Integrand size = 29, antiderivative size = 89 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{a+b \cos (c+d x)} \, dx=\frac {(A b-a B) x}{b^2}-\frac {2 a (A b-a B) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^2 \sqrt {a+b} d}+\frac {B \sin (c+d x)}{b d} \] Output:
(A*b-B*a)*x/b^2-2*a*(A*b-B*a)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^ (1/2))/(a-b)^(1/2)/b^2/(a+b)^(1/2)/d+B*sin(d*x+c)/b/d
Time = 0.87 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.96 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{a+b \cos (c+d x)} \, dx=\frac {(A b-a B) (c+d x)-\frac {2 a (-A b+a B) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}+b B \sin (c+d x)}{b^2 d} \] Input:
Integrate[(Cos[c + d*x]*(A + B*Cos[c + d*x]))/(a + b*Cos[c + d*x]),x]
Output:
((A*b - a*B)*(c + d*x) - (2*a*(-(A*b) + a*B)*ArcTanh[((a - b)*Tan[(c + d*x )/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] + b*B*Sin[c + d*x])/(b^2*d)
Time = 0.50 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.97, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.345, Rules used = {3042, 3447, 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.
\(\displaystyle \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{a+b \cos (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 3447 |
\(\displaystyle \int \frac {A \cos (c+d x)+B \cos ^2(c+d x)}{a+b \cos (c+d x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A \sin \left (c+d x+\frac {\pi }{2}\right )+B \sin \left (c+d x+\frac {\pi }{2}\right )^2}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 3502 |
\(\displaystyle \frac {\int \frac {(A b-a B) \cos (c+d x)}{a+b \cos (c+d x)}dx}{b}+\frac {B \sin (c+d x)}{b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(A b-a B) \int \frac {\cos (c+d x)}{a+b \cos (c+d x)}dx}{b}+\frac {B \sin (c+d x)}{b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(A b-a B) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}+\frac {B \sin (c+d x)}{b d}\) |
\(\Big \downarrow \) 3214 |
\(\displaystyle \frac {(A b-a B) \left (\frac {x}{b}-\frac {a \int \frac {1}{a+b \cos (c+d x)}dx}{b}\right )}{b}+\frac {B \sin (c+d x)}{b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(A b-a B) \left (\frac {x}{b}-\frac {a \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}\right )}{b}+\frac {B \sin (c+d x)}{b d}\) |
\(\Big \downarrow \) 3138 |
\(\displaystyle \frac {(A b-a B) \left (\frac {x}{b}-\frac {2 a \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{b d}\right )}{b}+\frac {B \sin (c+d x)}{b d}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {(A b-a B) \left (\frac {x}{b}-\frac {2 a \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b d \sqrt {a-b} \sqrt {a+b}}\right )}{b}+\frac {B \sin (c+d x)}{b d}\) |
Input:
Int[(Cos[c + d*x]*(A + B*Cos[c + d*x]))/(a + b*Cos[c + d*x]),x]
Output:
((A*b - a*B)*(x/b - (2*a*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b] ])/(Sqrt[a - b]*b*Sqrt[a + b]*d)))/b + (B*Sin[c + d*x])/(b*d)
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_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, 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]
Time = 1.38 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.24
method | result | size |
derivativedivides | \(\frac {\frac {\frac {2 B b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+2 \left (A b -B a \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2}}-\frac {2 a \left (A b -B a \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(110\) |
default | \(\frac {\frac {\frac {2 B b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+2 \left (A b -B a \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2}}-\frac {2 a \left (A b -B a \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(110\) |
risch | \(\frac {x A}{b}-\frac {a B x}{b^{2}}-\frac {i {\mathrm e}^{i \left (d x +c \right )} B}{2 d b}+\frac {i B \,{\mathrm e}^{-i \left (d x +c \right )}}{2 b d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, d b}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) B}{\sqrt {-a^{2}+b^{2}}\, d \,b^{2}}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, d b}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) B}{\sqrt {-a^{2}+b^{2}}\, d \,b^{2}}\) | \(352\) |
Input:
int(cos(d*x+c)*(A+B*cos(d*x+c))/(a+cos(d*x+c)*b),x,method=_RETURNVERBOSE)
Output:
1/d*(2/b^2*(B*b*tan(1/2*d*x+1/2*c)/(1+tan(1/2*d*x+1/2*c)^2)+(A*b-B*a)*arct an(tan(1/2*d*x+1/2*c)))-2*a*(A*b-B*a)/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 = 0.11 (sec) , antiderivative size = 322, normalized size of antiderivative = 3.62 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{a+b \cos (c+d x)} \, dx=\left [-\frac {2 \, {\left (B a^{3} - A a^{2} b - B a b^{2} + A b^{3}\right )} d x - {\left (B a^{2} - A a b\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - 2 \, {\left (B a^{2} b - B b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{2} b^{2} - b^{4}\right )} d}, -\frac {{\left (B a^{3} - A a^{2} b - B a b^{2} + A b^{3}\right )} d x - {\left (B a^{2} - A a b\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (B a^{2} b - B b^{3}\right )} \sin \left (d x + c\right )}{{\left (a^{2} b^{2} - b^{4}\right )} d}\right ] \] Input:
integrate(cos(d*x+c)*(A+B*cos(d*x+c))/(a+b*cos(d*x+c)),x, algorithm="frica s")
Output:
[-1/2*(2*(B*a^3 - A*a^2*b - B*a*b^2 + A*b^3)*d*x - (B*a^2 - A*a*b)*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*(B*a^2*b - B*b^3)*sin(d*x + c))/(( a^2*b^2 - b^4)*d), -((B*a^3 - A*a^2*b - B*a*b^2 + A*b^3)*d*x - (B*a^2 - A* a*b)*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) - (B*a^2*b - B*b^3)*sin(d*x + c))/((a^2*b^2 - b^4)*d)]
Leaf count of result is larger than twice the leaf count of optimal. 3225 vs. \(2 (76) = 152\).
Time = 56.42 (sec) , antiderivative size = 3225, normalized size of antiderivative = 36.24 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{a+b \cos (c+d x)} \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)*(A+B*cos(d*x+c))/(a+b*cos(d*x+c)),x)
Output:
Piecewise((zoo*x*(A + B*cos(c)), Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (A*d*x*t an(c/2 + d*x/2)**2/(b*d*tan(c/2 + d*x/2)**2 + b*d) + A*d*x/(b*d*tan(c/2 + d*x/2)**2 + b*d) - A*tan(c/2 + d*x/2)**3/(b*d*tan(c/2 + d*x/2)**2 + b*d) - A*tan(c/2 + d*x/2)/(b*d*tan(c/2 + d*x/2)**2 + b*d) - B*d*x*tan(c/2 + d*x/ 2)**2/(b*d*tan(c/2 + d*x/2)**2 + b*d) - B*d*x/(b*d*tan(c/2 + d*x/2)**2 + b *d) + B*tan(c/2 + d*x/2)**3/(b*d*tan(c/2 + d*x/2)**2 + b*d) + 3*B*tan(c/2 + d*x/2)/(b*d*tan(c/2 + d*x/2)**2 + b*d), Eq(a, b)), (A*d*x*tan(c/2 + d*x/ 2)**3/(b*d*tan(c/2 + d*x/2)**3 + b*d*tan(c/2 + d*x/2)) + A*d*x*tan(c/2 + d *x/2)/(b*d*tan(c/2 + d*x/2)**3 + b*d*tan(c/2 + d*x/2)) + A*tan(c/2 + d*x/2 )**2/(b*d*tan(c/2 + d*x/2)**3 + b*d*tan(c/2 + d*x/2)) + A/(b*d*tan(c/2 + d *x/2)**3 + b*d*tan(c/2 + d*x/2)) + B*d*x*tan(c/2 + d*x/2)**3/(b*d*tan(c/2 + d*x/2)**3 + b*d*tan(c/2 + d*x/2)) + B*d*x*tan(c/2 + d*x/2)/(b*d*tan(c/2 + d*x/2)**3 + b*d*tan(c/2 + d*x/2)) + 3*B*tan(c/2 + d*x/2)**2/(b*d*tan(c/2 + d*x/2)**3 + b*d*tan(c/2 + d*x/2)) + B/(b*d*tan(c/2 + d*x/2)**3 + b*d*ta n(c/2 + d*x/2)), Eq(a, -b)), ((A*sin(c + d*x)/d + 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*d))/a, Eq(b, 0)), ( x*(A + B*cos(c))*cos(c)/(a + b*cos(c)), Eq(d, 0)), (A*a*b*d*x*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2/(a*b**2*d*sqrt(-a/(a - b) - b/(a - b) )*tan(c/2 + d*x/2)**2 + a*b**2*d*sqrt(-a/(a - b) - b/(a - b)) - b**3*d*sqr t(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 - b**3*d*sqrt(-a/(a - b) ...
Exception generated. \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{a+b \cos (c+d x)} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(cos(d*x+c)*(A+B*cos(d*x+c))/(a+b*cos(d*x+c)),x, algorithm="maxim a")
Output:
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 = 0.14 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.60 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{a+b \cos (c+d x)} \, dx=-\frac {\frac {{\left (B a - A b\right )} {\left (d x + c\right )}}{b^{2}} - \frac {2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} b} + \frac {2 \, {\left (B a^{2} - A a b\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} b^{2}}}{d} \] Input:
integrate(cos(d*x+c)*(A+B*cos(d*x+c))/(a+b*cos(d*x+c)),x, algorithm="giac" )
Output:
-((B*a - A*b)*(d*x + c)/b^2 - 2*B*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2 *c)^2 + 1)*b) + 2*(B*a^2 - A*a*b)*(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^2))/d
Time = 42.30 (sec) , antiderivative size = 541, normalized size of antiderivative = 6.08 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{a+b \cos (c+d x)} \, dx=\frac {2\,B\,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 )}{d\,\left (a^2-b^2\right )}-\frac {2\,A\,b\,\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 )}{d\,\left (a^2-b^2\right )}-\frac {B\,b\,\sin \left (c+d\,x\right )}{d\,\left (a^2-b^2\right )}+\frac {2\,A\,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\,d\,\left (a^2-b^2\right )}-\frac {2\,B\,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^2\,d\,\left (a^2-b^2\right )}+\frac {A\,a\,\ln \left (\frac {a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b\,d\,\sqrt {b^2-a^2}}-\frac {A\,a\,\ln \left (\frac {b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b\,d\,\sqrt {b^2-a^2}}-\frac {B\,a^2\,\ln \left (\frac {a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b^2\,d\,\sqrt {b^2-a^2}}+\frac {B\,a^2\,\ln \left (\frac {b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b^2\,d\,\sqrt {b^2-a^2}}+\frac {B\,a^2\,\sin \left (c+d\,x\right )}{b\,d\,\left (a^2-b^2\right )} \] Input:
int((cos(c + d*x)*(A + B*cos(c + d*x)))/(a + b*cos(c + d*x)),x)
Output:
(2*B*a*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(d*(a^2 - b^2)) - (2*A *b*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(d*(a^2 - b^2)) - (B*b*sin (c + d*x))/(d*(a^2 - b^2)) + (2*A*a^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d *x)/2)))/(b*d*(a^2 - b^2)) - (2*B*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d *x)/2)))/(b^2*d*(a^2 - b^2)) + (A*a*log((a*sin(c/2 + (d*x)/2) - b*sin(c/2 + (d*x)/2) + cos(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2))/cos(c/2 + (d*x)/2)))/(b *d*(b^2 - a^2)^(1/2)) - (A*a*log((b*sin(c/2 + (d*x)/2) - a*sin(c/2 + (d*x) /2) + cos(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2))/cos(c/2 + (d*x)/2)))/(b*d*(b^2 - a^2)^(1/2)) - (B*a^2*log((a*sin(c/2 + (d*x)/2) - b*sin(c/2 + (d*x)/2) + cos(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2))/cos(c/2 + (d*x)/2)))/(b^2*d*(b^2 - a^2)^(1/2)) + (B*a^2*log((b*sin(c/2 + (d*x)/2) - a*sin(c/2 + (d*x)/2) + co s(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2))/cos(c/2 + (d*x)/2)))/(b^2*d*(b^2 - a^2 )^(1/2)) + (B*a^2*sin(c + d*x))/(b*d*(a^2 - b^2))
Time = 0.16 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.11 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{a+b \cos (c+d x)} \, dx=\frac {\sin \left (d x +c \right )}{d} \] Input:
int(cos(d*x+c)*(A+B*cos(d*x+c))/(a+b*cos(d*x+c)),x)
Output:
sin(c + d*x)/d