Integrand size = 30, antiderivative size = 140 \[ \int \frac {a^2-b^2 \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\frac {2 a \left (a^2+2 b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} d}-\frac {a b \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac {b \left (2 a^2+b^2\right ) \sin (c+d x)}{\left (a^2-b^2\right )^2 d (a+b \cos (c+d x))} \]
2*a*(a^2+2*b^2)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^( 5/2)/(a+b)^(5/2)/d-a*b*sin(d*x+c)/(a^2-b^2)/d/(a+b*cos(d*x+c))^2-b*(2*a^2+ b^2)*sin(d*x+c)/(a^2-b^2)^2/d/(a+b*cos(d*x+c))
Time = 0.85 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.83 \[ \int \frac {a^2-b^2 \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=-\frac {\frac {2 a \left (a^2+2 b^2\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{5/2}}+\frac {b \left (3 a^3+b \left (2 a^2+b^2\right ) \cos (c+d x)\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (a+b \cos (c+d x))^2}}{d} \]
-(((2*a*(a^2 + 2*b^2)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]] )/(-a^2 + b^2)^(5/2) + (b*(3*a^3 + b*(2*a^2 + b^2)*Cos[c + d*x])*Sin[c + d *x])/((a - b)^2*(a + b)^2*(a + b*Cos[c + d*x])^2))/d)
Time = 0.63 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.17, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.433, Rules used = {3042, 3495, 25, 3042, 3233, 27, 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.
| \(\displaystyle \int \frac {a^2-b^2 \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {a^2-b^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^4}dx\) |
| \(\Big \downarrow \) 3495 |
| \(\displaystyle -\int -\frac {a-b \cos (c+d x)}{(a+b \cos (c+d x))^3}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \int \frac {a-b \cos (c+d x)}{(a+b \cos (c+d x))^3}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {a-b \sin \left (c+d x+\frac {\pi }{2}\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3}dx\) |
| \(\Big \downarrow \) 3233 |
| \(\displaystyle -\frac {\int -\frac {2 \left (a^2-b \cos (c+d x) a+b^2\right )}{(a+b \cos (c+d x))^2}dx}{2 \left (a^2-b^2\right )}-\frac {a b \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a^2-b \cos (c+d x) a+b^2}{(a+b \cos (c+d x))^2}dx}{a^2-b^2}-\frac {a b \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a^2-b \sin \left (c+d x+\frac {\pi }{2}\right ) a+b^2}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx}{a^2-b^2}-\frac {a b \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 3233 |
| \(\displaystyle \frac {-\frac {\int -\frac {a \left (a^2+2 b^2\right )}{a+b \cos (c+d x)}dx}{a^2-b^2}-\frac {b \left (2 a^2+b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{a^2-b^2}-\frac {a b \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {a \left (a^2+2 b^2\right )}{a+b \cos (c+d x)}dx}{a^2-b^2}-\frac {b \left (2 a^2+b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{a^2-b^2}-\frac {a b \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {a \left (a^2+2 b^2\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{a^2-b^2}-\frac {b \left (2 a^2+b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{a^2-b^2}-\frac {a b \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a \left (a^2+2 b^2\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2-b^2}-\frac {b \left (2 a^2+b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{a^2-b^2}-\frac {a b \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {\frac {2 a \left (a^2+2 b^2\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{d \left (a^2-b^2\right )}-\frac {b \left (2 a^2+b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{a^2-b^2}-\frac {a b \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {2 a \left (a^2+2 b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b} \left (a^2-b^2\right )}-\frac {b \left (2 a^2+b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{a^2-b^2}-\frac {a b \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
-((a*b*Sin[c + d*x])/((a^2 - b^2)*d*(a + b*Cos[c + d*x])^2)) + ((2*a*(a^2 + 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) - (b*(2*a^2 + b^2)*Sin[c + d*x])/((a^2 - b^2)*d *(a + b*Cos[c + d*x])))/(a^2 - b^2)
3.7.22.3.1 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 = 1.63 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.41
| method | result | size |
| derivativedivides | \(\frac {\frac {-\frac {2 \left (3 a^{2}+a b +b^{2}\right ) b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {2 \left (3 a^{2}-a b +b^{2}\right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}^{2}}+\frac {2 a \left (a^{2}+2 b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(197\) |
| default | \(\frac {\frac {-\frac {2 \left (3 a^{2}+a b +b^{2}\right ) b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {2 \left (3 a^{2}-a b +b^{2}\right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}^{2}}+\frac {2 a \left (a^{2}+2 b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(197\) |
| risch | \(-\frac {2 i \left (b \,a^{3} {\mathrm e}^{3 i \left (d x +c \right )}+2 b^{3} a \,{\mathrm e}^{3 i \left (d x +c \right )}+4 a^{4} {\mathrm e}^{2 i \left (d x +c \right )}+4 b^{2} a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+b^{4} {\mathrm e}^{2 i \left (d x +c \right )}+7 b \,a^{3} {\mathrm e}^{i \left (d x +c \right )}+2 \,{\mathrm e}^{i \left (d x +c \right )} b^{3} a +2 a^{2} b^{2}+b^{4}\right )}{\left (a^{2}-b^{2}\right )^{2} d \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )^{2}}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}-\frac {2 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) b^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}+\frac {2 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) b^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}\) | \(498\) |
1/d*(2*(-(3*a^2+a*b+b^2)*b/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3-(3*a ^2-a*b+b^2)*b/(a+b)/(a^2-2*a*b+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+2*a*(a^2+2*b^2)/(a^4-2*a^2*b^2+b^4)/(( 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. 264 vs. \(2 (131) = 262\).
Time = 0.31 (sec) , antiderivative size = 598, normalized size of antiderivative = 4.27 \[ \int \frac {a^2-b^2 \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\left [-\frac {{\left (a^{5} + 2 \, a^{3} b^{2} + {\left (a^{3} b^{2} + 2 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{4} b + 2 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) + 2 \, {\left (3 \, a^{5} b - 3 \, a^{3} b^{3} + {\left (2 \, a^{4} b^{2} - a^{2} b^{4} - b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} d \cos \left (d x + c\right ) + {\left (a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6}\right )} d\right )}}, \frac {{\left (a^{5} + 2 \, a^{3} b^{2} + {\left (a^{3} b^{2} + 2 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{4} b + 2 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (3 \, a^{5} b - 3 \, a^{3} b^{3} + {\left (2 \, a^{4} b^{2} - a^{2} b^{4} - b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{{\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} d \cos \left (d x + c\right ) + {\left (a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6}\right )} d}\right ] \]
[-1/2*((a^5 + 2*a^3*b^2 + (a^3*b^2 + 2*a*b^4)*cos(d*x + c)^2 + 2*(a^4*b + 2*a^2*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)) + 2*(3 *a^5*b - 3*a^3*b^3 + (2*a^4*b^2 - a^2*b^4 - b^6)*cos(d*x + c))*sin(d*x + c ))/((a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*d*cos(d*x + c)^2 + 2*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*d*cos(d*x + c) + (a^8 - 3*a^6*b^2 + 3*a^4*b ^4 - a^2*b^6)*d), ((a^5 + 2*a^3*b^2 + (a^3*b^2 + 2*a*b^4)*cos(d*x + c)^2 + 2*(a^4*b + 2*a^2*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))) - (3*a^5*b - 3*a^3*b^3 + (2*a^4*b^ 2 - a^2*b^4 - b^6)*cos(d*x + c))*sin(d*x + c))/((a^6*b^2 - 3*a^4*b^4 + 3*a ^2*b^6 - b^8)*d*cos(d*x + c)^2 + 2*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7) *d*cos(d*x + c) + (a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6)*d)]
Timed out. \[ \int \frac {a^2-b^2 \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {a^2-b^2 \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\text {Exception raised: ValueError} \]
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 = 0.35 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.82 \[ \int \frac {a^2-b^2 \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=-\frac {2 \, {\left (\frac {{\left (a^{3} + 2 \, a b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {3 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}^{2}}\right )}}{d} \]
-2*((a^3 + 2*a*b^2)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + ar ctan(-(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(a^2 - b^2)))/ ((a^4 - 2*a^2*b^2 + b^4)*sqrt(a^2 - b^2)) + (3*a^3*b*tan(1/2*d*x + 1/2*c)^ 3 - 2*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - b^4*tan(1/2*d*x + 1/2*c)^3 + 3*a^3* b*tan(1/2*d*x + 1/2*c) + 2*a^2*b^2*tan(1/2*d*x + 1/2*c) + b^4*tan(1/2*d*x + 1/2*c))/((a^4 - 2*a^2*b^2 + b^4)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d *x + 1/2*c)^2 + a + b)^2))/d
Time = 4.55 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.73 \[ \int \frac {a^2-b^2 \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\frac {2\,a\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2+2\,b^2\right )\,\left (2\,a-2\,b\right )\,\left (a^2-2\,a\,b+b^2\right )}{2\,\sqrt {a+b}\,{\left (a-b\right )}^{5/2}\,\left (a^3+2\,a\,b^2\right )}\right )\,\left (a^2+2\,b^2\right )}{d\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}-\frac {\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (3\,a^2\,b+a\,b^2+b^3\right )}{{\left (a+b\right )}^2\,\left (a-b\right )}+\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,a^2\,b-a\,b^2+b^3\right )}{\left (a+b\right )\,\left (a^2-2\,a\,b+b^2\right )}}{d\,\left (2\,a\,b+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^2-2\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a^2-2\,a\,b+b^2\right )+a^2+b^2\right )} \]
(2*a*atan((a*tan(c/2 + (d*x)/2)*(a^2 + 2*b^2)*(2*a - 2*b)*(a^2 - 2*a*b + b ^2))/(2*(a + b)^(1/2)*(a - b)^(5/2)*(2*a*b^2 + a^3)))*(a^2 + 2*b^2))/(d*(a + b)^(5/2)*(a - b)^(5/2)) - ((2*tan(c/2 + (d*x)/2)^3*(a*b^2 + 3*a^2*b + b ^3))/((a + b)^2*(a - b)) + (2*tan(c/2 + (d*x)/2)*(3*a^2*b - a*b^2 + b^3))/ ((a + b)*(a^2 - 2*a*b + b^2)))/(d*(2*a*b + tan(c/2 + (d*x)/2)^2*(2*a^2 - 2 *b^2) + tan(c/2 + (d*x)/2)^4*(a^2 - 2*a*b + b^2) + a^2 + b^2))