Integrand size = 43, antiderivative size = 294 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx=-\frac {2 \left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 a^3 d}-\frac {2 (A b-a B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a^2 d}-\frac {2 b \left (A b^2-a (b B-a C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a^3 (a+b) d}+\frac {2 \left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 a^3 d}-\frac {2 (A b-a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d}+\frac {2 A \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 a d} \]
-2/3*(A*b-B*a)*sec(d*x+c)^(3/2)*sin(d*x+c)/a^2/d+2/5*A*sec(d*x+c)^(5/2)*si n(d*x+c)/a/d+2/5*(5*A*b^2-5*B*a*b+a^2*(3*A+5*C))*sin(d*x+c)*sec(d*x+c)^(1/ 2)/a^3/d-2/5*(5*A*b^2-5*B*a*b+a^2*(3*A+5*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)/a^3/d-2/3*(A*b-B*a)*(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)/a^2/d-2*b*(A*b^2-a*(B*b-C*a))*(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^3/(a+b)/d
Leaf count is larger than twice the leaf count of optimal. \(692\) vs. \(2(294)=588\).
Time = 8.89 (sec) , antiderivative size = 692, normalized size of antiderivative = 2.35 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx=-\frac {\frac {2 \left (19 a^2 A b+45 A b^3-10 a^3 B-45 a b^2 B+45 a^2 b C\right ) \cos ^2(c+d x) \left (\operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )-\operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )\right ) (b+a \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{a (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {2 \left (18 a^3 A+40 a A b^2-40 a^2 b B+30 a^3 C\right ) \cos ^2(c+d x) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) (b+a \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{b (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {\left (9 a^2 A b+15 A b^3-15 a b^2 B+15 a^2 b C\right ) \cos (2 (c+d x)) (b+a \sec (c+d x)) \left (-4 a b+4 a b \sec ^2(c+d x)-4 a b E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}+2 (2 a-b) b \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}-4 a^2 \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}+2 b^2 \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}\right ) \sin (c+d x)}{a b^2 (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \left (2-\sec ^2(c+d x)\right )}}{30 a^3 d}+\frac {\sqrt {\sec (c+d x)} \left (\frac {2 \left (3 a^2 A+5 A b^2-5 a b B+5 a^2 C\right ) \sin (c+d x)}{5 a^3}+\frac {2 \sec (c+d x) (-A b \sin (c+d x)+a B \sin (c+d x))}{3 a^2}+\frac {2 A \sec (c+d x) \tan (c+d x)}{5 a}\right )}{d} \]
Integrate[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(7/2))/(a + b*Cos[c + d*x]),x]
-1/30*((2*(19*a^2*A*b + 45*A*b^3 - 10*a^3*B - 45*a*b^2*B + 45*a^2*b*C)*Cos [c + d*x]^2*(EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1] - EllipticPi[-(a/b) , ArcSin[Sqrt[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*(1 8*a^3*A + 40*a*A*b^2 - 40*a^2*b*B + 30*a^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]*Sin[c + d*x])/(b*(a + b*Cos[c + d*x])*(1 - Cos[c + d*x]^2)) + (( 9*a^2*A*b + 15*A*b^3 - 15*a*b^2*B + 15*a^2*b*C)*Cos[2*(c + d*x)]*(b + a*Se c[c + d*x])*(-4*a*b + 4*a*b*Sec[c + d*x]^2 - 4*a*b*EllipticE[ArcSin[Sqrt[S ec[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 - Sec[c + d*x]^2] - 4*a^2*EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1 ]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] + 2*b^2*EllipticPi[-(a/b), A rcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] )*Sin[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)))/(a^3*d) + (Sqrt[Sec[c + d*x]]*((2*(3*a^2* A + 5*A*b^2 - 5*a*b*B + 5*a^2*C)*Sin[c + d*x])/(5*a^3) + (2*Sec[c + d*x]*( -(A*b*Sin[c + d*x]) + a*B*Sin[c + d*x]))/(3*a^2) + (2*A*Sec[c + d*x]*Tan[c + d*x])/(5*a)))/d
Time = 2.41 (sec) , antiderivative size = 272, normalized size of antiderivative = 0.93, number of steps used = 20, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.465, Rules used = {3042, 4709, 3042, 3534, 27, 3042, 3534, 27, 3042, 3534, 27, 3042, 3538, 27, 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.
| \(\displaystyle \int \frac {\sec ^{\frac {7}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sec (c+d x)^{7/2} \left (A+B \cos (c+d x)+C \cos (c+d x)^2\right )}{a+b \cos (c+d x)}dx\) |
| \(\Big \downarrow \) 4709 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \cos ^2(c+d x)+B \cos (c+d x)+A}{\cos ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \int -\frac {-3 A b \cos ^2(c+d x)-a (3 A+5 C) \cos (c+d x)+5 (A b-a B)}{2 \cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))}dx}{5 a}+\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\int \frac {-3 A b \cos ^2(c+d x)-a (3 A+5 C) \cos (c+d x)+5 (A b-a B)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))}dx}{5 a}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\int \frac {-3 A b \sin \left (c+d x+\frac {\pi }{2}\right )^2-a (3 A+5 C) \sin \left (c+d x+\frac {\pi }{2}\right )+5 (A b-a B)}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{5 a}\right )\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {2 \int -\frac {-5 b (A b-a B) \cos ^2(c+d x)+a (4 A b+5 a B) \cos (c+d x)+3 \left ((3 A+5 C) a^2-5 b B a+5 A b^2\right )}{2 \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}dx}{3 a}+\frac {10 (A b-a B) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\int \frac {-5 b (A b-a B) \cos ^2(c+d x)+a (4 A b+5 a B) \cos (c+d x)+3 \left ((3 A+5 C) a^2-5 b B a+5 A b^2\right )}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}dx}{3 a}}{5 a}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\int \frac {-5 b (A b-a B) \sin \left (c+d x+\frac {\pi }{2}\right )^2+a (4 A b+5 a B) \sin \left (c+d x+\frac {\pi }{2}\right )+3 \left ((3 A+5 C) a^2-5 b B a+5 A b^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{3 a}}{5 a}\right )\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 \int -\frac {3 b \left ((3 A+5 C) a^2-5 b B a+5 A b^2\right ) \cos ^2(c+d x)+a \left (3 (3 A+5 C) a^2-20 b B a+20 A b^2\right ) \cos (c+d x)+5 \left (-B a^3+b (A+3 C) a^2-3 b^2 B a+3 A b^3\right )}{2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{a}+\frac {6 \sin (c+d x) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{a d \sqrt {\cos (c+d x)}}}{3 a}}{5 a}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \sin (c+d x) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{a d \sqrt {\cos (c+d x)}}-\frac {\int \frac {3 b \left ((3 A+5 C) a^2-5 b B a+5 A b^2\right ) \cos ^2(c+d x)+a \left (3 (3 A+5 C) a^2-20 b B a+20 A b^2\right ) \cos (c+d x)+5 \left (-B a^3+b (A+3 C) a^2-3 b^2 B a+3 A b^3\right )}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{a}}{3 a}}{5 a}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \sin (c+d x) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{a d \sqrt {\cos (c+d x)}}-\frac {\int \frac {3 b \left ((3 A+5 C) a^2-5 b B a+5 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+a \left (3 (3 A+5 C) a^2-20 b B a+20 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+5 \left (-B a^3+b (A+3 C) a^2-3 b^2 B a+3 A b^3\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{3 a}}{5 a}\right )\) |
| \(\Big \downarrow \) 3538 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \sin (c+d x) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{a d \sqrt {\cos (c+d x)}}-\frac {3 \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right ) \int \sqrt {\cos (c+d x)}dx-\frac {\int -\frac {5 \left (a (A b-a B) \cos (c+d x) b^2+\left (-B a^3+b (A+3 C) a^2-3 b^2 B a+3 A b^3\right ) b\right )}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{a}}{3 a}}{5 a}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \sin (c+d x) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{a d \sqrt {\cos (c+d x)}}-\frac {3 \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right ) \int \sqrt {\cos (c+d x)}dx+\frac {5 \int \frac {a (A b-a B) \cos (c+d x) b^2+\left (-B a^3+b (A+3 C) a^2-3 b^2 B a+3 A b^3\right ) b}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{a}}{3 a}}{5 a}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \sin (c+d x) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{a d \sqrt {\cos (c+d x)}}-\frac {3 \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {5 \int \frac {a (A b-a B) \sin \left (c+d x+\frac {\pi }{2}\right ) b^2+\left (-B a^3+b (A+3 C) a^2-3 b^2 B a+3 A b^3\right ) b}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}}{a}}{3 a}}{5 a}\right )\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \sin (c+d x) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {5 \int \frac {a (A b-a B) \sin \left (c+d x+\frac {\pi }{2}\right ) b^2+\left (-B a^3+b (A+3 C) a^2-3 b^2 B a+3 A b^3\right ) b}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}+\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{d}}{a}}{3 a}}{5 a}\right )\) |
| \(\Big \downarrow \) 3481 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \sin (c+d x) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {5 \left (3 b^2 \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx+a b (A b-a B) \int \frac {1}{\sqrt {\cos (c+d x)}}dx\right )}{b}+\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{d}}{a}}{3 a}}{5 a}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \sin (c+d x) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {5 \left (3 b^2 \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx+a b (A b-a B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx\right )}{b}+\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{d}}{a}}{3 a}}{5 a}\right )\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \sin (c+d x) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {5 \left (3 b^2 \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx+\frac {2 a b (A b-a B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )}{b}+\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{d}}{a}}{3 a}}{5 a}\right )\) |
| \(\Big \downarrow \) 3284 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \sin (c+d x) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{d}+\frac {5 \left (\frac {6 b^2 \left (A b^2-a (b B-a C)\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{d (a+b)}+\frac {2 a b (A b-a B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )}{b}}{a}}{3 a}}{5 a}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*A*Sin[c + d*x])/(5*a*d*Cos[c + d *x]^(5/2)) - ((10*(A*b - a*B)*Sin[c + d*x])/(3*a*d*Cos[c + d*x]^(3/2)) - ( -(((6*(5*A*b^2 - 5*a*b*B + a^2*(3*A + 5*C))*EllipticE[(c + d*x)/2, 2])/d + (5*((2*a*b*(A*b - a*B)*EllipticF[(c + d*x)/2, 2])/d + (6*b^2*(A*b^2 - a*( b*B - a*C))*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/((a + b)*d)))/b)/a) + (6*(5*A*b^2 - 5*a*b*B + a^2*(3*A + 5*C))*Sin[c + d*x])/(a*d*Sqrt[Cos[c + d*x]]))/(3*a))/(5*a))
3.15.82.3.1 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[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b* c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int [(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b*c - a* d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A *b^2 - a*b*B + a^2*C)*(m + n + 3)*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] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ [n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) | | EqQ[a, 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. \(774\) vs. \(2(344)=688\).
Time = 164.54 (sec) , antiderivative size = 775, normalized size of antiderivative = 2.64
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2)/(a+b*cos(d*x+c)),x,me thod=_RETURNVERBOSE)
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2/5*A/a/(8*sin (1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/ 2*d*x+1/2*c)^2*(24*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-12*(2*sin(1/2*d *x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/ 2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2 *c)+12*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellip ticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2+8*sin(1/2*d*x+1/2*c) ^2*cos(1/2*d*x+1/2*c)-3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c) ^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*(-2*sin(1/2*d*x+1/2*c)^ 4+sin(1/2*d*x+1/2*c)^2)^(1/2)+2*(-A*b+B*a)/a^2*(-1/6*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)/(cos(1/2*d*x+1/2*c)^2-1 /2)^2+1/3*(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)*EllipticF(cos(1/2*d*x+ 1/2*c),2^(1/2)))+2*(A*b^2-B*a*b+C*a^2)/a^3/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2 *d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*s in(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin (1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+4*(A*b^2 -B*a*b+C*a^2)*b^2/a^3/(-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)))/sin(1/2*d*x+1/2...
Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx=\text {Timed out} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2)/(a+b*cos(d*x+c) ),x, algorithm="fricas")
Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx=\text {Timed out} \]
\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {7}{2}}}{b \cos \left (d x + c\right ) + a} \,d x } \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2)/(a+b*cos(d*x+c) ),x, algorithm="maxima")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sec(d*x + c)^(7/2)/(b*co s(d*x + c) + a), x)
\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {7}{2}}}{b \cos \left (d x + c\right ) + a} \,d x } \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2)/(a+b*cos(d*x+c) ),x, algorithm="giac")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sec(d*x + c)^(7/2)/(b*co s(d*x + c) + a), x)
Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx=\int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{a+b\,\cos \left (c+d\,x\right )} \,d x \]