Integrand size = 21, antiderivative size = 68 \[ \int \frac {\sec ^4(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 \sqrt {b \sec (c+d x)} \sin (c+d x)}{b^3 d} \]
-2*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1 /2*c),2^(1/2))/b^2/d/cos(d*x+c)^(1/2)/(b*sec(d*x+c))^(1/2)+2*sin(d*x+c)*(b *sec(d*x+c))^(1/2)/b^3/d
Time = 0.24 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.75 \[ \int \frac {\sec ^4(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\frac {-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{\sqrt {\cos (c+d x)}}+2 \tan (c+d x)}{b^2 d \sqrt {b \sec (c+d x)}} \]
((-2*EllipticE[(c + d*x)/2, 2])/Sqrt[Cos[c + d*x]] + 2*Tan[c + d*x])/(b^2* d*Sqrt[b*Sec[c + d*x]])
Time = 0.35 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2030, 3042, 4255, 3042, 4258, 3042, 3119}
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.
\(\displaystyle \int \frac {\sec ^4(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 2030 |
\(\displaystyle \frac {\int (b \sec (c+d x))^{3/2}dx}{b^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}dx}{b^4}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {\frac {2 b \sin (c+d x) \sqrt {b \sec (c+d x)}}{d}-b^2 \int \frac {1}{\sqrt {b \sec (c+d x)}}dx}{b^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {2 b \sin (c+d x) \sqrt {b \sec (c+d x)}}{d}-b^2 \int \frac {1}{\sqrt {b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b^4}\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle \frac {\frac {2 b \sin (c+d x) \sqrt {b \sec (c+d x)}}{d}-\frac {b^2 \int \sqrt {\cos (c+d x)}dx}{\sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}}{b^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {2 b \sin (c+d x) \sqrt {b \sec (c+d x)}}{d}-\frac {b^2 \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{\sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}}{b^4}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {\frac {2 b \sin (c+d x) \sqrt {b \sec (c+d x)}}{d}-\frac {2 b^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}}{b^4}\) |
((-2*b^2*EllipticE[(c + d*x)/2, 2])/(d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d *x]]) + (2*b*Sqrt[b*Sec[c + d*x]]*Sin[c + d*x])/d)/b^4
3.2.24.3.1 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[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[(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 = 0.58 (sec) , antiderivative size = 392, normalized size of antiderivative = 5.76
method | result | size |
default | \(-\frac {2 \left (i \operatorname {EllipticE}\left (i \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )-i \operatorname {EllipticF}\left (i \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )+2 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticE}\left (i \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ), i\right )-2 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ), i\right )+i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticE}\left (i \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ), i\right ) \sec \left (d x +c \right )-i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ), i\right ) \sec \left (d x +c \right )-\tan \left (d x +c \right )\right )}{d \left (\cos \left (d x +c \right )+1\right ) \sqrt {b \sec \left (d x +c \right )}\, b^{2}}\) | \(392\) |
-2/d/(cos(d*x+c)+1)/(b*sec(d*x+c))^(1/2)/b^2*(I*EllipticE(I*(-cot(d*x+c)+c sc(d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*c os(d*x+c)-I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*Ell ipticF(I*(-cot(d*x+c)+csc(d*x+c)),I)*cos(d*x+c)+2*I*(1/(cos(d*x+c)+1))^(1/ 2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(-cot(d*x+c)+csc(d*x+c)), I)-2*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*Elliptic F(I*(-cot(d*x+c)+csc(d*x+c)),I)+I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(co s(d*x+c)+1))^(1/2)*EllipticE(I*(-cot(d*x+c)+csc(d*x+c)),I)*sec(d*x+c)-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)-tan(d*x+c))
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.26 \[ \int \frac {\sec ^4(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\frac {-i \, \sqrt {2} \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + i \, \sqrt {2} \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{b^{3} d} \]
(-I*sqrt(2)*sqrt(b)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos( d*x + c) + I*sin(d*x + c))) + I*sqrt(2)*sqrt(b)*weierstrassZeta(-4, 0, wei erstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*sqrt(b/cos(d*x + c))*sin(d*x + c))/(b^3*d)
\[ \int \frac {\sec ^4(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\int \frac {\sec ^{4}{\left (c + d x \right )}}{\left (b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {\sec ^4(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{4}}{\left (b \sec \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {\sec ^4(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{4}}{\left (b \sec \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sec ^4(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^4\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]