Integrand size = 37, antiderivative size = 589 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=-\frac {4 b \left (8 A b^4+a^4 (4 A-3 C)-a^2 b^2 (14 A-C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^5 \sqrt {a+b} \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}-\frac {2 \left (12 a A b^3+16 A b^4-2 a^2 b^2 (8 A-C)-a^4 (A+3 C)-a^3 (9 A b-3 b C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^4 \sqrt {a+b} \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}+\frac {2 \left (A b^2+a^2 C\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac {4 \left (5 a^2 A b^2-3 A b^4+2 a^4 C\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (8 A b^4+a^4 (A-5 C)-a^2 b^2 (13 A-C)\right ) \sqrt {a+b \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^3 \left (a^2-b^2\right )^2 d} \]
2/3*(A*b^2+C*a^2)*sec(d*x+c)^(3/2)*sin(d*x+c)/a/(a^2-b^2)/d/(a+b*cos(d*x+c ))^(3/2)+4/3*(5*A*a^2*b^2-3*A*b^4+2*C*a^4)*sec(d*x+c)^(3/2)*sin(d*x+c)/a^2 /(a^2-b^2)^2/d/(a+b*cos(d*x+c))^(1/2)+2/3*(8*A*b^4+a^4*(A-5*C)-a^2*b^2*(13 *A-C))*sec(d*x+c)^(3/2)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/a^3/(a^2-b^2)^2/ d-4/3*b*(8*A*b^4+a^4*(4*A-3*C)-a^2*b^2*(14*A-C))*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))*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^5/(a^2-b^2)/d/(a+b)^(1/2)/sec(d*x+c)^(1/2)-2/3*(12*a*A*b^3+16*A*b^4-2*a ^2*b^2*(8*A-C)-a^4*(A+3*C)-a^3*(9*A*b-3*C*b))*csc(d*x+c)*EllipticF((a+b*co s(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),((-a-b)/(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^ 4/(a^2-b^2)/d/(a+b)^(1/2)/sec(d*x+c)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(3973\) vs. \(2(589)=1178\).
Time = 24.10 (sec) , antiderivative size = 3973, normalized size of antiderivative = 6.75 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\text {Result too large to show} \]
(Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((-4*b*(4*a^4*A - 14*a^2*A*b^ 2 + 8*A*b^4 - 3*a^4*C + a^2*b^2*C)*Sin[c + d*x])/(3*a^4*(a^2 - b^2)^2) - ( 2*(A*b^3*Sin[c + d*x] + a^2*b*C*Sin[c + d*x]))/(3*a^2*(a^2 - b^2)*(a + b*C os[c + d*x])^2) - (2*(11*a^2*A*b^3*Sin[c + d*x] - 7*A*b^5*Sin[c + d*x] + 5 *a^4*b*C*Sin[c + d*x] - a^2*b^3*C*Sin[c + d*x]))/(3*a^3*(a^2 - b^2)^2*(a + b*Cos[c + d*x])) + (2*A*Tan[c + d*x])/(3*a^3)))/d + (4*((8*a*A*b)/(3*(a^2 - b^2)^2*Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (28*A*b^3)/(3*a*( a^2 - b^2)^2*Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (16*A*b^5)/(3* a^3*(a^2 - b^2)^2*Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (2*a*b*C) /((a^2 - b^2)^2*Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (2*b^3*C)/( 3*a*(a^2 - b^2)^2*Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (a^2*A*Sq rt[Sec[c + d*x]])/(3*(a^2 - b^2)^2*Sqrt[a + b*Cos[c + d*x]]) + (5*A*b^2*Sq rt[Sec[c + d*x]])/((a^2 - b^2)^2*Sqrt[a + b*Cos[c + d*x]]) - (32*A*b^4*Sqr t[Sec[c + d*x]])/(3*a^2*(a^2 - b^2)^2*Sqrt[a + b*Cos[c + d*x]]) + (16*A*b^ 6*Sqrt[Sec[c + d*x]])/(3*a^4*(a^2 - b^2)^2*Sqrt[a + b*Cos[c + d*x]]) + (a^ 2*C*Sqrt[Sec[c + d*x]])/((a^2 - b^2)^2*Sqrt[a + b*Cos[c + d*x]]) - (5*b^2* C*Sqrt[Sec[c + d*x]])/(3*(a^2 - b^2)^2*Sqrt[a + b*Cos[c + d*x]]) + (2*b^4* C*Sqrt[Sec[c + d*x]])/(3*a^2*(a^2 - b^2)^2*Sqrt[a + b*Cos[c + d*x]]) + (8* A*b^2*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(3*(a^2 - b^2)^2*Sqrt[a + b*Cos [c + d*x]]) - (28*A*b^4*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(3*a^2*(a^...
Time = 2.69 (sec) , antiderivative size = 581, normalized size of antiderivative = 0.99, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.432, Rules used = {3042, 4709, 3042, 3535, 27, 3042, 3534, 27, 3042, 3534, 27, 3042, 3477, 3042, 3295, 3473}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sec (c+d x)^{5/2} \left (A+C \cos (c+d x)^2\right )}{(a+b \cos (c+d x))^{5/2}}dx\) |
| \(\Big \downarrow \) 4709 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \cos ^2(c+d x)+A}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \sin \left (c+d x+\frac {\pi }{2}\right )^2+A}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 3535 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \int -\frac {-4 \left (C a^2+A b^2\right ) \cos ^2(c+d x)+3 a b (A+C) \cos (c+d x)+3 \left (2 A b^2-a^2 (A-C)\right )}{2 \cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}dx}{3 a \left (a^2-b^2\right )}+\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}-\frac {\int \frac {-4 \left (C a^2+A b^2\right ) \cos ^2(c+d x)+3 a b (A+C) \cos (c+d x)+3 \left (2 A b^2-a^2 (A-C)\right )}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}dx}{3 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}-\frac {\int \frac {-4 \left (C a^2+A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+3 a b (A+C) \sin \left (c+d x+\frac {\pi }{2}\right )+3 \left (2 A b^2-a^2 (A-C)\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{3 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}-\frac {\frac {2 \int -\frac {4 \left (2 C a^4+5 A b^2 a^2-3 A b^4\right ) \cos ^2(c+d x)+2 a b \left (A b^2-a^2 (3 A+2 C)\right ) \cos (c+d x)+3 \left ((A-5 C) a^4-b^2 (13 A-C) a^2+8 A b^4\right )}{2 \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{a \left (a^2-b^2\right )}-\frac {4 \left (2 a^4 C+5 a^2 A b^2-3 A b^4\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}-\frac {-\frac {\int \frac {4 \left (2 C a^4+5 A b^2 a^2-3 A b^4\right ) \cos ^2(c+d x)+2 a b \left (A b^2-a^2 (3 A+2 C)\right ) \cos (c+d x)+3 \left ((A-5 C) a^4-b^2 (13 A-C) a^2+8 A b^4\right )}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{a \left (a^2-b^2\right )}-\frac {4 \left (2 a^4 C+5 a^2 A b^2-3 A b^4\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}-\frac {-\frac {\int \frac {4 \left (2 C a^4+5 A b^2 a^2-3 A b^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+2 a b \left (A b^2-a^2 (3 A+2 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+3 \left ((A-5 C) a^4-b^2 (13 A-C) a^2+8 A b^4\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}-\frac {4 \left (2 a^4 C+5 a^2 A b^2-3 A b^4\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}-\frac {-\frac {\frac {2 \int -\frac {3 \left (2 b \left ((4 A-3 C) a^4-b^2 (14 A-C) a^2+8 A b^4\right )+a \left (-\left ((A+3 C) a^4\right )-b^2 (7 A+C) a^2+4 A b^4\right ) \cos (c+d x)\right )}{2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{3 a}+\frac {2 \left (a^4 (A-5 C)-a^2 b^2 (13 A-C)+8 A b^4\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{a d \cos ^{\frac {3}{2}}(c+d x)}}{a \left (a^2-b^2\right )}-\frac {4 \left (2 a^4 C+5 a^2 A b^2-3 A b^4\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}-\frac {-\frac {\frac {2 \left (a^4 (A-5 C)-a^2 b^2 (13 A-C)+8 A b^4\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\int \frac {2 b \left ((4 A-3 C) a^4-b^2 (14 A-C) a^2+8 A b^4\right )+a \left (-\left ((A+3 C) a^4\right )-b^2 (7 A+C) a^2+4 A b^4\right ) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{a}}{a \left (a^2-b^2\right )}-\frac {4 \left (2 a^4 C+5 a^2 A b^2-3 A b^4\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}-\frac {-\frac {\frac {2 \left (a^4 (A-5 C)-a^2 b^2 (13 A-C)+8 A b^4\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\int \frac {2 b \left ((4 A-3 C) a^4-b^2 (14 A-C) a^2+8 A b^4\right )+a \left (-\left ((A+3 C) a^4\right )-b^2 (7 A+C) a^2+4 A b^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{a \left (a^2-b^2\right )}-\frac {4 \left (2 a^4 C+5 a^2 A b^2-3 A b^4\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 3477 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}-\frac {-\frac {\frac {2 \left (a^4 (A-5 C)-a^2 b^2 (13 A-C)+8 A b^4\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {2 b \left (a^4 (4 A-3 C)-a^2 b^2 (14 A-C)+8 A b^4\right ) \int \frac {\cos (c+d x)+1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx+(a-b) \left (-\left (a^4 (A+3 C)\right )-a^3 (9 A b-3 b C)-2 a^2 b^2 (8 A-C)+12 a A b^3+16 A b^4\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx}{a}}{a \left (a^2-b^2\right )}-\frac {4 \left (2 a^4 C+5 a^2 A b^2-3 A b^4\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}-\frac {-\frac {\frac {2 \left (a^4 (A-5 C)-a^2 b^2 (13 A-C)+8 A b^4\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {2 b \left (a^4 (4 A-3 C)-a^2 b^2 (14 A-C)+8 A b^4\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+(a-b) \left (-\left (a^4 (A+3 C)\right )-a^3 (9 A b-3 b C)-2 a^2 b^2 (8 A-C)+12 a A b^3+16 A b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{a \left (a^2-b^2\right )}-\frac {4 \left (2 a^4 C+5 a^2 A b^2-3 A b^4\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 3295 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}-\frac {-\frac {\frac {2 \left (a^4 (A-5 C)-a^2 b^2 (13 A-C)+8 A b^4\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {2 b \left (a^4 (4 A-3 C)-a^2 b^2 (14 A-C)+8 A b^4\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (a-b) \sqrt {a+b} \left (-\left (a^4 (A+3 C)\right )-a^3 (9 A b-3 b C)-2 a^2 b^2 (8 A-C)+12 a A b^3+16 A b^4\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}}{a}}{a \left (a^2-b^2\right )}-\frac {4 \left (2 a^4 C+5 a^2 A b^2-3 A b^4\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 3473 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}-\frac {-\frac {4 \left (2 a^4 C+5 a^2 A b^2-3 A b^4\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {2 \left (a^4 (A-5 C)-a^2 b^2 (13 A-C)+8 A b^4\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {4 b (a-b) \sqrt {a+b} \left (a^4 (4 A-3 C)-a^2 b^2 (14 A-C)+8 A b^4\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a^2 d}+\frac {2 (a-b) \sqrt {a+b} \left (-\left (a^4 (A+3 C)\right )-a^3 (9 A b-3 b C)-2 a^2 b^2 (8 A-C)+12 a A b^3+16 A b^4\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}}{a}}{a \left (a^2-b^2\right )}}{3 a \left (a^2-b^2\right )}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*(A*b^2 + a^2*C)*Sin[c + d*x])/(3 *a*(a^2 - b^2)*d*Cos[c + d*x]^(3/2)*(a + b*Cos[c + d*x])^(3/2)) - ((-4*(5* a^2*A*b^2 - 3*A*b^4 + 2*a^4*C)*Sin[c + d*x])/(a*(a^2 - b^2)*d*Cos[c + d*x] ^(3/2)*Sqrt[a + b*Cos[c + d*x]]) - (-(((4*(a - b)*b*Sqrt[a + b]*(8*A*b^4 + a^4*(4*A - 3*C) - a^2*b^2*(14*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^2*d) + (2*(a - b)*Sqrt[a + b]*(12*a*A*b^3 + 16*A*b^4 - 2*a^2*b^2*(8*A - C) - a^4*(A + 3*C) - a^3*(9*A*b - 3*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)])/(a*d))/a) + (2*(8*A*b^4 + a^4*(A - 5*C) - a^2*b^2*(13*A - C))*Sqrt [a + b*Cos[c + d*x]]*Sin[c + d*x])/(a*d*Cos[c + d*x]^(3/2)))/(a*(a^2 - b^2 )))/(3*a*(a^2 - b^2)))
3.15.46.3.1 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[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[(-(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_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 + a^2*C))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*S in[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[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d *(A*b^2 + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, 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[(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. \(8732\) vs. \(2(543)=1086\).
Time = 20.78 (sec) , antiderivative size = 8733, normalized size of antiderivative = 14.83
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(8733\) |
| parts | \(\text {Expression too large to display}\) | \(8742\) |
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{\frac {5}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
integral((C*cos(d*x + c)^2 + A)*sqrt(b*cos(d*x + c) + a)*sec(d*x + c)^(5/2 )/(b^3*cos(d*x + c)^3 + 3*a*b^2*cos(d*x + c)^2 + 3*a^2*b*cos(d*x + c) + a^ 3), x)
Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{\frac {5}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{\frac {5}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]