Integrand size = 33, antiderivative size = 315 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx=\frac {\left (a A b-3 a^2 B+2 b^2 B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{b^2 \left (a^2-b^2\right ) d}+\frac {(A b-a B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{b \left (a^2-b^2\right ) d}+\frac {\left (a^2 A b-3 A b^3-3 a^3 B+5 a b^2 B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{(a-b) b^2 (a+b)^2 d}-\frac {\left (a A b-3 a^2 B+2 b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d}+\frac {a (A b-a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))} \]
a*(A*b-B*a)*sec(d*x+c)^(3/2)*sin(d*x+c)/b/(a^2-b^2)/d/(a+b*sec(d*x+c))-(A* a*b-3*B*a^2+2*B*b^2)*sin(d*x+c)*sec(d*x+c)^(1/2)/b^2/(a^2-b^2)/d+(A*a*b-3* B*a^2+2*B*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(s in(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/b^2/(a^2-b^2) /d+(A*b-B*a)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin (1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/b/(a^2-b^2)/d+( A*a^2*b-3*A*b^3-3*B*a^3+5*B*a*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d* x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2*a/(a+b),2^(1/2))*cos(d*x+c)^(1/2) *sec(d*x+c)^(1/2)/(a-b)/b^2/(a+b)^2/d
Leaf count is larger than twice the leaf count of optimal. \(680\) vs. \(2(315)=630\).
Time = 7.29 (sec) , antiderivative size = 680, normalized size of antiderivative = 2.16 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx=-\frac {\frac {2 \left (-3 a^2 A b+4 A b^3+9 a^3 B-10 a b^2 B\right ) \cos ^2(c+d x) \left (\operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )-\operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )\right ) (a+b \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{b (b+a \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {2 \left (-4 a A b^2+8 a^2 b B-4 b^3 B\right ) \cos ^2(c+d x) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) (a+b \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{a (b+a \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {\left (-a^2 A b+3 a^3 B-2 a b^2 B\right ) \cos (2 (c+d x)) (a+b \sec (c+d x)) \left (-4 a b+4 a b \sec ^2(c+d x)-4 a b E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}-2 a (a-2 b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}+2 a^2 \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}-4 b^2 \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}\right ) \sin (c+d x)}{a^2 b (b+a \cos (c+d x)) \left (1-\cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \left (2-\sec ^2(c+d x)\right )}}{4 (a-b) b^2 (a+b) d}+\frac {\sqrt {\sec (c+d x)} \left (\frac {\left (a A b-3 a^2 B+2 b^2 B\right ) \sin (c+d x)}{b^2 \left (-a^2+b^2\right )}+\frac {-a A b \sin (c+d x)+a^2 B \sin (c+d x)}{b \left (-a^2+b^2\right ) (b+a \cos (c+d x))}\right )}{d} \]
-1/4*((2*(-3*a^2*A*b + 4*A*b^3 + 9*a^3*B - 10*a*b^2*B)*Cos[c + d*x]^2*(Ell ipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1] - EllipticPi[-(b/a), ArcSin[Sqrt[Se c[c + d*x]]], -1])*(a + b*Sec[c + d*x])*Sqrt[1 - Sec[c + d*x]^2]*Sin[c + d *x])/(b*(b + a*Cos[c + d*x])*(1 - Cos[c + d*x]^2)) + (2*(-4*a*A*b^2 + 8*a^ 2*b*B - 4*b^3*B)*Cos[c + d*x]^2*EllipticPi[-(b/a), ArcSin[Sqrt[Sec[c + d*x ]]], -1]*(a + b*Sec[c + d*x])*Sqrt[1 - Sec[c + d*x]^2]*Sin[c + d*x])/(a*(b + a*Cos[c + d*x])*(1 - Cos[c + d*x]^2)) + ((-(a^2*A*b) + 3*a^3*B - 2*a*b^ 2*B)*Cos[2*(c + d*x)]*(a + b*Sec[c + d*x])*(-4*a*b + 4*a*b*Sec[c + d*x]^2 - 4*a*b*EllipticE[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[ 1 - Sec[c + d*x]^2] - 2*a*(a - 2*b)*EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] + 2*a^2*EllipticPi[-(b/a), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^ 2] - 4*b^2*EllipticPi[-(b/a), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2])*Sin[c + d*x])/(a^2*b*(b + a*Cos[c + d*x]) *(1 - Cos[c + d*x]^2)*Sqrt[Sec[c + d*x]]*(2 - Sec[c + d*x]^2)))/((a - b)*b ^2*(a + b)*d) + (Sqrt[Sec[c + d*x]]*(((a*A*b - 3*a^2*B + 2*b^2*B)*Sin[c + d*x])/(b^2*(-a^2 + b^2)) + (-(a*A*b*Sin[c + d*x]) + a^2*B*Sin[c + d*x])/(b *(-a^2 + b^2)*(b + a*Cos[c + d*x]))))/d
Time = 2.19 (sec) , antiderivative size = 305, normalized size of antiderivative = 0.97, number of steps used = 18, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {3042, 4517, 27, 3042, 4590, 27, 3042, 4594, 3042, 4274, 3042, 4258, 3042, 3119, 3120, 4336, 3042, 3284}
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.
| \(\displaystyle \int \frac {\sec ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 4517 |
| \(\displaystyle \frac {\int \frac {\sqrt {\sec (c+d x)} \left (-\left (\left (-3 B a^2+A b a+2 b^2 B\right ) \sec ^2(c+d x)\right )-2 b (A b-a B) \sec (c+d x)+a (A b-a B)\right )}{2 (a+b \sec (c+d x))}dx}{b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\sqrt {\sec (c+d x)} \left (-\left (\left (-3 B a^2+A b a+2 b^2 B\right ) \sec ^2(c+d x)\right )-2 b (A b-a B) \sec (c+d x)+a (A b-a B)\right )}{a+b \sec (c+d x)}dx}{2 b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (\left (3 B a^2-A b a-2 b^2 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2-2 b (A b-a B) \csc \left (c+d x+\frac {\pi }{2}\right )+a (A b-a B)\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{2 b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 4590 |
| \(\displaystyle \frac {\frac {2 \int \frac {\left (-3 B a^3+A b a^2+4 b^2 B a-2 A b^3\right ) \sec ^2(c+d x)+2 b \left (-2 B a^2+A b a+b^2 B\right ) \sec (c+d x)+a \left (-3 B a^2+A b a+2 b^2 B\right )}{2 \sqrt {\sec (c+d x)} (a+b \sec (c+d x))}dx}{b}-\frac {2 \left (-3 a^2 B+a A b+2 b^2 B\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{b d}}{2 b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {\left (-3 B a^3+A b a^2+4 b^2 B a-2 A b^3\right ) \sec ^2(c+d x)+2 b \left (-2 B a^2+A b a+b^2 B\right ) \sec (c+d x)+a \left (-3 B a^2+A b a+2 b^2 B\right )}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))}dx}{b}-\frac {2 \left (-3 a^2 B+a A b+2 b^2 B\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{b d}}{2 b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {\left (-3 B a^3+A b a^2+4 b^2 B a-2 A b^3\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2+2 b \left (-2 B a^2+A b a+b^2 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+a \left (-3 B a^2+A b a+2 b^2 B\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}-\frac {2 \left (-3 a^2 B+a A b+2 b^2 B\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{b d}}{2 b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 4594 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {\left (-3 B a^2+A b a+2 b^2 B\right ) a^2+b (A b-a B) \sec (c+d x) a^2}{\sqrt {\sec (c+d x)}}dx}{a^2}+\left (-3 a^3 B+a^2 A b+5 a b^2 B-3 A b^3\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{a+b \sec (c+d x)}dx}{b}-\frac {2 \left (-3 a^2 B+a A b+2 b^2 B\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{b d}}{2 b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {\left (-3 B a^2+A b a+2 b^2 B\right ) a^2+b (A b-a B) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}+\left (-3 a^3 B+a^2 A b+5 a b^2 B-3 A b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {2 \left (-3 a^2 B+a A b+2 b^2 B\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{b d}}{2 b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {\frac {\frac {a^2 \left (-3 a^2 B+a A b+2 b^2 B\right ) \int \frac {1}{\sqrt {\sec (c+d x)}}dx+a^2 b (A b-a B) \int \sqrt {\sec (c+d x)}dx}{a^2}+\left (-3 a^3 B+a^2 A b+5 a b^2 B-3 A b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {2 \left (-3 a^2 B+a A b+2 b^2 B\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{b d}}{2 b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {a^2 \left (-3 a^2 B+a A b+2 b^2 B\right ) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+a^2 b (A b-a B) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\left (-3 a^3 B+a^2 A b+5 a b^2 B-3 A b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {2 \left (-3 a^2 B+a A b+2 b^2 B\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{b d}}{2 b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {\frac {\frac {a^2 \left (-3 a^2 B+a A b+2 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx+a^2 b (A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{a^2}+\left (-3 a^3 B+a^2 A b+5 a b^2 B-3 A b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {2 \left (-3 a^2 B+a A b+2 b^2 B\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{b d}}{2 b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {a^2 \left (-3 a^2 B+a A b+2 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+a^2 b (A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}+\left (-3 a^3 B+a^2 A b+5 a b^2 B-3 A b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {2 \left (-3 a^2 B+a A b+2 b^2 B\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{b d}}{2 b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {\frac {\frac {a^2 b (A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a^2 \left (-3 a^2 B+a A b+2 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a^2}+\left (-3 a^3 B+a^2 A b+5 a b^2 B-3 A b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {2 \left (-3 a^2 B+a A b+2 b^2 B\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{b d}}{2 b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {\frac {\left (-3 a^3 B+a^2 A b+5 a b^2 B-3 A b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {\frac {2 a^2 \left (-3 a^2 B+a A b+2 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 a^2 b (A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{a^2}}{b}-\frac {2 \left (-3 a^2 B+a A b+2 b^2 B\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{b d}}{2 b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 4336 |
| \(\displaystyle \frac {\frac {\left (-3 a^3 B+a^2 A b+5 a b^2 B-3 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx+\frac {\frac {2 a^2 \left (-3 a^2 B+a A b+2 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 a^2 b (A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{a^2}}{b}-\frac {2 \left (-3 a^2 B+a A b+2 b^2 B\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{b d}}{2 b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\left (-3 a^3 B+a^2 A b+5 a b^2 B-3 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx+\frac {\frac {2 a^2 \left (-3 a^2 B+a A b+2 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 a^2 b (A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{a^2}}{b}-\frac {2 \left (-3 a^2 B+a A b+2 b^2 B\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{b d}}{2 b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3284 |
| \(\displaystyle \frac {a (A b-a B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac {\frac {\frac {\frac {2 a^2 \left (-3 a^2 B+a A b+2 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 a^2 b (A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{a^2}+\frac {2 \left (-3 a^3 B+a^2 A b+5 a b^2 B-3 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{d (a+b)}}{b}-\frac {2 \left (-3 a^2 B+a A b+2 b^2 B\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{b d}}{2 b \left (a^2-b^2\right )}\) |
(a*(A*b - a*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(b*(a^2 - b^2)*d*(a + b*Se c[c + d*x])) + ((((2*a^2*(a*A*b - 3*a^2*B + 2*b^2*B)*Sqrt[Cos[c + d*x]]*El lipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (2*a^2*b*(A*b - a*B)*Sqrt[ Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d)/a^2 + (2*(a ^2*A*b - 3*A*b^3 - 3*a^3*B + 5*a*b^2*B)*Sqrt[Cos[c + d*x]]*EllipticPi[(2*a )/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/((a + b)*d))/b - (2*(a*A*b - 3*a^2*B + 2*b^2*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(b*d))/(2*b*(a^2 - b ^2))
3.5.23.3.1 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[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c , d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] )^n*Sin[c + d*x]^n Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[a Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d In t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[d*Sqrt[d*Sin[e + f*x]]*Sqrt[d*Csc[e + f*x]] Int[ 1/(Sqrt[d*Sin[e + f*x]]*(b + a*Sin[e + f*x])), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[a*d^2*( A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^(n - 2)/(b*f*(m + 1)*(a^2 - b^2))), x] - Simp[d/(b*(m + 1)*(a^2 - b^2)) Int[( a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 2)*Simp[a*d*(A*b - a*B)*( n - 2) + b*d*(A*b - a*B)*(m + 1)*Csc[e + f*x] - (a*A*b*d*(m + n) - d*B*(a^2 *(n - 1) + b^2*(m + 1)))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f , A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[ n, 1]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[(-C)*d*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1 )*((d*Csc[e + f*x])^(n - 1)/(b*f*(m + n + 1))), x] + Simp[d/(b*(m + n + 1)) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 1)*Simp[a*C*(n - 1) + ( A*b*(m + n + 1) + b*C*(m + n))*Csc[e + f*x] + (b*B*(m + n + 1) - a*C*n)*Csc [e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && GtQ[n, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))), x_Symbol] :> Simp[(A*b^2 - a*b*B + a^2*C)/(a^2*d^2) Int[(d*Csc[e + f*x])^(3/2)/(a + b*Csc[e + f*x]), x], x] + Simp[1/a^2 Int[(a*A - (A*b - a *B)*Csc[e + f*x])/Sqrt[d*Csc[e + f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(849\) vs. \(2(377)=754\).
Time = 52.01 (sec) , antiderivative size = 850, normalized size of antiderivative = 2.70
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*B/b^2/sin(1/ 2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2 *d*x+1/2*c)^2)^(1/2)*(2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-(sin(1/2*d *x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/ 2*c)^2-1)^(1/2))+2*B*a^2/b^2/(a^2-a*b)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*co s(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2) ^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))+2*(A*b-B*a)/b*(a^2 /b/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c )^2)^(1/2)/(2*a*cos(1/2*d*x+1/2*c)^2-a+b)-1/2/(a+b)/b*(sin(1/2*d*x+1/2*c)^ 2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/ 2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+1/2*a/b/(a^2-b ^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin (1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c) ,2^(1/2))-1/2*a/b/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1 /2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*Elli pticE(cos(1/2*d*x+1/2*c),2^(1/2))-1/2/b/(a^2-b^2)/(a^2-a*b)*a^3*(sin(1/2*d *x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c )^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^ (1/2))+3/2*b/(a^2-b^2)/(a^2-a*b)*a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/ 2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^...
Timed out. \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )}\right ) \sec ^{\frac {5}{2}}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \]
Timed out. \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {5}{2}}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx=\int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^2} \,d x \]