Integrand size = 31, antiderivative size = 109 \[ \int \frac {\cos (c+d x) \left (1-\cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=-\frac {\left (2 a^2-b^2\right ) x}{2 b^3}+\frac {2 a \sqrt {a-b} \sqrt {a+b} \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^3 d}+\frac {a \sin (c+d x)}{b^2 d}-\frac {\cos (c+d x) \sin (c+d x)}{2 b d} \]
-1/2*(2*a^2-b^2)*x/b^3+a*sin(d*x+c)/b^2/d-1/2*cos(d*x+c)*sin(d*x+c)/b/d+2* a*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^3/d
Time = 0.32 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.90 \[ \int \frac {\cos (c+d x) \left (1-\cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\frac {-2 \left (2 a^2-b^2\right ) (c+d x)+8 a \sqrt {-a^2+b^2} \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )+4 a b \sin (c+d x)-b^2 \sin (2 (c+d x))}{4 b^3 d} \]
(-2*(2*a^2 - b^2)*(c + d*x) + 8*a*Sqrt[-a^2 + b^2]*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]] + 4*a*b*Sin[c + d*x] - b^2*Sin[2*(c + d*x)]) /(4*b^3*d)
Time = 0.61 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.18, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.323, Rules used = {3042, 3529, 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.
| \(\displaystyle \int \frac {\cos (c+d x) \left (1-\cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \left (1-\sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 3529 |
| \(\displaystyle \frac {\int -\frac {-2 a \cos ^2(c+d x)-b \cos (c+d x)+a}{a+b \cos (c+d x)}dx}{2 b}-\frac {\sin (c+d x) \cos (c+d x)}{2 b d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\int \frac {-2 a \cos ^2(c+d x)-b \cos (c+d x)+a}{a+b \cos (c+d x)}dx}{2 b}-\frac {\sin (c+d x) \cos (c+d x)}{2 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {-2 a \sin \left (c+d x+\frac {\pi }{2}\right )^2-b \sin \left (c+d x+\frac {\pi }{2}\right )+a}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 b}-\frac {\sin (c+d x) \cos (c+d x)}{2 b d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle -\frac {\frac {\int \frac {a b+\left (2 a^2-b^2\right ) \cos (c+d x)}{a+b \cos (c+d x)}dx}{b}-\frac {2 a \sin (c+d x)}{b d}}{2 b}-\frac {\sin (c+d x) \cos (c+d x)}{2 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {\int \frac {a b+\left (2 a^2-b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {2 a \sin (c+d x)}{b d}}{2 b}-\frac {\sin (c+d x) \cos (c+d x)}{2 b d}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle -\frac {\frac {\frac {x \left (2 a^2-b^2\right )}{b}-\frac {2 a \left (a^2-b^2\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{b}}{b}-\frac {2 a \sin (c+d x)}{b d}}{2 b}-\frac {\sin (c+d x) \cos (c+d x)}{2 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {\frac {x \left (2 a^2-b^2\right )}{b}-\frac {2 a \left (a^2-b^2\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}}{b}-\frac {2 a \sin (c+d x)}{b d}}{2 b}-\frac {\sin (c+d x) \cos (c+d x)}{2 b d}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle -\frac {\frac {\frac {x \left (2 a^2-b^2\right )}{b}-\frac {4 a \left (a^2-b^2\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{b d}}{b}-\frac {2 a \sin (c+d x)}{b d}}{2 b}-\frac {\sin (c+d x) \cos (c+d x)}{2 b d}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle -\frac {\frac {\frac {x \left (2 a^2-b^2\right )}{b}-\frac {4 a \left (a^2-b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b d \sqrt {a-b} \sqrt {a+b}}}{b}-\frac {2 a \sin (c+d x)}{b d}}{2 b}-\frac {\sin (c+d x) \cos (c+d x)}{2 b d}\) |
-1/2*(Cos[c + d*x]*Sin[c + d*x])/(b*d) - ((((2*a^2 - b^2)*x)/b - (4*a*(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*a*Sin[c + d*x])/(b*d))/(2*b)
3.6.95.3.1 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_)])^(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 = 1.44 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.34
| method | result | size |
| derivativedivides | \(\frac {-\frac {2 \left (\frac {\left (-a b -\frac {1}{2} b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-a b +\frac {1}{2} b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {\left (2 a^{2}-b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{b^{3}}+\frac {2 \left (a -b \right ) a \left (a +b \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(146\) |
| default | \(\frac {-\frac {2 \left (\frac {\left (-a b -\frac {1}{2} b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-a b +\frac {1}{2} b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {\left (2 a^{2}-b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{b^{3}}+\frac {2 \left (a -b \right ) a \left (a +b \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(146\) |
| risch | \(-\frac {x \,a^{2}}{b^{3}}+\frac {x}{2 b}-\frac {i a \,{\mathrm e}^{i \left (d x +c \right )}}{2 b^{2} d}+\frac {i a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 b^{2} d}+\frac {\sqrt {-a^{2}+b^{2}}\, a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i \sqrt {-a^{2}+b^{2}}-a}{b}\right )}{d \,b^{3}}-\frac {\sqrt {-a^{2}+b^{2}}\, a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}+a}{b}\right )}{d \,b^{3}}-\frac {\sin \left (2 d x +2 c \right )}{4 d b}\) | \(176\) |
1/d*(-2/b^3*(((-a*b-1/2*b^2)*tan(1/2*d*x+1/2*c)^3+(-a*b+1/2*b^2)*tan(1/2*d *x+1/2*c))/(1+tan(1/2*d*x+1/2*c)^2)^2+1/2*(2*a^2-b^2)*arctan(tan(1/2*d*x+1 /2*c)))+2*(a-b)*a*(a+b)/b^3/((a-b)*(a+b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1 /2*c)/((a-b)*(a+b))^(1/2)))
Time = 0.29 (sec) , antiderivative size = 251, normalized size of antiderivative = 2.30 \[ \int \frac {\cos (c+d x) \left (1-\cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\left [-\frac {{\left (2 \, a^{2} - b^{2}\right )} d x - \sqrt {-a^{2} + b^{2}} a \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) + {\left (b^{2} \cos \left (d x + c\right ) - 2 \, a b\right )} \sin \left (d x + c\right )}{2 \, b^{3} d}, -\frac {{\left (2 \, a^{2} - b^{2}\right )} d x - 2 \, \sqrt {a^{2} - b^{2}} a \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) + {\left (b^{2} \cos \left (d x + c\right ) - 2 \, a b\right )} \sin \left (d x + c\right )}{2 \, b^{3} d}\right ] \]
[-1/2*((2*a^2 - b^2)*d*x - sqrt(-a^2 + b^2)*a*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)) + (b^2*cos(d*x + c) - 2*a*b)*sin(d*x + c))/(b^3*d), -1/2*((2*a^2 - b^2)*d*x - 2*sqrt(a^2 - b^2)*a*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d* x + c))) + (b^2*cos(d*x + c) - 2*a*b)*sin(d*x + c))/(b^3*d)]
Leaf count of result is larger than twice the leaf count of optimal. 2958 vs. \(2 (92) = 184\).
Time = 159.03 (sec) , antiderivative size = 2958, normalized size of antiderivative = 27.14 \[ \int \frac {\cos (c+d x) \left (1-\cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\text {Too large to display} \]
Piecewise((zoo*x*(1 - cos(c)**2), Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (-x*sin (c + d*x)**2/(2*b) - x*cos(c + d*x)**2/(2*b) - sin(c + d*x)*cos(c + d*x)/( 2*b*d) + sin(c + d*x)/(b*d), Eq(a, b)), (-x*sin(c + d*x)**2/(2*b) - x*cos( c + d*x)**2/(2*b) - sin(c + d*x)*cos(c + d*x)/(2*b*d) - sin(c + d*x)/(b*d) , Eq(a, -b)), ((-2*sin(c + d*x)**3/(3*d) - sin(c + d*x)*cos(c + d*x)**2/d + sin(c + d*x)/d)/a, Eq(b, 0)), (x*(1 - cos(c)**2)*cos(c)/(a + b*cos(c)), Eq(d, 0)), (-2*a**2*d*x*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**4/( 2*b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**4 + 4*b**3*d*sqrt( -a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + 2*b**3*d*sqrt(-a/(a - b) - b /(a - b))) - 4*a**2*d*x*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2/( 2*b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**4 + 4*b**3*d*sqrt( -a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + 2*b**3*d*sqrt(-a/(a - b) - b /(a - b))) - 2*a**2*d*x*sqrt(-a/(a - b) - b/(a - b))/(2*b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**4 + 4*b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + 2*b**3*d*sqrt(-a/(a - b) - b/(a - b))) + 2*a**2* log(-sqrt(-a/(a - b) - b/(a - b)) + tan(c/2 + d*x/2))*tan(c/2 + d*x/2)**4/ (2*b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**4 + 4*b**3*d*sqrt (-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + 2*b**3*d*sqrt(-a/(a - b) - b/(a - b))) + 4*a**2*log(-sqrt(-a/(a - b) - b/(a - b)) + tan(c/2 + d*x/2)) *tan(c/2 + d*x/2)**2/(2*b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d...
Exception generated. \[ \int \frac {\cos (c+d x) \left (1-\cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\text {Exception raised: ValueError} \]
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 = 0.31 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.71 \[ \int \frac {\cos (c+d x) \left (1-\cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=-\frac {\frac {{\left (2 \, a^{2} - b^{2}\right )} {\left (d x + c\right )}}{b^{3}} + \frac {4 \, {\left (a^{3} - a b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} b^{3}} - \frac {2 \, {\left (2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} b^{2}}}{2 \, d} \]
-1/2*((2*a^2 - b^2)*(d*x + c)/b^3 + 4*(a^3 - 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)))/(sqrt(a^2 - b^2)*b^3) - 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))/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*b^2))/d
Time = 1.89 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.35 \[ \int \frac {\cos (c+d x) \left (1-\cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\frac {\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-\frac {\sin \left (2\,c+2\,d\,x\right )}{4}}{b\,d}-\frac {2\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b^3\,d}+\frac {a\,\sin \left (c+d\,x\right )}{b^2\,d}-\frac {2\,a\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a+b\right )}\right )\,\sqrt {b^2-a^2}}{b^3\,d} \]