Integrand size = 23, antiderivative size = 263 \[ \int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=-\frac {4 a \left (a^2-3 b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 \left (2 a^2-3 b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{3 b^2 \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}-\frac {2 a^2 \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac {4 a \left (a^2-3 b^2\right ) \sin (c+d x)}{3 b \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}} \]
-2/3*a^2*sin(d*x+c)/b/(a^2-b^2)/d/(a+b*cos(d*x+c))^(3/2)+4/3*a*(a^2-3*b^2) *sin(d*x+c)/b/(a^2-b^2)^2/d/(a+b*cos(d*x+c))^(1/2)-4/3*a*(a^2-3*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)^2/d/((a+b*cos( d*x+c))/(a+b))^(1/2)+2/3*(2*a^2-3*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/ 2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2)*(b/(a+b))^(1/2))*((a+b*c os(d*x+c))/(a+b))^(1/2)/b^2/(a^2-b^2)/d/(a+b*cos(d*x+c))^(1/2)
Time = 0.92 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.67 \[ \int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\frac {2 \left (-\frac {\left (\frac {a+b \cos (c+d x)}{a+b}\right )^{3/2} \left (2 \left (a^3-3 a b^2\right ) E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )+\left (-2 a^3+2 a^2 b+3 a b^2-3 b^3\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )\right )}{(a-b)^2}+\frac {a b \left (a^3-5 a b^2+2 b \left (a^2-3 b^2\right ) \cos (c+d x)\right ) \sin (c+d x)}{\left (a^2-b^2\right )^2}\right )}{3 b^2 d (a+b \cos (c+d x))^{3/2}} \]
(2*(-((((a + b*Cos[c + d*x])/(a + b))^(3/2)*(2*(a^3 - 3*a*b^2)*EllipticE[( c + d*x)/2, (2*b)/(a + b)] + (-2*a^3 + 2*a^2*b + 3*a*b^2 - 3*b^3)*Elliptic F[(c + d*x)/2, (2*b)/(a + b)]))/(a - b)^2) + (a*b*(a^3 - 5*a*b^2 + 2*b*(a^ 2 - 3*b^2)*Cos[c + d*x])*Sin[c + d*x])/(a^2 - b^2)^2))/(3*b^2*d*(a + b*Cos [c + d*x])^(3/2))
Time = 1.24 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.03, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.652, Rules used = {3042, 3269, 27, 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.
| \(\displaystyle \int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 3269 |
| \(\displaystyle \frac {2 \int \frac {3 a b+\left (2 a^2-3 b^2\right ) \cos (c+d x)}{2 (a+b \cos (c+d x))^{3/2}}dx}{3 b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {3 a b+\left (2 a^2-3 b^2\right ) \cos (c+d x)}{(a+b \cos (c+d x))^{3/2}}dx}{3 b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {3 a b+\left (2 a^2-3 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{3 b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3233 |
| \(\displaystyle \frac {\frac {4 a \left (a^2-3 b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {2 \int -\frac {b \left (a^2+3 b^2\right )-2 a \left (a^2-3 b^2\right ) \cos (c+d x)}{2 \sqrt {a+b \cos (c+d x)}}dx}{a^2-b^2}}{3 b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {b \left (a^2+3 b^2\right )-2 a \left (a^2-3 b^2\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{a^2-b^2}+\frac {4 a \left (a^2-3 b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {b \left (a^2+3 b^2\right )-2 a \left (a^2-3 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2-b^2}+\frac {4 a \left (a^2-3 b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3231 |
| \(\displaystyle \frac {\frac {\frac {\left (2 a^4-5 a^2 b^2+3 b^4\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}}dx}{b}-\frac {2 a \left (a^2-3 b^2\right ) \int \sqrt {a+b \cos (c+d x)}dx}{b}}{a^2-b^2}+\frac {4 a \left (a^2-3 b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\left (2 a^4-5 a^2 b^2+3 b^4\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {2 a \left (a^2-3 b^2\right ) \int \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}}{a^2-b^2}+\frac {4 a \left (a^2-3 b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3134 |
| \(\displaystyle \frac {\frac {\frac {\left (2 a^4-5 a^2 b^2+3 b^4\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {2 a \left (a^2-3 b^2\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}dx}{b \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a^2-b^2}+\frac {4 a \left (a^2-3 b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\left (2 a^4-5 a^2 b^2+3 b^4\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {2 a \left (a^2-3 b^2\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{b \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a^2-b^2}+\frac {4 a \left (a^2-3 b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3132 |
| \(\displaystyle \frac {\frac {\frac {\left (2 a^4-5 a^2 b^2+3 b^4\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {4 a \left (a^2-3 b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a^2-b^2}+\frac {4 a \left (a^2-3 b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3142 |
| \(\displaystyle \frac {\frac {\frac {\left (2 a^4-5 a^2 b^2+3 b^4\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{b \sqrt {a+b \cos (c+d x)}}-\frac {4 a \left (a^2-3 b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a^2-b^2}+\frac {4 a \left (a^2-3 b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\left (2 a^4-5 a^2 b^2+3 b^4\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{b \sqrt {a+b \cos (c+d x)}}-\frac {4 a \left (a^2-3 b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a^2-b^2}+\frac {4 a \left (a^2-3 b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3140 |
| \(\displaystyle \frac {\frac {4 a \left (a^2-3 b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}+\frac {\frac {2 \left (2 a^4-5 a^2 b^2+3 b^4\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{b d \sqrt {a+b \cos (c+d x)}}-\frac {4 a \left (a^2-3 b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a^2-b^2}}{3 b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\) |
(-2*a^2*Sin[c + d*x])/(3*b*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^(3/2)) + ((( -4*a*(a^2 - 3*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)]) + (2*(2*a^4 - 5*a^2*b^2 + 3*b^4)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/( a + b)])/(b*d*Sqrt[a + b*Cos[c + d*x]]))/(a^2 - b^2) + (4*a*(a^2 - 3*b^2)* Sin[c + d*x])/((a^2 - b^2)*d*Sqrt[a + b*Cos[c + d*x]]))/(3*b*(a^2 - b^2))
3.6.41.3.1 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_)]), 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_)*((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. \(849\) vs. \(2(301)=602\).
Time = 8.48 (sec) , antiderivative size = 850, normalized size of antiderivative = 3.23
-(-(-2*b*cos(1/2*d*x+1/2*c)^2-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2/b^2*(sin (1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2* sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2 *d*x+1/2*c),(-2*b/(a-b))^(1/2))+2*a^2/b^2*(1/6/b/(a-b)/(a+b)*cos(1/2*d*x+1 /2*c)*(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/ 2*d*x+1/2*c)^2+1/2*(a-b)/b)^2+8/3*sin(1/2*d*x+1/2*c)^2*b/(a-b)^2/(a+b)^2*c os(1/2*d*x+1/2*c)*a/(-(-2*b*cos(1/2*d*x+1/2*c)^2-a+b)*sin(1/2*d*x+1/2*c)^2 )^(1/2)+(3*a-b)/(3*a^3+3*a^2*b-3*a*b^2-3*b^3)*(sin(1/2*d*x+1/2*c)^2)^(1/2) *((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+( a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b)) ^(1/2))-4/3*a/(a-b)/(a+b)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x +1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1 /2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-EllipticE (cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))))+4*a/b^2/sin(1/2*d*x+1/2*c)^2/(2* b*sin(1/2*d*x+1/2*c)^2-a-b)/(a^2-b^2)*(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin (1/2*d*x+1/2*c)^2)^(1/2)*(2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*b+Elli pticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*(-2*b/(a-b)*sin(1/2*d*x+1/2*c )^2+(a+b)/(a-b))^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a-EllipticE(cos(1/2*d* x+1/2*c),(-2*b/(a-b))^(1/2))*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b)) ^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*b))/sin(1/2*d*x+1/2*c)/(-2*b*sin(1/...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.12 (sec) , antiderivative size = 814, normalized size of antiderivative = 3.10 \[ \int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\text {Too large to display} \]
1/9*(6*(a^4*b^2 - 5*a^2*b^4 + 2*(a^3*b^3 - 3*a*b^5)*cos(d*x + c))*sqrt(b*c os(d*x + c) + a)*sin(d*x + c) + (sqrt(2)*(-4*I*a^4*b^2 + 9*I*a^2*b^4 - 9*I *b^6)*cos(d*x + c)^2 - 2*sqrt(2)*(4*I*a^5*b - 9*I*a^3*b^3 + 9*I*a*b^5)*cos (d*x + c) + sqrt(2)*(-4*I*a^6 + 9*I*a^4*b^2 - 9*I*a^2*b^4))*sqrt(b)*weiers trassPInverse(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^4*b^2 - 9 *I*a^2*b^4 + 9*I*b^6)*cos(d*x + c)^2 - 2*sqrt(2)*(-4*I*a^5*b + 9*I*a^3*b^3 - 9*I*a*b^5)*cos(d*x + c) + sqrt(2)*(4*I*a^6 - 9*I*a^4*b^2 + 9*I*a^2*b^4) )*sqrt(b)*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) - 6*(sqrt(2 )*(I*a^3*b^3 - 3*I*a*b^5)*cos(d*x + c)^2 + 2*sqrt(2)*(I*a^4*b^2 - 3*I*a^2* b^4)*cos(d*x + c) + sqrt(2)*(I*a^5*b - 3*I*a^3*b^3))*sqrt(b)*weierstrassZe ta(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, weierstrassPInver se(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)) - 6*(sqrt(2)*(-I*a^3*b^3 + 3*I*a*b^5) *cos(d*x + c)^2 + 2*sqrt(2)*(-I*a^4*b^2 + 3*I*a^2*b^4)*cos(d*x + c) + sqrt (2)*(-I*a^5*b + 3*I*a^3*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^4*b^5 - 2*a^2*b^7 + b^9)*d*cos(d*x + c)^2 + 2*(a^5*b...
Timed out. \[ \int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\int { \frac {\cos \left (d x + c\right )^{2}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\int { \frac {\cos \left (d x + c\right )^{2}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]