Integrand size = 30, antiderivative size = 93 \[ \int \frac {a^2-b^2 \cos ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\frac {2 \left (a^2+b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2} d}-\frac {2 a b \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \cos (c+d x))} \]
2*(a^2+b^2)*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-2*a*b*sin(d*x+c)/(a^2-b^2)/d/(a+b*cos(d*x+c))
Time = 0.48 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.98 \[ \int \frac {a^2-b^2 \cos ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\frac {2 \left (\frac {\left (a^2+b^2\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{3/2}}-\frac {a b \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))}\right )}{d} \]
(2*(((a^2 + b^2)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/(-a ^2 + b^2)^(3/2) - (a*b*Sin[c + d*x])/((a - b)*(a + b)*(a + b*Cos[c + d*x]) )))/d
Time = 0.43 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.12, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3495, 25, 3042, 3233, 25, 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.
| \(\displaystyle \int \frac {a^2-b^2 \cos ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {a^2-b^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3}dx\) |
| \(\Big \downarrow \) 3495 |
| \(\displaystyle -\int -\frac {a-b \cos (c+d x)}{(a+b \cos (c+d x))^2}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \int \frac {a-b \cos (c+d x)}{(a+b \cos (c+d x))^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {a-b \sin \left (c+d x+\frac {\pi }{2}\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 3233 |
| \(\displaystyle -\frac {\int -\frac {a^2+b^2}{a+b \cos (c+d x)}dx}{a^2-b^2}-\frac {2 a b \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {a^2+b^2}{a+b \cos (c+d x)}dx}{a^2-b^2}-\frac {2 a b \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\left (a^2+b^2\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{a^2-b^2}-\frac {2 a b \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\left (a^2+b^2\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2-b^2}-\frac {2 a b \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {2 \left (a^2+b^2\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{d \left (a^2-b^2\right )}-\frac {2 a b \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {2 \left (a^2+b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b} \left (a^2-b^2\right )}-\frac {2 a b \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
(2*(a^2 + b^2)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*Sqrt[a + b]*(a^2 - b^2)*d) - (2*a*b*Sin[c + d*x])/((a^2 - b^2)*d*(a + b*Cos[c + d*x]))
3.7.21.3.1 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_)])^(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]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C/b^2 Int[(a + b*Sin[e + f*x])^(m + 1) *Simp[-a + b*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C, m}, x] && EqQ[A*b^2 + a^2*C, 0]
Time = 1.57 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.30
| method | result | size |
| derivativedivides | \(\frac {-\frac {4 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}+\frac {2 \left (a^{2}+b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(121\) |
| default | \(\frac {-\frac {4 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}+\frac {2 \left (a^{2}+b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(121\) |
| risch | \(\frac {4 i a \left (a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}{\left (-a^{2}+b^{2}\right ) d \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) a^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) b^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) a^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) b^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}\) | \(394\) |
1/d*(-4*a*b/(a^2-b^2)*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)+2*(a^2+b^2)/(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 = 0.31 (sec) , antiderivative size = 346, normalized size of antiderivative = 3.72 \[ \int \frac {a^2-b^2 \cos ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\left [\frac {{\left (a^{3} + a b^{2} + {\left (a^{2} b + b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - 4 \, {\left (a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \cos \left (d x + c\right ) + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d\right )}}, \frac {{\left (a^{3} + a b^{2} + {\left (a^{2} b + b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - 2 \, {\left (a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )}{{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \cos \left (d x + c\right ) + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d}\right ] \]
[1/2*((a^3 + a*b^2 + (a^2*b + b^3)*cos(d*x + c))*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)) - 4*(a^3*b - a*b^3)*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), ((a^3 + a*b^2 + (a^2*b + b^3)*cos(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a ^2 - b^2)*sin(d*x + c))) - 2*(a^3*b - a*b^3)*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)]
Timed out. \[ \int \frac {a^2-b^2 \cos ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {a^2-b^2 \cos ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Exception raised: ValueError} \]
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 = 0.32 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.54 \[ \int \frac {a^2-b^2 \cos ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=-\frac {2 \, {\left (\frac {2 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )} {\left (a^{2} - b^{2}\right )}} - \frac {{\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )} {\left (a^{2} + b^{2}\right )}}{{\left (a^{2} - b^{2}\right )}^{\frac {3}{2}}}\right )}}{d} \]
-2*(2*a*b*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)) - (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^2 + b^2)/(a^2 - b^2)^(3/2))/d
Time = 1.84 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.14 \[ \int \frac {a^2-b^2 \cos ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\frac {2\,\mathrm {atan}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a-2\,b\right )}{2\,\sqrt {a+b}\,\sqrt {a-b}}\right )\,\left (a^2+b^2\right )}{d\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}-\frac {4\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a+b\right )\,\left (a-b\right )\,\left (\left (a-b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a+b\right )} \]