} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) + 3 \, {\left (2 \, a^{3} - a b^{2}\right )} d x - {\left (2 \, b^{3} \cos \left (d x + c\right )^{2} - 3 \, a b^{2} \cos \left (d x + c\right ) + 6 \, a^{2} b - 2 \, b^{3}\right )} \sin \left (d x + c\right )}{6 \, b^{4} d}, -\frac {6 \, \sqrt {a^{2} - b^{2}} a^{2} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - 3 \, {\left (2 \, a^{3} - a b^{2}\right )} d x + {\left (2 \, b^{3} \cos \left (d x + c\right )^{2} - 3 \, a b^{2} \cos \left (d x + c\right ) + 6 \, a^{2} b - 2 \, b^{3}\right )} \sin \left (d x + c\right )}{6 \, b^{4} d}\right ] \]
[1/6*(3*sqrt(-a^2 + b^2)*a^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)) + 3*(2*a^3 - a*b^2) *d*x - (2*b^3*cos(d*x + c)^2 - 3*a*b^2*cos(d*x + c) + 6*a^2*b - 2*b^3)*sin (d*x + c))/(b^4*d), -1/6*(6*sqrt(a^2 - b^2)*a^2*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) - 3*(2*a^3 - a*b^2)*d*x + (2*b^3*cos(d* x + c)^2 - 3*a*b^2*cos(d*x + c) + 6*a^2*b - 2*b^3)*sin(d*x + c))/(b^4*d)]
Timed out. \[ \int \frac {\cos ^2(c+d x) \left (1-\cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\cos ^2(c+d x) \left (1-\cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\text {Exception raised: ValueError} \]
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
Time = 0.30 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.53 \[ \int \frac {\cos ^2(c+d x) \left (1-\cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\frac {\frac {3 \, {\left (2 \, a^{3} - a b^{2}\right )} {\left (d x + c\right )}}{b^{4}} + \frac {12 \, {\left (a^{4} - a^{2} b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} b^{4}} - \frac {2 \, {\left (6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} b^{3}}}{6 \, d} \]
1/6*(3*(2*a^3 - a*b^2)*(d*x + c)/b^4 + 12*(a^4 - a^2*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*ta n(1/2*d*x + 1/2*c))/sqrt(a^2 - b^2)))/(sqrt(a^2 - b^2)*b^4) - 2*(6*a^2*tan (1/2*d*x + 1/2*c)^5 + 3*a*b*tan(1/2*d*x + 1/2*c)^5 + 12*a^2*tan(1/2*d*x + 1/2*c)^3 - 8*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*a^2*tan(1/2*d*x + 1/2*c) - 3*a *b*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*b^3))/d
Time = 2.25 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.22 \[ \int \frac {\cos ^2(c+d x) \left (1-\cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\frac {\frac {\sin \left (c+d\,x\right )}{4}-\frac {\sin \left (3\,c+3\,d\,x\right )}{12}}{b\,d}-\frac {a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-\frac {a\,\sin \left (2\,c+2\,d\,x\right )}{4}}{b^2\,d}-\frac {a^2\,\sin \left (c+d\,x\right )}{b^3\,d}+\frac {2\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b^4\,d}+\frac {2\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a+b\right )}\right )\,\sqrt {b^2-a^2}}{b^4\,d} \]
(sin(c + d*x)/4 - sin(3*c + 3*d*x)/12)/(b*d) - (a*atan(sin(c/2 + (d*x)/2)/ cos(c/2 + (d*x)/2)) - (a*sin(2*c + 2*d*x))/4)/(b^2*d) - (a^2*sin(c + d*x)) /(b^3*d) + (2*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(b^4*d) + ( 2*a^2*atanh((sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2))/(cos(c/2 + (d*x)/2)*(a + b)))*(b^2 - a^2)^(1/2))/(b^4*d)