Integrand size = 44, antiderivative size = 479 \[ \int \frac {\sqrt {\cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx=-\frac {(a-b) \sqrt {a+b} (4 b B-3 a C) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{4 a b^2 d}-\frac {\sqrt {a+b} (3 a C-2 b (2 B+C)) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{4 b^2 d}+\frac {\sqrt {a+b} \left (4 a b B-3 a^2 C-4 b^2 C\right ) \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{4 b^3 d}+\frac {(4 b B-3 a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{4 b^2 d \sqrt {\cos (c+d x)}}+\frac {C \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{2 b d} \]
1/4*(4*B*b-3*C*a)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/b^2/d/cos(d*x+c)^(1/2) +1/2*C*sin(d*x+c)*cos(d*x+c)^(1/2)*(a+b*cos(d*x+c))^(1/2)/b/d-1/4*(a-b)*(4 *B*b-3*C*a)*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^2/d-1/4*(3*a*C-2*b*(2*B+C))*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+sec(d*x+c))/( a-b))^(1/2)/b^2/d+1/4*(4*B*a*b-3*C*a^2-4*C*b^2)*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^3/d
Result contains complex when optimal does not.
Time = 19.41 (sec) , antiderivative size = 1175, normalized size of antiderivative = 2.45 \[ \int \frac {\sqrt {\cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx =\text {Too large to display} \]
Integrate[(Sqrt[Cos[c + d*x]]*(B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Sqrt[a + b*Cos[c + d*x]],x]
(C*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(2*b*d) + ((- 4*a*(4*b*B - a*C)*Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Cos[c + d*x]*Csc[(c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]*Csc[c + d*x]*EllipticF[ArcSin[Sqrt[((a + b*Cos[c + d*x])* Csc[(c + d*x)/2]^2)/a]/Sqrt[2]], (-2*a)/(-a + b)]*Sin[(c + d*x)/2]^4)/((a + b)*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]) - 16*a*b*C*((Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Cos[c + d*x]*Csc[(c + d*x )/2]^2)/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]*Csc[c + d*x] *EllipticF[ArcSin[Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]/Sqrt[2 ]], (-2*a)/(-a + b)]*Sin[(c + d*x)/2]^4)/((a + b)*Sqrt[Cos[c + d*x]]*Sqrt[ a + b*Cos[c + d*x]]) - (Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*Sqrt[- (((a + b)*Cos[c + d*x]*Csc[(c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos[c + d*x])* Csc[(c + d*x)/2]^2)/a]*Csc[c + d*x]*EllipticPi[-(a/b), ArcSin[Sqrt[((a + b *Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]/Sqrt[2]], (-2*a)/(-a + b)]*Sin[(c + d*x)/2]^4)/(b*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]])) + 2*(4*b*B - 3 *a*C)*((I*Cos[(c + d*x)/2]*Sqrt[a + b*Cos[c + d*x]]*EllipticE[I*ArcSinh[Si n[(c + d*x)/2]/Sqrt[Cos[c + d*x]]], (-2*a)/(-a - b)]*Sec[c + d*x])/(b*Sqrt [Cos[(c + d*x)/2]^2*Sec[c + d*x]]*Sqrt[((a + b*Cos[c + d*x])*Sec[c + d*x]) /(a + b)]) + (2*a*((a*Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(( (a + b)*Cos[c + d*x]*Csc[(c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos[c + d*x])...
Time = 2.26 (sec) , antiderivative size = 480, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3042, 3508, 3042, 3469, 27, 3042, 3540, 25, 3042, 3532, 3042, 3288, 3477, 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.
\(\displaystyle \int \frac {\sqrt {\cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 3508 |
\(\displaystyle \int \frac {\cos ^{\frac {3}{2}}(c+d x) (B+C \cos (c+d x))}{\sqrt {a+b \cos (c+d x)}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (B+C \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 3469 |
\(\displaystyle \frac {\int \frac {(4 b B-3 a C) \cos ^2(c+d x)+2 b C \cos (c+d x)+a C}{2 \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx}{2 b}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {(4 b B-3 a C) \cos ^2(c+d x)+2 b C \cos (c+d x)+a C}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx}{4 b}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {(4 b B-3 a C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+2 b C \sin \left (c+d x+\frac {\pi }{2}\right )+a C}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{4 b}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}\) |
\(\Big \downarrow \) 3540 |
\(\displaystyle \frac {\frac {\int -\frac {\left (-3 C a^2+4 b B a-4 b^2 C\right ) \cos ^2(c+d x)-2 a b C \cos (c+d x)+a (4 b B-3 a C)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{2 b}+\frac {(4 b B-3 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{4 b}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {(4 b B-3 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {\int \frac {\left (-3 C a^2+4 b B a-4 b^2 C\right ) \cos ^2(c+d x)-2 a b C \cos (c+d x)+a (4 b B-3 a C)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{2 b}}{4 b}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {(4 b B-3 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {\int \frac {\left (-3 C a^2+4 b B a-4 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-2 a b C \sin \left (c+d x+\frac {\pi }{2}\right )+a (4 b B-3 a C)}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 b}}{4 b}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}\) |
\(\Big \downarrow \) 3532 |
\(\displaystyle \frac {\frac {(4 b B-3 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {\left (-3 a^2 C+4 a b B-4 b^2 C\right ) \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {a+b \cos (c+d x)}}dx+\int \frac {a (4 b B-3 a C)-2 a b C \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{2 b}}{4 b}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {(4 b B-3 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {\left (-3 a^2 C+4 a b B-4 b^2 C\right ) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\int \frac {a (4 b B-3 a C)-2 a b C \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 b}}{4 b}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}\) |
\(\Big \downarrow \) 3288 |
\(\displaystyle \frac {\frac {(4 b B-3 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {\int \frac {a (4 b B-3 a C)-2 a b C \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} \left (-3 a^2 C+4 a b B-4 b^2 C\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}}{2 b}}{4 b}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}\) |
\(\Big \downarrow \) 3477 |
\(\displaystyle \frac {\frac {(4 b B-3 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {a (4 b B-3 a C) \int \frac {\cos (c+d x)+1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx+a (3 a C-2 b (2 B+C)) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx-\frac {2 \sqrt {a+b} \left (-3 a^2 C+4 a b B-4 b^2 C\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}}{2 b}}{4 b}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {(4 b B-3 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {a (3 a C-2 b (2 B+C)) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+a (4 b B-3 a C) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} \left (-3 a^2 C+4 a b B-4 b^2 C\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}}{2 b}}{4 b}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}\) |
\(\Big \downarrow \) 3295 |
\(\displaystyle \frac {\frac {(4 b B-3 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {a (4 b B-3 a C) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} \left (-3 a^2 C+4 a b B-4 b^2 C\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}+\frac {2 \sqrt {a+b} (3 a C-2 b (2 B+C)) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}}{2 b}}{4 b}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}\) |
\(\Big \downarrow \) 3473 |
\(\displaystyle \frac {\frac {(4 b B-3 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {-\frac {2 \sqrt {a+b} \left (-3 a^2 C+4 a b B-4 b^2 C\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}+\frac {2 \sqrt {a+b} (3 a C-2 b (2 B+C)) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}+\frac {2 (a-b) \sqrt {a+b} (4 b B-3 a C) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a d}}{2 b}}{4 b}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}\) |
(C*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(2*b*d) + (-1 /2*((2*(a - b)*Sqrt[a + b]*(4*b*B - 3*a*C)*Cot[c + d*x]*EllipticE[ArcSin[S qrt[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*Sqrt[a + b]*(3*a*C - 2*b*(2*B + C))*Cot[c + d*x]*Elliptic F[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)])/d - (2*Sqrt[a + b]*(4*a*b*B - 3*a^2*C - 4*b^2*C)*Cot[c + d *x]*EllipticPi[(a + b)/b, ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqr t[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)])/(b*d))/b + ((4*b*B - 3*a*C)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(b*d*Sqrt[Cos[c + d*x]]))/(4*b)
3.10.21.3.1 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[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_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[(-b)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^( n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1)) Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a^2*A*d*(m + n + 1) + b*B*(b*c*( m - 1) + a*d*(n + 1)) + (a*d*(2*A*b + a*B)*(m + n + 1) - b*B*(a*c - b*d*(m + n)))*Sin[e + f*x] + b*(A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin [e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] && !(IGt Q[n, 1] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 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[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ .)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> S imp[(A - B)/(a - b) Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f* x]]), x], x] - Simp[(A*b - a*B)/(a - b) Int[(1 + Sin[e + f*x])/((a + b*Si n[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*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] && NeQ[A, B]
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[1/b^2 Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 2)/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)*Sqrt[(c_.) + (d_.)*sin[(e _.) + (f_.)*(x_)]]), x_Symbol] :> Simp[C/b^2 Int[Sqrt[a + b*Sin[e + f*x]] /Sqrt[c + d*Sin[e + f*x]], x], x] + Simp[1/b^2 Int[(A*b^2 - a^2*C + b*(b* B - 2*a*C)*Sin[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, A, B, C}, 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_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(Sqrt[c + d*Sin[e + f *x]]/(d*f*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[1/(2*d) Int[(1/((a + b*Si n[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x]]))*Simp[2*a*A*d - C*(b*c - a*d) - 2*(a*c*C - d*(A*b + a*B))*Sin[e + f*x] + (2*b*B*d - C*(b*c + a*d))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{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. \(2522\) vs. \(2(437)=874\).
Time = 16.83 (sec) , antiderivative size = 2523, normalized size of antiderivative = 5.27
method | result | size |
parts | \(\text {Expression too large to display}\) | \(2523\) |
default | \(\text {Expression too large to display}\) | \(2535\) |
int((B*cos(d*x+c)+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2)/(a+b*cos(d*x+c))^(1/2), x,method=_RETURNVERBOSE)
-B/d/b*((cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*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*co s(d*x+c)^2+(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*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))*b *cos(d*x+c)^2-2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c) )/(1+cos(d*x+c)))^(1/2)*EllipticPi(cot(d*x+c)-csc(d*x+c),-1,(-(a-b)/(a+b)) ^(1/2))*a*cos(d*x+c)^2+2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*c os(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)+2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+ b*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))*b*cos(d*x+c)-4*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)* (a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticPi(cot(d*x+c)-csc(d*x+c),-1 ,(-(a-b)/(a+b))^(1/2))*a*cos(d*x+c)+(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/( a+b)*(a+b*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)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b* 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))*b-2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+ c))/(1+cos(d*x+c)))^(1/2)*EllipticPi(cot(d*x+c)-csc(d*x+c),-1,(-(a-b)/(a+b ))^(1/2))*a-b*cos(d*x+c)^2*sin(d*x+c)-a*cos(d*x+c)*sin(d*x+c))/(1+cos(d*x+ c))/(a+b*cos(d*x+c))^(1/2)/cos(d*x+c)^(1/2)-1/4*C/d/b^2*(2*(cos(d*x+c)/...
\[ \int \frac {\sqrt {\cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )}}{\sqrt {b \cos \left (d x + c\right ) + a}} \,d x } \]
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2)/(a+b*cos(d*x+c))^ (1/2),x, algorithm="fricas")
integral((C*cos(d*x + c)^2 + B*cos(d*x + c))*sqrt(cos(d*x + c))/sqrt(b*cos (d*x + c) + a), x)
Timed out. \[ \int \frac {\sqrt {\cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx=\text {Timed out} \]
\[ \int \frac {\sqrt {\cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )}}{\sqrt {b \cos \left (d x + c\right ) + a}} \,d x } \]
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2)/(a+b*cos(d*x+c))^ (1/2),x, algorithm="maxima")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*sqrt(cos(d*x + c))/sqrt(b*co s(d*x + c) + a), x)
\[ \int \frac {\sqrt {\cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )}}{\sqrt {b \cos \left (d x + c\right ) + a}} \,d x } \]
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2)/(a+b*cos(d*x+c))^ (1/2),x, algorithm="giac")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*sqrt(cos(d*x + c))/sqrt(b*co s(d*x + c) + a), x)
Timed out. \[ \int \frac {\sqrt {\cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx=\int \frac {\sqrt {\cos \left (c+d\,x\right )}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )\right )}{\sqrt {a+b\,\cos \left (c+d\,x\right )}} \,d x \]