Integrand size = 12, antiderivative size = 100 \[ \int \frac {1}{(b \sec (c+d x))^{7/2}} \, dx=\frac {10 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{21 b^4 d}+\frac {2 \sin (c+d x)}{7 b d (b \sec (c+d x))^{5/2}}+\frac {10 \sin (c+d x)}{21 b^3 d \sqrt {b \sec (c+d x)}} \]
2/7*sin(d*x+c)/b/d/(b*sec(d*x+c))^(5/2)+10/21*sin(d*x+c)/b^3/d/(b*sec(d*x+ c))^(1/2)+10/21*(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)*(b*sec(d*x+c))^(1/2)/b^4/d
Time = 0.02 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.66 \[ \int \frac {1}{(b \sec (c+d x))^{7/2}} \, dx=\frac {\sqrt {b \sec (c+d x)} \left (40 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+26 \sin (2 (c+d x))+3 \sin (4 (c+d x))\right )}{84 b^4 d} \]
(Sqrt[b*Sec[c + d*x]]*(40*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + 2 6*Sin[2*(c + d*x)] + 3*Sin[4*(c + d*x)]))/(84*b^4*d)
Time = 0.43 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.08, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 4256, 3042, 4256, 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.
\(\displaystyle \int \frac {1}{(b \sec (c+d x))^{7/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{7/2}}dx\) |
\(\Big \downarrow \) 4256 |
\(\displaystyle \frac {5 \int \frac {1}{(b \sec (c+d x))^{3/2}}dx}{7 b^2}+\frac {2 \sin (c+d x)}{7 b d (b \sec (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5 \int \frac {1}{\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{7 b^2}+\frac {2 \sin (c+d x)}{7 b d (b \sec (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 4256 |
\(\displaystyle \frac {5 \left (\frac {\int \sqrt {b \sec (c+d x)}dx}{3 b^2}+\frac {2 \sin (c+d x)}{3 b d \sqrt {b \sec (c+d x)}}\right )}{7 b^2}+\frac {2 \sin (c+d x)}{7 b d (b \sec (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5 \left (\frac {\int \sqrt {b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{3 b^2}+\frac {2 \sin (c+d x)}{3 b d \sqrt {b \sec (c+d x)}}\right )}{7 b^2}+\frac {2 \sin (c+d x)}{7 b d (b \sec (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle \frac {5 \left (\frac {\sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 b^2}+\frac {2 \sin (c+d x)}{3 b d \sqrt {b \sec (c+d x)}}\right )}{7 b^2}+\frac {2 \sin (c+d x)}{7 b d (b \sec (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5 \left (\frac {\sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b^2}+\frac {2 \sin (c+d x)}{3 b d \sqrt {b \sec (c+d x)}}\right )}{7 b^2}+\frac {2 \sin (c+d x)}{7 b d (b \sec (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {5 \left (\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 b^2 d}+\frac {2 \sin (c+d x)}{3 b d \sqrt {b \sec (c+d x)}}\right )}{7 b^2}+\frac {2 \sin (c+d x)}{7 b d (b \sec (c+d x))^{5/2}}\) |
(2*Sin[c + d*x])/(7*b*d*(b*Sec[c + d*x])^(5/2)) + (5*((2*Sqrt[Cos[c + d*x] ]*EllipticF[(c + d*x)/2, 2]*Sqrt[b*Sec[c + d*x]])/(3*b^2*d) + (2*Sin[c + d *x])/(3*b*d*Sqrt[b*Sec[c + d*x]])))/(7*b^2)
3.2.31.3.1 Defintions of rubi rules used
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[Cos[c + d*x]*(( b*Csc[c + d*x])^(n + 1)/(b*d*n)), x] + Simp[(n + 1)/(b^2*n) Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[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 = 0.50 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.60
method | result | size |
default | \(-\frac {2 \left (5 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ), i\right )+5 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ), i\right ) \sec \left (d x +c \right )-3 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )-5 \sin \left (d x +c \right )\right )}{21 d \sqrt {b \sec \left (d x +c \right )}\, b^{3}}\) | \(160\) |
-2/21/d/(b*sec(d*x+c))^(1/2)/b^3*(5*I*(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*(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)*sec(d*x+c)-3*cos(d*x+c)^2*sin(d*x+c)-5*sin(d*x+c))
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(b \sec (c+d x))^{7/2}} \, dx=\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{3} + 5 \, \cos \left (d x + c\right )\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right ) - 5 i \, \sqrt {2} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )}{21 \, b^{4} d} \]
1/21*(2*(3*cos(d*x + c)^3 + 5*cos(d*x + c))*sqrt(b/cos(d*x + c))*sin(d*x + c) - 5*I*sqrt(2)*sqrt(b)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin( d*x + c)) + 5*I*sqrt(2)*sqrt(b)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)))/(b^4*d)
\[ \int \frac {1}{(b \sec (c+d x))^{7/2}} \, dx=\int \frac {1}{\left (b \sec {\left (c + d x \right )}\right )^{\frac {7}{2}}}\, dx \]
\[ \int \frac {1}{(b \sec (c+d x))^{7/2}} \, dx=\int { \frac {1}{\left (b \sec \left (d x + c\right )\right )^{\frac {7}{2}}} \,d x } \]
\[ \int \frac {1}{(b \sec (c+d x))^{7/2}} \, dx=\int { \frac {1}{\left (b \sec \left (d x + c\right )\right )^{\frac {7}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{(b \sec (c+d x))^{7/2}} \, dx=\int \frac {1}{{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{7/2}} \,d x \]