Integrand size = 19, antiderivative size = 85 \[ \int \frac {\cos (c+d x)}{(a+b \cos (c+d x))^2} \, dx=-\frac {2 b \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2} d}+\frac {a \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \cos (c+d x))} \] Output:
-2*b*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(3/2)/(a+b)^ (3/2)/d+a*sin(d*x+c)/(a^2-b^2)/d/(a+b*cos(d*x+c))
Time = 0.41 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.98 \[ \int \frac {\cos (c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {-\frac {2 b \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{3/2}}+\frac {a \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))}}{d} \] Input:
Integrate[Cos[c + d*x]/(a + b*Cos[c + d*x])^2,x]
Output:
((-2*b*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^ (3/2) + (a*Sin[c + d*x])/((a - b)*(a + b)*(a + b*Cos[c + d*x])))/d
Time = 0.33 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.13, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {3042, 3233, 27, 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.
\(\displaystyle \int \frac {\cos (c+d x)}{(a+b \cos (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 3233 |
\(\displaystyle \frac {a \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\int \frac {b}{a+b \cos (c+d x)}dx}{a^2-b^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {b \int \frac {1}{a+b \cos (c+d x)}dx}{a^2-b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {a \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {b \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2-b^2}\) |
\(\Big \downarrow \) 3138 |
\(\displaystyle \frac {a \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {2 b \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{d \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {a \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {2 b \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b} \left (a^2-b^2\right )}\) |
Input:
Int[Cos[c + d*x]/(a + b*Cos[c + d*x])^2,x]
Output:
(-2*b*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*Sqr t[a + b]*(a^2 - b^2)*d) + (a*Sin[c + d*x])/((a^2 - b^2)*d*(a + b*Cos[c + d *x]))
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_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(b*c - a*d))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*( m + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]
Time = 0.90 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.34
method | result | size |
derivativedivides | \(\frac {\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}-\frac {2 b \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(114\) |
default | \(\frac {\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}-\frac {2 b \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(114\) |
risch | \(-\frac {2 i a \left (a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}{b \left (-a^{2}+b^{2}\right ) d \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}\) | \(226\) |
Input:
int(cos(d*x+c)/(a+cos(d*x+c)*b)^2,x,method=_RETURNVERBOSE)
Output:
1/d*(2*a/(a^2-b^2)*tan(1/2*d*x+1/2*c)/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+ 1/2*c)^2*b+a+b)-2*b/(a-b)/(a+b)/((a-b)*(a+b))^(1/2)*arctan((a-b)*tan(1/2*d *x+1/2*c)/((a-b)*(a+b))^(1/2)))
Time = 0.10 (sec) , antiderivative size = 321, normalized size of antiderivative = 3.78 \[ \int \frac {\cos (c+d x)}{(a+b \cos (c+d x))^2} \, dx=\left [\frac {{\left (b^{2} \cos \left (d x + c\right ) + a b\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) + 2 \, {\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \cos \left (d x + c\right ) + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d\right )}}, -\frac {{\left (b^{2} \cos \left (d x + c\right ) + a b\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )}{{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \cos \left (d x + c\right ) + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d}\right ] \] Input:
integrate(cos(d*x+c)/(a+b*cos(d*x+c))^2,x, algorithm="fricas")
Output:
[1/2*((b^2*cos(d*x + c) + 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*(a^3 - a*b^2)*sin(d*x + c))/((a^4*b - 2*a^2*b^3 + b^5)*d*cos(d*x + c) + (a^5 - 2*a^3*b^2 + a*b^4)*d), -((b^2*cos(d*x + c) + a*b)*sqrt(a^2 - b^2) *arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) - (a^3 - a*b ^2)*sin(d*x + c))/((a^4*b - 2*a^2*b^3 + b^5)*d*cos(d*x + c) + (a^5 - 2*a^3 *b^2 + a*b^4)*d)]
Leaf count of result is larger than twice the leaf count of optimal. 2470 vs. \(2 (70) = 140\).
Time = 148.79 (sec) , antiderivative size = 2470, normalized size of antiderivative = 29.06 \[ \int \frac {\cos (c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)/(a+b*cos(d*x+c))**2,x)
Output:
Piecewise((zoo*x/cos(c), Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (-tan(c/2 + d*x/ 2)**3/(6*b**2*d) + tan(c/2 + d*x/2)/(2*b**2*d), Eq(a, b)), (1/(2*b**2*d*ta n(c/2 + d*x/2)) - 1/(6*b**2*d*tan(c/2 + d*x/2)**3), Eq(a, -b)), (x*cos(c)/ (a + b*cos(c))**2, Eq(d, 0)), (2*a**2*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)/(a**4*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + a**4* d*sqrt(-a/(a - b) - b/(a - b)) - 2*a**3*b*d*sqrt(-a/(a - b) - b/(a - b))*t an(c/2 + d*x/2)**2 - 2*a**2*b**2*d*sqrt(-a/(a - b) - b/(a - b)) + 2*a*b**3 *d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 - b**4*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + b**4*d*sqrt(-a/(a - b) - b/(a - b))) - 2*a*b*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)/(a**4*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + a**4*d*sqrt(-a/(a - b) - b/(a - b) ) - 2*a**3*b*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 - 2*a**2*b **2*d*sqrt(-a/(a - b) - b/(a - b)) + 2*a*b**3*d*sqrt(-a/(a - b) - b/(a - b ))*tan(c/2 + d*x/2)**2 - b**4*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x /2)**2 + b**4*d*sqrt(-a/(a - b) - b/(a - b))) - a*b*log(-sqrt(-a/(a - b) - b/(a - b)) + tan(c/2 + d*x/2))*tan(c/2 + d*x/2)**2/(a**4*d*sqrt(-a/(a - b ) - b/(a - b))*tan(c/2 + d*x/2)**2 + a**4*d*sqrt(-a/(a - b) - b/(a - b)) - 2*a**3*b*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 - 2*a**2*b**2 *d*sqrt(-a/(a - b) - b/(a - b)) + 2*a*b**3*d*sqrt(-a/(a - b) - b/(a - b))* tan(c/2 + d*x/2)**2 - b**4*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x...
Exception generated. \[ \int \frac {\cos (c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(cos(d*x+c)/(a+b*cos(d*x+c))^2,x, algorithm="maxima")
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 = 0.56 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.59 \[ \int \frac {\cos (c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {2 \, {\left (\frac {{\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )} b}{{\left (a^{2} - b^{2}\right )}^{\frac {3}{2}}} + \frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )} {\left (a^{2} - b^{2}\right )}}\right )}}{d} \] Input:
integrate(cos(d*x+c)/(a+b*cos(d*x+c))^2,x, algorithm="giac")
Output:
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)))*b/(a^2 - b^2)^(3/ 2) + a*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1 /2*c)^2 + a + b)*(a^2 - b^2)))/d
Time = 42.54 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.16 \[ \int \frac {\cos (c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a+b\right )\,\left (a-b\right )\,\left (\left (a-b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a+b\right )}-\frac {2\,b\,\mathrm {atan}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a-2\,b\right )}{2\,\sqrt {a+b}\,\sqrt {a-b}}\right )}{d\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}} \] Input:
int(cos(c + d*x)/(a + b*cos(c + d*x))^2,x)
Output:
(2*a*tan(c/2 + (d*x)/2))/(d*(a + b)*(a - b)*(a + b + tan(c/2 + (d*x)/2)^2* (a - b))) - (2*b*atan((tan(c/2 + (d*x)/2)*(2*a - 2*b))/(2*(a + b)^(1/2)*(a - b)^(1/2))))/(d*(a + b)^(3/2)*(a - b)^(3/2))
Time = 0.16 (sec) , antiderivative size = 192, normalized size of antiderivative = 2.26 \[ \int \frac {\cos (c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {-2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) \cos \left (d x +c \right ) b^{2}-2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) a b +\sin \left (d x +c \right ) a^{3}-\sin \left (d x +c \right ) a \,b^{2}}{d \left (\cos \left (d x +c \right ) a^{4} b -2 \cos \left (d x +c \right ) a^{2} b^{3}+\cos \left (d x +c \right ) b^{5}+a^{5}-2 a^{3} b^{2}+a \,b^{4}\right )} \] Input:
int(cos(d*x+c)/(a+b*cos(d*x+c))^2,x)
Output:
( - 2*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqr t(a**2 - b**2))*cos(c + d*x)*b**2 - 2*sqrt(a**2 - b**2)*atan((tan((c + d*x )/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*a*b + sin(c + d*x)*a**3 - sin(c + d*x)*a*b**2)/(d*(cos(c + d*x)*a**4*b - 2*cos(c + d*x)*a**2*b**3 + cos(c + d*x)*b**5 + a**5 - 2*a**3*b**2 + a*b**4))