Integrand size = 35, antiderivative size = 607 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=-\frac {2 \left (8 a^4 A b-28 a^2 A b^3+16 A b^5-3 a^5 B+15 a^3 b^2 B-8 a b^4 B\right ) \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^5 (a-b) (a+b)^{3/2} d \sqrt {\sec (c+d x)}}-\frac {2 \left (16 A b^4-a^4 (A-3 B)+4 a b^3 (3 A-2 B)-9 a^3 b (A-B)-2 a^2 b^2 (8 A+3 B)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^4 \sqrt {a+b} \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}+\frac {2 b (A b-a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac {2 b \left (10 a^2 A b-6 A b^3-7 a^3 B+3 a b^2 B\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (a^4 A-13 a^2 A b^2+8 A b^4+8 a^3 b B-4 a b^3 B\right ) \sqrt {a+b \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^3 \left (a^2-b^2\right )^2 d} \]
2/3*b*(A*b-B*a)*sec(d*x+c)^(3/2)*sin(d*x+c)/a/(a^2-b^2)/d/(a+b*cos(d*x+c)) ^(3/2)+2/3*b*(10*A*a^2*b-6*A*b^3-7*B*a^3+3*B*a*b^2)*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*(A*a^4-13*A*a^2*b^2+8*A *b^4+8*B*a^3*b-4*B*a*b^3)*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-2/3*(8*A*a^4*b-28*A*a^2*b^3+16*A*b^5-3*B*a^5+15*B*a^3 *b^2-8*B*a*b^4)*csc(d*x+c)*EllipticE((a+b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/co s(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*(16*A*b^4-a^4*(A-3*B)+4*a*b^3*(3*A-2*B)-9*a^3*b*(A-B)-2*a^2*b ^2*(8*A+3*B))*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^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. \(4316\) vs. \(2(607)=1214\).
Time = 24.30 (sec) , antiderivative size = 4316, normalized size of antiderivative = 7.11 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\text {Result too large to show} \]
(Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*(-8*a^4*A*b + 28*a^2*A*b^ 3 - 16*A*b^5 + 3*a^5*B - 15*a^3*b^2*B + 8*a*b^4*B)*Sin[c + d*x])/(3*a^4*(a ^2 - b^2)^2) + (2*(-(A*b^3*Sin[c + d*x]) + a*b^2*B*Sin[c + d*x]))/(3*a^2*( a^2 - b^2)*(a + b*Cos[c + d*x])^2) + (2*(-11*a^2*A*b^3*Sin[c + d*x] + 7*A* b^5*Sin[c + d*x] + 8*a^3*b^2*B*Sin[c + d*x] - 4*a*b^4*B*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 + (2*((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]]) - (a^2*B)/((a^2 - b^2)^2*Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (5*b^2*B)/((a^2 - b^2)^2*Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d* x]]) - (8*b^4*B)/(3*a^2*(a^2 - b^2)^2*Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (a^2*A*Sqrt[Sec[c + d*x]])/(3*(a^2 - b^2)^2*Sqrt[a + b*Cos[c + d*x]]) + (5*A*b^2*Sqrt[Sec[c + d*x]])/((a^2 - b^2)^2*Sqrt[a + b*Cos[c + d* x]]) - (32*A*b^4*Sqrt[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]]) - (3*a*b*B*Sqrt[Sec[c + d*x]])/((a^2 - b^2)^2*Sqrt[a + b*Co s[c + d*x]]) + (17*b^3*B*Sqrt[Sec[c + d*x]])/(3*a*(a^2 - b^2)^2*Sqrt[a + b *Cos[c + d*x]]) - (8*b^5*B*Sqrt[Sec[c + d*x]])/(3*a^3*(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*...
Time = 2.90 (sec) , antiderivative size = 602, normalized size of antiderivative = 0.99, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.457, Rules used = {3042, 3440, 3042, 3479, 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.
| \(\displaystyle \int \frac {\sec ^{\frac {5}{2}}(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 3440 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 3479 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \int \frac {4 b (A b-a B) \cos ^2(c+d x)-3 a (A b-a B) \cos (c+d x)+3 \left (A a^2+b B a-2 A b^2\right )}{2 \cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}dx}{3 a \left (a^2-b^2\right )}+\frac {2 b (A b-a B) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {4 b (A b-a B) \cos ^2(c+d x)-3 a (A b-a B) \cos (c+d x)+3 \left (A a^2+b B a-2 A b^2\right )}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}dx}{3 a \left (a^2-b^2\right )}+\frac {2 b (A b-a B) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {4 b (A b-a B) \sin \left (c+d x+\frac {\pi }{2}\right )^2-3 a (A b-a B) \sin \left (c+d x+\frac {\pi }{2}\right )+3 \left (A a^2+b B a-2 A b^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{3 a \left (a^2-b^2\right )}+\frac {2 b (A b-a B) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}\right )\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 \int \frac {2 b \left (-7 B a^3+10 A b a^2+3 b^2 B a-6 A b^3\right ) \cos ^2(c+d x)-a \left (-3 B a^3+6 A b a^2-b^2 B a-2 A b^3\right ) \cos (c+d x)+3 \left (A a^4+8 b B a^3-13 A b^2 a^2-4 b^3 B a+8 A b^4\right )}{2 \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{a \left (a^2-b^2\right )}+\frac {2 b \left (-7 a^3 B+10 a^2 A b+3 a b^2 B-6 A b^3\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 b (A b-a B) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {2 b \left (-7 B a^3+10 A b a^2+3 b^2 B a-6 A b^3\right ) \cos ^2(c+d x)-a \left (-3 B a^3+6 A b a^2-b^2 B a-2 A b^3\right ) \cos (c+d x)+3 \left (A a^4+8 b B a^3-13 A b^2 a^2-4 b^3 B a+8 A b^4\right )}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{a \left (a^2-b^2\right )}+\frac {2 b \left (-7 a^3 B+10 a^2 A b+3 a b^2 B-6 A b^3\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 b (A b-a B) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {2 b \left (-7 B a^3+10 A b a^2+3 b^2 B a-6 A b^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-a \left (-3 B a^3+6 A b a^2-b^2 B a-2 A b^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+3 \left (A a^4+8 b B a^3-13 A b^2 a^2-4 b^3 B a+8 A b^4\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}+\frac {2 b \left (-7 a^3 B+10 a^2 A b+3 a b^2 B-6 A b^3\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 b (A b-a B) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}\right )\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 \int -\frac {3 \left (-3 B a^5+8 A b a^4+15 b^2 B a^3-28 A b^3 a^2-8 b^4 B a-\left (A a^4-6 b B a^3+7 A b^2 a^2+2 b^3 B a-4 A b^4\right ) \cos (c+d x) a+16 A b^5\right )}{2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{3 a}+\frac {2 \left (a^4 A+8 a^3 b B-13 a^2 A b^2-4 a b^3 B+8 A b^4\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{a d \cos ^{\frac {3}{2}}(c+d x)}}{a \left (a^2-b^2\right )}+\frac {2 b \left (-7 a^3 B+10 a^2 A b+3 a b^2 B-6 A b^3\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 b (A b-a B) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 \left (a^4 A+8 a^3 b B-13 a^2 A b^2-4 a b^3 B+8 A b^4\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\int \frac {-3 B a^5+8 A b a^4+15 b^2 B a^3-28 A b^3 a^2-8 b^4 B a-\left (A a^4-6 b B a^3+7 A b^2 a^2+2 b^3 B a-4 A b^4\right ) \cos (c+d x) a+16 A b^5}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{a}}{a \left (a^2-b^2\right )}+\frac {2 b \left (-7 a^3 B+10 a^2 A b+3 a b^2 B-6 A b^3\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 b (A b-a B) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 \left (a^4 A+8 a^3 b B-13 a^2 A b^2-4 a b^3 B+8 A b^4\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\int \frac {-3 B a^5+8 A b a^4+15 b^2 B a^3-28 A b^3 a^2-8 b^4 B a-\left (A a^4-6 b B a^3+7 A b^2 a^2+2 b^3 B a-4 A b^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a+16 A b^5}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{a \left (a^2-b^2\right )}+\frac {2 b \left (-7 a^3 B+10 a^2 A b+3 a b^2 B-6 A b^3\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 b (A b-a B) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}\right )\) |
| \(\Big \downarrow \) 3477 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 \left (a^4 A+8 a^3 b B-13 a^2 A b^2-4 a b^3 B+8 A b^4\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {(a-b) \left (-\left (a^4 (A-3 B)\right )-9 a^3 b (A-B)-2 a^2 b^2 (8 A+3 B)+4 a b^3 (3 A-2 B)+16 A b^4\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx+\left (-3 a^5 B+8 a^4 A b+15 a^3 b^2 B-28 a^2 A b^3-8 a b^4 B+16 A b^5\right ) \int \frac {\cos (c+d x)+1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{a}}{a \left (a^2-b^2\right )}+\frac {2 b \left (-7 a^3 B+10 a^2 A b+3 a b^2 B-6 A b^3\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 b (A b-a B) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 \left (a^4 A+8 a^3 b B-13 a^2 A b^2-4 a b^3 B+8 A b^4\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {(a-b) \left (-\left (a^4 (A-3 B)\right )-9 a^3 b (A-B)-2 a^2 b^2 (8 A+3 B)+4 a b^3 (3 A-2 B)+16 A b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\left (-3 a^5 B+8 a^4 A b+15 a^3 b^2 B-28 a^2 A b^3-8 a b^4 B+16 A b^5\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{a \left (a^2-b^2\right )}+\frac {2 b \left (-7 a^3 B+10 a^2 A b+3 a b^2 B-6 A b^3\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 b (A b-a B) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}\right )\) |
| \(\Big \downarrow \) 3295 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 \left (a^4 A+8 a^3 b B-13 a^2 A b^2-4 a b^3 B+8 A b^4\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\left (-3 a^5 B+8 a^4 A b+15 a^3 b^2 B-28 a^2 A b^3-8 a b^4 B+16 A b^5\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (a-b) \sqrt {a+b} \left (-\left (a^4 (A-3 B)\right )-9 a^3 b (A-B)-2 a^2 b^2 (8 A+3 B)+4 a b^3 (3 A-2 B)+16 A b^4\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}}{a}}{a \left (a^2-b^2\right )}+\frac {2 b \left (-7 a^3 B+10 a^2 A b+3 a b^2 B-6 A b^3\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 b (A b-a B) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}\right )\) |
| \(\Big \downarrow \) 3473 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 b (A b-a B) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}+\frac {\frac {2 b \left (-7 a^3 B+10 a^2 A b+3 a b^2 B-6 A b^3\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}+\frac {\frac {2 \left (a^4 A+8 a^3 b B-13 a^2 A b^2-4 a b^3 B+8 A b^4\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (a-b) \sqrt {a+b} \left (-\left (a^4 (A-3 B)\right )-9 a^3 b (A-B)-2 a^2 b^2 (8 A+3 B)+4 a b^3 (3 A-2 B)+16 A b^4\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}+\frac {2 (a-b) \sqrt {a+b} \left (-3 a^5 B+8 a^4 A b+15 a^3 b^2 B-28 a^2 A b^3-8 a b^4 B+16 A b^5\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a^2 d}}{a}}{a \left (a^2-b^2\right )}}{3 a \left (a^2-b^2\right )}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*b*(A*b - a*B)*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)) + ((2*b*(10* a^2*A*b - 6*A*b^3 - 7*a^3*B + 3*a*b^2*B)*Sin[c + d*x])/(a*(a^2 - b^2)*d*Co s[c + d*x]^(3/2)*Sqrt[a + b*Cos[c + d*x]]) + (-(((2*(a - b)*Sqrt[a + b]*(8 *a^4*A*b - 28*a^2*A*b^3 + 16*A*b^5 - 3*a^5*B + 15*a^3*b^2*B - 8*a*b^4*B)*C ot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Co s[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sq rt[(a*(1 + Sec[c + d*x]))/(a - b)])/(a^2*d) + (2*(a - b)*Sqrt[a + b]*(16*A *b^4 - a^4*(A - 3*B) + 4*a*b^3*(3*A - 2*B) - 9*a^3*b*(A - B) - 2*a^2*b^2*( 8*A + 3*B))*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*(a^4*A - 1 3*a^2*A*b^2 + 8*A*b^4 + 8*a^3*b*B - 4*a*b^3*B)*Sqrt[a + b*Cos[c + d*x]]*Si n[c + d*x])/(a*d*Cos[c + d*x]^(3/2)))/(a*(a^2 - b^2)))/(3*a*(a^2 - b^2)))
3.7.25.3.1 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[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)* (x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Sim p[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p Int[(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(g*Sin[e + f*x])^p), x], x] /; FreeQ[{a, b, c, d, e, f, g , m, n, p}, x] && NeQ[b*c - a*d, 0] && !IntegerQ[p] && !(IntegerQ[m] && I ntegerQ[n])
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_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[(-(A*b^2 - a*b*B))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin [e + f*x])^(1 + n)/(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*A - b*B)*(b*c - a*d)*(m + 1) + b*d*(A*b - a*B)*(m + n + 2) + (A*b - a*B)*(a*d*(m + 1) - b*c*(m + 2))*Sin[e + f*x] - b*d*(A*b - a*B) *(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n }, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Rat ionalQ[m] && m < -1 && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(I ntegerQ[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_.) + (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])))
Leaf count of result is larger than twice the leaf count of optimal. \(9527\) vs. \(2(561)=1122\).
Time = 25.71 (sec) , antiderivative size = 9528, normalized size of antiderivative = 15.70
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(9528\) |
| default | \(\text {Expression too large to display}\) | \(9947\) |
\[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {5}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
integral((B*cos(d*x + c) + 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. \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {5}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {5}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\int \frac {\left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]