Integrand size = 35, antiderivative size = 733 \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^{5/2} \sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {\left (6 a^3 A b-14 a A b^3-15 a^4 B+26 a^2 b^2 B-3 b^4 B\right ) \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a (a-b) b^3 (a+b)^{3/2} d \sqrt {\sec (c+d x)}}+\frac {\left (3 b^3 (4 A-B)+15 a^3 B-a b^2 (2 A+21 B)-a^2 (6 A b-5 b B)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 (a-b) b^3 (a+b)^{3/2} d \sqrt {\sec (c+d x)}}-\frac {\sqrt {a+b} (2 A b-5 a B) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{b^4 d \sqrt {\sec (c+d x)}}+\frac {2 a (A b-a B) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 a \left (2 a^2 A b-6 A b^3-5 a^3 B+9 a b^2 B\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)}}-\frac {\left (6 a^3 A b-14 a A b^3-15 a^4 B+26 a^2 b^2 B-3 b^4 B\right ) \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d} \]
2/3*a*(A*b-B*a)*sin(d*x+c)/b/(a^2-b^2)/d/(a+b*cos(d*x+c))^(3/2)/sec(d*x+c) ^(3/2)+2/3*a*(2*A*a^2*b-6*A*b^3-5*B*a^3+9*B*a*b^2)*sin(d*x+c)/b^2/(a^2-b^2 )^2/d/(a+b*cos(d*x+c))^(1/2)/sec(d*x+c)^(1/2)-1/3*(6*A*a^3*b-14*A*a*b^3-15 *B*a^4+26*B*a^2*b^2-3*B*b^4)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)*sec(d*x+c)^ (1/2)/b^3/(a^2-b^2)^2/d+1/3*(6*A*a^3*b-14*A*a*b^3-15*B*a^4+26*B*a^2*b^2-3* B*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/(a-b)/b^3/(a+b)^(3/2)/d/sec(d*x+c)^(1/2 )+1/3*(3*b^3*(4*A-B)+15*B*a^3-a*b^2*(2*A+21*B)-a^2*(6*A*b-5*B*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-b)/b^3/(a+b)^(3/2)/d/sec(d*x+c)^(1/2)-(2*A*b-5*B*a)*c sc(d*x+c)*EllipticPi((a+b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),( a+b)/b,((-a-b)/(a-b))^(1/2))*(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)/b^4/d/sec(d*x+c)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(2318\) vs. \(2(733)=1466\).
Time = 20.14 (sec) , antiderivative size = 2318, normalized size of antiderivative = 3.16 \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^{5/2} \sec ^{\frac {5}{2}}(c+d x)} \, dx=\text {Result too large to show} \]
(Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((-2*a*(-3*a^2*A*b + 7*A*b^3 + 6*a^3*B - 10*a*b^2*B)*Sin[c + d*x])/(3*b^3*(a^2 - b^2)^2) + (2*(-(a^3*A* b*Sin[c + d*x]) + a^4*B*Sin[c + d*x]))/(3*b^3*(-a^2 + b^2)*(a + b*Cos[c + d*x])^2) + (2*(-4*a^4*A*b*Sin[c + d*x] + 8*a^2*A*b^3*Sin[c + d*x] + 7*a^5* B*Sin[c + d*x] - 11*a^3*b^2*B*Sin[c + d*x]))/(3*b^3*(-a^2 + b^2)^2*(a + b* Cos[c + d*x]))))/d + (Sqrt[(1 - Tan[(c + d*x)/2]^2)^(-1)]*Sqrt[(a + b + a* Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2)]*(6*a^ 4*A*b*Tan[(c + d*x)/2] + 6*a^3*A*b^2*Tan[(c + d*x)/2] - 14*a^2*A*b^3*Tan[( c + d*x)/2] - 14*a*A*b^4*Tan[(c + d*x)/2] - 15*a^5*B*Tan[(c + d*x)/2] - 15 *a^4*b*B*Tan[(c + d*x)/2] + 26*a^3*b^2*B*Tan[(c + d*x)/2] + 26*a^2*b^3*B*T an[(c + d*x)/2] - 3*a*b^4*B*Tan[(c + d*x)/2] - 3*b^5*B*Tan[(c + d*x)/2] - 12*a^3*A*b^2*Tan[(c + d*x)/2]^3 + 28*a*A*b^4*Tan[(c + d*x)/2]^3 + 30*a^4*b *B*Tan[(c + d*x)/2]^3 - 52*a^2*b^3*B*Tan[(c + d*x)/2]^3 + 6*b^5*B*Tan[(c + d*x)/2]^3 - 6*a^4*A*b*Tan[(c + d*x)/2]^5 + 6*a^3*A*b^2*Tan[(c + d*x)/2]^5 + 14*a^2*A*b^3*Tan[(c + d*x)/2]^5 - 14*a*A*b^4*Tan[(c + d*x)/2]^5 + 15*a^ 5*B*Tan[(c + d*x)/2]^5 - 15*a^4*b*B*Tan[(c + d*x)/2]^5 - 26*a^3*b^2*B*Tan[ (c + d*x)/2]^5 + 26*a^2*b^3*B*Tan[(c + d*x)/2]^5 + 3*a*b^4*B*Tan[(c + d*x) /2]^5 - 3*b^5*B*Tan[(c + d*x)/2]^5 - 12*a^4*A*b*EllipticPi[-1, ArcSin[Tan[ (c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] + 24*a^2*A*b^3*...
Time = 3.65 (sec) , antiderivative size = 719, normalized size of antiderivative = 0.98, number of steps used = 18, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.514, Rules used = {3042, 3440, 3042, 3468, 27, 3042, 3526, 27, 3042, 3540, 3042, 3532, 3042, 3288, 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.
\(\displaystyle \int \frac {A+B \cos (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
\(\Big \downarrow \) 3440 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^{5/2}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
\(\Big \downarrow \) 3468 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 a (A b-a B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}-\frac {2 \int -\frac {\sqrt {\cos (c+d x)} \left (-\left (\left (-5 B a^2+2 A b a+3 b^2 B\right ) \cos ^2(c+d x)\right )-3 b (A b-a B) \cos (c+d x)+3 a (A b-a B)\right )}{2 (a+b \cos (c+d x))^{3/2}}dx}{3 b \left (a^2-b^2\right )}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\sqrt {\cos (c+d x)} \left (-\left (\left (-5 B a^2+2 A b a+3 b^2 B\right ) \cos ^2(c+d x)\right )-3 b (A b-a B) \cos (c+d x)+3 a (A b-a B)\right )}{(a+b \cos (c+d x))^{3/2}}dx}{3 b \left (a^2-b^2\right )}+\frac {2 a (A b-a B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (\left (5 B a^2-2 A b a-3 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-3 b (A b-a B) \sin \left (c+d x+\frac {\pi }{2}\right )+3 a (A b-a B)\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{3 b \left (a^2-b^2\right )}+\frac {2 a (A b-a B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\right )\) |
\(\Big \downarrow \) 3526 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a \left (-5 a^3 B+2 a^2 A b+9 a b^2 B-6 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {2 \int -\frac {-\left (\left (-15 B a^4+6 A b a^3+26 b^2 B a^2-14 A b^3 a-3 b^4 B\right ) \cos ^2(c+d x)\right )+b \left (2 B a^3+A b a^2-6 b^2 B a+3 A b^3\right ) \cos (c+d x)+a \left (-5 B a^3+2 A b a^2+9 b^2 B a-6 A b^3\right )}{2 \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx}{b \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}+\frac {2 a (A b-a B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {-\left (\left (-15 B a^4+6 A b a^3+26 b^2 B a^2-14 A b^3 a-3 b^4 B\right ) \cos ^2(c+d x)\right )+b \left (2 B a^3+A b a^2-6 b^2 B a+3 A b^3\right ) \cos (c+d x)+a \left (-5 B a^3+2 A b a^2+9 b^2 B a-6 A b^3\right )}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx}{b \left (a^2-b^2\right )}+\frac {2 a \left (-5 a^3 B+2 a^2 A b+9 a b^2 B-6 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 b \left (a^2-b^2\right )}+\frac {2 a (A b-a B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {\left (15 B a^4-6 A b a^3-26 b^2 B a^2+14 A b^3 a+3 b^4 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+b \left (2 B a^3+A b a^2-6 b^2 B a+3 A b^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+a \left (-5 B a^3+2 A b a^2+9 b^2 B a-6 A b^3\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b \left (a^2-b^2\right )}+\frac {2 a \left (-5 a^3 B+2 a^2 A b+9 a b^2 B-6 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 b \left (a^2-b^2\right )}+\frac {2 a (A b-a B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\right )\) |
\(\Big \downarrow \) 3540 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {\int \frac {3 \left (a^2-b^2\right )^2 (2 A b-5 a B) \cos ^2(c+d x)+2 a b \left (-5 B a^3+2 A b a^2+9 b^2 B a-6 A b^3\right ) \cos (c+d x)+a \left (-15 B a^4+6 A b a^3+26 b^2 B a^2-14 A b^3 a-3 b^4 B\right )}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{2 b}-\frac {\left (-15 a^4 B+6 a^3 A b+26 a^2 b^2 B-14 a A b^3-3 b^4 B\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{b \left (a^2-b^2\right )}+\frac {2 a \left (-5 a^3 B+2 a^2 A b+9 a b^2 B-6 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 b \left (a^2-b^2\right )}+\frac {2 a (A b-a B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {\int \frac {3 \left (a^2-b^2\right )^2 (2 A b-5 a B) \sin \left (c+d x+\frac {\pi }{2}\right )^2+2 a b \left (-5 B a^3+2 A b a^2+9 b^2 B a-6 A b^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+a \left (-15 B a^4+6 A b a^3+26 b^2 B a^2-14 A b^3 a-3 b^4 B\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 b}-\frac {\left (-15 a^4 B+6 a^3 A b+26 a^2 b^2 B-14 a A b^3-3 b^4 B\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{b \left (a^2-b^2\right )}+\frac {2 a \left (-5 a^3 B+2 a^2 A b+9 a b^2 B-6 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 b \left (a^2-b^2\right )}+\frac {2 a (A b-a B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\right )\) |
\(\Big \downarrow \) 3532 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {3 \left (a^2-b^2\right )^2 (2 A b-5 a B) \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {a+b \cos (c+d x)}}dx+\int \frac {a \left (-15 B a^4+6 A b a^3+26 b^2 B a^2-14 A b^3 a-3 b^4 B\right )+2 a b \left (-5 B a^3+2 A b a^2+9 b^2 B a-6 A b^3\right ) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{2 b}-\frac {\left (-15 a^4 B+6 a^3 A b+26 a^2 b^2 B-14 a A b^3-3 b^4 B\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{b \left (a^2-b^2\right )}+\frac {2 a \left (-5 a^3 B+2 a^2 A b+9 a b^2 B-6 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 b \left (a^2-b^2\right )}+\frac {2 a (A b-a B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {3 \left (a^2-b^2\right )^2 (2 A b-5 a B) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\int \frac {a \left (-15 B a^4+6 A b a^3+26 b^2 B a^2-14 A b^3 a-3 b^4 B\right )+2 a b \left (-5 B a^3+2 A b a^2+9 b^2 B a-6 A b^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 b}-\frac {\left (-15 a^4 B+6 a^3 A b+26 a^2 b^2 B-14 a A b^3-3 b^4 B\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{b \left (a^2-b^2\right )}+\frac {2 a \left (-5 a^3 B+2 a^2 A b+9 a b^2 B-6 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 b \left (a^2-b^2\right )}+\frac {2 a (A b-a B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\right )\) |
\(\Big \downarrow \) 3288 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {\int \frac {a \left (-15 B a^4+6 A b a^3+26 b^2 B a^2-14 A b^3 a-3 b^4 B\right )+2 a b \left (-5 B a^3+2 A b a^2+9 b^2 B a-6 A b^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 \sqrt {a+b} \left (a^2-b^2\right )^2 (2 A b-5 a B) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}}{2 b}-\frac {\left (-15 a^4 B+6 a^3 A b+26 a^2 b^2 B-14 a A b^3-3 b^4 B\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{b \left (a^2-b^2\right )}+\frac {2 a \left (-5 a^3 B+2 a^2 A b+9 a b^2 B-6 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 b \left (a^2-b^2\right )}+\frac {2 a (A b-a B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\right )\) |
\(\Big \downarrow \) 3477 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {a (a-b) \left (15 a^3 B-a^2 (6 A b-5 b B)-a b^2 (2 A+21 B)+3 b^3 (4 A-B)\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx+a \left (-15 a^4 B+6 a^3 A b+26 a^2 b^2 B-14 a A b^3-3 b^4 B\right ) \int \frac {\cos (c+d x)+1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx-\frac {6 \sqrt {a+b} \left (a^2-b^2\right )^2 (2 A b-5 a B) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}}{2 b}-\frac {\left (-15 a^4 B+6 a^3 A b+26 a^2 b^2 B-14 a A b^3-3 b^4 B\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{b \left (a^2-b^2\right )}+\frac {2 a \left (-5 a^3 B+2 a^2 A b+9 a b^2 B-6 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 b \left (a^2-b^2\right )}+\frac {2 a (A b-a B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {a (a-b) \left (15 a^3 B-a^2 (6 A b-5 b B)-a b^2 (2 A+21 B)+3 b^3 (4 A-B)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+a \left (-15 a^4 B+6 a^3 A b+26 a^2 b^2 B-14 a A b^3-3 b^4 B\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 \sqrt {a+b} \left (a^2-b^2\right )^2 (2 A b-5 a B) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}}{2 b}-\frac {\left (-15 a^4 B+6 a^3 A b+26 a^2 b^2 B-14 a A b^3-3 b^4 B\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{b \left (a^2-b^2\right )}+\frac {2 a \left (-5 a^3 B+2 a^2 A b+9 a b^2 B-6 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 b \left (a^2-b^2\right )}+\frac {2 a (A b-a B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\right )\) |
\(\Big \downarrow \) 3295 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {a \left (-15 a^4 B+6 a^3 A b+26 a^2 b^2 B-14 a A b^3-3 b^4 B\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 \sqrt {a+b} \left (a^2-b^2\right )^2 (2 A b-5 a B) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}+\frac {2 (a-b) \sqrt {a+b} \left (15 a^3 B-a^2 (6 A b-5 b B)-a b^2 (2 A+21 B)+3 b^3 (4 A-B)\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}}{2 b}-\frac {\left (-15 a^4 B+6 a^3 A b+26 a^2 b^2 B-14 a A b^3-3 b^4 B\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{b \left (a^2-b^2\right )}+\frac {2 a \left (-5 a^3 B+2 a^2 A b+9 a b^2 B-6 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 b \left (a^2-b^2\right )}+\frac {2 a (A b-a B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\right )\) |
\(\Big \downarrow \) 3473 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 a (A b-a B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}+\frac {\frac {2 a \left (-5 a^3 B+2 a^2 A b+9 a b^2 B-6 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}+\frac {\frac {-\frac {6 \sqrt {a+b} \left (a^2-b^2\right )^2 (2 A b-5 a B) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}+\frac {2 (a-b) \sqrt {a+b} \left (15 a^3 B-a^2 (6 A b-5 b B)-a b^2 (2 A+21 B)+3 b^3 (4 A-B)\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}+\frac {2 (a-b) \sqrt {a+b} \left (-15 a^4 B+6 a^3 A b+26 a^2 b^2 B-14 a A b^3-3 b^4 B\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a d}}{2 b}-\frac {\left (-15 a^4 B+6 a^3 A b+26 a^2 b^2 B-14 a A b^3-3 b^4 B\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{b \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*a*(A*b - a*B)*Cos[c + d*x]^(3/2) *Sin[c + d*x])/(3*b*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^(3/2)) + ((2*a*(2*a ^2*A*b - 6*A*b^3 - 5*a^3*B + 9*a*b^2*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/( b*(a^2 - b^2)*d*Sqrt[a + b*Cos[c + d*x]]) + (((2*(a - b)*Sqrt[a + b]*(6*a^ 3*A*b - 14*a*A*b^3 - 15*a^4*B + 26*a^2*b^2*B - 3*b^4*B)*Cot[c + d*x]*Ellip ticE[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) + (2*(a - b)*Sqrt[a + b]*(3*b^3*(4*A - B) + 15*a^ 3*B - a*b^2*(2*A + 21*B) - a^2*(6*A*b - 5*b*B))*Cot[c + d*x]*EllipticF[Arc Sin[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)])/d - (6*Sqrt[a + b]*(a^2 - b^2)^2*(2*A*b - 5*a*B)*Cot[c + d*x]*E llipticPi[(a + b)/b, 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)]*Sqr t[(a*(1 + Sec[c + d*x]))/(a - b)])/(b*d))/(2*b) - ((6*a^3*A*b - 14*a*A*b^3 - 15*a^4*B + 26*a^2*b^2*B - 3*b^4*B)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x ])/(b*d*Sqrt[Cos[c + d*x]]))/(b*(a^2 - b^2)))/(3*b*(a^2 - b^2)))
3.7.30.3.1 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[(b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.) *(x_)]], x_Symbol] :> Simp[2*b*(Tan[e + f*x]/(d*f))*Rt[(c + d)/b, 2]*Sqrt[c *((1 + Csc[e + f*x])/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*Ellipti cPi[(c + d)/d, 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}, x] && NeQ[c^2 - d^2, 0] && PosQ[(c + d)/b]
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_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[(-(b*c - a*d))*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1)*Simp[b*(b*c - a*d)*(B*c - A*d)*(m - 1) + a*d*(a*A*c + b*B*c - (A*b + a *B)*d)*(n + 1) + (b*(b*d*(B*c - A*d) + a*(A*c*d + B*(c^2 - 2*d^2)))*(n + 1) - a*(b*c - a*d)*(B*c - A*d)*(n + 2))*Sin[e + f*x] + b*(d*(A*b*c + a*B*c - a*A*d)*(m + n + 1) - b*B*(c^2*m + d^2*(n + 1)))*Sin[e + f*x]^2, 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] && GtQ[m, 1] && LtQ[n, -1]
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[(-(c^2*C - B*c*d + A*d^2))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + (c*C - B* d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 1) - a*c*(n + 2))) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x ] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f *x]^2, x], x], x] /; FreeQ[{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] && GtQ[m, 0] && LtQ[n, -1]
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 2)/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)*Sqrt[(c_.) + (d_.)*sin[(e _.) + (f_.)*(x_)]]), x_Symbol] :> Simp[C/b^2 Int[Sqrt[a + b*Sin[e + f*x]] /Sqrt[c + d*Sin[e + f*x]], x], x] + Simp[1/b^2 Int[(A*b^2 - a^2*C + b*(b* B - 2*a*C)*Sin[e + f*x])/((a + b*Sin[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x ]]), x], x] /; FreeQ[{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[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(Sqrt[c + d*Sin[e + f *x]]/(d*f*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[1/(2*d) Int[(1/((a + b*Si n[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x]]))*Simp[2*a*A*d - C*(b*c - a*d) - 2*(a*c*C - d*(A*b + a*B))*Sin[e + f*x] + (2*b*B*d - C*(b*c + a*d))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{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]
Leaf count of result is larger than twice the leaf count of optimal. \(10352\) vs. \(2(675)=1350\).
Time = 21.29 (sec) , antiderivative size = 10353, normalized size of antiderivative = 14.12
method | result | size |
parts | \(\text {Expression too large to display}\) | \(10353\) |
default | \(\text {Expression too large to display}\) | \(10874\) |
Timed out. \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^{5/2} \sec ^{\frac {5}{2}}(c+d x)} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^{5/2} \sec ^{\frac {5}{2}}(c+d x)} \, dx=\text {Timed out} \]
\[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^{5/2} \sec ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^{5/2} \sec ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^{5/2} \sec ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {A+B\,\cos \left (c+d\,x\right )}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]