\(\int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\sqrt {a+b \cos (c+d x)}} \, dx\) [1137]

Optimal result

Integrand size = 45, antiderivative size = 485 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx=-\frac {(a-b) \sqrt {a+b} (4 b B-3 a C) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{4 a b^2 d}-\frac {\sqrt {a+b} (3 a C-2 b (2 B+C)) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{4 b^2 d}-\frac {\sqrt {a+b} \left (8 A b^2-4 a b B+3 a^2 C+4 b^2 C\right ) \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{4 b^3 d}+\frac {(4 b B-3 a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{4 b^2 d \sqrt {\cos (c+d x)}}+\frac {C \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{2 b d} \] Output:

-1/4*(a-b)*(a+b)^(1/2)*(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*(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*(a+b)^(1/2)*(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)/co 
s(d*x+c)^(1/2),(-(a+b)/(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*(a+b)^(1/2)*(8*A*b^2-4*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*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+ 
sec(d*x+c))/(a-b))^(1/2)/b^3/d+1/4*(4*B*b-3*C*a)*(a+b*cos(d*x+c))^(1/2)*si 
n(d*x+c)/b^2/d/cos(d*x+c)^(1/2)+1/2*C*cos(d*x+c)^(1/2)*(a+b*cos(d*x+c))^(1 
/2)*sin(d*x+c)/b/d
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 19.79 (sec) , antiderivative size = 1182, normalized size of antiderivative = 2.44 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx =\text {Too large to display} \] Input:

Integrate[(Sqrt[Cos[c + d*x]]*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Sqr 
t[a + b*Cos[c + d*x]],x]
 

Output:

(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]]) - 4*a*(8*A*b + 4*b*C)*(( 
Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Cos[c + d*x]*C 
sc[(c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]*C 
sc[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[S 
qrt[((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[Sin[(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])*Se 
c[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*Co...
 

Rubi [A] (verified)

Time = 2.16 (sec) , antiderivative size = 486, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.311, Rules used = {3042, 3528, 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.

\(\displaystyle \int \frac {\sqrt {\cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\)

\(\Big \downarrow \) 3528

\(\displaystyle \frac {\int \frac {(4 b B-3 a C) \cos ^2(c+d x)+2 b (2 A+C) \cos (c+d x)+a C}{2 \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx}{2 b}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(4 b B-3 a C) \cos ^2(c+d x)+2 b (2 A+C) \cos (c+d x)+a C}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx}{4 b}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {(4 b B-3 a C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+2 b (2 A+C) \sin \left (c+d x+\frac {\pi }{2}\right )+a C}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{4 b}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}\)

\(\Big \downarrow \) 3540

\(\displaystyle \frac {\frac {\int -\frac {-\left (\left (3 C a^2-4 b B a+8 A b^2+4 b^2 C\right ) \cos ^2(c+d x)\right )-2 a b C \cos (c+d x)+a (4 b B-3 a C)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{2 b}+\frac {(4 b B-3 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{4 b}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {(4 b B-3 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {\int \frac {-\left (\left (3 C a^2-4 b B a+8 A b^2+4 b^2 C\right ) \cos ^2(c+d x)\right )-2 a b C \cos (c+d x)+a (4 b B-3 a C)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{2 b}}{4 b}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {(4 b B-3 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {\int \frac {\left (-3 C a^2+4 b B a-8 A b^2-4 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-2 a b C \sin \left (c+d x+\frac {\pi }{2}\right )+a (4 b B-3 a C)}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 b}}{4 b}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}\)

\(\Big \downarrow \) 3532

\(\displaystyle \frac {\frac {(4 b B-3 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {\int \frac {a (4 b B-3 a C)-2 a b C \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx-\left (3 a^2 C-4 a b B+8 A b^2+4 b^2 C\right ) \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {a+b \cos (c+d x)}}dx}{2 b}}{4 b}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {(4 b B-3 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {\int \frac {a (4 b B-3 a C)-2 a b C \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\left (3 a^2 C-4 a b B+8 A b^2+4 b^2 C\right ) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 b}}{4 b}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}\)

\(\Big \downarrow \) 3288

\(\displaystyle \frac {\frac {(4 b B-3 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {\int \frac {a (4 b B-3 a C)-2 a b C \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sqrt {a+b} \cot (c+d x) \left (3 a^2 C-4 a b B+8 A b^2+4 b^2 C\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}}{2 b}}{4 b}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}\)

\(\Big \downarrow \) 3477

\(\displaystyle \frac {\frac {(4 b B-3 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {a (4 b B-3 a C) \int \frac {\cos (c+d x)+1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx+a (3 a C-2 b (2 B+C)) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx+\frac {2 \sqrt {a+b} \cot (c+d x) \left (3 a^2 C-4 a b B+8 A b^2+4 b^2 C\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}}{2 b}}{4 b}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {(4 b B-3 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {a (3 a C-2 b (2 B+C)) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+a (4 b B-3 a C) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sqrt {a+b} \cot (c+d x) \left (3 a^2 C-4 a b B+8 A b^2+4 b^2 C\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}}{2 b}}{4 b}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}\)

\(\Big \downarrow \) 3295

\(\displaystyle \frac {\frac {(4 b B-3 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {a (4 b B-3 a C) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sqrt {a+b} \cot (c+d x) \left (3 a^2 C-4 a b B+8 A b^2+4 b^2 C\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}+\frac {2 \sqrt {a+b} (3 a C-2 b (2 B+C)) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}}{2 b}}{4 b}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}\)

\(\Big \downarrow \) 3473

\(\displaystyle \frac {\frac {(4 b B-3 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {\frac {2 \sqrt {a+b} \cot (c+d x) \left (3 a^2 C-4 a b B+8 A b^2+4 b^2 C\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}+\frac {2 \sqrt {a+b} (3 a C-2 b (2 B+C)) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}+\frac {2 (a-b) \sqrt {a+b} (4 b B-3 a C) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a d}}{2 b}}{4 b}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}\)

Input:

Int[(Sqrt[Cos[c + d*x]]*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Sqrt[a + 
b*Cos[c + d*x]],x]
 

Output:

(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]*(8*A*b^2 - 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]*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)])/(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)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3288

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]
 

rule 3295

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]
 

rule 3473

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]
 

rule 3477

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]
 

rule 3528

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])))
 

rule 3532

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]
 

rule 3540

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1044\) vs. \(2(434)=868\).

Time = 5.82 (sec) , antiderivative size = 1045, normalized size of antiderivative = 2.15

method	result	size
default	\(\text {Expression too large to display}\)	\(1045\)
parts	\(\text {Expression too large to display}\)	\(1271\)

Input:

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))^(1/2 
),x,method=_RETURNVERBOSE)
 

Output:

1/4/d*((-16*cos(d*x+c)-16)*A*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/ 
2)*b^2*EllipticPi(-csc(d*x+c)+cot(d*x+c),-1,(-(a-b)/(a+b))^(1/2))+(8*cos(d 
*x+c)+8)*B*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a*b*EllipticPi( 
-csc(d*x+c)+cot(d*x+c),-1,(-(a-b)/(a+b))^(1/2))+(-6*cos(d*x+c)-6)*C*(1/(a+ 
b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a^2*EllipticPi(-csc(d*x+c)+cot(d 
*x+c),-1,(-(a-b)/(a+b))^(1/2))+(-8*cos(d*x+c)-8)*C*(1/(a+b)*(a+b*cos(d*x+c 
))/(1+cos(d*x+c)))^(1/2)*b^2*EllipticPi(-csc(d*x+c)+cot(d*x+c),-1,(-(a-b)/ 
(a+b))^(1/2))+(-4*cos(d*x+c)-4)*B*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)) 
)^(1/2)*a*b*EllipticE(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))+(-4*cos 
(d*x+c)-4)*B*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*b^2*EllipticE 
(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))+(3*cos(d*x+c)+3)*C*(1/(a+b)* 
(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a^2*EllipticE(-csc(d*x+c)+cot(d*x+c 
),(-(a-b)/(a+b))^(1/2))+(3*cos(d*x+c)+3)*C*(1/(a+b)*(a+b*cos(d*x+c))/(1+co 
s(d*x+c)))^(1/2)*a*b*EllipticE(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2) 
)+(8*cos(d*x+c)+8)*A*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*b^2*E 
llipticF(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))+(-2*cos(d*x+c)-2)*C* 
(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a*b*EllipticF(-csc(d*x+c)+ 
cot(d*x+c),(-(a-b)/(a+b))^(1/2))+(4*cos(d*x+c)+4)*C*(1/(a+b)*(a+b*cos(d*x+ 
c))/(1+cos(d*x+c)))^(1/2)*b^2*EllipticF(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+ 
b))^(1/2))+4*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*b^2*cos(d*x+c)*sin(d*x...
 

Fricas [F]

\[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sqrt {\cos \left (d x + c\right )}}{\sqrt {b \cos \left (d x + c\right ) + a}} \,d x } \] Input:

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) 
)^(1/2),x, algorithm="fricas")
 

Output:

integral((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sqrt(cos(d*x + c))/sqrt(b 
*cos(d*x + c) + a), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx=\text {Timed out} \] Input:

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))**(1/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sqrt {\cos \left (d x + c\right )}}{\sqrt {b \cos \left (d x + c\right ) + a}} \,d x } \] Input:

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) 
)^(1/2),x, algorithm="maxima")
 

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sqrt(cos(d*x + c))/sqrt( 
b*cos(d*x + c) + a), x)
 

Giac [F]

\[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sqrt {\cos \left (d x + c\right )}}{\sqrt {b \cos \left (d x + c\right ) + a}} \,d x } \] Input:

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) 
)^(1/2),x, algorithm="giac")
 

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sqrt(cos(d*x + c))/sqrt( 
b*cos(d*x + c) + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx=\int \frac {\sqrt {\cos \left (c+d\,x\right )}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{\sqrt {a+b\,\cos \left (c+d\,x\right )}} \,d x \] Input:

int((cos(c + d*x)^(1/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/(a + b*co 
s(c + d*x))^(1/2),x)
 

Output:

int((cos(c + d*x)^(1/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/(a + b*co 
s(c + d*x))^(1/2), x)
 

Reduce [F]

\[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx=\left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )}{\cos \left (d x +c \right ) b +a}d x \right ) b +\left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}}{\cos \left (d x +c \right ) b +a}d x \right ) c +\left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right ) b +a}d x \right ) a \] Input:

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))^(1/2 
),x)
 

Output:

int((sqrt(cos(c + d*x)*b + a)*sqrt(cos(c + d*x))*cos(c + d*x))/(cos(c + d* 
x)*b + a),x)*b + int((sqrt(cos(c + d*x)*b + a)*sqrt(cos(c + d*x))*cos(c + 
d*x)**2)/(cos(c + d*x)*b + a),x)*c + int((sqrt(cos(c + d*x)*b + a)*sqrt(co 
s(c + d*x)))/(cos(c + d*x)*b + a),x)*a