Integrand size = 43, antiderivative size = 147 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\frac {2 (b B-a C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 d}+\frac {2 \left (b^2 (3 A+C)-3 a (b B-a C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 b^3 d}-\frac {2 a \left (A b^2-a (b B-a C)\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b^3 (a+b) d}+\frac {2 C \sqrt {\cos (c+d x)} \sin (c+d x)}{3 b d} \]
2*(B*b-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^2/d+2/3*(b^2*(3*A+C)-3*a*(B*b-C*a))*(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^3/d-2*a*(A*b^2-a*(B*b-C*a))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2 *c)*EllipticPi(sin(1/2*d*x+1/2*c),2*b/(a+b),2^(1/2))/b^3/(a+b)/d+2/3*C*sin (d*x+c)*cos(d*x+c)^(1/2)/b/d
Time = 2.46 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.46 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\frac {-\frac {2 (-3 b B+a C) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {4 (3 A+C) \left ((a+b) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )\right )}{a+b}+4 C \sqrt {\cos (c+d x)} \sin (c+d x)+\frac {6 (b B-a C) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a b^2 \sqrt {\sin ^2(c+d x)}}}{6 b d} \]
Integrate[(Sqrt[Cos[c + d*x]]*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(a + b*Cos[c + d*x]),x]
((-2*(-3*b*B + a*C)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) + ( 4*(3*A + C)*((a + b)*EllipticF[(c + d*x)/2, 2] - a*EllipticPi[(2*b)/(a + b ), (c + d*x)/2, 2]))/(a + b) + 4*C*Sqrt[Cos[c + d*x]]*Sin[c + d*x] + (6*(b *B - a*C)*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*a*(a + b)* EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + (-2*a^2 + b^2)*EllipticPi[-(b/ a), ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(a*b^2*Sqrt[Sin[c + d*x ]^2]))/(6*b*d)
Time = 1.15 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.07, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.279, Rules used = {3042, 3528, 27, 3042, 3538, 25, 3042, 3119, 3481, 3042, 3120, 3284}
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.
\(\displaystyle \int \frac {\sqrt {\cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 3528 |
\(\displaystyle \frac {2 \int \frac {3 (b B-a C) \cos ^2(c+d x)+b (3 A+C) \cos (c+d x)+a C}{2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{3 b}+\frac {2 C \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {3 (b B-a C) \cos ^2(c+d x)+b (3 A+C) \cos (c+d x)+a C}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{3 b}+\frac {2 C \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {3 (b B-a C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+b (3 A+C) \sin \left (c+d x+\frac {\pi }{2}\right )+a C}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{3 b}+\frac {2 C \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}\) |
\(\Big \downarrow \) 3538 |
\(\displaystyle \frac {\frac {3 (b B-a C) \int \sqrt {\cos (c+d x)}dx}{b}-\frac {\int -\frac {a b C+\left (b^2 (3 A+C)-3 a (b B-a C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{3 b}+\frac {2 C \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {\int \frac {a b C+\left (b^2 (3 A+C)-3 a (b B-a C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}+\frac {3 (b B-a C) \int \sqrt {\cos (c+d x)}dx}{b}}{3 b}+\frac {2 C \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {a b C+\left (b^2 (3 A+C)-3 a (b B-a C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}+\frac {3 (b B-a C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}}{3 b}+\frac {2 C \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {\frac {\int \frac {a b C+\left (b^2 (3 A+C)-3 a (b B-a C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}+\frac {6 (b B-a C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{3 b}+\frac {2 C \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}\) |
\(\Big \downarrow \) 3481 |
\(\displaystyle \frac {\frac {\frac {\left (b^2 (3 A+C)-3 a (b B-a C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{b}-\frac {3 a \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{b}+\frac {6 (b B-a C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{3 b}+\frac {2 C \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {\left (b^2 (3 A+C)-3 a (b B-a C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {3 a \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}}{b}+\frac {6 (b B-a C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{3 b}+\frac {2 C \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {\frac {\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (b^2 (3 A+C)-3 a (b B-a C)\right )}{b d}-\frac {3 a \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}}{b}+\frac {6 (b B-a C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{3 b}+\frac {2 C \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}\) |
\(\Big \downarrow \) 3284 |
\(\displaystyle \frac {\frac {\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (b^2 (3 A+C)-3 a (b B-a C)\right )}{b d}-\frac {6 a \left (A b^2-a (b B-a C)\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b d (a+b)}}{b}+\frac {6 (b B-a C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{3 b}+\frac {2 C \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}\) |
((6*(b*B - a*C)*EllipticE[(c + d*x)/2, 2])/(b*d) + ((2*(b^2*(3*A + C) - 3* a*(b*B - a*C))*EllipticF[(c + d*x)/2, 2])/(b*d) - (6*a*(A*b^2 - a*(b*B - a *C))*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(b*(a + b)*d))/b)/(3*b) + (2*C*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*b*d)
3.11.97.3.1 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[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c , d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]
Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[ B/d Int[(a + b*Sin[e + f*x])^m, x], x] - Simp[(B*c - A*d)/d Int[(a + b* Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_ .) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x ])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Simp[1/(d*(m + n + 2)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A* d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2) - C*(a *c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n} , x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[ m, 0] && !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[C/(b*d) Int[Sqrt[a + b*Sin[e + f*x]], x] , x] - Simp[1/(b*d) Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[ e + f*x], x]/(Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])), x], x] /; Fre eQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0 ] && NeQ[c^2 - d^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(980\) vs. \(2(219)=438\).
Time = 3.34 (sec) , antiderivative size = 981, normalized size of antiderivative = 6.67
int(cos(d*x+c)^(1/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c)),x,me thod=_RETURNVERBOSE)
2/3*((-1+2*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-4*C*cos(1/2 *d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4*a*b^2+4*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+ 1/2*c)^4*b^3+3*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^( 1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))*a*b^2-3*a*A*b^2*(si n(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1 /2*d*x+1/2*c),2^(1/2))+3*A*b^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x +1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*B*(sin(1/2*d*x+ 1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticPi(cos(1/2*d*x+1/ 2*c),-2*b/(a-b),2^(1/2))*a^2*b+3*B*a^2*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*s in(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*B*a*b ^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF (cos(1/2*d*x+1/2*c),2^(1/2))+3*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d *x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^2-3*B*(sin( 1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2 *d*x+1/2*c),2^(1/2))*b^3+2*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*a*b^2 -2*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*b^3+3*C*(sin(1/2*d*x+1/2*c)^2 )^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2* b/(a-b),2^(1/2))*a^3-3*a^3*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1 /2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3*a^2*b*C*(sin(1/2* d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*...
Timed out. \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\text {Timed out} \]
integrate(cos(d*x+c)^(1/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c) ),x, algorithm="fricas")
Timed out. \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\text {Timed out} \]
\[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sqrt {\cos \left (d x + c\right )}}{b \cos \left (d x + c\right ) + a} \,d x } \]
integrate(cos(d*x+c)^(1/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c) ),x, algorithm="maxima")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sqrt(cos(d*x + c))/(b*co s(d*x + c) + a), x)
\[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sqrt {\cos \left (d x + c\right )}}{b \cos \left (d x + c\right ) + a} \,d x } \]
integrate(cos(d*x+c)^(1/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c) ),x, algorithm="giac")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sqrt(cos(d*x + c))/(b*co s(d*x + c) + a), x)
Timed out. \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\int \frac {\sqrt {\cos \left (c+d\,x\right )}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{a+b\,\cos \left (c+d\,x\right )} \,d x \]