Integrand size = 21, antiderivative size = 98 \[ \int \frac {\sec ^2(c+d x)}{(b \cos (c+d x))^{5/2}} \, dx=\frac {10 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 b^2 d \sqrt {b \cos (c+d x)}}+\frac {2 b \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac {10 \sin (c+d x)}{21 b d (b \cos (c+d x))^{3/2}} \]
2/7*b*sin(d*x+c)/d/(b*cos(d*x+c))^(7/2)+10/21*sin(d*x+c)/b/d/(b*cos(d*x+c) )^(3/2)+10/21*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(si n(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)/b^2/d/(b*cos(d*x+c))^(1/2)
Time = 0.17 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.67 \[ \int \frac {\sec ^2(c+d x)}{(b \cos (c+d x))^{5/2}} \, dx=\frac {10 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+2 \left (5+3 \sec ^2(c+d x)\right ) \tan (c+d x)}{21 b^2 d \sqrt {b \cos (c+d x)}} \]
(10*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + 2*(5 + 3*Sec[c + d*x]^2 )*Tan[c + d*x])/(21*b^2*d*Sqrt[b*Cos[c + d*x]])
Time = 0.46 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.14, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 2030, 3116, 3042, 3116, 3042, 3121, 3042, 3120}
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.
| \(\displaystyle \int \frac {\sec ^2(c+d x)}{(b \cos (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 2030 |
| \(\displaystyle b^2 \int \frac {1}{\left (b \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{9/2}}dx\) |
| \(\Big \downarrow \) 3116 |
| \(\displaystyle b^2 \left (\frac {5 \int \frac {1}{(b \cos (c+d x))^{5/2}}dx}{7 b^2}+\frac {2 \sin (c+d x)}{7 b d (b \cos (c+d x))^{7/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle b^2 \left (\frac {5 \int \frac {1}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx}{7 b^2}+\frac {2 \sin (c+d x)}{7 b d (b \cos (c+d x))^{7/2}}\right )\) |
| \(\Big \downarrow \) 3116 |
| \(\displaystyle b^2 \left (\frac {5 \left (\frac {\int \frac {1}{\sqrt {b \cos (c+d x)}}dx}{3 b^2}+\frac {2 \sin (c+d x)}{3 b d (b \cos (c+d x))^{3/2}}\right )}{7 b^2}+\frac {2 \sin (c+d x)}{7 b d (b \cos (c+d x))^{7/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle b^2 \left (\frac {5 \left (\frac {\int \frac {1}{\sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b^2}+\frac {2 \sin (c+d x)}{3 b d (b \cos (c+d x))^{3/2}}\right )}{7 b^2}+\frac {2 \sin (c+d x)}{7 b d (b \cos (c+d x))^{7/2}}\right )\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle b^2 \left (\frac {5 \left (\frac {\sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 b^2 \sqrt {b \cos (c+d x)}}+\frac {2 \sin (c+d x)}{3 b d (b \cos (c+d x))^{3/2}}\right )}{7 b^2}+\frac {2 \sin (c+d x)}{7 b d (b \cos (c+d x))^{7/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle b^2 \left (\frac {5 \left (\frac {\sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b^2 \sqrt {b \cos (c+d x)}}+\frac {2 \sin (c+d x)}{3 b d (b \cos (c+d x))^{3/2}}\right )}{7 b^2}+\frac {2 \sin (c+d x)}{7 b d (b \cos (c+d x))^{7/2}}\right )\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle b^2 \left (\frac {5 \left (\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 b^2 d \sqrt {b \cos (c+d x)}}+\frac {2 \sin (c+d x)}{3 b d (b \cos (c+d x))^{3/2}}\right )}{7 b^2}+\frac {2 \sin (c+d x)}{7 b d (b \cos (c+d x))^{7/2}}\right )\) |
b^2*((2*Sin[c + d*x])/(7*b*d*(b*Cos[c + d*x])^(7/2)) + (5*((2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(3*b^2*d*Sqrt[b*Cos[c + d*x]]) + (2*Sin[ c + d*x])/(3*b*d*(b*Cos[c + d*x])^(3/2))))/(7*b^2))
3.2.37.3.1 Defintions of rubi rules used
Int[(Fx_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Simp[1/b^m Int[(b*v) ^(m + n)*Fx, x], x] /; FreeQ[{b, n}, x] && IntegerQ[m]
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1)) I nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && IntegerQ[2*n]
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[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) ^n/Sin[c + d*x]^n Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt Q[-1, n, 1] && IntegerQ[2*n]
Leaf count of result is larger than twice the leaf count of optimal. \(397\) vs. \(2(110)=220\).
Time = 2.23 (sec) , antiderivative size = 398, normalized size of antiderivative = 4.06
| method | result | size |
| default | \(-\frac {2 \left (-40 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-40 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+60 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+40 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-30 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-16 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+5 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}{21 b^{2} \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}\) | \(398\) |
-2/21*(-40*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*E llipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^6-40*cos(1/2*d*x+1 /2*c)*sin(1/2*d*x+1/2*c)^6+60*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+ 1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c) ^4+40*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-30*(sin(1/2*d*x+1/2*c)^2)^(1 /2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)) *sin(1/2*d*x+1/2*c)^2-16*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+5*(sin(1/ 2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d *x+1/2*c),2^(1/2)))/b^2*((2*cos(1/2*d*x+1/2*c)^2-1)*b*sin(1/2*d*x+1/2*c)^2 )^(1/2)/(-b*(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)^2-1)^3/sin(1/2*d*x+1/2*c)/((2*cos(1/2*d*x+1/2*c)^2-1)*b)^(1/2 )/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.17 \[ \int \frac {\sec ^2(c+d x)}{(b \cos (c+d x))^{5/2}} \, dx=\frac {-5 i \, \sqrt {2} \sqrt {b} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} \sqrt {b} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, \sqrt {b \cos \left (d x + c\right )} {\left (5 \, \cos \left (d x + c\right )^{2} + 3\right )} \sin \left (d x + c\right )}{21 \, b^{3} d \cos \left (d x + c\right )^{4}} \]
1/21*(-5*I*sqrt(2)*sqrt(b)*cos(d*x + c)^4*weierstrassPInverse(-4, 0, cos(d *x + c) + I*sin(d*x + c)) + 5*I*sqrt(2)*sqrt(b)*cos(d*x + c)^4*weierstrass PInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*sqrt(b*cos(d*x + c))*(5 *cos(d*x + c)^2 + 3)*sin(d*x + c))/(b^3*d*cos(d*x + c)^4)
Timed out. \[ \int \frac {\sec ^2(c+d x)}{(b \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\sec ^2(c+d x)}{(b \cos (c+d x))^{5/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{2}}{\left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {\sec ^2(c+d x)}{(b \cos (c+d x))^{5/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{2}}{\left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sec ^2(c+d x)}{(b \cos (c+d x))^{5/2}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^2\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]