Integrand size = 34, antiderivative size = 204 \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=-\frac {2 \left (a b B-2 a^2 C+b^2 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b^2 \left (a^2-b^2\right ) d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 (b B-2 a C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{b^2 d \sqrt {a+b \cos (c+d x)}}+\frac {2 a (b B-a C) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}} \] Output:
-2*(B*a*b-2*C*a^2+C*b^2)*(a+b*cos(d*x+c))^(1/2)*EllipticE(sin(1/2*d*x+1/2* c),2^(1/2)*(b/(a+b))^(1/2))/b^2/(a^2-b^2)/d/((a+b*cos(d*x+c))/(a+b))^(1/2) +2*(B*b-2*C*a)*((a+b*cos(d*x+c))/(a+b))^(1/2)*InverseJacobiAM(1/2*d*x+1/2* c,2^(1/2)*(b/(a+b))^(1/2))/b^2/d/(a+b*cos(d*x+c))^(1/2)+2*a*(B*b-C*a)*sin( d*x+c)/b/(a^2-b^2)/d/(a+b*cos(d*x+c))^(1/2)
Time = 1.40 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.83 \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=-\frac {2 \left (-\left ((a+b) \left (-a b B+2 a^2 C-b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )\right )+\left (a^2-b^2\right ) (-b B+2 a C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )+a b (-b B+a C) \sin (c+d x)\right )}{(a-b) b^2 (a+b) d \sqrt {a+b \cos (c+d x)}} \] Input:
Integrate[(B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(a + b*Cos[c + d*x])^(3/2),x ]
Output:
(-2*(-((a + b)*(-(a*b*B) + 2*a^2*C - b^2*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticE[(c + d*x)/2, (2*b)/(a + b)]) + (a^2 - b^2)*(-(b*B) + 2*a*C) *Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)] + a*b*(-(b*B) + a*C)*Sin[c + d*x]))/((a - b)*b^2*(a + b)*d*Sqrt[a + b*Cos[ c + d*x]])
Time = 1.02 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.07, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {3042, 3500, 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.
| \(\displaystyle \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {2 a (b B-a C) \sin (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {2 \int \frac {b (b B-a C)+\left (-2 C a^2+b B a+b^2 C\right ) \cos (c+d x)}{2 \sqrt {a+b \cos (c+d x)}}dx}{b \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 a (b B-a C) \sin (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {\int \frac {b (b B-a C)+\left (-2 C a^2+b B a+b^2 C\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a (b B-a C) \sin (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {\int \frac {b (b B-a C)+\left (-2 C a^2+b B a+b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3231 |
| \(\displaystyle \frac {2 a (b B-a C) \sin (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {\left (-2 a^2 C+a b B+b^2 C\right ) \int \sqrt {a+b \cos (c+d x)}dx}{b}-\frac {\left (a^2-b^2\right ) (b B-2 a C) \int \frac {1}{\sqrt {a+b \cos (c+d x)}}dx}{b}}{b \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a (b B-a C) \sin (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {\left (-2 a^2 C+a b B+b^2 C\right ) \int \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {\left (a^2-b^2\right ) (b B-2 a C) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}}{b \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3134 |
| \(\displaystyle \frac {2 a (b B-a C) \sin (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {\left (-2 a^2 C+a b B+b^2 C\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}dx}{b \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (a^2-b^2\right ) (b B-2 a C) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}}{b \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a (b B-a C) \sin (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {\left (-2 a^2 C+a b B+b^2 C\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{b \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (a^2-b^2\right ) (b B-2 a C) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}}{b \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3132 |
| \(\displaystyle \frac {2 a (b B-a C) \sin (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {2 \left (-2 a^2 C+a b B+b^2 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (a^2-b^2\right ) (b B-2 a C) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}}{b \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3142 |
| \(\displaystyle \frac {2 a (b B-a C) \sin (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {2 \left (-2 a^2 C+a b B+b^2 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (a^2-b^2\right ) (b B-2 a C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{b \sqrt {a+b \cos (c+d x)}}}{b \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a (b B-a C) \sin (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {2 \left (-2 a^2 C+a b B+b^2 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (a^2-b^2\right ) (b B-2 a C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{b \sqrt {a+b \cos (c+d x)}}}{b \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3140 |
| \(\displaystyle \frac {2 a (b B-a C) \sin (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {2 \left (-2 a^2 C+a b B+b^2 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 \left (a^2-b^2\right ) (b B-2 a C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{b d \sqrt {a+b \cos (c+d x)}}}{b \left (a^2-b^2\right )}\) |
Input:
Int[(B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(a + b*Cos[c + d*x])^(3/2),x]
Output:
-(((2*(a*b*B - 2*a^2*C + 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)*(b*B - 2*a*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]]))/(b*(a^2 - b^2))) + (2* a*(b*B - a*C)*Sin[c + d*x])/(b*(a^2 - b^2)*d*Sqrt[a + b*Cos[c + d*x]])
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_)*((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. \(518\) vs. \(2(203)=406\).
Time = 2.80 (sec) , antiderivative size = 519, normalized size of antiderivative = 2.54
| method | result | size |
| default | \(-\frac {\sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a +b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\frac {2 \sqrt {-\frac {2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a -b}+\frac {a +b}{a -b}}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (B b \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right )-2 C \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) a +C \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) a -C \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) b \right )}{b^{2} \sqrt {-2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (a +b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}+\frac {2 a \left (B b -C a \right ) \sqrt {-2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (a +b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b \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 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a -b}+\frac {a +b}{a -b}}\, \operatorname {EllipticE}\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 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a -b}+\frac {a +b}{a -b}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) b \right )}{b^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right ) \left (a^{2}-b^{2}\right )}\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b}\, d}\) | \(519\) |
| parts | \(\text {Expression too large to display}\) | \(905\) |
Input:
int((B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(3/2),x,method=_RETURNV ERBOSE)
Output:
-(-(-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/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/(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*b*Ellip ticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-2*C*EllipticF(cos(1/2*d*x+1/2* c),(-2*b/(a-b))^(1/2))*a+C*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2) )*a-C*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b)+2*a*(B*b-C*a)/b^ 2/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2*b-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*sin(1/2*d*x+1/2*c)^2 *b*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)*EllipticE(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)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b))/sin(1/2* d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*b+a+b)^(1/2)/d
Result contains complex when optimal does not.
Time = 0.13 (sec) , antiderivative size = 663, normalized size of antiderivative = 3.25 \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx =\text {Too large to display} \] Input:
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(3/2),x, algorith m="fricas")
Output:
2/3*(sqrt(1/2)*(4*I*C*a^4 - 2*I*B*a^3*b - 5*I*C*a^2*b^2 + 3*I*B*a*b^3 + (4 *I*C*a^3*b - 2*I*B*a^2*b^2 - 5*I*C*a*b^3 + 3*I*B*b^4)*cos(d*x + c))*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(1/2)*(-4*I*C *a^4 + 2*I*B*a^3*b + 5*I*C*a^2*b^2 - 3*I*B*a*b^3 + (-4*I*C*a^3*b + 2*I*B*a ^2*b^2 + 5*I*C*a*b^3 - 3*I*B*b^4)*cos(d*x + c))*sqrt(b)*weierstrassPInvers e(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(1/2)*(2*I*C*a^3*b - I*B*a^2*b^ 2 - I*C*a*b^3 + (2*I*C*a^2*b^2 - I*B*a*b^3 - I*C*b^4)*cos(d*x + c))*sqrt(b )*weierstrassZeta(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, we ierstrassPInverse(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(1/2)*(-2*I*C* a^3*b + I*B*a^2*b^2 + I*C*a*b^3 + (-2*I*C*a^2*b^2 + I*B*a*b^3 + I*C*b^4)*c os(d*x + c))*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 *(C*a^2*b^2 - B*a*b^3)*sqrt(b*cos(d*x + c) + a)*sin(d*x + c))/((a^2*b^4 - b^6)*d*cos(d*x + c) + (a^3*b^3 - a*b^5)*d)
Timed out. \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \] Input:
integrate((B*cos(d*x+c)+C*cos(d*x+c)**2)/(a+b*cos(d*x+c))**(3/2),x)
Output:
Timed out
\[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(3/2),x, algorith m="maxima")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))/(b*cos(d*x + c) + a)^(3/2), x)
\[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(3/2),x, algorith m="giac")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))/(b*cos(d*x + c) + a)^(3/2), x)
Timed out. \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \] Input:
int((B*cos(c + d*x) + C*cos(c + d*x)^2)/(a + b*cos(c + d*x))^(3/2),x)
Output:
int((B*cos(c + d*x) + C*cos(c + d*x)^2)/(a + b*cos(c + d*x))^(3/2), x)
\[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )}{\cos \left (d x +c \right )^{2} b^{2}+2 \cos \left (d x +c \right ) a b +a^{2}}d x \right ) b +\left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )^{2}}{\cos \left (d x +c \right )^{2} b^{2}+2 \cos \left (d x +c \right ) a b +a^{2}}d x \right ) c \] Input:
int((B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(3/2),x)
Output:
int((sqrt(cos(c + d*x)*b + a)*cos(c + d*x))/(cos(c + d*x)**2*b**2 + 2*cos( c + d*x)*a*b + a**2),x)*b + int((sqrt(cos(c + d*x)*b + a)*cos(c + d*x)**2) /(cos(c + d*x)**2*b**2 + 2*cos(c + d*x)*a*b + a**2),x)*c