Integrand size = 25, antiderivative size = 513 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=-\frac {8 b \left (2 a^4-7 a^2 b^2+4 b^4\right ) \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^5 (a-b) (a+b)^{3/2} d \sqrt {\sec (c+d x)}}+\frac {2 \left (a^4+9 a^3 b+16 a^2 b^2-12 a b^3-16 b^4\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^4 (a-b) (a+b)^{3/2} d \sqrt {\sec (c+d x)}}+\frac {2 b^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac {4 b^2 \left (5 a^2-3 b^2\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (a^4-13 a^2 b^2+8 b^4\right ) \sqrt {a+b \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^3 \left (a^2-b^2\right )^2 d} \]
2/3*b^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*b^2*(5*a^2-3*b^2)*sec(d*x+c)^(3/2)*sin(d*x+c)/a^2/(a^2-b^2)^2/d/(a+b*co s(d*x+c))^(1/2)+2/3*(a^4-13*a^2*b^2+8*b^4)*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-8/3*b*(2*a^4-7*a^2*b^2+4*b^4)*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-b)/(a+b)^(3/2)/d/sec(d*x+c)^(1/2)+2/3*(a^4+9*a^ 3*b+16*a^2*b^2-12*a*b^3-16*b^4)*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))*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-b)/(a+b)^ (3/2)/d/sec(d*x+c)^(1/2)
Time = 13.24 (sec) , antiderivative size = 546, normalized size of antiderivative = 1.06 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\frac {4 \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (4 b \left (2 a^5+2 a^4 b-7 a^3 b^2-7 a^2 b^3+4 a b^4+4 b^5\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {-a+b}{a+b}\right )+a \left (a^5-8 a^4 b+7 a^3 b^2+28 a^2 b^3-4 a b^4-16 b^5\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right )+2 b \left (2 a^4-7 a^2 b^2+4 b^4\right ) \cos (c+d x) (a+b \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 a^4 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)} \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )}}+\frac {\sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {8 b \left (2 a^4-7 a^2 b^2+4 b^4\right ) \sin (c+d x)}{3 a^4 \left (a^2-b^2\right )^2}-\frac {2 b^3 \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {2 \left (11 a^2 b^3 \sin (c+d x)-7 b^5 \sin (c+d x)\right )}{3 a^3 \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}+\frac {2 \tan (c+d x)}{3 a^3}\right )}{d} \]
(4*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(4*b*(2*a^5 + 2*a^4*b - 7*a^3*b^2 - 7*a^2*b^3 + 4*a*b^4 + 4*b^5)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt [(a + b*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[( c + d*x)/2]], (-a + b)/(a + b)] + a*(a^5 - 8*a^4*b + 7*a^3*b^2 + 28*a^2*b^ 3 - 4*a*b^4 - 16*b^5)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(a + b*Co s[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2 ]], (-a + b)/(a + b)] + 2*b*(2*a^4 - 7*a^2*b^2 + 4*b^4)*Cos[c + d*x]*(a + b*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(3*a^4*(a^2 - b^2)^2 *d*Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[(c + d*x)/2]^2]) + (Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((-8*b*(2*a^4 - 7*a^2*b^2 + 4*b^4)*Sin[c + d*x] )/(3*a^4*(a^2 - b^2)^2) - (2*b^3*Sin[c + d*x])/(3*a^2*(a^2 - b^2)*(a + b*C os[c + d*x])^2) - (2*(11*a^2*b^3*Sin[c + d*x] - 7*b^5*Sin[c + d*x]))/(3*a^ 3*(a^2 - b^2)^2*(a + b*Cos[c + d*x])) + (2*Tan[c + d*x])/(3*a^3)))/d
Time = 2.17 (sec) , antiderivative size = 512, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.640, Rules used = {3042, 4710, 3042, 3281, 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.
\(\displaystyle \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
\(\Big \downarrow \) 4710 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^{5/2}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
\(\Big \downarrow \) 3281 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \int \frac {4 b^2 \cos ^2(c+d x)-3 a b \cos (c+d x)+3 \left (a^2-2 b^2\right )}{2 \cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}dx}{3 a \left (a^2-b^2\right )}+\frac {2 b^2 \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {4 b^2 \cos ^2(c+d x)-3 a b \cos (c+d x)+3 \left (a^2-2 b^2\right )}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}dx}{3 a \left (a^2-b^2\right )}+\frac {2 b^2 \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {4 b^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2-3 a b \sin \left (c+d x+\frac {\pi }{2}\right )+3 \left (a^2-2 b^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{3 a \left (a^2-b^2\right )}+\frac {2 b^2 \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}\right )\) |
\(\Big \downarrow \) 3534 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 \int \frac {4 b^2 \left (5 a^2-3 b^2\right ) \cos ^2(c+d x)-2 a b \left (3 a^2-b^2\right ) \cos (c+d x)+3 \left (a^4-13 b^2 a^2+8 b^4\right )}{2 \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{a \left (a^2-b^2\right )}+\frac {4 b^2 \left (5 a^2-3 b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 b^2 \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {4 b^2 \left (5 a^2-3 b^2\right ) \cos ^2(c+d x)-2 a b \left (3 a^2-b^2\right ) \cos (c+d x)+3 \left (a^4-13 b^2 a^2+8 b^4\right )}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{a \left (a^2-b^2\right )}+\frac {4 b^2 \left (5 a^2-3 b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 b^2 \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {4 b^2 \left (5 a^2-3 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-2 a b \left (3 a^2-b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+3 \left (a^4-13 b^2 a^2+8 b^4\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}+\frac {4 b^2 \left (5 a^2-3 b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 b^2 \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}\right )\) |
\(\Big \downarrow \) 3534 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 \int -\frac {3 \left (4 b \left (2 a^4-7 b^2 a^2+4 b^4\right )-a \left (a^4+7 b^2 a^2-4 b^4\right ) \cos (c+d x)\right )}{2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{3 a}+\frac {2 \left (a^4-13 a^2 b^2+8 b^4\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{a d \cos ^{\frac {3}{2}}(c+d x)}}{a \left (a^2-b^2\right )}+\frac {4 b^2 \left (5 a^2-3 b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 b^2 \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 \left (a^4-13 a^2 b^2+8 b^4\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\int \frac {4 b \left (2 a^4-7 b^2 a^2+4 b^4\right )-a \left (a^4+7 b^2 a^2-4 b^4\right ) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{a}}{a \left (a^2-b^2\right )}+\frac {4 b^2 \left (5 a^2-3 b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 b^2 \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 \left (a^4-13 a^2 b^2+8 b^4\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\int \frac {4 b \left (2 a^4-7 b^2 a^2+4 b^4\right )-a \left (a^4+7 b^2 a^2-4 b^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{a \left (a^2-b^2\right )}+\frac {4 b^2 \left (5 a^2-3 b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 b^2 \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}\right )\) |
\(\Big \downarrow \) 3477 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 \left (a^4-13 a^2 b^2+8 b^4\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {4 b \left (2 a^4-7 a^2 b^2+4 b^4\right ) \int \frac {\cos (c+d x)+1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx-(a-b) \left (a^4+9 a^3 b+16 a^2 b^2-12 a b^3-16 b^4\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx}{a}}{a \left (a^2-b^2\right )}+\frac {4 b^2 \left (5 a^2-3 b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 b^2 \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 \left (a^4-13 a^2 b^2+8 b^4\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {4 b \left (2 a^4-7 a^2 b^2+4 b^4\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-(a-b) \left (a^4+9 a^3 b+16 a^2 b^2-12 a b^3-16 b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{a \left (a^2-b^2\right )}+\frac {4 b^2 \left (5 a^2-3 b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 b^2 \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}\right )\) |
\(\Big \downarrow \) 3295 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 \left (a^4-13 a^2 b^2+8 b^4\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {4 b \left (2 a^4-7 a^2 b^2+4 b^4\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (a-b) \sqrt {a+b} \left (a^4+9 a^3 b+16 a^2 b^2-12 a b^3-16 b^4\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}}{a}}{a \left (a^2-b^2\right )}+\frac {4 b^2 \left (5 a^2-3 b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 b^2 \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}\right )\) |
\(\Big \downarrow \) 3473 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 b^2 \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}+\frac {\frac {4 b^2 \left (5 a^2-3 b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}+\frac {\frac {2 \left (a^4-13 a^2 b^2+8 b^4\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {8 b (a-b) \sqrt {a+b} \left (2 a^4-7 a^2 b^2+4 b^4\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a^2 d}-\frac {2 (a-b) \sqrt {a+b} \left (a^4+9 a^3 b+16 a^2 b^2-12 a b^3-16 b^4\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}}{a}}{a \left (a^2-b^2\right )}}{3 a \left (a^2-b^2\right )}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*b^2*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*b^2*(5*a^2 - 3*b ^2)*Sin[c + d*x])/(a*(a^2 - b^2)*d*Cos[c + d*x]^(3/2)*Sqrt[a + b*Cos[c + d *x]]) + (-(((8*(a - b)*b*Sqrt[a + b]*(2*a^4 - 7*a^2*b^2 + 4*b^4)*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]*(a^4 + 9*a^3* b + 16*a^2*b^2 - 12*a*b^3 - 16*b^4)*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))]*Sq rt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/( a*d))/a) + (2*(a^4 - 13*a^2*b^2 + 8*b^4)*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)))
3.8.63.3.1 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[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*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[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n + 3)*Si n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a* d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[ 2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(IntegerQ[2* n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) || EqQ[a, 0])))
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[(csc[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m Int[ActivateTrig[u]/(c*Sin[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. \(5848\) vs. \(2(467)=934\).
Time = 13.99 (sec) , antiderivative size = 5849, normalized size of antiderivative = 11.40
\[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{\frac {5}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
integral(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. \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{\frac {5}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{\frac {5}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]