Integrand size = 50, antiderivative size = 180 \[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\frac {2 (b B-2 a C) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{\left (a^2-b^2\right ) d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 C \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}-\frac {2 b (b B-2 a C) \sin (c+d x)}{\left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}} \]
-2*b*(B*b-2*C*a)*sin(d*x+c)/(a^2-b^2)/d/(a+b*cos(d*x+c))^(1/2)+2*(B*b-2*C* a)*(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)/(a^2-b^2)/d/((a+b*co s(d*x+c))/(a+b))^(1/2)+2*C*(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)/d/(a+b*cos(d*x+c))^(1/2)
Time = 0.83 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.83 \[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\frac {2 \left (-\left ((a+b) (-b B+2 a C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )\right )+\left (a^2-b^2\right ) C \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )+b (-b B+2 a C) \sin (c+d x)\right )}{(a-b) (a+b) d \sqrt {a+b \cos (c+d x)}} \]
Integrate[(a*b*B - a^2*C + b^2*B*Cos[c + d*x] + b^2*C*Cos[c + d*x]^2)/(a + b*Cos[c + d*x])^(5/2),x]
(2*(-((a + b)*(-(b*B) + 2*a*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*Elliptic E[(c + d*x)/2, (2*b)/(a + b)]) + (a^2 - b^2)*C*Sqrt[(a + b*Cos[c + d*x])/( a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)] + b*(-(b*B) + 2*a*C)*Sin[c + d*x]))/((a - b)*(a + b)*d*Sqrt[a + b*Cos[c + d*x]])
Time = 0.99 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.13, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.260, Rules used = {2014, 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.
| \(\displaystyle \int \frac {a^2 (-C)+a b B+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 2014 |
| \(\displaystyle \frac {\int \frac {C \cos (c+d x) b^3+(b B-a C) b^2}{(a+b \cos (c+d x))^{3/2}}dx}{b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {C \sin \left (c+d x+\frac {\pi }{2}\right ) b^3+(b B-a C) b^2}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{b^2}\) |
| \(\Big \downarrow \) 3233 |
| \(\displaystyle \frac {-\frac {2 \int -\frac {(b B-2 a C) \cos (c+d x) b^3+\left (-C a^2+b B a-b^2 C\right ) b^2}{2 \sqrt {a+b \cos (c+d x)}}dx}{a^2-b^2}-\frac {2 b^3 (b B-2 a C) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{b^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {(b B-2 a C) \cos (c+d x) b^3+\left (-C a^2+b B a-b^2 C\right ) b^2}{\sqrt {a+b \cos (c+d x)}}dx}{a^2-b^2}-\frac {2 b^3 (b B-2 a C) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {(b B-2 a C) \sin \left (c+d x+\frac {\pi }{2}\right ) b^3+\left (-C a^2+b B a-b^2 C\right ) b^2}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2-b^2}-\frac {2 b^3 (b B-2 a C) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{b^2}\) |
| \(\Big \downarrow \) 3231 |
| \(\displaystyle \frac {\frac {b^2 C \left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}}dx+b^2 (b B-2 a C) \int \sqrt {a+b \cos (c+d x)}dx}{a^2-b^2}-\frac {2 b^3 (b B-2 a C) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {b^2 C \left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+b^2 (b B-2 a C) \int \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2-b^2}-\frac {2 b^3 (b B-2 a C) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{b^2}\) |
| \(\Big \downarrow \) 3134 |
| \(\displaystyle \frac {\frac {b^2 C \left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {b^2 (b B-2 a C) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a^2-b^2}-\frac {2 b^3 (b B-2 a C) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {b^2 C \left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {b^2 (b B-2 a C) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a^2-b^2}-\frac {2 b^3 (b B-2 a C) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{b^2}\) |
| \(\Big \downarrow \) 3132 |
| \(\displaystyle \frac {\frac {b^2 C \left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b^2 (b B-2 a C) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a^2-b^2}-\frac {2 b^3 (b B-2 a C) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{b^2}\) |
| \(\Big \downarrow \) 3142 |
| \(\displaystyle \frac {\frac {\frac {b^2 C \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}+\frac {2 b^2 (b B-2 a C) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a^2-b^2}-\frac {2 b^3 (b B-2 a C) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {b^2 C \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}+\frac {2 b^2 (b B-2 a C) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a^2-b^2}-\frac {2 b^3 (b B-2 a C) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{b^2}\) |
| \(\Big \downarrow \) 3140 |
| \(\displaystyle \frac {\frac {\frac {2 b^2 C \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}+\frac {2 b^2 (b B-2 a C) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a^2-b^2}-\frac {2 b^3 (b B-2 a C) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{b^2}\) |
Int[(a*b*B - a^2*C + b^2*B*Cos[c + d*x] + b^2*C*Cos[c + d*x]^2)/(a + b*Cos [c + d*x])^(5/2),x]
(((2*b^2*(b*B - 2*a*C)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2* b)/(a + b)])/(d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) + (2*b^2*(a^2 - b^2)*C *Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)]) /(d*Sqrt[a + b*Cos[c + d*x]]))/(a^2 - b^2) - (2*b^3*(b*B - 2*a*C)*Sin[c + d*x])/((a^2 - b^2)*d*Sqrt[a + b*Cos[c + d*x]]))/b^2
3.11.63.3.1 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[(u_.)*((a_) + (b_.)*(v_))^(m_)*((A_.) + (B_.)*(v_) + (C_.)*(v_)^2), x_S ymbol] :> Simp[1/b^2 Int[u*(a + b*v)^(m + 1)*Simp[b*B - a*C + b*C*v, x], x], x] /; FreeQ[{a, b, A, B, C}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0] && LeQ [m, -1]
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]
Time = 4.87 (sec) , antiderivative size = 426, normalized size of antiderivative = 2.37
| method | result | size |
| default | \(-\frac {\sqrt {-\left (-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -a +b \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\frac {2 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a -b}{a -b}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right )}{\sqrt {-2 b \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (a +b \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}-\frac {2 \left (B b -2 C a \right ) \sqrt {-2 b \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (a +b \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (2 b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-\frac {2 b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a -b}+\frac {a +b}{a -b}}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) a -\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-\frac {2 b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a -b}+\frac {a +b}{a -b}}\, b E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right )\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (2 b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a -b \right ) \left (a^{2}-b^{2}\right )}\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b}\, d}\) | \(426\) |
| parts | \(\text {Expression too large to display}\) | \(2101\) |
int((B*a*b-a^2*C+b^2*B*cos(d*x+c)+b^2*C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(5/ 2),x,method=_RETURNVERBOSE)
-(-(-2*cos(1/2*d*x+1/2*c)^2*b-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*C*(sin(1 /2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b* sin(1/2*d*x+1/2*c)^4+(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*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*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d *x+1/2*c)^2)^(1/2)*(2*b*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+(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)*Ell ipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a-(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)*b*EllipticE(cos(1/2 *d*x+1/2*c),(-2*b/(a-b))^(1/2))))/sin(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 = 0.12 (sec) , antiderivative size = 609, normalized size of antiderivative = 3.38 \[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\frac {6 \, {\left (2 \, C a b^{2} - B b^{3}\right )} \sqrt {b \cos \left (d x + c\right ) + a} \sin \left (d x + c\right ) - {\left (\sqrt {2} {\left (i \, C a^{2} b + i \, B a b^{2} - 3 i \, C b^{3}\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, C a^{3} + i \, B a^{2} b - 3 i \, C a 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 (-i \, C a^{2} b - i \, B a b^{2} + 3 i \, C b^{3}\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, C a^{3} - i \, B a^{2} b + 3 i \, C a 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 \, C a b^{2} + i \, B b^{3}\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-2 i \, C a^{2} b + i \, B a b^{2}\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 \, C a b^{2} - i \, B b^{3}\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (2 i \, C a^{2} b - i \, B a b^{2}\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^{2} - b^{4}\right )} d \cos \left (d x + c\right ) + {\left (a^{3} b - a b^{3}\right )} d\right )}} \]
integrate((B*a*b-a^2*C+b^2*B*cos(d*x+c)+b^2*C*cos(d*x+c)^2)/(a+b*cos(d*x+c ))^(5/2),x, algorithm="fricas")
1/3*(6*(2*C*a*b^2 - B*b^3)*sqrt(b*cos(d*x + c) + a)*sin(d*x + c) - (sqrt(2 )*(I*C*a^2*b + I*B*a*b^2 - 3*I*C*b^3)*cos(d*x + c) + sqrt(2)*(I*C*a^3 + I* B*a^2*b - 3*I*C*a*b^2))*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)*(-I*C*a^2*b - I*B*a*b^2 + 3*I*C*b^3)*cos(d*x + c) + sqrt(2)*(-I*C*a^3 - I*B*a^2*b + 3*I*C*a*b^2))*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) + 3*(sqrt(2)*(-2*I*C*a*b^2 + I*B*b^3)*cos (d*x + c) + sqrt(2)*(-2*I*C*a^2*b + I*B*a*b^2))*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)) + 3*(sqrt(2)*(2*I*C*a*b^2 - I*B*b^3)*cos(d *x + c) + sqrt(2)*(2*I*C*a^2*b - I*B*a*b^2))*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^2 - b^4)*d*cos(d*x + c) + (a^3*b - a *b^3)*d)
Timed out. \[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\int { \frac {C b^{2} \cos \left (d x + c\right )^{2} + B b^{2} \cos \left (d x + c\right ) - C a^{2} + B a b}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
integrate((B*a*b-a^2*C+b^2*B*cos(d*x+c)+b^2*C*cos(d*x+c)^2)/(a+b*cos(d*x+c ))^(5/2),x, algorithm="maxima")
integrate((C*b^2*cos(d*x + c)^2 + B*b^2*cos(d*x + c) - C*a^2 + B*a*b)/(b*c os(d*x + c) + a)^(5/2), x)
\[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\int { \frac {C b^{2} \cos \left (d x + c\right )^{2} + B b^{2} \cos \left (d x + c\right ) - C a^{2} + B a b}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
integrate((B*a*b-a^2*C+b^2*B*cos(d*x+c)+b^2*C*cos(d*x+c)^2)/(a+b*cos(d*x+c ))^(5/2),x, algorithm="giac")
integrate((C*b^2*cos(d*x + c)^2 + B*b^2*cos(d*x + c) - C*a^2 + B*a*b)/(b*c os(d*x + c) + a)^(5/2), x)
Timed out. \[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\int \frac {-C\,a^2+B\,a\,b+C\,b^2\,{\cos \left (c+d\,x\right )}^2+B\,b^2\,\cos \left (c+d\,x\right )}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]
int((C*b^2*cos(c + d*x)^2 - C*a^2 + B*a*b + B*b^2*cos(c + d*x))/(a + b*cos (c + d*x))^(5/2),x)