Integrand size = 45, antiderivative size = 509 \[ \int \sqrt {a+b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\frac {(a-b) \sqrt {a+b} (2 A-C) \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{a d \sqrt {\sec (c+d x)}}+\frac {\sqrt {a+b} (2 A b-a (2 A-2 B-C)) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{a d \sqrt {\sec (c+d x)}}-\frac {\sqrt {a+b} (2 b B+a C) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{b d \sqrt {\sec (c+d x)}}+\frac {2 A \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{d}-\frac {(2 A-C) \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{d} \] Output:
(a-b)*(a+b)^(1/2)*(2*A-C)*cos(d*x+c)^(1/2)*csc(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/d/sec(d*x+c)^(1/2)+( a+b)^(1/2)*(2*A*b-a*(2*A-2*B-C))*cos(d*x+c)^(1/2)*csc(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/d/sec(d*x+c)^ (1/2)-(a+b)^(1/2)*(2*B*b+C*a)*cos(d*x+c)^(1/2)*csc(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/d/sec(d *x+c)^(1/2)+2*A*(a+b*cos(d*x+c))^(1/2)*sec(d*x+c)^(1/2)*sin(d*x+c)/d-(2*A- C)*(a+b*cos(d*x+c))^(1/2)*sec(d*x+c)^(1/2)*sin(d*x+c)/d
Time = 15.10 (sec) , antiderivative size = 896, normalized size of antiderivative = 1.76 \[ \int \sqrt {a+b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx =\text {Too large to display} \] Input:
Integrate[Sqrt[a + b*Cos[c + d*x]]*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2) *Sec[c + d*x]^(3/2),x]
Output:
(2*A*Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d + (Sqrt[( 1 - Tan[(c + d*x)/2]^2)^(-1)]*(-2*a*A*Tan[(c + d*x)/2] - 2*A*b*Tan[(c + d* x)/2] + a*C*Tan[(c + d*x)/2] + b*C*Tan[(c + d*x)/2] + 4*A*b*Tan[(c + d*x)/ 2]^3 - 2*b*C*Tan[(c + d*x)/2]^3 + 2*a*A*Tan[(c + d*x)/2]^5 - 2*A*b*Tan[(c + d*x)/2]^5 - a*C*Tan[(c + d*x)/2]^5 + b*C*Tan[(c + d*x)/2]^5 + 4*b*B*Elli pticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[1 - Tan[(c + d *x)/2]^2]*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] + 2*a*C*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sqr t[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] + 4*b*B*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] + 2*a*C*EllipticPi[ -1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x) /2]^2)/(a + b)] - (a + b)*(2*A - C)*EllipticE[ArcSin[Tan[(c + d*x)/2]], (- a + b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*(1 + Tan[(c + d*x)/2]^2)*Sqrt [(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] + 2*(b*(A - B) + a*(A + B - C))*EllipticF[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b) ]*Sqrt[1 - Tan[(c + d*x)/2]^2]*(1 + Tan[(c + d*x)/2]^2)*Sqrt[(a + b + a*Ta n[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)]))/(d*(1 + Tan[(c + d*...
Time = 2.34 (sec) , antiderivative size = 479, normalized size of antiderivative = 0.94, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4709, 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.
| \(\displaystyle \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sec (c+d x)^{3/2} \sqrt {a+b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos (c+d x)^2\right )dx\) |
| \(\Big \downarrow \) 4709 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sqrt {a+b \cos (c+d x)} \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right )}{\cos ^{\frac {3}{2}}(c+d x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )} \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 3526 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (2 \int \frac {-b (2 A-C) \cos ^2(c+d x)+(b B-a (A-C)) \cos (c+d x)+A b+a B}{2 \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx+\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\int \frac {-b (2 A-C) \cos ^2(c+d x)+(b B-a (A-C)) \cos (c+d x)+A b+a B}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx+\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\int \frac {-b (2 A-C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+(b B-a (A-C)) \sin \left (c+d x+\frac {\pi }{2}\right )+A b+a B}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3540 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {b (2 b B+a C) \cos ^2(c+d x)+2 b (A b+a B) \cos (c+d x)+a b (2 A-C)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{2 b}-\frac {(2 A-C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}+\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {b (2 b B+a C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+2 b (A b+a B) \sin \left (c+d x+\frac {\pi }{2}\right )+a b (2 A-C)}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 b}-\frac {(2 A-C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}+\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3532 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a b (2 A-C)+2 b (A b+a B) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx+b (a C+2 b B) \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {a+b \cos (c+d x)}}dx}{2 b}-\frac {(2 A-C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}+\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a b (2 A-C)+2 b (A b+a B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+b (a C+2 b B) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 b}-\frac {(2 A-C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}+\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3288 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a b (2 A-C)+2 b (A b+a B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} (a C+2 b B) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}}{2 b}-\frac {(2 A-C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}+\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3477 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {b (2 A b-a (2 A-2 B-C)) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx+a b (2 A-C) \int \frac {\cos (c+d x)+1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx-\frac {2 \sqrt {a+b} (a C+2 b B) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}}{2 b}-\frac {(2 A-C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}+\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {b (2 A b-a (2 A-2 B-C)) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+a b (2 A-C) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} (a C+2 b B) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}}{2 b}-\frac {(2 A-C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}+\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3295 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {a b (2 A-C) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b \sqrt {a+b} \cot (c+d x) (2 A b-a (2 A-2 B-C)) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}-\frac {2 \sqrt {a+b} (a C+2 b B) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}}{2 b}-\frac {(2 A-C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}+\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3473 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 b \sqrt {a+b} \cot (c+d x) (2 A b-a (2 A-2 B-C)) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}+\frac {2 b (a-b) \sqrt {a+b} (2 A-C) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a d}-\frac {2 \sqrt {a+b} (a C+2 b B) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}}{2 b}-\frac {(2 A-C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}+\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\right )\) |
Input:
Int[Sqrt[a + b*Cos[c + d*x]]*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(3/2),x]
Output:
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(((2*(a - b)*b*Sqrt[a + b]*(2*A - 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*b*Sqrt[a + b]*(2*A*b - a* (2*A - 2*B - C))*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(S qrt[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]*(2*b*B + a*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)])/d)/(2*b) + (2*A*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]) - (( 2*A - C)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]))
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]
Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Simp[(c*Sec[a + b*x])^m*(c*Cos[a + b*x])^m Int[ActivateTrig[u]/(c*Cos[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSineIntegrandQ[u, x]
Leaf count of result is larger than twice the leaf count of optimal. \(1154\) vs. \(2(456)=912\).
Time = 14.27 (sec) , antiderivative size = 1155, normalized size of antiderivative = 2.27
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(1155\) |
| default | \(\text {Expression too large to display}\) | \(1221\) |
Input:
int((a+b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2 ),x,method=_RETURNVERBOSE)
Output:
2*A/d*((cos(d*x+c)^2+2*cos(d*x+c)+1)*(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)*a*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)*(cos(d*x+c)/(1+c os(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*b*Ellipt icE(-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)*(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)*a*EllipticF(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))+(-c os(d*x+c)^2-2*cos(d*x+c)-1)*(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)*b*EllipticF(-csc(d*x+c)+cot(d*x+c),(-( a-b)/(a+b))^(1/2))+cos(d*x+c)*sin(d*x+c)*b+a*sin(d*x+c))*(a+b*cos(d*x+c))^ (1/2)*cos(d*x+c)*sec(d*x+c)^(3/2)/(b*cos(d*x+c)^2+a*cos(d*x+c)+b*cos(d*x+c )+a)-2*B/d*sec(d*x+c)^(3/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/(a+b*cos(d*x +c))^(1/2)*(EllipticF(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))*a-Ellip ticF(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))*b+2*b*EllipticPi(-csc(d* x+c)+cot(d*x+c),-1,(-(a-b)/(a+b))^(1/2)))*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos (d*x+c)))^(1/2)*(cos(d*x+c)^2+cos(d*x+c))-C/d*((2*cos(d*x+c)^2+4*cos(d*x+c )+2)*(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*EllipticPi(-csc(d*x+c)+cot(d*x+c),-1,(-(a-b)/(a+b))^(1/2))+ (cos(d*x+c)^2+2*cos(d*x+c)+1)*(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)*a*EllipticE(-csc(d*x+c)+cot(d*x+c...
\[ \int \sqrt {a+b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sqrt {b \cos \left (d x + c\right ) + a} \sec \left (d x + c\right )^{\frac {3}{2}} \,d x } \] Input:
integrate((a+b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(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)* sec(d*x + c)^(3/2), x)
Timed out. \[ \int \sqrt {a+b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\text {Timed out} \] Input:
integrate((a+b*cos(d*x+c))**(1/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x +c)**(3/2),x)
Output:
Timed out
\[ \int \sqrt {a+b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sqrt {b \cos \left (d x + c\right ) + a} \sec \left (d x + c\right )^{\frac {3}{2}} \,d x } \] Input:
integrate((a+b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c )^(3/2),x, algorithm="maxima")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sqrt(b*cos(d*x + c) + a) *sec(d*x + c)^(3/2), x)
\[ \int \sqrt {a+b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sqrt {b \cos \left (d x + c\right ) + a} \sec \left (d x + c\right )^{\frac {3}{2}} \,d x } \] Input:
integrate((a+b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c )^(3/2),x, algorithm="giac")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sqrt(b*cos(d*x + c) + a) *sec(d*x + c)^(3/2), x)
Timed out. \[ \int \sqrt {a+b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\int {\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,\sqrt {a+b\,\cos \left (c+d\,x\right )}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right ) \,d x \] Input:
int((1/cos(c + d*x))^(3/2)*(a + b*cos(c + d*x))^(1/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2),x)
Output:
int((1/cos(c + d*x))^(3/2)*(a + b*cos(c + d*x))^(1/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2), x)
\[ \int \sqrt {a+b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right ) b +a}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right ) b +a}\, \sec \left (d x +c \right )d x \right ) a \] Input:
int((a+b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2 ),x)
Output:
int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x)*b + a)*cos(c + d*x)*sec(c + d*x), x)*b + int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x)*b + a)*cos(c + d*x)**2*sec (c + d*x),x)*c + int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x)*b + a)*sec(c + d *x),x)*a