Integrand size = 43, antiderivative size = 403 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{b^3 \left (a^2-b^2\right ) d}-\frac {\left (9 a^3 b B-12 a b^3 B-a^2 b^2 (3 A-16 C)-15 a^4 C+2 b^4 (3 A+C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 b^4 \left (a^2-b^2\right ) d}+\frac {a \left (3 A b^4+3 a^3 b B-5 a b^3 B-a^2 b^2 (A-7 C)-5 a^4 C\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{(a-b) b^4 (a+b)^2 d}-\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}+\frac {\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}} \]
-(A*b^2-a*(B*b-C*a))*sin(d*x+c)/b/(a^2-b^2)/d/(a+b*cos(d*x+c))/sec(d*x+c)^ (3/2)+1/3*(3*A*b^2-3*B*a*b+5*C*a^2-2*C*b^2)*sin(d*x+c)/b^2/(a^2-b^2)/d/sec (d*x+c)^(1/2)+(3*B*a^2*b-2*B*b^3-a*b^2*(A-4*C)-5*a^3*C)*(cos(1/2*d*x+1/2*c )^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d* x+c)^(1/2)*sec(d*x+c)^(1/2)/b^3/(a^2-b^2)/d-1/3*(9*B*a^3*b-12*B*a*b^3-a^2* b^2*(3*A-16*C)-15*a^4*C+2*b^4*(3*A+C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/ 2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d* x+c)^(1/2)/b^4/(a^2-b^2)/d+a*(3*A*b^4+3*B*a^3*b-5*B*a*b^3-a^2*b^2*(A-7*C)- 5*a^4*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin(1/ 2*d*x+1/2*c),2*b/(a+b),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/(a-b)/b^ 4/(a+b)^2/d
Time = 7.94 (sec) , antiderivative size = 742, normalized size of antiderivative = 1.84 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\frac {2 \left (-3 a A b^2-3 a^2 b B+6 b^3 B+5 a^3 C-8 a b^2 C\right ) \cos ^2(c+d x) \left (\operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )-\operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )\right ) (b+a \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{a (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {2 \left (12 A b^3-12 a b^2 B+8 a^2 b C+4 b^3 C\right ) \cos ^2(c+d x) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) (b+a \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{b (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {\left (3 a A b^2-9 a^2 b B+6 b^3 B+15 a^3 C-12 a b^2 C\right ) \cos (2 (c+d x)) (b+a \sec (c+d x)) \left (-4 a b+4 a b \sec ^2(c+d x)-4 a b E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}+2 (2 a-b) b \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}-4 a^2 \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}+2 b^2 \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}\right ) \sin (c+d x)}{a b^2 (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \left (2-\sec ^2(c+d x)\right )}}{12 b^2 (-a+b) (a+b) d}+\frac {\sqrt {\sec (c+d x)} \left (\frac {a \left (A b^2-a b B+a^2 C\right ) \sin (c+d x)}{b^3 \left (a^2-b^2\right )}-\frac {-a^2 A b^2 \sin (c+d x)+a^3 b B \sin (c+d x)-a^4 C \sin (c+d x)}{b^3 \left (-a^2+b^2\right ) (a+b \cos (c+d x))}+\frac {C \sin (2 (c+d x))}{3 b^2}\right )}{d} \]
Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/((a + b*Cos[c + d*x])^2* Sec[c + d*x]^(3/2)),x]
((2*(-3*a*A*b^2 - 3*a^2*b*B + 6*b^3*B + 5*a^3*C - 8*a*b^2*C)*Cos[c + d*x]^ 2*(EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1] - EllipticPi[-(a/b), ArcSin[S qrt[Sec[c + d*x]]], -1])*(b + a*Sec[c + d*x])*Sqrt[1 - Sec[c + d*x]^2]*Sin [c + d*x])/(a*(a + b*Cos[c + d*x])*(1 - Cos[c + d*x]^2)) + (2*(12*A*b^3 - 12*a*b^2*B + 8*a^2*b*C + 4*b^3*C)*Cos[c + d*x]^2*EllipticPi[-(a/b), ArcSin [Sqrt[Sec[c + d*x]]], -1]*(b + a*Sec[c + d*x])*Sqrt[1 - Sec[c + d*x]^2]*Si n[c + d*x])/(b*(a + b*Cos[c + d*x])*(1 - Cos[c + d*x]^2)) + ((3*a*A*b^2 - 9*a^2*b*B + 6*b^3*B + 15*a^3*C - 12*a*b^2*C)*Cos[2*(c + d*x)]*(b + a*Sec[c + d*x])*(-4*a*b + 4*a*b*Sec[c + d*x]^2 - 4*a*b*EllipticE[ArcSin[Sqrt[Sec[ c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] + 2*(2*a - b)* b*EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Se c[c + d*x]^2] - 4*a^2*EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1]*S qrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] + 2*b^2*EllipticPi[-(a/b), ArcS in[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2])*S in[c + d*x])/(a*b^2*(a + b*Cos[c + d*x])*(1 - Cos[c + d*x]^2)*Sqrt[Sec[c + d*x]]*(2 - Sec[c + d*x]^2)))/(12*b^2*(-a + b)*(a + b)*d) + (Sqrt[Sec[c + d*x]]*((a*(A*b^2 - a*b*B + a^2*C)*Sin[c + d*x])/(b^3*(a^2 - b^2)) - (-(a^2 *A*b^2*Sin[c + d*x]) + a^3*b*B*Sin[c + d*x] - a^4*C*Sin[c + d*x])/(b^3*(-a ^2 + b^2)*(a + b*Cos[c + d*x])) + (C*Sin[2*(c + d*x)])/(3*b^2)))/d
Time = 2.38 (sec) , antiderivative size = 356, normalized size of antiderivative = 0.88, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.395, Rules used = {3042, 4709, 3042, 3526, 27, 3042, 3528, 27, 3042, 3538, 25, 3042, 3119, 3481, 3042, 3120, 3284}
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.
\(\displaystyle \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \cos (c+d x)+C \cos (c+d x)^2}{\sec (c+d x)^{3/2} (a+b \cos (c+d x))^2}dx\) |
\(\Big \downarrow \) 4709 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right )}{(a+b \cos (c+d x))^2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 3526 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\int \frac {\sqrt {\cos (c+d x)} \left (-\left (\left (5 C a^2-3 b B a+3 A b^2-2 b^2 C\right ) \cos ^2(c+d x)\right )+2 b (b B-a (A+C)) \cos (c+d x)+3 \left (A b^2-a (b B-a C)\right )\right )}{2 (a+b \cos (c+d x))}dx}{b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\int \frac {\sqrt {\cos (c+d x)} \left (-\left (\left (5 C a^2-3 b B a+3 A b^2-2 b^2 C\right ) \cos ^2(c+d x)\right )+2 b (b B-a (A+C)) \cos (c+d x)+3 \left (A b^2-a (b B-a C)\right )\right )}{a+b \cos (c+d x)}dx}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (\left (-5 C a^2+3 b B a-3 A b^2+2 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+2 b (b B-a (A+C)) \sin \left (c+d x+\frac {\pi }{2}\right )+3 \left (A b^2-a (b B-a C)\right )\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\) |
\(\Big \downarrow \) 3528 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {2 \int -\frac {3 \left (-5 C a^3+3 b B a^2-b^2 (A-4 C) a-2 b^3 B\right ) \cos ^2(c+d x)-2 b \left (2 C a^2-3 b B a+3 A b^2+b^2 C\right ) \cos (c+d x)+a \left (5 C a^2-3 b B a+3 A b^2-2 b^2 C\right )}{2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{3 b}-\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {\int \frac {3 \left (-5 C a^3+3 b B a^2-b^2 (A-4 C) a-2 b^3 B\right ) \cos ^2(c+d x)-2 b \left (2 C a^2-3 b B a+3 A b^2+b^2 C\right ) \cos (c+d x)+a \left (5 C a^2-3 b B a+3 A b^2-2 b^2 C\right )}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{3 b}-\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {\int \frac {3 \left (-5 C a^3+3 b B a^2-b^2 (A-4 C) a-2 b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-2 b \left (2 C a^2-3 b B a+3 A b^2+b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+a \left (5 C a^2-3 b B a+3 A b^2-2 b^2 C\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{3 b}-\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\) |
\(\Big \downarrow \) 3538 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {\frac {3 \left (-5 a^3 C+3 a^2 b B-a b^2 (A-4 C)-2 b^3 B\right ) \int \sqrt {\cos (c+d x)}dx}{b}-\frac {\int -\frac {a b \left (5 C a^2-3 b B a+3 A b^2-2 b^2 C\right )-\left (-15 C a^4+9 b B a^3-b^2 (3 A-16 C) a^2-12 b^3 B a+2 b^4 (3 A+C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{3 b}-\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {\frac {3 \left (-5 a^3 C+3 a^2 b B-a b^2 (A-4 C)-2 b^3 B\right ) \int \sqrt {\cos (c+d x)}dx}{b}+\frac {\int \frac {a b \left (5 C a^2-3 b B a+3 A b^2-2 b^2 C\right )-\left (-15 C a^4+9 b B a^3-b^2 (3 A-16 C) a^2-12 b^3 B a+2 b^4 (3 A+C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{3 b}-\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {\frac {3 \left (-5 a^3 C+3 a^2 b B-a b^2 (A-4 C)-2 b^3 B\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}+\frac {\int \frac {a b \left (5 C a^2-3 b B a+3 A b^2-2 b^2 C\right )+\left (15 C a^4-9 b B a^3+b^2 (3 A-16 C) a^2+12 b^3 B a-2 b^4 (3 A+C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}}{3 b}-\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {\frac {\int \frac {a b \left (5 C a^2-3 b B a+3 A b^2-2 b^2 C\right )+\left (15 C a^4-9 b B a^3+b^2 (3 A-16 C) a^2+12 b^3 B a-2 b^4 (3 A+C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}+\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-5 a^3 C+3 a^2 b B-a b^2 (A-4 C)-2 b^3 B\right )}{b d}}{3 b}-\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\) |
\(\Big \downarrow \) 3481 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {\frac {\frac {3 a \left (-5 a^4 C+3 a^3 b B-a^2 b^2 (A-7 C)-5 a b^3 B+3 A b^4\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}-\frac {\left (-15 a^4 C+9 a^3 b B-a^2 b^2 (3 A-16 C)-12 a b^3 B+2 b^4 (3 A+C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{b}}{b}+\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-5 a^3 C+3 a^2 b B-a b^2 (A-4 C)-2 b^3 B\right )}{b d}}{3 b}-\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {\frac {\frac {3 a \left (-5 a^4 C+3 a^3 b B-a^2 b^2 (A-7 C)-5 a b^3 B+3 A b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}-\frac {\left (-15 a^4 C+9 a^3 b B-a^2 b^2 (3 A-16 C)-12 a b^3 B+2 b^4 (3 A+C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}}{b}+\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-5 a^3 C+3 a^2 b B-a b^2 (A-4 C)-2 b^3 B\right )}{b d}}{3 b}-\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {\frac {\frac {3 a \left (-5 a^4 C+3 a^3 b B-a^2 b^2 (A-7 C)-5 a b^3 B+3 A b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}-\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (-15 a^4 C+9 a^3 b B-a^2 b^2 (3 A-16 C)-12 a b^3 B+2 b^4 (3 A+C)\right )}{b d}}{b}+\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-5 a^3 C+3 a^2 b B-a b^2 (A-4 C)-2 b^3 B\right )}{b d}}{3 b}-\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\) |
\(\Big \downarrow \) 3284 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {-\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b d}-\frac {\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-5 a^3 C+3 a^2 b B-a b^2 (A-4 C)-2 b^3 B\right )}{b d}+\frac {\frac {6 a \left (-5 a^4 C+3 a^3 b B-a^2 b^2 (A-7 C)-5 a b^3 B+3 A b^4\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b d (a+b)}-\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (-15 a^4 C+9 a^3 b B-a^2 b^2 (3 A-16 C)-12 a b^3 B+2 b^4 (3 A+C)\right )}{b d}}{b}}{3 b}}{2 b \left (a^2-b^2\right )}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-(((A*b^2 - a*(b*B - a*C))*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(b*(a^2 - b^2)*d*(a + b*Cos[c + d*x]))) - (-1/3*( (6*(3*a^2*b*B - 2*b^3*B - a*b^2*(A - 4*C) - 5*a^3*C)*EllipticE[(c + d*x)/2 , 2])/(b*d) + ((-2*(9*a^3*b*B - 12*a*b^3*B - a^2*b^2*(3*A - 16*C) - 15*a^4 *C + 2*b^4*(3*A + C))*EllipticF[(c + d*x)/2, 2])/(b*d) + (6*a*(3*A*b^4 + 3 *a^3*b*B - 5*a*b^3*B - a^2*b^2*(A - 7*C) - 5*a^4*C)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(b*(a + b)*d))/b)/b - (2*(3*A*b^2 - 3*a*b*B + 5*a^2*C - 2*b^2*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*b*d))/(2*b*(a^2 - b^2)))
3.15.93.3.1 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[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c , d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]
Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[ B/d Int[(a + b*Sin[e + f*x])^m, x], x] - Simp[(B*c - A*d)/d Int[(a + b* Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, 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_)])^(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)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[C/(b*d) Int[Sqrt[a + b*Sin[e + f*x]], x] , x] - Simp[1/(b*d) Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[ e + f*x], x]/(Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])), x], x] /; Fre eQ[{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. \(1128\) vs. \(2(461)=922\).
Time = 6.40 (sec) , antiderivative size = 1129, normalized size of antiderivative = 2.80
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2/sec(d*x+c)^(3/2),x, method=_RETURNVERBOSE)
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*a^2*(A*b^2-B *a*b+C*a^2)/b^4*(-b^2/a/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c )^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2*b+a-b)-1/2/(a+b)/a *(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/ 2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^ (1/2))-1/2*b/a/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2* c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellipti cF(cos(1/2*d*x+1/2*c),2^(1/2))+1/2*b/a/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(1 /2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x +1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-3*a/(a^2-b^2)/(-2*a *b+2*b^2)*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2) /(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d *x+1/2*c),-2*b/(a-b),2^(1/2))+1/a/(a^2-b^2)/(-2*a*b+2*b^2)*b^3*(sin(1/2*d* x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c) ^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^ (1/2)))+4*a/b^3*(2*A*b^2-3*B*a*b+4*C*a^2)/(-2*a*b+2*b^2)*(sin(1/2*d*x+1/2* c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin (1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2)) +2/3/b^4*(4*C*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+3*A*b^2*(sin(1/2 *d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2...
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2/sec(d*x+c)^(3 /2),x, algorithm="fricas")
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \]
\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*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)^2* sec(d*x + c)^(3/2)), x)
\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*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)^2* sec(d*x + c)^(3/2)), x)
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]
int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/((1/cos(c + d*x))^(3/2)*(a + b *cos(c + d*x))^2),x)