\(\int \frac {1}{x \sec ^{\frac {5}{2}}(a+b \log (c x^n))} \, dx\) [277]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [B] (verified)
   Fricas [C] (verification not implemented)
   Sympy [F(-1)]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 19, antiderivative size = 93 \[ \int \frac {1}{x \sec ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\frac {6 \sqrt {\cos \left (a+b \log \left (c x^n\right )\right )} E\left (\left .\frac {1}{2} \left (a+b \log \left (c x^n\right )\right )\right |2\right ) \sqrt {\sec \left (a+b \log \left (c x^n\right )\right )}}{5 b n}+\frac {2 \sin \left (a+b \log \left (c x^n\right )\right )}{5 b n \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )} \]

[Out]

2/5*sin(a+b*ln(c*x^n))/b/n/sec(a+b*ln(c*x^n))^(3/2)+6/5*(cos(1/2*a+1/2*b*ln(c*x^n))^2)^(1/2)/cos(1/2*a+1/2*b*l
n(c*x^n))*EllipticE(sin(1/2*a+1/2*b*ln(c*x^n)),2^(1/2))*cos(a+b*ln(c*x^n))^(1/2)*sec(a+b*ln(c*x^n))^(1/2)/b/n

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3854, 3856, 2719} \[ \int \frac {1}{x \sec ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\frac {2 \sin \left (a+b \log \left (c x^n\right )\right )}{5 b n \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}+\frac {6 \sqrt {\sec \left (a+b \log \left (c x^n\right )\right )} \sqrt {\cos \left (a+b \log \left (c x^n\right )\right )} E\left (\left .\frac {1}{2} \left (a+b \log \left (c x^n\right )\right )\right |2\right )}{5 b n} \]

[In]

Int[1/(x*Sec[a + b*Log[c*x^n]]^(5/2)),x]

[Out]

(6*Sqrt[Cos[a + b*Log[c*x^n]]]*EllipticE[(a + b*Log[c*x^n])/2, 2]*Sqrt[Sec[a + b*Log[c*x^n]]])/(5*b*n) + (2*Si
n[a + b*Log[c*x^n]])/(5*b*n*Sec[a + b*Log[c*x^n]]^(3/2))

Rule 2719

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]

Rule 3854

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
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{\sec ^{\frac {5}{2}}(a+b x)} \, dx,x,\log \left (c x^n\right )\right )}{n} \\ & = \frac {2 \sin \left (a+b \log \left (c x^n\right )\right )}{5 b n \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}+\frac {3 \text {Subst}\left (\int \frac {1}{\sqrt {\sec (a+b x)}} \, dx,x,\log \left (c x^n\right )\right )}{5 n} \\ & = \frac {2 \sin \left (a+b \log \left (c x^n\right )\right )}{5 b n \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}+\frac {\left (3 \sqrt {\cos \left (a+b \log \left (c x^n\right )\right )} \sqrt {\sec \left (a+b \log \left (c x^n\right )\right )}\right ) \text {Subst}\left (\int \sqrt {\cos (a+b x)} \, dx,x,\log \left (c x^n\right )\right )}{5 n} \\ & = \frac {6 \sqrt {\cos \left (a+b \log \left (c x^n\right )\right )} E\left (\left .\frac {1}{2} \left (a+b \log \left (c x^n\right )\right )\right |2\right ) \sqrt {\sec \left (a+b \log \left (c x^n\right )\right )}}{5 b n}+\frac {2 \sin \left (a+b \log \left (c x^n\right )\right )}{5 b n \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x \sec ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\frac {\sqrt {\sec \left (a+b \log \left (c x^n\right )\right )} \left (12 \sqrt {\cos \left (a+b \log \left (c x^n\right )\right )} E\left (\left .\frac {1}{2} \left (a+b \log \left (c x^n\right )\right )\right |2\right )+\sin \left (a+b \log \left (c x^n\right )\right )+\sin \left (3 \left (a+b \log \left (c x^n\right )\right )\right )\right )}{10 b n} \]

[In]

Integrate[1/(x*Sec[a + b*Log[c*x^n]]^(5/2)),x]

[Out]

(Sqrt[Sec[a + b*Log[c*x^n]]]*(12*Sqrt[Cos[a + b*Log[c*x^n]]]*EllipticE[(a + b*Log[c*x^n])/2, 2] + Sin[a + b*Lo
g[c*x^n]] + Sin[3*(a + b*Log[c*x^n])]))/(10*b*n)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(279\) vs. \(2(119)=238\).

Time = 3.02 (sec) , antiderivative size = 280, normalized size of antiderivative = 3.01

method result size
derivativedivides \(-\frac {2 \sqrt {\left (2 {\cos \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{2}-1\right ) {\sin \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{2}}\, \left (-8 \cos \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right ) {\sin \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{6}+8 \cos \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right ) {\sin \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{4}-2 {\sin \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{2} \cos \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )-3 \sqrt {\frac {1}{2}-\frac {\cos \left (a +2 b \ln \left (\sqrt {c \,x^{n}}\right )\right )}{2}}\, \sqrt {-1+2 {\sin \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right ), \sqrt {2}\right )\right )}{5 n \sqrt {-2 {\sin \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{4}+{\sin \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{2}}\, \sin \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right ) \sqrt {2 {\cos \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{2}-1}\, b}\) \(280\)
default \(-\frac {2 \sqrt {\left (2 {\cos \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{2}-1\right ) {\sin \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{2}}\, \left (-8 \cos \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right ) {\sin \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{6}+8 \cos \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right ) {\sin \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{4}-2 {\sin \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{2} \cos \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )-3 \sqrt {\frac {1}{2}-\frac {\cos \left (a +2 b \ln \left (\sqrt {c \,x^{n}}\right )\right )}{2}}\, \sqrt {-1+2 {\sin \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right ), \sqrt {2}\right )\right )}{5 n \sqrt {-2 {\sin \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{4}+{\sin \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{2}}\, \sin \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right ) \sqrt {2 {\cos \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{2}-1}\, b}\) \(280\)

[In]

int(1/x/sec(a+b*ln(c*x^n))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-2/5/n*((2*cos(1/2*a+1/2*b*ln(c*x^n))^2-1)*sin(1/2*a+1/2*b*ln(c*x^n))^2)^(1/2)*(-8*cos(1/2*a+1/2*b*ln(c*x^n))*
sin(1/2*a+1/2*b*ln(c*x^n))^6+8*cos(1/2*a+1/2*b*ln(c*x^n))*sin(1/2*a+1/2*b*ln(c*x^n))^4-2*sin(1/2*a+1/2*b*ln(c*
x^n))^2*cos(1/2*a+1/2*b*ln(c*x^n))-3*(sin(1/2*a+1/2*b*ln(c*x^n))^2)^(1/2)*(-1+2*sin(1/2*a+1/2*b*ln(c*x^n))^2)^
(1/2)*EllipticE(cos(1/2*a+1/2*b*ln(c*x^n)),2^(1/2)))/(-2*sin(1/2*a+1/2*b*ln(c*x^n))^4+sin(1/2*a+1/2*b*ln(c*x^n
))^2)^(1/2)/sin(1/2*a+1/2*b*ln(c*x^n))/(2*cos(1/2*a+1/2*b*ln(c*x^n))^2-1)^(1/2)/b

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.09 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.22 \[ \int \frac {1}{x \sec ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\frac {2 \, \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{\frac {3}{2}} \sin \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) + 3 i \, \sqrt {2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) + i \, \sin \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )\right )\right ) - 3 i \, \sqrt {2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) - i \, \sin \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )\right )\right )}{5 \, b n} \]

[In]

integrate(1/x/sec(a+b*log(c*x^n))^(5/2),x, algorithm="fricas")

[Out]

1/5*(2*cos(b*n*log(x) + b*log(c) + a)^(3/2)*sin(b*n*log(x) + b*log(c) + a) + 3*I*sqrt(2)*weierstrassZeta(-4, 0
, weierstrassPInverse(-4, 0, cos(b*n*log(x) + b*log(c) + a) + I*sin(b*n*log(x) + b*log(c) + a))) - 3*I*sqrt(2)
*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(b*n*log(x) + b*log(c) + a) - I*sin(b*n*log(x) + b*log(c
) + a))))/(b*n)

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{x \sec ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\text {Timed out} \]

[In]

integrate(1/x/sec(a+b*ln(c*x**n))**(5/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {1}{x \sec ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\int { \frac {1}{x \sec \left (b \log \left (c x^{n}\right ) + a\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate(1/x/sec(a+b*log(c*x^n))^(5/2),x, algorithm="maxima")

[Out]

integrate(1/(x*sec(b*log(c*x^n) + a)^(5/2)), x)

Giac [F]

\[ \int \frac {1}{x \sec ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\int { \frac {1}{x \sec \left (b \log \left (c x^{n}\right ) + a\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate(1/x/sec(a+b*log(c*x^n))^(5/2),x, algorithm="giac")

[Out]

integrate(1/(x*sec(b*log(c*x^n) + a)^(5/2)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x \sec ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\int \frac {1}{x\,{\left (\frac {1}{\cos \left (a+b\,\ln \left (c\,x^n\right )\right )}\right )}^{5/2}} \,d x \]

[In]

int(1/(x*(1/cos(a + b*log(c*x^n)))^(5/2)),x)

[Out]

int(1/(x*(1/cos(a + b*log(c*x^n)))^(5/2)), x)