Integrand size = 35, antiderivative size = 427 \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{\sqrt {a+b \cos (c+d x)}} \, dx=-\frac {(a-b) \sqrt {a+b} B \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{a b d}+\frac {\sqrt {a+b} B \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{b d}-\frac {\sqrt {a+b} (2 A b-a B) \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{b^2 d}+\frac {a B \sin (c+d x)}{b d \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}+\frac {B \sqrt {\cos (c+d x)} \sin (c+d x)}{d \sqrt {a+b \cos (c+d x)}} \]
a*B*sin(d*x+c)/b/d/cos(d*x+c)^(1/2)/(a+b*cos(d*x+c))^(1/2)+B*sin(d*x+c)*co s(d*x+c)^(1/2)/d/(a+b*cos(d*x+c))^(1/2)-(a-b)*B*cot(d*x+c)*EllipticE((a+b* cos(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),((-a-b)/(a-b))^(1/2))*(a+b) ^(1/2)*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/a/b/d +B*cot(d*x+c)*EllipticF((a+b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2 ),((-a-b)/(a-b))^(1/2))*(a+b)^(1/2)*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+s ec(d*x+c))/(a-b))^(1/2)/b/d-(2*A*b-B*a)*cot(d*x+c)*EllipticPi((a+b*cos(d*x +c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),(a+b)/b,((-a-b)/(a-b))^(1/2))*(a+b )^(1/2)*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/b^2/ d
Time = 7.04 (sec) , antiderivative size = 402, normalized size of antiderivative = 0.94 \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{\sqrt {a+b \cos (c+d x)}} \, dx=\frac {\sqrt {\cos (c+d x)} \left (2 (a+b) B \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {-a+b}{a+b}\right )-4 A b \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right )+8 A b \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right )-4 a B \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right )+b B \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sec \left (\frac {1}{2} (c+d x)\right ) \sin \left (\frac {3}{2} (c+d x)\right )+2 a B \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \tan \left (\frac {1}{2} (c+d x)\right )-b B \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \tan \left (\frac {1}{2} (c+d x)\right )\right )}{2 b d \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {a+b \cos (c+d x)}} \]
(Sqrt[Cos[c + d*x]]*(2*(a + b)*B*Sqrt[(a + b*Cos[c + d*x])/((a + b)*(1 + C os[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)] - 4*A *b*Sqrt[(a + b*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSi n[Tan[(c + d*x)/2]], (-a + b)/(a + b)] + 8*A*b*Sqrt[(a + b*Cos[c + d*x])/( (a + b)*(1 + Cos[c + d*x]))]*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)] - 4*a*B*Sqrt[(a + b*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x] ))]*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)] + b*B*Sqrt[ Cos[c + d*x]/(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sin[(3*(c + d*x))/2] + 2 *a*B*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Tan[(c + d*x)/2] - b*B*Sqrt[Cos [c + d*x]/(1 + Cos[c + d*x])]*Tan[(c + d*x)/2]))/(2*b*d*Sqrt[Cos[c + d*x]/ (1 + Cos[c + d*x])]*Sqrt[a + b*Cos[c + d*x]])
Time = 1.79 (sec) , antiderivative size = 436, normalized size of antiderivative = 1.02, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3482, 3042, 3530, 3042, 3288, 3472, 25, 27, 3042, 3280, 3042, 3295, 3473}
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.
\(\displaystyle \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{\sqrt {a+b \cos (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 3482 |
\(\displaystyle \frac {1}{2} \int \frac {(2 A b-a B) \cos ^2(c+d x)+2 a A \cos (c+d x)+a B}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2}}dx+\frac {B \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a+b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \int \frac {(2 A b-a B) \sin \left (c+d x+\frac {\pi }{2}\right )^2+2 a A \sin \left (c+d x+\frac {\pi }{2}\right )+a B}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx+\frac {B \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a+b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3530 |
\(\displaystyle \frac {1}{2} \left (\frac {\int \frac {B \cos (c+d x) a^2+b B a}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2}}dx}{b}+\frac {(2 A b-a B) \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {a+b \cos (c+d x)}}dx}{b}\right )+\frac {B \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a+b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (\frac {\int \frac {B \sin \left (c+d x+\frac {\pi }{2}\right ) a^2+b B a}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{b}+\frac {(2 A b-a B) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}\right )+\frac {B \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a+b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3288 |
\(\displaystyle \frac {1}{2} \left (\frac {\int \frac {B \sin \left (c+d x+\frac {\pi }{2}\right ) a^2+b B a}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{b}-\frac {2 \sqrt {a+b} (2 A b-a B) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b^2 d}\right )+\frac {B \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a+b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3472 |
\(\displaystyle \frac {1}{2} \left (\frac {\frac {\int -\frac {a \left (a^2-b^2\right ) B}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{a^2-b^2}+\frac {2 a B \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 \sqrt {a+b} (2 A b-a B) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b^2 d}\right )+\frac {B \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a+b \cos (c+d x)}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (\frac {\frac {2 a B \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}-\frac {\int \frac {a \left (a^2-b^2\right ) B}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{a^2-b^2}}{b}-\frac {2 \sqrt {a+b} (2 A b-a B) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b^2 d}\right )+\frac {B \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a+b \cos (c+d x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \left (\frac {\frac {2 a B \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}-a B \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{b}-\frac {2 \sqrt {a+b} (2 A b-a B) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b^2 d}\right )+\frac {B \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a+b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (\frac {\frac {2 a B \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}-a B \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {2 \sqrt {a+b} (2 A b-a B) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b^2 d}\right )+\frac {B \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a+b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3280 |
\(\displaystyle \frac {1}{2} \left (\frac {\frac {2 a B \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}-a B \left (\int \frac {\cos (c+d x)+1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx-\int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx\right )}{b}-\frac {2 \sqrt {a+b} (2 A b-a B) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b^2 d}\right )+\frac {B \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a+b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (\frac {\frac {2 a B \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}-a B \left (\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\right )}{b}-\frac {2 \sqrt {a+b} (2 A b-a B) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b^2 d}\right )+\frac {B \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a+b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3295 |
\(\displaystyle \frac {1}{2} \left (\frac {\frac {2 a B \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}-a B \left (\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}\right )}{b}-\frac {2 \sqrt {a+b} (2 A b-a B) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b^2 d}\right )+\frac {B \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a+b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3473 |
\(\displaystyle \frac {1}{2} \left (\frac {\frac {2 a B \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}-a B \left (\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a^2 d}-\frac {2 \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}\right )}{b}-\frac {2 \sqrt {a+b} (2 A b-a B) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b^2 d}\right )+\frac {B \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a+b \cos (c+d x)}}\) |
(B*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + b*Cos[c + d*x]]) + ((-2*Sq rt[a + b]*(2*A*b - a*B)*Cot[c + d*x]*EllipticPi[(a + b)/b, ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sq rt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/( b^2*d) + (-(a*B*((2*(a - b)*Sqrt[a + b]*Cot[c + d*x]*EllipticE[ArcSin[Sqrt [a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b)) ]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b) ])/(a^2*d) - (2*Sqrt[a + b]*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(a*d))) + (2*a*B*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]))/b)/ 2
3.5.21.3.1 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[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)*Sqrt[(c_.) + (d_.)*sin [(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[1/(a - b) Int[1/(Sqrt[a + b*Sin [e + f*x]]*Sqrt[c + d*Sin[e + f*x]]), x], x] - Simp[b/(a - b) Int[(1 + Si n[e + f*x])/((a + b*Sin[e + f*x])^(3/2)*Sqrt[c + d*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] && NeQ[c^2 - d^2, 0]
Int[Sqrt[(b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.) *(x_)]], x_Symbol] :> Simp[2*b*(Tan[e + f*x]/(d*f))*Rt[(c + d)/b, 2]*Sqrt[c *((1 + Csc[e + f*x])/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*Ellipti cPi[(c + d)/d, ArcSin[Sqrt[c + d*Sin[e + f*x]]/Sqrt[b*Sin[e + f*x]]/Rt[(c + d)/b, 2]], -(c + d)/(c - d)], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2, 0] && PosQ[(c + d)/b]
Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f _.)*(x_)]]), x_Symbol] :> Simp[-2*(Tan[e + f*x]/(a*f))*Rt[(a + b)/d, 2]*Sqr t[a*((1 - Csc[e + f*x])/(a + b))]*Sqrt[a*((1 + Csc[e + f*x])/(a - b))]*Elli pticF[ArcSin[Sqrt[a + b*Sin[e + f*x]]/Sqrt[d*Sin[e + f*x]]/Rt[(a + b)/d, 2] ], -(a + b)/(a - b)], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && PosQ[(a + b)/d]
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(d_.)*sin[(e_.) + (f_.)*( x_)]]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)), x_Symbol] :> Simp[2*(A *b - a*B)*(Cos[e + f*x]/(f*(a^2 - b^2)*Sqrt[a + b*Sin[e + f*x]]*Sqrt[d*Sin[ e + f*x]])), x] + Simp[d/(a^2 - b^2) Int[(A*b - a*B + (a*A - b*B)*Sin[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*(d*Sin[e + f*x])^(3/2)), x], x] /; FreeQ[{ a, b, d, e, f, A, B}, x] && NeQ[a^2 - b^2, 0]
Int[((A_) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((b_.)*sin[(e_.) + (f_.)*(x_)]) ^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*A* (c - d)*(Tan[e + f*x]/(f*b*c^2))*Rt[(c + d)/b, 2]*Sqrt[c*((1 + Csc[e + f*x] )/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*EllipticE[ArcSin[Sqrt[c + d*Sin[e + f*x]]/Sqrt[b*Sin[e + f*x]]/Rt[(c + d)/b, 2]], -(c + d)/(c - d)], x] /; FreeQ[{b, c, d, e, f, A, B}, x] && NeQ[c^2 - d^2, 0] && EqQ[A, B] && PosQ[(c + d)/b]
Int[Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[-2*B*Cos[e + f*x]*Sqrt[a + b*Sin[e + f*x]]*((c + d*Sin[e + f*x])^n/(f*(2* n + 3))), x] + Simp[1/(2*n + 3) Int[((c + d*Sin[e + f*x])^(n - 1)/Sqrt[a + b*Sin[e + f*x]])*Simp[a*A*c*(2*n + 3) + B*(b*c + 2*a*d*n) + (B*(a*c + b*d )*(2*n + 1) + A*(b*c + a*d)*(2*n + 3))*Sin[e + f*x] + (A*b*d*(2*n + 3) + B* (a*d + 2*b*c*n))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Eq Q[n^2, 1/4]
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 2)/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_ )])^(3/2)), x_Symbol] :> Simp[C/(b*d) Int[Sqrt[d*Sin[e + f*x]]/Sqrt[a + b *Sin[e + f*x]], x], x] + Simp[1/b Int[(A*b + (b*B - a*C)*Sin[e + f*x])/(( a + b*Sin[e + f*x])^(3/2)*Sqrt[d*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(980\) vs. \(2(395)=790\).
Time = 11.54 (sec) , antiderivative size = 981, normalized size of antiderivative = 2.30
method | result | size |
parts | \(\text {Expression too large to display}\) | \(981\) |
default | \(\text {Expression too large to display}\) | \(1469\) |
2*A/d*(EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))-2*EllipticPi( cot(d*x+c)-csc(d*x+c),-1,(-(a-b)/(a+b))^(1/2)))*(cos(d*x+c)/(1+cos(d*x+c)) )^(1/2)*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)/(a+cos(d*x+c)*b)^(1/ 2)*(1+cos(d*x+c))/cos(d*x+c)^(1/2)+B/d*(-EllipticE(cot(d*x+c)-csc(d*x+c),( -(a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+cos(d*x+c)*b)/( 1+cos(d*x+c))/(a+b))^(1/2)*a*cos(d*x+c)^2-EllipticE(cot(d*x+c)-csc(d*x+c), (-(a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+cos(d*x+c)*b)/ (1+cos(d*x+c))/(a+b))^(1/2)*b*cos(d*x+c)^2+2*EllipticPi(cot(d*x+c)-csc(d*x +c),-1,(-(a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+cos(d*x +c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*a*cos(d*x+c)^2-2*EllipticE(cot(d*x+c)-c sc(d*x+c),(-(a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+cos( d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*a*cos(d*x+c)-2*EllipticE(cot(d*x+c)- csc(d*x+c),(-(a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+cos (d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*b*cos(d*x+c)+4*EllipticPi(cot(d*x+c )-csc(d*x+c),-1,(-(a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(( a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*a*cos(d*x+c)+b*cos(d*x+c)^2*si n(d*x+c)-(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c) ,(-(a-b)/(a+b))^(1/2))*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*a-(co s(d*x+c)/(1+cos(d*x+c)))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+ b))^(1/2))*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*b+2*(cos(d*x+c...
\[ \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{\sqrt {a+b \cos (c+d x)}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sqrt {\cos \left (d x + c\right )}}{\sqrt {b \cos \left (d x + c\right ) + a}} \,d x } \]
\[ \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{\sqrt {a+b \cos (c+d x)}} \, dx=\int \frac {\left (A + B \cos {\left (c + d x \right )}\right ) \sqrt {\cos {\left (c + d x \right )}}}{\sqrt {a + b \cos {\left (c + d x \right )}}}\, dx \]
\[ \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{\sqrt {a+b \cos (c+d x)}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sqrt {\cos \left (d x + c\right )}}{\sqrt {b \cos \left (d x + c\right ) + a}} \,d x } \]
\[ \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{\sqrt {a+b \cos (c+d x)}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sqrt {\cos \left (d x + c\right )}}{\sqrt {b \cos \left (d x + c\right ) + a}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{\sqrt {a+b \cos (c+d x)}} \, dx=\int \frac {\sqrt {\cos \left (c+d\,x\right )}\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{\sqrt {a+b\,\cos \left (c+d\,x\right )}} \,d x \]