Integrand size = 34, antiderivative size = 307 \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=-\frac {2 \left (a^2 b B+3 b^3 B+2 a^3 C-6 a b^2 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 \left (a b B+2 a^2 C-3 b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{3 b^2 \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}+\frac {2 a (b B-a C) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac {2 \left (a^2 b B+3 b^3 B+2 a^3 C-6 a b^2 C\right ) \sin (c+d x)}{3 b \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}} \]
2/3*a*(B*b-C*a)*sin(d*x+c)/b/(a^2-b^2)/d/(a+b*cos(d*x+c))^(3/2)+2/3*(B*a^2 *b+3*B*b^3+2*C*a^3-6*C*a*b^2)*sin(d*x+c)/b/(a^2-b^2)^2/d/(a+b*cos(d*x+c))^ (1/2)-2/3*(B*a^2*b+3*B*b^3+2*C*a^3-6*C*a*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*(B*a*b+2*C*a^2-3*C*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*cos(d*x+c))/(a +b))^(1/2)/b^2/(a^2-b^2)/d/(a+b*cos(d*x+c))^(1/2)
Time = 2.59 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.73 \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\frac {2 \left (-\frac {\left (\frac {a+b \cos (c+d x)}{a+b}\right )^{3/2} \left (\left (a^2 b B+3 b^3 B+2 a^3 C-6 a b^2 C\right ) E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-(a-b) \left (a b B+2 a^2 C-3 b^2 C\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )\right )}{(a-b)^2}+\frac {b \left (a \left (2 a^2 b B+2 b^3 B+a^3 C-5 a b^2 C\right )+b \left (a^2 b B+3 b^3 B+2 a^3 C-6 a b^2 C\right ) \cos (c+d x)\right ) \sin (c+d x)}{\left (a^2-b^2\right )^2}\right )}{3 b^2 d (a+b \cos (c+d x))^{3/2}} \]
(2*(-((((a + b*Cos[c + d*x])/(a + b))^(3/2)*((a^2*b*B + 3*b^3*B + 2*a^3*C - 6*a*b^2*C)*EllipticE[(c + d*x)/2, (2*b)/(a + b)] - (a - b)*(a*b*B + 2*a^ 2*C - 3*b^2*C)*EllipticF[(c + d*x)/2, (2*b)/(a + b)]))/(a - b)^2) + (b*(a* (2*a^2*b*B + 2*b^3*B + a^3*C - 5*a*b^2*C) + b*(a^2*b*B + 3*b^3*B + 2*a^3*C - 6*a*b^2*C)*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 = 1.48 (sec) , antiderivative size = 317, normalized size of antiderivative = 1.03, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.441, Rules used = {3042, 3500, 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.
| \(\displaystyle \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {2 a (b B-a C) \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}-\frac {2 \int \frac {3 b (b B-a C)-\left (2 C a^2+b B a-3 b^2 C\right ) \cos (c+d x)}{2 (a+b \cos (c+d x))^{3/2}}dx}{3 b \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 a (b B-a C) \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}-\frac {\int \frac {3 b (b B-a C)-\left (2 C a^2+b B a-3 b^2 C\right ) \cos (c+d x)}{(a+b \cos (c+d x))^{3/2}}dx}{3 b \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a (b B-a C) \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}-\frac {\int \frac {3 b (b B-a C)+\left (-2 C a^2-b B a+3 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{3 b \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3233 |
| \(\displaystyle \frac {2 a (b B-a C) \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}-\frac {-\frac {2 \int -\frac {b \left (-C a^2+4 b B a-3 b^2 C\right )+\left (2 C a^3+b B a^2-6 b^2 C a+3 b^3 B\right ) \cos (c+d x)}{2 \sqrt {a+b \cos (c+d x)}}dx}{a^2-b^2}-\frac {2 \left (2 a^3 C+a^2 b B-6 a b^2 C+3 b^3 B\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 b \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 a (b B-a C) \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}-\frac {\frac {\int \frac {b \left (-C a^2+4 b B a-3 b^2 C\right )+\left (2 C a^3+b B a^2-6 b^2 C a+3 b^3 B\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{a^2-b^2}-\frac {2 \left (2 a^3 C+a^2 b B-6 a b^2 C+3 b^3 B\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 b \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a (b B-a C) \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}-\frac {\frac {\int \frac {b \left (-C a^2+4 b B a-3 b^2 C\right )+\left (2 C a^3+b B a^2-6 b^2 C a+3 b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2-b^2}-\frac {2 \left (2 a^3 C+a^2 b B-6 a b^2 C+3 b^3 B\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 b \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3231 |
| \(\displaystyle \frac {2 a (b B-a C) \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}-\frac {\frac {\frac {\left (2 a^3 C+a^2 b B-6 a b^2 C+3 b^3 B\right ) \int \sqrt {a+b \cos (c+d x)}dx}{b}-\frac {\left (a^2-b^2\right ) \left (2 a^2 C+a b B-3 b^2 C\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}}dx}{b}}{a^2-b^2}-\frac {2 \left (2 a^3 C+a^2 b B-6 a b^2 C+3 b^3 B\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 b \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a (b B-a C) \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}-\frac {\frac {\frac {\left (2 a^3 C+a^2 b B-6 a b^2 C+3 b^3 B\right ) \int \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {\left (a^2-b^2\right ) \left (2 a^2 C+a b B-3 b^2 C\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}}{a^2-b^2}-\frac {2 \left (2 a^3 C+a^2 b B-6 a b^2 C+3 b^3 B\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 b \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3134 |
| \(\displaystyle \frac {2 a (b B-a C) \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}-\frac {\frac {\frac {\left (2 a^3 C+a^2 b B-6 a b^2 C+3 b^3 B\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}dx}{b \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (a^2-b^2\right ) \left (2 a^2 C+a b B-3 b^2 C\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}}{a^2-b^2}-\frac {2 \left (2 a^3 C+a^2 b B-6 a b^2 C+3 b^3 B\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 b \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a (b B-a C) \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}-\frac {\frac {\frac {\left (2 a^3 C+a^2 b B-6 a b^2 C+3 b^3 B\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{b \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (a^2-b^2\right ) \left (2 a^2 C+a b B-3 b^2 C\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}}{a^2-b^2}-\frac {2 \left (2 a^3 C+a^2 b B-6 a b^2 C+3 b^3 B\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 b \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3132 |
| \(\displaystyle \frac {2 a (b B-a C) \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}-\frac {\frac {\frac {2 \left (2 a^3 C+a^2 b B-6 a b^2 C+3 b^3 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (a^2-b^2\right ) \left (2 a^2 C+a b B-3 b^2 C\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}}{a^2-b^2}-\frac {2 \left (2 a^3 C+a^2 b B-6 a b^2 C+3 b^3 B\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 b \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3142 |
| \(\displaystyle \frac {2 a (b B-a C) \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}-\frac {\frac {\frac {2 \left (2 a^3 C+a^2 b B-6 a b^2 C+3 b^3 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (a^2-b^2\right ) \left (2 a^2 C+a b B-3 b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{b \sqrt {a+b \cos (c+d x)}}}{a^2-b^2}-\frac {2 \left (2 a^3 C+a^2 b B-6 a b^2 C+3 b^3 B\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 b \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a (b B-a C) \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}-\frac {\frac {\frac {2 \left (2 a^3 C+a^2 b B-6 a b^2 C+3 b^3 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (a^2-b^2\right ) \left (2 a^2 C+a b B-3 b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{b \sqrt {a+b \cos (c+d x)}}}{a^2-b^2}-\frac {2 \left (2 a^3 C+a^2 b B-6 a b^2 C+3 b^3 B\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 b \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3140 |
| \(\displaystyle \frac {2 a (b B-a C) \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}-\frac {\frac {\frac {2 \left (2 a^3 C+a^2 b B-6 a b^2 C+3 b^3 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 \left (a^2-b^2\right ) \left (2 a^2 C+a b B-3 b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{b d \sqrt {a+b \cos (c+d x)}}}{a^2-b^2}-\frac {2 \left (2 a^3 C+a^2 b B-6 a b^2 C+3 b^3 B\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 b \left (a^2-b^2\right )}\) |
(2*a*(b*B - a*C)*Sin[c + d*x])/(3*b*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^(3/ 2)) - (((2*(a^2*b*B + 3*b^3*B + 2*a^3*C - 6*a*b^2*C)*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*(a^2 - b^2)*(a*b*B + 2*a^2*C - 3*b^2*C)*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) - (2*(a^2*b*B + 3*b^3*B + 2*a^3*C - 6*a*b^2*C)*Si n[c + d*x])/((a^2 - b^2)*d*Sqrt[a + b*Cos[c + d*x]]))/(3*b*(a^2 - b^2))
3.9.49.3.1 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_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*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*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A *b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(863\) vs. \(2(345)=690\).
Time = 21.92 (sec) , antiderivative size = 864, normalized size of antiderivative = 2.81
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(864\) |
| parts | \(\text {Expression too large to display}\) | \(1598\) |
-(-(-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*C/b^2*(s in(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/b^2*(B*b-2*C*a)/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)*s in(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+El lipticE(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)-2*a*(B*b-C*a)/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*cos(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)))))/sin(1/2*d*x...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.16 (sec) , antiderivative size = 1076, normalized size of antiderivative = 3.50 \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\text {Too large to display} \]
1/9*(6*(C*a^4*b^2 + 2*B*a^3*b^3 - 5*C*a^2*b^4 + 2*B*a*b^5 + (2*C*a^3*b^3 + B*a^2*b^4 - 6*C*a*b^5 + 3*B*b^6)*cos(d*x + c))*sqrt(b*cos(d*x + c) + a)*s in(d*x + c) + (sqrt(2)*(-4*I*C*a^4*b^2 - 2*I*B*a^3*b^3 + 9*I*C*a^2*b^4 + 6 *I*B*a*b^5 - 9*I*C*b^6)*cos(d*x + c)^2 - 2*sqrt(2)*(4*I*C*a^5*b + 2*I*B*a^ 4*b^2 - 9*I*C*a^3*b^3 - 6*I*B*a^2*b^4 + 9*I*C*a*b^5)*cos(d*x + c) + sqrt(2 )*(-4*I*C*a^6 - 2*I*B*a^5*b + 9*I*C*a^4*b^2 + 6*I*B*a^3*b^3 - 9*I*C*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) + (sqrt(2 )*(4*I*C*a^4*b^2 + 2*I*B*a^3*b^3 - 9*I*C*a^2*b^4 - 6*I*B*a*b^5 + 9*I*C*b^6 )*cos(d*x + c)^2 - 2*sqrt(2)*(-4*I*C*a^5*b - 2*I*B*a^4*b^2 + 9*I*C*a^3*b^3 + 6*I*B*a^2*b^4 - 9*I*C*a*b^5)*cos(d*x + c) + sqrt(2)*(4*I*C*a^6 + 2*I*B* a^5*b - 9*I*C*a^4*b^2 - 6*I*B*a^3*b^3 + 9*I*C*a^2*b^4))*sqrt(b)*weierstras sPInverse(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*c os(d*x + c) - 3*I*b*sin(d*x + c) + 2*a)/b) - 3*(sqrt(2)*(2*I*C*a^3*b^3 + I *B*a^2*b^4 - 6*I*C*a*b^5 + 3*I*B*b^6)*cos(d*x + c)^2 + 2*sqrt(2)*(2*I*C*a^ 4*b^2 + I*B*a^3*b^3 - 6*I*C*a^2*b^4 + 3*I*B*a*b^5)*cos(d*x + c) + sqrt(2)* (2*I*C*a^5*b + I*B*a^4*b^2 - 6*I*C*a^3*b^3 + 3*I*B*a^2*b^4))*sqrt(b)*weier strassZeta(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, weierstra ssPInverse(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*C*a^3*b...
Timed out. \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]