Integrand size = 45, antiderivative size = 535 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{3/2}} \, dx=-\frac {\left (2 A b^2-2 a b B+3 a^2 C-b^2 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{a b^2 \sqrt {a+b} d}+\frac {\left (2 A b^2-a (b (2 B-C)-3 a C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{a b^2 \sqrt {a+b} d}-\frac {\sqrt {a+b} (2 b B-3 a C) \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{b^3 d}-\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}+\frac {\left (2 A b^2-2 a b B+3 a^2 C-b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}} \] Output:
-(2*A*b^2-2*B*a*b+3*C*a^2-C*b^2)*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/(a+b)^(1/2)/d+(2*A*b^2-a*( b*(2*B-C)-3*a*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*(1-sec(d*x+c))/(a+b))^(1/2)*(a*( 1+sec(d*x+c))/(a-b))^(1/2)/a/b^2/(a+b)^(1/2)/d-(a+b)^(1/2)*(2*B*b-3*C*a)*c ot(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-2*(A*b^2-a*(B*b-C*a))*cos(d*x+c)^(1/2)*sin(d*x+c)/b /(a^2-b^2)/d/(a+b*cos(d*x+c))^(1/2)+(2*A*b^2-2*B*a*b+3*C*a^2-C*b^2)*(a+b*c os(d*x+c))^(1/2)*sin(d*x+c)/b^2/(a^2-b^2)/d/cos(d*x+c)^(1/2)
Result contains complex when optimal does not.
Time = 7.72 (sec) , antiderivative size = 1256, normalized size of antiderivative = 2.35 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{3/2}} \, dx =\text {Too large to display} \] Input:
Integrate[(Sqrt[Cos[c + d*x]]*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(a + b*Cos[c + d*x])^(3/2),x]
Output:
(2*Sqrt[Cos[c + d*x]]*(A*b^2*Sin[c + d*x] - a*b*B*Sin[c + d*x] + a^2*C*Sin [c + d*x]))/(b*(-a^2 + b^2)*d*Sqrt[a + b*Cos[c + d*x]]) + ((-4*a*(a^2*C - b^2*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[C os[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]) - 4*a*(2*a*A*b - 2*b^2*B + 2*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), ArcSi n[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*(2*A*b^2 - 2*a*b*B + 3*a^2*C - b^2*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])*Sec[c + d*x])/(a + b)]) + (2*a*((a*Sqrt[((a + b)...
Time = 2.57 (sec) , antiderivative size = 564, normalized size of antiderivative = 1.05, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.289, Rules used = {3042, 3526, 27, 3042, 3540, 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.
| \(\displaystyle \int \frac {\sqrt {\cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 3526 |
| \(\displaystyle -\frac {2 \int \frac {A b^2+(b B-a (A+C)) \cos (c+d x) b-\left (3 C a^2-2 b B a+2 A b^2-b^2 C\right ) \cos ^2(c+d x)-a (b B-a C)}{2 \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx}{b \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {A b^2+(b B-a (A+C)) \cos (c+d x) b-\left (3 C a^2-2 b B a+2 A b^2-b^2 C\right ) \cos ^2(c+d x)-a (b B-a C)}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx}{b \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {A b^2+(b B-a (A+C)) \sin \left (c+d x+\frac {\pi }{2}\right ) b+\left (-3 C a^2+2 b B a-2 A b^2+b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-a (b B-a C)}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3540 |
| \(\displaystyle -\frac {\frac {\int \frac {-\left (\left (a^2-b^2\right ) (2 b B-3 a C) \cos ^2(c+d x)\right )+2 b \left (A b^2-a (b B-a C)\right ) \cos (c+d x)+a \left (3 C a^2-2 b B a+2 A b^2-b^2 C\right )}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{2 b}-\frac {\sin (c+d x) \left (3 a^2 C-2 a b B+2 A b^2-b^2 C\right ) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{b \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {\int \frac {-\left (\left (a^2-b^2\right ) (2 b B-3 a C) \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )+2 b \left (A b^2-a (b B-a C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+a \left (3 C a^2-2 b B a+2 A b^2-b^2 C\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 b}-\frac {\sin (c+d x) \left (3 a^2 C-2 a b B+2 A b^2-b^2 C\right ) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{b \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3532 |
| \(\displaystyle -\frac {\frac {\int \frac {a \left (3 C a^2-2 b B a+2 A b^2-b^2 C\right )+2 b \left (A b^2-a (b B-a C)\right ) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx-\left (a^2-b^2\right ) (2 b B-3 a C) \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {a+b \cos (c+d x)}}dx}{2 b}-\frac {\sin (c+d x) \left (3 a^2 C-2 a b B+2 A b^2-b^2 C\right ) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{b \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {\int \frac {a \left (3 C a^2-2 b B a+2 A b^2-b^2 C\right )+2 b \left (A b^2-a (b B-a C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\left (a^2-b^2\right ) (2 b B-3 a C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 b}-\frac {\sin (c+d x) \left (3 a^2 C-2 a b B+2 A b^2-b^2 C\right ) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{b \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3288 |
| \(\displaystyle -\frac {\frac {\int \frac {a \left (3 C a^2-2 b B a+2 A b^2-b^2 C\right )+2 b \left (A b^2-a (b B-a C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sqrt {a+b} \left (a^2-b^2\right ) (2 b B-3 a C) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}}{2 b}-\frac {\sin (c+d x) \left (3 a^2 C-2 a b B+2 A b^2-b^2 C\right ) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{b \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3477 |
| \(\displaystyle -\frac {\frac {a \left (3 a^2 C-2 a b B+2 A b^2-b^2 C\right ) \int \frac {\cos (c+d x)+1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx-(a-b) \left (2 A b^2-a (b (2 B-C)-3 a C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx+\frac {2 \sqrt {a+b} \left (a^2-b^2\right ) (2 b B-3 a C) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}}{2 b}-\frac {\sin (c+d x) \left (3 a^2 C-2 a b B+2 A b^2-b^2 C\right ) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{b \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {a \left (3 a^2 C-2 a b B+2 A b^2-b^2 C\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-(a-b) \left (2 A b^2-a (b (2 B-C)-3 a C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sqrt {a+b} \left (a^2-b^2\right ) (2 b B-3 a C) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}}{2 b}-\frac {\sin (c+d x) \left (3 a^2 C-2 a b B+2 A b^2-b^2 C\right ) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{b \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3295 |
| \(\displaystyle -\frac {\frac {a \left (3 a^2 C-2 a b B+2 A b^2-b^2 C\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sqrt {a+b} \left (a^2-b^2\right ) (2 b B-3 a C) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}-\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (2 A b^2-a (b (2 B-C)-3 a C)\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}}{2 b}-\frac {\sin (c+d x) \left (3 a^2 C-2 a b B+2 A b^2-b^2 C\right ) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{b \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3473 |
| \(\displaystyle -\frac {\frac {\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (3 a^2 C-2 a b B+2 A b^2-b^2 C\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a d}+\frac {2 \sqrt {a+b} \left (a^2-b^2\right ) (2 b B-3 a C) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}-\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (2 A b^2-a (b (2 B-C)-3 a C)\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}}{2 b}-\frac {\sin (c+d x) \left (3 a^2 C-2 a b B+2 A b^2-b^2 C\right ) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{b \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\) |
Input:
Int[(Sqrt[Cos[c + d*x]]*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(a + b*Co s[c + d*x])^(3/2),x]
Output:
(-2*(A*b^2 - a*(b*B - a*C))*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(b*(a^2 - b^2 )*d*Sqrt[a + b*Cos[c + d*x]]) - (((2*(a - b)*Sqrt[a + b]*(2*A*b^2 - 2*a*b* B + 3*a^2*C - b^2*C)*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*d) - (2*(a - b)*Sqrt[a + b]*(2*A*b^2 - a*(b*(2*B - C) - 3*a*C))*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*Sqrt[a + b]*(a^2 - b^2)*(2*b*B - 3*a*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))/(2*b) - ((2*A*b^2 - 2*a*b*B + 3*a^2*C - b^2*C)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(b*d*Sqrt[Cos[c + d*x]]))/(b*(a^2 - b^2))
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_)])/(((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[(-(c^2*C - B*c*d + A*d^2))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + (c*C - B* d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 1) - a*c*(n + 2))) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x ] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)))*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] && GtQ[m, 0] && LtQ[n, -1]
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. \(2079\) vs. \(2(494)=988\).
Time = 0.66 (sec) , antiderivative size = 2080, normalized size of antiderivative = 3.89
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(2080\) |
| parts | \(\text {Expression too large to display}\) | \(2159\) |
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))^(3/2 ),x,method=_RETURNVERBOSE)
Output:
-1/d*(C*b^3*cos(d*x+c)^2*sin(d*x+c)+2*A*b^3*cos(d*x+c)*sin(d*x+c)+sin(d*x+ c)*cos(d*x+c)*(-cos(d*x+c)+2)*a^2*b*C-2*A*a*b^2*cos(d*x+c)*sin(d*x+c)+2*B* a^2*b*cos(d*x+c)*sin(d*x+c)-3*C*a^3*cos(d*x+c)*sin(d*x+c)+(-4*cos(d*x+c)^2 -8*cos(d*x+c)-4)*B*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(cos(d* x+c)/(1+cos(d*x+c)))^(1/2)*b^3*EllipticPi(-csc(d*x+c)+cot(d*x+c),-1,(-(a-b )/(a+b))^(1/2))+(-6*cos(d*x+c)^2-12*cos(d*x+c)-6)*C*(1/(a+b)*(a+b*cos(d*x+ c))/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*a^3*EllipticPi (-csc(d*x+c)+cot(d*x+c),-1,(-(a-b)/(a+b))^(1/2))+(2*cos(d*x+c)^2+4*cos(d*x +c)+2)*A*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+co s(d*x+c)))^(1/2)*b^3*EllipticE(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2) )+(3*cos(d*x+c)^2+6*cos(d*x+c)+3)*C*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c )))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*a^3*EllipticE(-csc(d*x+c)+cot( d*x+c),(-(a-b)/(a+b))^(1/2))+(-cos(d*x+c)^2-2*cos(d*x+c)-1)*C*(1/(a+b)*(a+ b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*b^3* EllipticE(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))+(-2*cos(d*x+c)^2-4* cos(d*x+c)-2)*A*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c )/(1+cos(d*x+c)))^(1/2)*b^3*EllipticF(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b) )^(1/2))+(2*cos(d*x+c)^2+4*cos(d*x+c)+2)*B*(1/(a+b)*(a+b*cos(d*x+c))/(1+co s(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*b^3*EllipticF(-csc(d*x+ c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))-2*B*a*b^2*cos(d*x+c)*sin(d*x+c)+C*a...
\[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sqrt {\cos \left (d x + c\right )}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] 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) )^(3/2),x, algorithm="fricas")
Output:
integral((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sqrt(b*cos(d*x + c) + a)* sqrt(cos(d*x + c))/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2), x)
Timed out. \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \] 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))**(3/2),x)
Output:
Timed out
\[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sqrt {\cos \left (d x + c\right )}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] 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) )^(3/2),x, algorithm="maxima")
Output:
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)^(3/2), x)
\[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sqrt {\cos \left (d x + c\right )}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] 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) )^(3/2),x, algorithm="giac")
Output:
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)^(3/2), x)
Timed out. \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{3/2}} \, dx=\int \frac {\sqrt {\cos \left (c+d\,x\right )}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \] 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))^(3/2),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))^(3/2), x)
\[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{3/2}} \, dx=\left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )}{\cos \left (d x +c \right )^{2} b^{2}+2 \cos \left (d x +c \right ) a b +a^{2}}d x \right ) b +\left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}}{\cos \left (d x +c \right )^{2} b^{2}+2 \cos \left (d x +c \right ) a b +a^{2}}d x \right ) c +\left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2} b^{2}+2 \cos \left (d x +c \right ) a b +a^{2}}d x \right ) a \] 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))^(3/2 ),x)
Output:
int((sqrt(cos(c + d*x)*b + a)*sqrt(cos(c + d*x))*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)*sqrt(cos(c + d*x))*cos(c + d*x)**2)/(cos(c + d*x)**2*b**2 + 2*cos(c + d *x)*a*b + a**2),x)*c + int((sqrt(cos(c + d*x)*b + a)*sqrt(cos(c + d*x)))/( cos(c + d*x)**2*b**2 + 2*cos(c + d*x)*a*b + a**2),x)*a