Integrand size = 23, antiderivative size = 186 \[ \int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\frac {2 \left (2 a^2-b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b^2 \left (a^2-b^2\right ) d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {4 a \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{b^2 d \sqrt {a+b \cos (c+d x)}}-\frac {2 a^2 \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}} \]
-2*a^2*sin(d*x+c)/b/(a^2-b^2)/d/(a+b*cos(d*x+c))^(1/2)+2*(2*a^2-b^2)*(cos( 1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^ (1/2)*(b/(a+b))^(1/2))*(a+b*cos(d*x+c))^(1/2)/b^2/(a^2-b^2)/d/((a+b*cos(d* x+c))/(a+b))^(1/2)-4*a*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ell ipticF(sin(1/2*d*x+1/2*c),2^(1/2)*(b/(a+b))^(1/2))*((a+b*cos(d*x+c))/(a+b) )^(1/2)/b^2/d/(a+b*cos(d*x+c))^(1/2)
Time = 0.59 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.85 \[ \int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\frac {2 \left (2 a^3+2 a^2 b-a b^2-b^3\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-2 a \left (2 \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )+a b \sin (c+d x)\right )}{(a-b) b^2 (a+b) d \sqrt {a+b \cos (c+d x)}} \]
(2*(2*a^3 + 2*a^2*b - a*b^2 - b^3)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*Elli pticE[(c + d*x)/2, (2*b)/(a + b)] - 2*a*(2*(a^2 - b^2)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)] + a*b*Sin[c + d*x])) /((a - b)*b^2*(a + b)*d*Sqrt[a + b*Cos[c + d*x]])
Time = 0.95 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.08, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.522, Rules used = {3042, 3269, 27, 3042, 3231, 3042, 3134, 3042, 3132, 3142, 3042, 3140}
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.
| \(\displaystyle \int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 3269 |
| \(\displaystyle \frac {2 \int \frac {a b+\left (2 a^2-b^2\right ) \cos (c+d x)}{2 \sqrt {a+b \cos (c+d x)}}dx}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a b+\left (2 a^2-b^2\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a b+\left (2 a^2-b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3231 |
| \(\displaystyle \frac {\frac {\left (2 a^2-b^2\right ) \int \sqrt {a+b \cos (c+d x)}dx}{b}-\frac {2 a \left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}}dx}{b}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\left (2 a^2-b^2\right ) \int \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {2 a \left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3134 |
| \(\displaystyle \frac {\frac {\left (2 a^2-b^2\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}dx}{b \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 a \left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\left (2 a^2-b^2\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{b \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 a \left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3132 |
| \(\displaystyle \frac {\frac {2 \left (2 a^2-b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 a \left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3142 |
| \(\displaystyle \frac {\frac {2 \left (2 a^2-b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 a \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{b \sqrt {a+b \cos (c+d x)}}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {2 \left (2 a^2-b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 a \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{b \sqrt {a+b \cos (c+d x)}}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3140 |
| \(\displaystyle \frac {\frac {2 \left (2 a^2-b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {4 a \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{b d \sqrt {a+b \cos (c+d x)}}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\) |
((2*(2*a^2 - b^2)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(b*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) - (4*a*(a^2 - b^2)*Sqrt[( a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(b*d*S qrt[a + b*Cos[c + d*x]]))/(b*(a^2 - b^2)) - (2*a^2*Sin[c + d*x])/(b*(a^2 - b^2)*d*Sqrt[a + b*Cos[c + d*x]])
3.6.32.3.1 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[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a + b]/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c + d*x])/(a + b)] Int[Sqrt[a/(a + b) + ( b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2 , 0] && !GtQ[a + b, 0]
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*S qrt[a + b]))*EllipticF[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[ {a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a + b*Sin[c + d*x]] Int[1/Sqrt[a/(a + b) + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && !GtQ[a + b, 0]
Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + ( f_.)*(x_)]], x_Symbol] :> Simp[(b*c - a*d)/b Int[1/Sqrt[a + b*Sin[e + f*x ]], x], x] + Simp[d/b Int[Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a, b , c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2))*Cos[e + f*x]*((a + b*Sin[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[b*(m + 1) *(2*b*c*d - a*(c^2 + d^2)) + (a^2*d^2 - 2*a*b*c*d*(m + 2) + b^2*(d^2*(m + 1 ) + c^2*(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]
Leaf count of result is larger than twice the leaf count of optimal. \(529\) vs. \(2(234)=468\).
Time = 5.25 (sec) , antiderivative size = 530, normalized size of antiderivative = 2.85
| method | result | size |
| default | \(-\frac {2 \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2} b -2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-\frac {2 b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a -b}+\frac {a +b}{a -b}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) a^{3}+2 a \,b^{2} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-\frac {2 b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a -b}+\frac {a +b}{a -b}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right )+2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-\frac {2 b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a -b}+\frac {a +b}{a -b}}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) a^{3}-2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-\frac {2 b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a -b}+\frac {a +b}{a -b}}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) a^{2} b -\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-\frac {2 b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a -b}+\frac {a +b}{a -b}}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) a \,b^{2}+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-\frac {2 b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a -b}+\frac {a +b}{a -b}}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) b^{3}\right )}{b^{2} \left (a -b \right ) \left (a +b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b}\, d}\) | \(530\) |
-2*(2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*a^2*b-2*(sin(1/2*d*x+1/2*c)^ 2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticF(cos (1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3+2*a*b^2*(sin(1/2*d*x+1/2*c)^2)^(1/ 2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticF(cos(1/2*d *x+1/2*c),(-2*b/(a-b))^(1/2))+2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*s in(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/ (a-b))^(1/2))*a^3-2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1 /2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2) )*a^2*b-(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b )/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b^2+(sin (1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/ 2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^3)/b^2/(a-b)/(a+b)/s in(1/2*d*x+1/2*c)/(-2*b*sin(1/2*d*x+1/2*c)^2+a+b)^(1/2)/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.13 (sec) , antiderivative size = 567, normalized size of antiderivative = 3.05 \[ \int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=-\frac {6 \, \sqrt {b \cos \left (d x + c\right ) + a} a^{2} b^{2} \sin \left (d x + c\right ) + {\left (\sqrt {2} {\left (-4 i \, a^{3} b + 5 i \, a b^{3}\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-4 i \, a^{4} + 5 i \, a^{2} b^{2}\right )}\right )} \sqrt {b} {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) + 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right ) + {\left (\sqrt {2} {\left (4 i \, a^{3} b - 5 i \, a b^{3}\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (4 i \, a^{4} - 5 i \, a^{2} b^{2}\right )}\right )} \sqrt {b} {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) - 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right ) - 3 \, {\left (\sqrt {2} {\left (2 i \, a^{2} b^{2} - i \, b^{4}\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (2 i \, a^{3} b - i \, a b^{3}\right )}\right )} \sqrt {b} {\rm weierstrassZeta}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) + 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right )\right ) - 3 \, {\left (\sqrt {2} {\left (-2 i \, a^{2} b^{2} + i \, b^{4}\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-2 i \, a^{3} b + i \, a b^{3}\right )}\right )} \sqrt {b} {\rm weierstrassZeta}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) - 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right )\right )}{3 \, {\left ({\left (a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right ) + {\left (a^{3} b^{3} - a b^{5}\right )} d\right )}} \]
-1/3*(6*sqrt(b*cos(d*x + c) + a)*a^2*b^2*sin(d*x + c) + (sqrt(2)*(-4*I*a^3 *b + 5*I*a*b^3)*cos(d*x + c) + sqrt(2)*(-4*I*a^4 + 5*I*a^2*b^2))*sqrt(b)*w eierstrassPInverse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1 /3*(3*b*cos(d*x + c) + 3*I*b*sin(d*x + c) + 2*a)/b) + (sqrt(2)*(4*I*a^3*b - 5*I*a*b^3)*cos(d*x + c) + sqrt(2)*(4*I*a^4 - 5*I*a^2*b^2))*sqrt(b)*weier strassPInverse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*( 3*b*cos(d*x + c) - 3*I*b*sin(d*x + c) + 2*a)/b) - 3*(sqrt(2)*(2*I*a^2*b^2 - I*b^4)*cos(d*x + c) + sqrt(2)*(2*I*a^3*b - I*a*b^3))*sqrt(b)*weierstrass Zeta(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, weierstrassPInv erse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d* x + c) + 3*I*b*sin(d*x + c) + 2*a)/b)) - 3*(sqrt(2)*(-2*I*a^2*b^2 + I*b^4) *cos(d*x + c) + sqrt(2)*(-2*I*a^3*b + I*a*b^3))*sqrt(b)*weierstrassZeta(4/ 3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, weierstrassPInverse(4/ 3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) - 3*I*b*sin(d*x + c) + 2*a)/b)))/((a^2*b^4 - b^6)*d*cos(d*x + c) + (a^3*b^ 3 - a*b^5)*d)
\[ \int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\int \frac {\cos ^{2}{\left (c + d x \right )}}{\left (a + b \cos {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\int { \frac {\cos \left (d x + c\right )^{2}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\int { \frac {\cos \left (d x + c\right )^{2}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]