Integrand size = 21, antiderivative size = 98 \[ \int \sec ^2(c+d x) (b \sec (c+d x))^{5/2} \, dx=\frac {10 b^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{21 d}+\frac {10 b (b \sec (c+d x))^{3/2} \sin (c+d x)}{21 d}+\frac {2 (b \sec (c+d x))^{7/2} \sin (c+d x)}{7 b d} \]
10/21*b*(b*sec(d*x+c))^(3/2)*sin(d*x+c)/d+2/7*(b*sec(d*x+c))^(7/2)*sin(d*x +c)/b/d+10/21*b^2*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Elliptic F(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*(b*sec(d*x+c))^(1/2)/d
Time = 0.37 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.62 \[ \int \sec ^2(c+d x) (b \sec (c+d x))^{5/2} \, dx=\frac {(b \sec (c+d x))^{5/2} \left (10 \cos ^{\frac {5}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+5 \sin (2 (c+d x))+6 \tan (c+d x)\right )}{21 d} \]
((b*Sec[c + d*x])^(5/2)*(10*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2] + 5*Sin[2*(c + d*x)] + 6*Tan[c + d*x]))/(21*d)
Time = 0.45 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.10, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {2030, 3042, 4255, 3042, 4255, 3042, 4258, 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.
\(\displaystyle \int \sec ^2(c+d x) (b \sec (c+d x))^{5/2} \, dx\) |
\(\Big \downarrow \) 2030 |
\(\displaystyle \frac {\int (b \sec (c+d x))^{9/2}dx}{b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{9/2}dx}{b^2}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {\frac {5}{7} b^2 \int (b \sec (c+d x))^{5/2}dx+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{7/2}}{7 d}}{b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {5}{7} b^2 \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}dx+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{7/2}}{7 d}}{b^2}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {\frac {5}{7} b^2 \left (\frac {1}{3} b^2 \int \sqrt {b \sec (c+d x)}dx+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{7/2}}{7 d}}{b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {5}{7} b^2 \left (\frac {1}{3} b^2 \int \sqrt {b \csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{7/2}}{7 d}}{b^2}\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle \frac {\frac {5}{7} b^2 \left (\frac {1}{3} b^2 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{7/2}}{7 d}}{b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {5}{7} b^2 \left (\frac {1}{3} b^2 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{7/2}}{7 d}}{b^2}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {\frac {5}{7} b^2 \left (\frac {2 b^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 d}+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{7/2}}{7 d}}{b^2}\) |
((2*b*(b*Sec[c + d*x])^(7/2)*Sin[c + d*x])/(7*d) + (5*b^2*((2*b^2*Sqrt[Cos [c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[b*Sec[c + d*x]])/(3*d) + (2*b*(b *Sec[c + d*x])^(3/2)*Sin[c + d*x])/(3*d)))/7)/b^2
3.1.90.3.1 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[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[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
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]
Result contains complex when optimal does not.
Time = 4.77 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.64
method | result | size |
default | \(-\frac {2 i b^{2} \sqrt {b \sec \left (d x +c \right )}\, \left (5 \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ), i\right ) \cos \left (d x +c \right )+5 \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ), i\right )+5 i \tan \left (d x +c \right )+3 i \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}\right )}{21 d}\) | \(161\) |
-2/21*I/d*b^2*(b*sec(d*x+c))^(1/2)*(5*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c) /(cos(d*x+c)+1))^(1/2)*EllipticF(I*(-cot(d*x+c)+csc(d*x+c)),I)*cos(d*x+c)+ 5*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*( -cot(d*x+c)+csc(d*x+c)),I)+5*I*tan(d*x+c)+3*I*tan(d*x+c)*sec(d*x+c)^2)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.23 \[ \int \sec ^2(c+d x) (b \sec (c+d x))^{5/2} \, dx=\frac {-5 i \, \sqrt {2} b^{\frac {5}{2}} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} b^{\frac {5}{2}} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (5 \, b^{2} \cos \left (d x + c\right )^{2} + 3 \, b^{2}\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{21 \, d \cos \left (d x + c\right )^{3}} \]
1/21*(-5*I*sqrt(2)*b^(5/2)*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d *x + c) + I*sin(d*x + c)) + 5*I*sqrt(2)*b^(5/2)*cos(d*x + c)^3*weierstrass PInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*(5*b^2*cos(d*x + c)^2 + 3*b^2)*sqrt(b/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^3)
\[ \int \sec ^2(c+d x) (b \sec (c+d x))^{5/2} \, dx=\int \left (b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}} \sec ^{2}{\left (c + d x \right )}\, dx \]
\[ \int \sec ^2(c+d x) (b \sec (c+d x))^{5/2} \, dx=\int { \left (b \sec \left (d x + c\right )\right )^{\frac {5}{2}} \sec \left (d x + c\right )^{2} \,d x } \]
\[ \int \sec ^2(c+d x) (b \sec (c+d x))^{5/2} \, dx=\int { \left (b \sec \left (d x + c\right )\right )^{\frac {5}{2}} \sec \left (d x + c\right )^{2} \,d x } \]
Timed out. \[ \int \sec ^2(c+d x) (b \sec (c+d x))^{5/2} \, dx=\int \frac {{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^2} \,d x \]