Integrand size = 45, antiderivative size = 595 \[ \int (a+b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\frac {(a-b) \sqrt {a+b} (8 a A-4 b B-5 a C) \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{4 a d \sqrt {\sec (c+d x)}}-\frac {\sqrt {a+b} (a (8 A-8 B-5 C)-2 b (8 A+2 B+C)) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{4 d \sqrt {\sec (c+d x)}}-\frac {\sqrt {a+b} \left (8 A b^2+12 a b B+3 a^2 C+4 b^2 C\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{4 b d \sqrt {\sec (c+d x)}}-\frac {b (4 A-C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{2 d \sqrt {\sec (c+d x)}}-\frac {(8 a A-4 b B-5 a C) \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{4 d}+\frac {2 A (a+b \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)} \sin (c+d x)}{d} \]
-1/2*b*(4*A-C)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/d/sec(d*x+c)^(1/2)+2*A*(a +b*cos(d*x+c))^(3/2)*sin(d*x+c)*sec(d*x+c)^(1/2)/d-1/4*(8*A*a-4*B*b-5*C*a) *sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)*sec(d*x+c)^(1/2)/d+1/4*(a-b)*(8*A*a-4*B *b-5*C*a)*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+b)^(1/2)*cos(d*x+c)^(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)-1/4*( a*(8*A-8*B-5*C)-2*b*(8*A+2*B+C))*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+b)^(1/2)*cos(d*x+ c)^(1/2)*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/d/s ec(d*x+c)^(1/2)-1/4*(8*A*b^2+12*B*a*b+3*C*a^2+4*C*b^2)*csc(d*x+c)*Elliptic Pi((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)*cos(d*x+c)^(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)
Leaf count is larger than twice the leaf count of optimal. \(1453\) vs. \(2(595)=1190\).
Time = 24.37 (sec) , antiderivative size = 1453, normalized size of antiderivative = 2.44 \[ \int (a+b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx =\text {Too large to display} \]
Integrate[(a + b*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^ 2)*Sec[c + d*x]^(3/2),x]
(Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(2*a*A*Sin[c + d*x] + (b*C*Si n[2*(c + d*x)])/4))/d + (8*a^2*A*Tan[(c + d*x)/2] + 8*a*A*b*Tan[(c + d*x)/ 2] - 4*a*b*B*Tan[(c + d*x)/2] - 4*b^2*B*Tan[(c + d*x)/2] - 5*a^2*C*Tan[(c + d*x)/2] - 5*a*b*C*Tan[(c + d*x)/2] - 16*a*A*b*Tan[(c + d*x)/2]^3 + 8*b^2 *B*Tan[(c + d*x)/2]^3 + 10*a*b*C*Tan[(c + d*x)/2]^3 - 8*a^2*A*Tan[(c + d*x )/2]^5 + 8*a*A*b*Tan[(c + d*x)/2]^5 + 4*a*b*B*Tan[(c + d*x)/2]^5 - 4*b^2*B *Tan[(c + d*x)/2]^5 + 5*a^2*C*Tan[(c + d*x)/2]^5 - 5*a*b*C*Tan[(c + d*x)/2 ]^5 - 16*A*b^2*EllipticPi[-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)] - 24*a*b*B*EllipticPi[-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)] - 6*a^2*C*EllipticPi[-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)] - 8*b^2*C*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)] - 16*A*b^2*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)] - 24*a*b*B*EllipticPi[-1, ArcSin [Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(...
Time = 2.99 (sec) , antiderivative size = 560, normalized size of antiderivative = 0.94, number of steps used = 18, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4709, 3042, 3526, 27, 3042, 3528, 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.
\(\displaystyle \int \sec ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sec (c+d x)^{3/2} (a+b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos (c+d x)^2\right )dx\) |
\(\Big \downarrow \) 4709 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {(a+b \cos (c+d x))^{3/2} \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right )}{\cos ^{\frac {3}{2}}(c+d x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx\) |
\(\Big \downarrow \) 3526 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (2 \int \frac {\sqrt {a+b \cos (c+d x)} \left (-b (4 A-C) \cos ^2(c+d x)+(b B-a (A-C)) \cos (c+d x)+3 A b+a B\right )}{2 \sqrt {\cos (c+d x)}}dx+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{d \sqrt {\cos (c+d x)}}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\int \frac {\sqrt {a+b \cos (c+d x)} \left (-b (4 A-C) \cos ^2(c+d x)+(b B-a (A-C)) \cos (c+d x)+3 A b+a B\right )}{\sqrt {\cos (c+d x)}}dx+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{d \sqrt {\cos (c+d x)}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\int \frac {\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )} \left (-b (4 A-C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+(b B-a (A-C)) \sin \left (c+d x+\frac {\pi }{2}\right )+3 A b+a B\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{d \sqrt {\cos (c+d x)}}\right )\) |
\(\Big \downarrow \) 3528 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{2} \int \frac {-b (8 a A-4 b B-5 a C) \cos ^2(c+d x)+2 \left (-2 (A-C) a^2+4 b B a+b^2 (2 A+C)\right ) \cos (c+d x)+a (8 A b+C b+4 a B)}{2 \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx-\frac {b (4 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{d \sqrt {\cos (c+d x)}}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} \int \frac {-b (8 a A-4 b B-5 a C) \cos ^2(c+d x)+2 \left (-2 (A-C) a^2+4 b B a+b^2 (2 A+C)\right ) \cos (c+d x)+a (8 A b+C b+4 a B)}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx-\frac {b (4 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{d \sqrt {\cos (c+d x)}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} \int \frac {-b (8 a A-4 b B-5 a C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+2 \left (-2 (A-C) a^2+4 b B a+b^2 (2 A+C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+a (8 A b+C b+4 a B)}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {b (4 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{d \sqrt {\cos (c+d x)}}\right )\) |
\(\Big \downarrow \) 3540 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} \left (\frac {\int \frac {b \left (3 C a^2+12 b B a+8 A b^2+4 b^2 C\right ) \cos ^2(c+d x)+2 a b (8 A b+C b+4 a B) \cos (c+d x)+a b (8 a A-4 b B-5 a C)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{2 b}-\frac {\sin (c+d x) (8 a A-5 a C-4 b B) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\right )-\frac {b (4 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{d \sqrt {\cos (c+d x)}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} \left (\frac {\int \frac {b \left (3 C a^2+12 b B a+8 A b^2+4 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+2 a b (8 A b+C b+4 a B) \sin \left (c+d x+\frac {\pi }{2}\right )+a b (8 a A-4 b B-5 a C)}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 b}-\frac {\sin (c+d x) (8 a A-5 a C-4 b B) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\right )-\frac {b (4 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{d \sqrt {\cos (c+d x)}}\right )\) |
\(\Big \downarrow \) 3532 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} \left (\frac {b \left (3 a^2 C+12 a b B+8 A b^2+4 b^2 C\right ) \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {a+b \cos (c+d x)}}dx+\int \frac {a b (8 a A-4 b B-5 a C)+2 a b (8 A b+C b+4 a B) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{2 b}-\frac {\sin (c+d x) (8 a A-5 a C-4 b B) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\right )-\frac {b (4 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{d \sqrt {\cos (c+d x)}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} \left (\frac {b \left (3 a^2 C+12 a b B+8 A b^2+4 b^2 C\right ) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\int \frac {a b (8 a A-4 b B-5 a C)+2 a b (8 A b+C b+4 a B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 b}-\frac {\sin (c+d x) (8 a A-5 a C-4 b B) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\right )-\frac {b (4 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{d \sqrt {\cos (c+d x)}}\right )\) |
\(\Big \downarrow \) 3288 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} \left (\frac {\int \frac {a b (8 a A-4 b B-5 a C)+2 a b (8 A b+C b+4 a B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} \cot (c+d x) \left (3 a^2 C+12 a b B+8 A b^2+4 b^2 C\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}}{2 b}-\frac {\sin (c+d x) (8 a A-5 a C-4 b B) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\right )-\frac {b (4 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{d \sqrt {\cos (c+d x)}}\right )\) |
\(\Big \downarrow \) 3477 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} \left (\frac {a b (8 a A-5 a C-4 b B) \int \frac {\cos (c+d x)+1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx-a b (a (8 A-8 B-5 C)-2 b (8 A+2 B+C)) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx-\frac {2 \sqrt {a+b} \cot (c+d x) \left (3 a^2 C+12 a b B+8 A b^2+4 b^2 C\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}}{2 b}-\frac {\sin (c+d x) (8 a A-5 a C-4 b B) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\right )-\frac {b (4 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{d \sqrt {\cos (c+d x)}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} \left (\frac {-a b (a (8 A-8 B-5 C)-2 b (8 A+2 B+C)) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+a b (8 a A-5 a C-4 b B) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} \cot (c+d x) \left (3 a^2 C+12 a b B+8 A b^2+4 b^2 C\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}}{2 b}-\frac {\sin (c+d x) (8 a A-5 a C-4 b B) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\right )-\frac {b (4 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{d \sqrt {\cos (c+d x)}}\right )\) |
\(\Big \downarrow \) 3295 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} \left (\frac {a b (8 a A-5 a C-4 b B) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} \cot (c+d x) \left (3 a^2 C+12 a b B+8 A b^2+4 b^2 C\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}-\frac {2 b \sqrt {a+b} \cot (c+d x) (a (8 A-8 B-5 C)-2 b (8 A+2 B+C)) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}}{2 b}-\frac {\sin (c+d x) (8 a A-5 a C-4 b B) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\right )-\frac {b (4 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{d \sqrt {\cos (c+d x)}}\right )\) |
\(\Big \downarrow \) 3473 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} \left (\frac {-\frac {2 \sqrt {a+b} \cot (c+d x) \left (3 a^2 C+12 a b B+8 A b^2+4 b^2 C\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}-\frac {2 b \sqrt {a+b} \cot (c+d x) (a (8 A-8 B-5 C)-2 b (8 A+2 B+C)) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}+\frac {2 b (a-b) \sqrt {a+b} \cot (c+d x) (8 a A-5 a C-4 b B) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a d}}{2 b}-\frac {\sin (c+d x) (8 a A-5 a C-4 b B) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\right )-\frac {b (4 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{d \sqrt {\cos (c+d x)}}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-1/2*(b*(4*A - C)*Sqrt[Cos[c + d*x] ]*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/d + (2*A*(a + b*Cos[c + d*x])^(3/ 2)*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]) + (((2*(a - b)*b*Sqrt[a + b]*(8*a* A - 4*b*B - 5*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]*(a*(8*A - 8*B - 5*C) - 2*b*(8*A + 2*B + 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)])/d - (2*Sqrt[a + b]*(8*A*b^2 + 12*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)])/d)/(2*b) - ((8*a*A - 4*b* B - 5*a*C)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]))/ 4)
3.16.14.3.1 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_)])/(((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_)])^(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])))
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. \(4337\) vs. \(2(537)=1074\).
Time = 14.42 (sec) , antiderivative size = 4338, normalized size of antiderivative = 7.29
method | result | size |
parts | \(\text {Expression too large to display}\) | \(4338\) |
default | \(\text {Expression too large to display}\) | \(4928\) |
int((a+b*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2 ),x,method=_RETURNVERBOSE)
-2*A/d*(-a^2*(-(1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*((a*(1-cos(d*x+c))^2 *csc(d*x+c)^2-b*(1-cos(d*x+c))^2*csc(d*x+c)^2+a+b)/(a+b))^(1/2)*EllipticF( cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))-2*b*(-(1-cos(d*x+c))^2*csc(d*x +c)^2+1)^(1/2)*((a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2*csc(d* x+c)^2+a+b)/(a+b))^(1/2)*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1 /2))*a+b^2*(-(1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*((a*(1-cos(d*x+c))^2*c sc(d*x+c)^2-b*(1-cos(d*x+c))^2*csc(d*x+c)^2+a+b)/(a+b))^(1/2)*EllipticF(co t(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))+(-(1-cos(d*x+c))^2*csc(d*x+c)^2+ 1)^(1/2)*((a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2*csc(d*x+c)^2 +a+b)/(a+b))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*a ^2+b*(-(1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*((a*(1-cos(d*x+c))^2*csc(d*x +c)^2-b*(1-cos(d*x+c))^2*csc(d*x+c)^2+a+b)/(a+b))^(1/2)*EllipticE(cot(d*x+ c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*a-2*b^2*(-(1-cos(d*x+c))^2*csc(d*x+c)^ 2+1)^(1/2)*((a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2*csc(d*x+c) ^2+a+b)/(a+b))^(1/2)*EllipticPi(cot(d*x+c)-csc(d*x+c),-1,(-(a-b)/(a+b))^(1 /2))+a^2*(1-cos(d*x+c))^3*csc(d*x+c)^3-a*b*(1-cos(d*x+c))^3*csc(d*x+c)^3+a ^2*(-cot(d*x+c)+csc(d*x+c))+b*a*(-cot(d*x+c)+csc(d*x+c)))*((a*(1-cos(d*x+c ))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2*csc(d*x+c)^2+a+b)/((1-cos(d*x+c))^2*c sc(d*x+c)^2+1))^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)*(-((1-cos(d*x+c))^ 2*csc(d*x+c)^2+1)/((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(3/2)/(a*(1-cos(d*...
\[ \int (a+b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{\frac {3}{2}} \,d x } \]
integrate((a+b*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c )^(3/2),x, algorithm="fricas")
integral((C*b*cos(d*x + c)^3 + (C*a + B*b)*cos(d*x + c)^2 + A*a + (B*a + A *b)*cos(d*x + c))*sqrt(b*cos(d*x + c) + a)*sec(d*x + c)^(3/2), x)
Timed out. \[ \int (a+b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\text {Timed out} \]
\[ \int (a+b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{\frac {3}{2}} \,d x } \]
integrate((a+b*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c )^(3/2),x, algorithm="maxima")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^(3/ 2)*sec(d*x + c)^(3/2), x)
\[ \int (a+b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{\frac {3}{2}} \,d x } \]
integrate((a+b*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c )^(3/2),x, algorithm="giac")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^(3/ 2)*sec(d*x + c)^(3/2), x)
Timed out. \[ \int (a+b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\int {\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{3/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right ) \,d x \]
int((1/cos(c + d*x))^(3/2)*(a + b*cos(c + d*x))^(3/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2),x)