Integrand size = 37, antiderivative size = 534 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=-\frac {2 \left (16 A b^4-2 a^2 b^2 (4 A-5 C)-a^4 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{5 a^5 \sqrt {a+b} d \sqrt {\sec (c+d x)}}-\frac {2 \left (12 a A b^2+16 A b^3+2 a^2 b (2 A+5 C)+a^3 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{5 a^4 \sqrt {a+b} d \sqrt {\sec (c+d x)}}+\frac {2 b \left (8 A b^2-a^2 (3 A-5 C)\right ) \sqrt {a+b \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 a^3 \left (a^2-b^2\right ) d}+\frac {2 \left (A b^2+a^2 C\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (6 A b^2-a^2 (A-5 C)\right ) \sqrt {a+b \cos (c+d x)} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d} \]
2*(A*b^2+C*a^2)*sec(d*x+c)^(5/2)*sin(d*x+c)/a/(a^2-b^2)/d/(a+b*cos(d*x+c)) ^(1/2)+2/5*b*(8*A*b^2-a^2*(3*A-5*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)/d-2/5*(6*A*b^2-a^2*(A-5*C))*sec(d*x+c)^(5/2)*s in(d*x+c)*(a+b*cos(d*x+c))^(1/2)/a^2/(a^2-b^2)/d-2/5*(16*A*b^4-2*a^2*b^2*( 4*A-5*C)-a^4*(3*A+5*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/d/(a+b)^(1/2)/sec(d* x+c)^(1/2)-2/5*(12*a*A*b^2+16*A*b^3+2*a^2*b*(2*A+5*C)+a^3*(3*A+5*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))*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/d/(a+b)^(1/2)/sec(d*x+c)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(3767\) vs. \(2(534)=1068\).
Time = 22.99 (sec) , antiderivative size = 3767, normalized size of antiderivative = 7.05 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\text {Result too large to show} \]
(Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*(3*a^4*A + 8*a^2*A*b^2 - 16*A*b^4 + 5*a^4*C - 10*a^2*b^2*C)*Sin[c + d*x])/(5*a^4*(a^2 - b^2)) + (2* (A*b^4*Sin[c + d*x] + a^2*b^2*C*Sin[c + d*x]))/(a^3*(a^2 - b^2)*(a + b*Cos [c + d*x])) - (6*A*b*Tan[c + d*x])/(5*a^3) + (2*A*Sec[c + d*x]*Tan[c + d*x ])/(5*a^2)))/d + (2*((-3*a*A)/(5*(a^2 - b^2)*Sqrt[a + b*Cos[c + d*x]]*Sqrt [Sec[c + d*x]]) - (8*A*b^2)/(5*a*(a^2 - b^2)*Sqrt[a + b*Cos[c + d*x]]*Sqrt [Sec[c + d*x]]) + (16*A*b^4)/(5*a^3*(a^2 - b^2)*Sqrt[a + b*Cos[c + d*x]]*S qrt[Sec[c + d*x]]) - (a*C)/((a^2 - b^2)*Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[ c + d*x]]) + (2*b^2*C)/(a*(a^2 - b^2)*Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (4*A*b*Sqrt[Sec[c + d*x]])/(5*(a^2 - b^2)*Sqrt[a + b*Cos[c + d* x]]) - (12*A*b^3*Sqrt[Sec[c + d*x]])/(5*a^2*(a^2 - b^2)*Sqrt[a + b*Cos[c + d*x]]) + (16*A*b^5*Sqrt[Sec[c + d*x]])/(5*a^4*(a^2 - b^2)*Sqrt[a + b*Cos[ c + d*x]]) - (2*b*C*Sqrt[Sec[c + d*x]])/((a^2 - b^2)*Sqrt[a + b*Cos[c + d* x]]) + (2*b^3*C*Sqrt[Sec[c + d*x]])/(a^2*(a^2 - b^2)*Sqrt[a + b*Cos[c + d* x]]) - (3*A*b*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(5*(a^2 - b^2)*Sqrt[a + b*Cos[c + d*x]]) - (8*A*b^3*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(5*a^2*( a^2 - b^2)*Sqrt[a + b*Cos[c + d*x]]) + (16*A*b^5*Cos[2*(c + d*x)]*Sqrt[Sec [c + d*x]])/(5*a^4*(a^2 - b^2)*Sqrt[a + b*Cos[c + d*x]]) - (b*C*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/((a^2 - b^2)*Sqrt[a + b*Cos[c + d*x]]) + (2*b^3 *C*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(a^2*(a^2 - b^2)*Sqrt[a + b*Cos...
Time = 2.53 (sec) , antiderivative size = 528, 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.
\(\displaystyle \int \frac {\sec ^{\frac {7}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (c+d x)^{7/2} \left (A+C \cos (c+d x)^2\right )}{(a+b \cos (c+d x))^{3/2}}dx\) |
\(\Big \downarrow \) 4709 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \cos ^2(c+d x)+A}{\cos ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \sin \left (c+d x+\frac {\pi }{2}\right )^2+A}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx\) |
\(\Big \downarrow \) 3535 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \int -\frac {-\left ((A-5 C) a^2\right )+b (A+C) \cos (c+d x) a+6 A b^2-4 \left (C a^2+A b^2\right ) \cos ^2(c+d x)}{2 \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{a \left (a^2-b^2\right )}+\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}-\frac {\int \frac {-\left ((A-5 C) a^2\right )+b (A+C) \cos (c+d x) a+6 A b^2-4 \left (C a^2+A b^2\right ) \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{a \left (a^2-b^2\right )}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}-\frac {\int \frac {-\left ((A-5 C) a^2\right )+b (A+C) \sin \left (c+d x+\frac {\pi }{2}\right ) a+6 A b^2-4 \left (C a^2+A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}\right )\) |
\(\Big \downarrow \) 3534 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {2 \int -\frac {-2 b \left (6 A b^2-a^2 (A-5 C)\right ) \cos ^2(c+d x)+a \left ((3 A+5 C) a^2+2 A b^2\right ) \cos (c+d x)+3 b \left (8 A b^2-a^2 (3 A-5 C)\right )}{2 \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{5 a}+\frac {2 \left (6 A b^2-a^2 (A-5 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 a d \cos ^{\frac {5}{2}}(c+d x)}}{a \left (a^2-b^2\right )}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {2 \left (6 A b^2-a^2 (A-5 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\int \frac {-2 b \left (6 A b^2-a^2 (A-5 C)\right ) \cos ^2(c+d x)+a \left ((3 A+5 C) a^2+2 A b^2\right ) \cos (c+d x)+3 b \left (8 A b^2-a^2 (3 A-5 C)\right )}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{5 a}}{a \left (a^2-b^2\right )}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {2 \left (6 A b^2-a^2 (A-5 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\int \frac {-2 b \left (6 A b^2-a^2 (A-5 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+a \left ((3 A+5 C) a^2+2 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+3 b \left (8 A b^2-a^2 (3 A-5 C)\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 a}}{a \left (a^2-b^2\right )}\right )\) |
\(\Big \downarrow \) 3534 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {2 \left (6 A b^2-a^2 (A-5 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {2 \int -\frac {3 \left (-\left ((3 A+5 C) a^4\right )-2 b^2 (4 A-5 C) a^2+b \left ((A+5 C) a^2+4 A b^2\right ) \cos (c+d x) a+16 A b^4\right )}{2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{3 a}+\frac {2 b \left (8 A b^2-a^2 (3 A-5 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{a d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}}{a \left (a^2-b^2\right )}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {2 \left (6 A b^2-a^2 (A-5 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {2 b \left (8 A b^2-a^2 (3 A-5 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\int \frac {-\left ((3 A+5 C) a^4\right )-2 b^2 (4 A-5 C) a^2+b \left ((A+5 C) a^2+4 A b^2\right ) \cos (c+d x) a+16 A b^4}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{a}}{5 a}}{a \left (a^2-b^2\right )}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {2 \left (6 A b^2-a^2 (A-5 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {2 b \left (8 A b^2-a^2 (3 A-5 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\int \frac {-\left ((3 A+5 C) a^4\right )-2 b^2 (4 A-5 C) a^2+b \left ((A+5 C) a^2+4 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a+16 A b^4}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{5 a}}{a \left (a^2-b^2\right )}\right )\) |
\(\Big \downarrow \) 3477 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {2 \left (6 A b^2-a^2 (A-5 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {2 b \left (8 A b^2-a^2 (3 A-5 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\left (-\left (a^4 (3 A+5 C)\right )-2 a^2 b^2 (4 A-5 C)+16 A b^4\right ) \int \frac {\cos (c+d x)+1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx+(a-b) \left (a^3 (3 A+5 C)+2 a^2 b (2 A+5 C)+12 a A b^2+16 A b^3\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx}{a}}{5 a}}{a \left (a^2-b^2\right )}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {2 \left (6 A b^2-a^2 (A-5 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {2 b \left (8 A b^2-a^2 (3 A-5 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\left (-\left (a^4 (3 A+5 C)\right )-2 a^2 b^2 (4 A-5 C)+16 A b^4\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+(a-b) \left (a^3 (3 A+5 C)+2 a^2 b (2 A+5 C)+12 a A b^2+16 A b^3\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{5 a}}{a \left (a^2-b^2\right )}\right )\) |
\(\Big \downarrow \) 3295 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {2 \left (6 A b^2-a^2 (A-5 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {2 b \left (8 A b^2-a^2 (3 A-5 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\left (-\left (a^4 (3 A+5 C)\right )-2 a^2 b^2 (4 A-5 C)+16 A b^4\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (a-b) \sqrt {a+b} \left (a^3 (3 A+5 C)+2 a^2 b (2 A+5 C)+12 a A b^2+16 A b^3\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}}{a}}{5 a}}{a \left (a^2-b^2\right )}\right )\) |
\(\Big \downarrow \) 3473 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {2 \left (6 A b^2-a^2 (A-5 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {2 b \left (8 A b^2-a^2 (3 A-5 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (a-b) \sqrt {a+b} \left (-\left (a^4 (3 A+5 C)\right )-2 a^2 b^2 (4 A-5 C)+16 A b^4\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a^2 d}+\frac {2 (a-b) \sqrt {a+b} \left (a^3 (3 A+5 C)+2 a^2 b (2 A+5 C)+12 a A b^2+16 A b^3\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}}{a}}{5 a}}{a \left (a^2-b^2\right )}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*(A*b^2 + a^2*C)*Sin[c + d*x])/(a *(a^2 - b^2)*d*Cos[c + d*x]^(5/2)*Sqrt[a + b*Cos[c + d*x]]) - ((2*(6*A*b^2 - a^2*(A - 5*C))*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(5*a*d*Cos[c + d* x]^(5/2)) - (-(((2*(a - b)*Sqrt[a + b]*(16*A*b^4 - 2*a^2*b^2*(4*A - 5*C) - a^4*(3*A + 5*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^2 + 16*A*b^3 + 2*a^2*b*(2*A + 5*C) + a^3*(3*A + 5* C))*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sq rt[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*b*(8*A*b^2 - a^2*( 3*A - 5*C))*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(a*d*Cos[c + d*x]^(3/2) ))/(5*a))/(a*(a^2 - b^2)))
3.15.40.3.1 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. \(4609\) vs. \(2(490)=980\).
Time = 22.28 (sec) , antiderivative size = 4610, normalized size of antiderivative = 8.63
method | result | size |
parts | \(\text {Expression too large to display}\) | \(4610\) |
default | \(\text {Expression too large to display}\) | \(5481\) |
2/5*A/d/(a+b)/(a-b)/a^4*(16*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+ b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b )/(a+b))^(1/2))*a*b^4*cos(d*x+c)^4+a^5*sin(d*x+c)+2*a^2*b^3*cos(d*x+c)^2*s in(d*x+c)-8*a*b^4*cos(d*x+c)^2*sin(d*x+c)-2*a^4*b*cos(d*x+c)*sin(d*x+c)-a^ 3*b^2*cos(d*x+c)*sin(d*x+c)+2*a^2*b^3*cos(d*x+c)*sin(d*x+c)-3*a^3*b^2*cos( d*x+c)^3*sin(d*x+c)+8*a^2*b^3*cos(d*x+c)^3*sin(d*x+c)+8*a*b^4*cos(d*x+c)^3 *sin(d*x+c)-2*a^4*b*cos(d*x+c)^2*sin(d*x+c)+5*a^3*b^2*cos(d*x+c)^2*sin(d*x +c)+3*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1 +cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a^4*b* cos(d*x+c)^2+8*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*(cos( d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/ 2)*a^3*b^2*cos(d*x+c)^2+8*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^( 1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d *x+c)))^(1/2)*a^2*b^3*cos(d*x+c)^2-16*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a -b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+ c))/(1+cos(d*x+c)))^(1/2)*a*b^4*cos(d*x+c)^2+(cos(d*x+c)/(1+cos(d*x+c)))^( 1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF(cot(d*x+c)- csc(d*x+c),(-(a-b)/(a+b))^(1/2))*a^4*b*cos(d*x+c)^2-8*(cos(d*x+c)/(1+cos(d *x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF(co t(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*a^3*b^2*cos(d*x+c)^2+4*(cos(d...
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{\frac {7}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
integral((C*cos(d*x + c)^2 + A)*sqrt(b*cos(d*x + c) + a)*sec(d*x + c)^(7/2 )/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2), x)
Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{\frac {7}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{\frac {7}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]