Integrand size = 23, antiderivative size = 219 \[ \int \frac {\cos ^2(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=-\frac {2 \cos (c+d x)}{3 b d (a+b \sin (c+d x))^{3/2}}+\frac {4 a \cos (c+d x)}{3 b \left (a^2-b^2\right ) d \sqrt {a+b \sin (c+d x)}}+\frac {4 a E\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )|\frac {2 b}{a+b}\right ) \sqrt {a+b \sin (c+d x)}}{3 b^2 \left (a^2-b^2\right ) d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {4 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}{3 b^2 d \sqrt {a+b \sin (c+d x)}} \]
-2/3*cos(d*x+c)/b/d/(a+b*sin(d*x+c))^(3/2)+4/3*a*cos(d*x+c)/b/(a^2-b^2)/d/ (a+b*sin(d*x+c))^(1/2)-4/3*a*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c +1/4*Pi+1/2*d*x)*EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2)*(b/(a+b))^(1/ 2))*(a+b*sin(d*x+c))^(1/2)/b^2/(a^2-b^2)/d/((a+b*sin(d*x+c))/(a+b))^(1/2)+ 4/3*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*Elliptic F(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2)*(b/(a+b))^(1/2))*((a+b*sin(d*x+c))/(a+ b))^(1/2)/b^2/d/(a+b*sin(d*x+c))^(1/2)
Time = 0.74 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.76 \[ \int \frac {\cos ^2(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=\frac {-4 a (a+b)^2 E\left (\frac {1}{4} (-2 c+\pi -2 d x)|\frac {2 b}{a+b}\right ) \left (\frac {a+b \sin (c+d x)}{a+b}\right )^{3/2}+4 (a-b) (a+b)^2 \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),\frac {2 b}{a+b}\right ) \left (\frac {a+b \sin (c+d x)}{a+b}\right )^{3/2}+2 b \cos (c+d x) \left (a^2+b^2+2 a b \sin (c+d x)\right )}{3 (a-b) b^2 (a+b) d (a+b \sin (c+d x))^{3/2}} \]
(-4*a*(a + b)^2*EllipticE[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*((a + b*Si n[c + d*x])/(a + b))^(3/2) + 4*(a - b)*(a + b)^2*EllipticF[(-2*c + Pi - 2* d*x)/4, (2*b)/(a + b)]*((a + b*Sin[c + d*x])/(a + b))^(3/2) + 2*b*Cos[c + d*x]*(a^2 + b^2 + 2*a*b*Sin[c + d*x]))/(3*(a - b)*b^2*(a + b)*d*(a + b*Sin [c + d*x])^(3/2))
Time = 1.04 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.05, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.609, Rules used = {3042, 3172, 3042, 3233, 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.
| \(\displaystyle \int \frac {\cos ^2(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^2}{(a+b \sin (c+d x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3172 |
| \(\displaystyle -\frac {2 \int \frac {\sin (c+d x)}{(a+b \sin (c+d x))^{3/2}}dx}{3 b}-\frac {2 \cos (c+d x)}{3 b d (a+b \sin (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {2 \int \frac {\sin (c+d x)}{(a+b \sin (c+d x))^{3/2}}dx}{3 b}-\frac {2 \cos (c+d x)}{3 b d (a+b \sin (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3233 |
| \(\displaystyle -\frac {2 \left (-\frac {2 \int \frac {b+a \sin (c+d x)}{2 \sqrt {a+b \sin (c+d x)}}dx}{a^2-b^2}-\frac {2 a \cos (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}\right )}{3 b}-\frac {2 \cos (c+d x)}{3 b d (a+b \sin (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {2 \left (-\frac {\int \frac {b+a \sin (c+d x)}{\sqrt {a+b \sin (c+d x)}}dx}{a^2-b^2}-\frac {2 a \cos (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}\right )}{3 b}-\frac {2 \cos (c+d x)}{3 b d (a+b \sin (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {2 \left (-\frac {\int \frac {b+a \sin (c+d x)}{\sqrt {a+b \sin (c+d x)}}dx}{a^2-b^2}-\frac {2 a \cos (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}\right )}{3 b}-\frac {2 \cos (c+d x)}{3 b d (a+b \sin (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3231 |
| \(\displaystyle -\frac {2 \left (-\frac {\frac {a \int \sqrt {a+b \sin (c+d x)}dx}{b}-\frac {\left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}}dx}{b}}{a^2-b^2}-\frac {2 a \cos (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}\right )}{3 b}-\frac {2 \cos (c+d x)}{3 b d (a+b \sin (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {2 \left (-\frac {\frac {a \int \sqrt {a+b \sin (c+d x)}dx}{b}-\frac {\left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}}dx}{b}}{a^2-b^2}-\frac {2 a \cos (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}\right )}{3 b}-\frac {2 \cos (c+d x)}{3 b d (a+b \sin (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3134 |
| \(\displaystyle -\frac {2 \left (-\frac {\frac {a \sqrt {a+b \sin (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}dx}{b \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {\left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}}dx}{b}}{a^2-b^2}-\frac {2 a \cos (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}\right )}{3 b}-\frac {2 \cos (c+d x)}{3 b d (a+b \sin (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {2 \left (-\frac {\frac {a \sqrt {a+b \sin (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}dx}{b \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {\left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}}dx}{b}}{a^2-b^2}-\frac {2 a \cos (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}\right )}{3 b}-\frac {2 \cos (c+d x)}{3 b d (a+b \sin (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3132 |
| \(\displaystyle -\frac {2 \left (-\frac {\frac {2 a \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {\left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}}dx}{b}}{a^2-b^2}-\frac {2 a \cos (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}\right )}{3 b}-\frac {2 \cos (c+d x)}{3 b d (a+b \sin (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3142 |
| \(\displaystyle -\frac {2 \left (-\frac {\frac {2 a \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {\left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}}dx}{b \sqrt {a+b \sin (c+d x)}}}{a^2-b^2}-\frac {2 a \cos (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}\right )}{3 b}-\frac {2 \cos (c+d x)}{3 b d (a+b \sin (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {2 \left (-\frac {\frac {2 a \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {\left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}}dx}{b \sqrt {a+b \sin (c+d x)}}}{a^2-b^2}-\frac {2 a \cos (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}\right )}{3 b}-\frac {2 \cos (c+d x)}{3 b d (a+b \sin (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3140 |
| \(\displaystyle -\frac {2 \left (-\frac {2 a \cos (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}-\frac {\frac {2 a \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {2 \left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),\frac {2 b}{a+b}\right )}{b d \sqrt {a+b \sin (c+d x)}}}{a^2-b^2}\right )}{3 b}-\frac {2 \cos (c+d x)}{3 b d (a+b \sin (c+d x))^{3/2}}\) |
(-2*Cos[c + d*x])/(3*b*d*(a + b*Sin[c + d*x])^(3/2)) - (2*((-2*a*Cos[c + d *x])/((a^2 - b^2)*d*Sqrt[a + b*Sin[c + d*x]]) - ((2*a*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(b*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) - (2*(a^2 - b^2)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(b*d*Sqrt[a + b*Sin[c + d*x]]))/ (a^2 - b^2)))/(3*b)
3.6.36.3.1 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[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x ])^(m + 1)/(b*f*(m + 1))), x] + Simp[g^2*((p - 1)/(b*(m + 1))) Int[(g*Cos [e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1)*Sin[e + f*x], x], x] /; Fre eQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && I ntegersQ[2*m, 2*p]
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_)]), 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]
Leaf count of result is larger than twice the leaf count of optimal. \(863\) vs. \(2(265)=530\).
Time = 1.43 (sec) , antiderivative size = 864, normalized size of antiderivative = 3.95
| method | result | size |
| default | \(\frac {\frac {4 a \,b^{3} \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+\frac {2 \left (a^{2} b^{2}+b^{4}\right ) \left (\cos ^{2}\left (d x +c \right )\right )}{3}+\frac {4 \sqrt {-\frac {b \sin \left (d x +c \right )}{a +b}+\frac {b}{a +b}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a -b}-\frac {b}{a -b}}\, \sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}\, b \left (F\left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{2} b -F\left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) b^{3}-E\left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{3}+E\left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a \,b^{2}\right ) \sin \left (d x +c \right )}{3}+\frac {4 \sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a +b}+\frac {b}{a +b}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a -b}-\frac {b}{a -b}}\, F\left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{3} b}{3}-\frac {4 \sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a +b}+\frac {b}{a +b}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a -b}-\frac {b}{a -b}}\, F\left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a \,b^{3}}{3}-\frac {4 \sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a +b}+\frac {b}{a +b}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a -b}-\frac {b}{a -b}}\, E\left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{4}}{3}+\frac {4 \sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a +b}+\frac {b}{a +b}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a -b}-\frac {b}{a -b}}\, E\left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{2} b^{2}}{3}}{\left (a^{2}-b^{2}\right ) \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}} b^{3} \cos \left (d x +c \right ) d}\) | \(864\) |
2/3*(2*a*b^3*cos(d*x+c)^2*sin(d*x+c)+(a^2*b^2+b^4)*cos(d*x+c)^2+2*(-b/(a+b )*sin(d*x+c)+b/(a+b))^(1/2)*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*(b/(a-b)*s in(d*x+c)+a/(a-b))^(1/2)*b*(EllipticF((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),( (a-b)/(a+b))^(1/2))*a^2*b-EllipticF((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a -b)/(a+b))^(1/2))*b^3-EllipticE((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b)/ (a+b))^(1/2))*a^3+EllipticE((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b)/(a+b ))^(1/2))*a*b^2)*sin(d*x+c)+2*(b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2)*(-b/(a+b) *sin(d*x+c)+b/(a+b))^(1/2)*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*EllipticF(( b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^3*b-2*(b/(a-b)*si n(d*x+c)+a/(a-b))^(1/2)*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*(-b/(a-b)*sin( d*x+c)-b/(a-b))^(1/2)*EllipticF((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b)/ (a+b))^(1/2))*a*b^3-2*(b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2)*(-b/(a+b)*sin(d*x +c)+b/(a+b))^(1/2)*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*EllipticE((b/(a-b)* sin(d*x+c)+a/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^4+2*(b/(a-b)*sin(d*x+c)+a /(a-b))^(1/2)*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*(-b/(a-b)*sin(d*x+c)-b/( a-b))^(1/2)*EllipticE((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b)/(a+b))^(1/ 2))*a^2*b^2)/(a^2-b^2)/(a+b*sin(d*x+c))^(3/2)/b^3/cos(d*x+c)/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.12 (sec) , antiderivative size = 711, normalized size of antiderivative = 3.25 \[ \int \frac {\cos ^2(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=-\frac {2 \, {\left ({\left (\sqrt {2} {\left (2 \, a^{2} b^{2} - 3 \, b^{4}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} {\left (2 \, a^{3} b - 3 \, a b^{3}\right )} \sin \left (d x + c\right ) - \sqrt {2} {\left (2 \, a^{4} - a^{2} b^{2} - 3 \, b^{4}\right )}\right )} \sqrt {i \, b} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 i \, a^{3} - 9 i \, 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 i \, a}{3 \, b}\right ) + {\left (\sqrt {2} {\left (2 \, a^{2} b^{2} - 3 \, b^{4}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} {\left (2 \, a^{3} b - 3 \, a b^{3}\right )} \sin \left (d x + c\right ) - \sqrt {2} {\left (2 \, a^{4} - a^{2} b^{2} - 3 \, b^{4}\right )}\right )} \sqrt {-i \, b} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (-8 i \, a^{3} + 9 i \, 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 i \, a}{3 \, b}\right ) + 3 \, {\left (i \, \sqrt {2} a b^{3} \cos \left (d x + c\right )^{2} - 2 i \, \sqrt {2} a^{2} b^{2} \sin \left (d x + c\right ) + \sqrt {2} {\left (-i \, a^{3} b - i \, a b^{3}\right )}\right )} \sqrt {i \, b} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 i \, a^{3} - 9 i \, 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 i \, a^{3} - 9 i \, 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 i \, a}{3 \, b}\right )\right ) + 3 \, {\left (-i \, \sqrt {2} a b^{3} \cos \left (d x + c\right )^{2} + 2 i \, \sqrt {2} a^{2} b^{2} \sin \left (d x + c\right ) + \sqrt {2} {\left (i \, a^{3} b + i \, a b^{3}\right )}\right )} \sqrt {-i \, b} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (-8 i \, a^{3} + 9 i \, 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 i \, a^{3} + 9 i \, 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 i \, a}{3 \, b}\right )\right ) + 3 \, {\left (2 \, a b^{3} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )\right )} \sqrt {b \sin \left (d x + c\right ) + a}\right )}}{9 \, {\left ({\left (a^{2} b^{5} - b^{7}\right )} d \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{3} b^{4} - a b^{6}\right )} d \sin \left (d x + c\right ) - {\left (a^{4} b^{3} - b^{7}\right )} d\right )}} \]
-2/9*((sqrt(2)*(2*a^2*b^2 - 3*b^4)*cos(d*x + c)^2 - 2*sqrt(2)*(2*a^3*b - 3 *a*b^3)*sin(d*x + c) - sqrt(2)*(2*a^4 - a^2*b^2 - 3*b^4))*sqrt(I*b)*weiers trassPInverse(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*I*a^3 - 9*I*a*b^2)/b^3, 1 /3*(3*b*cos(d*x + c) - 3*I*b*sin(d*x + c) - 2*I*a)/b) + (sqrt(2)*(2*a^2*b^ 2 - 3*b^4)*cos(d*x + c)^2 - 2*sqrt(2)*(2*a^3*b - 3*a*b^3)*sin(d*x + c) - s qrt(2)*(2*a^4 - a^2*b^2 - 3*b^4))*sqrt(-I*b)*weierstrassPInverse(-4/3*(4*a ^2 - 3*b^2)/b^2, -8/27*(-8*I*a^3 + 9*I*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) + 3*I*b*sin(d*x + c) + 2*I*a)/b) + 3*(I*sqrt(2)*a*b^3*cos(d*x + c)^2 - 2*I* sqrt(2)*a^2*b^2*sin(d*x + c) + sqrt(2)*(-I*a^3*b - I*a*b^3))*sqrt(I*b)*wei erstrassZeta(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*I*a^3 - 9*I*a*b^2)/b^3, we ierstrassPInverse(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*I*a^3 - 9*I*a*b^2)/b^ 3, 1/3*(3*b*cos(d*x + c) - 3*I*b*sin(d*x + c) - 2*I*a)/b)) + 3*(-I*sqrt(2) *a*b^3*cos(d*x + c)^2 + 2*I*sqrt(2)*a^2*b^2*sin(d*x + c) + sqrt(2)*(I*a^3* b + I*a*b^3))*sqrt(-I*b)*weierstrassZeta(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*( -8*I*a^3 + 9*I*a*b^2)/b^3, weierstrassPInverse(-4/3*(4*a^2 - 3*b^2)/b^2, - 8/27*(-8*I*a^3 + 9*I*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) + 3*I*b*sin(d*x + c ) + 2*I*a)/b)) + 3*(2*a*b^3*cos(d*x + c)*sin(d*x + c) + (a^2*b^2 + b^4)*co s(d*x + c))*sqrt(b*sin(d*x + c) + a))/((a^2*b^5 - b^7)*d*cos(d*x + c)^2 - 2*(a^3*b^4 - a*b^6)*d*sin(d*x + c) - (a^4*b^3 - b^7)*d)
\[ \int \frac {\cos ^2(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=\int \frac {\cos ^{2}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {\cos ^2(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=\int { \frac {\cos \left (d x + c\right )^{2}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\cos ^2(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\cos ^2(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2}{{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \]