Integrand size = 21, antiderivative size = 41 \[ \int \frac {\sec ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}} \] Output:
2*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)
Time = 0.14 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.93 \[ \int \frac {\sec ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \cos ^{\frac {5}{2}}(c+d x) (b \sec (c+d x))^{5/2}} \] Input:
Integrate[Sec[c + d*x]^2/(b*Sec[c + d*x])^(5/2),x]
Output:
(2*EllipticE[(c + d*x)/2, 2])/(d*Cos[c + d*x]^(5/2)*(b*Sec[c + d*x])^(5/2) )
Time = 0.26 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2030, 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.
\(\displaystyle \int \frac {\sec ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 2030 |
\(\displaystyle \frac {\int \frac {1}{\sqrt {b \sec (c+d x)}}dx}{b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {1}{\sqrt {b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b^2}\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle \frac {\int \sqrt {\cos (c+d x)}dx}{b^2 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}\) |
Input:
Int[Sec[c + d*x]^2/(b*Sec[c + d*x])^(5/2),x]
Output:
(2*EllipticE[(c + d*x)/2, 2])/(b^2*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x ]])
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*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 = 1.15 (sec) , antiderivative size = 178, normalized size of antiderivative = 4.34
method | result | size |
default | \(\frac {2 i \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \left (-\cos \left (d x +c \right )-2-\sec \left (d x +c \right )\right )+2 i \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \left (2+\cos \left (d x +c \right )+\sec \left (d x +c \right )\right ) \operatorname {EllipticE}\left (i \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ), i\right )+2 \sin \left (d x +c \right )}{b^{2} d \left (\cos \left (d x +c \right )+1\right ) \sqrt {b \sec \left (d x +c \right )}}\) | \(178\) |
risch | \(-\frac {i \sqrt {2}}{d \,b^{2} \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}-\frac {i \left (-\frac {2 \left ({\mathrm e}^{2 i \left (d x +c \right )} b +b \right )}{b \sqrt {{\mathrm e}^{i \left (d x +c \right )} \left ({\mathrm e}^{2 i \left (d x +c \right )} b +b \right )}}+\frac {i \sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}\, \sqrt {2}\, \sqrt {i \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}\, \sqrt {i {\mathrm e}^{i \left (d x +c \right )}}\, \left (-2 i \operatorname {EllipticE}\left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right )+i \operatorname {EllipticF}\left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right )\right )}{\sqrt {b \,{\mathrm e}^{3 i \left (d x +c \right )}+b \,{\mathrm e}^{i \left (d x +c \right )}}}\right ) \sqrt {2}\, \sqrt {b \,{\mathrm e}^{i \left (d x +c \right )} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}}{d \,b^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}\) | \(305\) |
Input:
int(sec(d*x+c)^2/(b*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
Output:
2/b^2/d/(cos(d*x+c)+1)/(b*sec(d*x+c))^(1/2)*(I*(cos(d*x+c)/(cos(d*x+c)+1)) ^(1/2)*EllipticF(I*(-cot(d*x+c)+csc(d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2)*(- cos(d*x+c)-2-sec(d*x+c))+I*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*( -cot(d*x+c)+csc(d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2)*(2+cos(d*x+c)+sec(d*x+ c))+sin(d*x+c))
Result contains complex when optimal does not.
Time = 0.08 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.61 \[ \int \frac {\sec ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\frac {i \, \sqrt {2} \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - i \, \sqrt {2} \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{b^{3} d} \] Input:
integrate(sec(d*x+c)^2/(b*sec(d*x+c))^(5/2),x, algorithm="fricas")
Output:
(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, weie rstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(b^3*d)
\[ \int \frac {\sec ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\int \frac {\sec ^{2}{\left (c + d x \right )}}{\left (b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(sec(d*x+c)**2/(b*sec(d*x+c))**(5/2),x)
Output:
Integral(sec(c + d*x)**2/(b*sec(c + d*x))**(5/2), x)
\[ \int \frac {\sec ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{2}}{\left (b \sec \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate(sec(d*x+c)^2/(b*sec(d*x+c))^(5/2),x, algorithm="maxima")
Output:
integrate(sec(d*x + c)^2/(b*sec(d*x + c))^(5/2), x)
\[ \int \frac {\sec ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{2}}{\left (b \sec \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate(sec(d*x+c)^2/(b*sec(d*x+c))^(5/2),x, algorithm="giac")
Output:
integrate(sec(d*x + c)^2/(b*sec(d*x + c))^(5/2), x)
Timed out. \[ \int \frac {\sec ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^2\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \] Input:
int(1/(cos(c + d*x)^2*(b/cos(c + d*x))^(5/2)),x)
Output:
int(1/(cos(c + d*x)^2*(b/cos(c + d*x))^(5/2)), x)
\[ \int \frac {\sec ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\frac {\sqrt {b}\, \left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )}d x \right )}{b^{3}} \] Input:
int(sec(d*x+c)^2/(b*sec(d*x+c))^(5/2),x)
Output:
(sqrt(b)*int(sqrt(sec(c + d*x))/sec(c + d*x),x))/b**3