Integrand size = 25, antiderivative size = 85 \[ \int \frac {1-\cos ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=-\frac {x}{b^2}+\frac {2 a \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^2 \sqrt {a+b} d}+\frac {\sin (c+d x)}{b d (a+b \cos (c+d x))} \]
-x/b^2+sin(d*x+c)/b/d/(a+b*cos(d*x+c))+2*a*arctan((a-b)^(1/2)*tan(1/2*d*x+ 1/2*c)/(a+b)^(1/2))/b^2/d/(a-b)^(1/2)/(a+b)^(1/2)
Time = 0.45 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.94 \[ \int \frac {1-\cos ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=-\frac {c+d x+\frac {2 a \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}-\frac {b \sin (c+d x)}{a+b \cos (c+d x)}}{b^2 d} \]
-((c + d*x + (2*a*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/Sq rt[-a^2 + b^2] - (b*Sin[c + d*x])/(a + b*Cos[c + d*x]))/(b^2*d))
Time = 0.41 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3501, 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.
\(\displaystyle \int \frac {1-\cos ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1-\sin \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 3501 |
\(\displaystyle \frac {\sin (c+d x)}{b d (a+b \cos (c+d x))}-\frac {\int \frac {\left (a^2-b^2\right ) \cos (c+d x)}{a+b \cos (c+d x)}dx}{b \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sin (c+d x)}{b d (a+b \cos (c+d x))}-\frac {\int \frac {\cos (c+d x)}{a+b \cos (c+d x)}dx}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sin (c+d x)}{b d (a+b \cos (c+d x))}-\frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}\) |
\(\Big \downarrow \) 3214 |
\(\displaystyle \frac {\sin (c+d x)}{b d (a+b \cos (c+d x))}-\frac {\frac {x}{b}-\frac {a \int \frac {1}{a+b \cos (c+d x)}dx}{b}}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sin (c+d x)}{b d (a+b \cos (c+d x))}-\frac {\frac {x}{b}-\frac {a \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}}{b}\) |
\(\Big \downarrow \) 3138 |
\(\displaystyle \frac {\sin (c+d x)}{b d (a+b \cos (c+d x))}-\frac {\frac {x}{b}-\frac {2 a \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{b d}}{b}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\sin (c+d x)}{b d (a+b \cos (c+d x))}-\frac {\frac {x}{b}-\frac {2 a \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b d \sqrt {a-b} \sqrt {a+b}}}{b}\) |
-((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) + Sin[c + d*x]/(b*d*(a + b*Cos[c + d*x]))
3.7.5.3.1 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_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 + a^2*C))*Cos[e + f*x]*((a + b*S in[e + f*x])^(m + 1)/(b*f*(m + 1)*(a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^ 2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*b*(A + C)*(m + 1) - (A* b^2 + a^2*C + b^2*(A + C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
Time = 1.62 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.29
method | result | size |
derivativedivides | \(\frac {-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2}}+\frac {\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b}+\frac {2 a \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}}{b^{2}}}{d}\) | \(110\) |
default | \(\frac {-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2}}+\frac {\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b}+\frac {2 a \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}}{b^{2}}}{d}\) | \(110\) |
risch | \(-\frac {x}{b^{2}}+\frac {2 i \left (a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}{b^{2} d \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,b^{2}}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+a \sqrt {-a^{2}+b^{2}}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,b^{2}}\) | \(200\) |
1/d*(-2/b^2*arctan(tan(1/2*d*x+1/2*c))+2/b^2*(b*tan(1/2*d*x+1/2*c)/(tan(1/ 2*d*x+1/2*c)^2*a-b*tan(1/2*d*x+1/2*c)^2+a+b)+a/((a-b)*(a+b))^(1/2)*arctan( (a-b)*tan(1/2*d*x+1/2*c)/((a-b)*(a+b))^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 153 vs. \(2 (76) = 152\).
Time = 0.31 (sec) , antiderivative size = 376, normalized size of antiderivative = 4.42 \[ \int \frac {1-\cos ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\left [-\frac {2 \, {\left (a^{2} b - b^{3}\right )} d x \cos \left (d x + c\right ) + 2 \, {\left (a^{3} - a b^{2}\right )} d x + {\left (a b \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - 2 \, {\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{2} b^{3} - b^{5}\right )} d \cos \left (d x + c\right ) + {\left (a^{3} b^{2} - a b^{4}\right )} d\right )}}, -\frac {{\left (a^{2} b - b^{3}\right )} d x \cos \left (d x + c\right ) + {\left (a^{3} - a b^{2}\right )} d x - {\left (a b \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )}{{\left (a^{2} b^{3} - b^{5}\right )} d \cos \left (d x + c\right ) + {\left (a^{3} b^{2} - a b^{4}\right )} d}\right ] \]
[-1/2*(2*(a^2*b - b^3)*d*x*cos(d*x + c) + 2*(a^3 - a*b^2)*d*x + (a*b*cos(d *x + c) + a^2)*sqrt(-a^2 + b^2)*log((2*a*b*cos(d*x + c) + (2*a^2 - b^2)*co s(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^2*b - b^3 )*sin(d*x + c))/((a^2*b^3 - b^5)*d*cos(d*x + c) + (a^3*b^2 - a*b^4)*d), -( (a^2*b - b^3)*d*x*cos(d*x + c) + (a^3 - a*b^2)*d*x - (a*b*cos(d*x + c) + a ^2)*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) - (a^2*b - b^3)*sin(d*x + c))/((a^2*b^3 - b^5)*d*cos(d*x + c) + (a^ 3*b^2 - a*b^4)*d)]
Timed out. \[ \int \frac {1-\cos ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {1-\cos ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Exception raised: ValueError} \]
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
Time = 0.30 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.65 \[ \int \frac {1-\cos ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=-\frac {\frac {2 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )} a}{\sqrt {a^{2} - b^{2}} b^{2}} + \frac {d x + c}{b^{2}} - \frac {2 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )} b}}{d} \]
-(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)))*a/(sqrt(a^2 - b^ 2)*b^2) + (d*x + c)/b^2 - 2*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)*b))/d
Time = 2.00 (sec) , antiderivative size = 277, normalized size of antiderivative = 3.26 \[ \int \frac {1-\cos ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {2\,\left (-a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\sqrt {b^2-a^2}+a^2\,\mathrm {atan}\left (\frac {\left (a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}\right )\,1{}\mathrm {i}\right )}{b^2\,d\,\sqrt {b^2-a^2}\,\left (a+b\,\cos \left (c+d\,x\right )\right )}+\frac {2\,\left (\frac {\sin \left (c+d\,x\right )\,\sqrt {b^2-a^2}}{2}-\cos \left (c+d\,x\right )\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\sqrt {b^2-a^2}+a\,\mathrm {atan}\left (\frac {\left (a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}\right )\,\cos \left (c+d\,x\right )\,1{}\mathrm {i}\right )}{b\,d\,\sqrt {b^2-a^2}\,\left (a+b\,\cos \left (c+d\,x\right )\right )} \]
(2*(a^2*atan(((a*sin(c/2 + (d*x)/2) - b*sin(c/2 + (d*x)/2))*1i)/(cos(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)))*1i - a*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d* x)/2))*(b^2 - a^2)^(1/2)))/(b^2*d*(b^2 - a^2)^(1/2)*(a + b*cos(c + d*x))) + (2*((sin(c + d*x)*(b^2 - a^2)^(1/2))/2 - cos(c + d*x)*atan(sin(c/2 + (d* x)/2)/cos(c/2 + (d*x)/2))*(b^2 - a^2)^(1/2) + a*atan(((a*sin(c/2 + (d*x)/2 ) - b*sin(c/2 + (d*x)/2))*1i)/(cos(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)))*cos( c + d*x)*1i))/(b*d*(b^2 - a^2)^(1/2)*(a + b*cos(c + d*x)))