\(\int \frac {\log ^2(\frac {1}{x})}{x^5} \, dx\) [262]

Optimal result

Integrand size = 10, antiderivative size = 32 \[ \int \frac {\log ^2\left (\frac {1}{x}\right )}{x^5} \, dx=-\frac {1}{32 x^4}+\frac {\log \left (\frac {1}{x}\right )}{8 x^4}-\frac {\log ^2\left (\frac {1}{x}\right )}{4 x^4} \]

[Out]

-1/32/x^4+1/8*ln(1/x)/x^4-1/4*ln(1/x)^2/x^4

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2342, 2341} \[ \int \frac {\log ^2\left (\frac {1}{x}\right )}{x^5} \, dx=-\frac {1}{32 x^4}-\frac {\log ^2\left (\frac {1}{x}\right )}{4 x^4}+\frac {\log \left (\frac {1}{x}\right )}{8 x^4} \]

[In]

Int[Log[x^(-1)]^2/x^5,x]

[Out]

-1/32*1/x^4 + Log[x^(-1)]/(8*x^4) - Log[x^(-1)]^2/(4*x^4)

Rule 2341

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Log[c*x^
n])/(d*(m + 1))), x] - Simp[b*n*((d*x)^(m + 1)/(d*(m + 1)^2)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2342

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Lo
g[c*x^n])^p/(d*(m + 1))), x] - Dist[b*n*(p/(m + 1)), Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{
a, b, c, d, m, n}, x] && NeQ[m, -1] && GtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {\log ^2\left (\frac {1}{x}\right )}{4 x^4}-\frac {1}{2} \int \frac {\log \left (\frac {1}{x}\right )}{x^5} \, dx \\ & = -\frac {1}{32 x^4}+\frac {\log \left (\frac {1}{x}\right )}{8 x^4}-\frac {\log ^2\left (\frac {1}{x}\right )}{4 x^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {\log ^2\left (\frac {1}{x}\right )}{x^5} \, dx=-\frac {1}{32 x^4}+\frac {\log \left (\frac {1}{x}\right )}{8 x^4}-\frac {\log ^2\left (\frac {1}{x}\right )}{4 x^4} \]

[In]

Integrate[Log[x^(-1)]^2/x^5,x]

[Out]

-1/32*1/x^4 + Log[x^(-1)]/(8*x^4) - Log[x^(-1)]^2/(4*x^4)

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.66

method	result	size
norman	\(\frac {-\frac {1}{32}-\frac {\ln \left (\frac {1}{x}\right )^{2}}{4}+\frac {\ln \left (\frac {1}{x}\right )}{8}}{x^{4}}\)	\(21\)
parallelrisch	\(\frac {-1-8 \ln \left (\frac {1}{x}\right )^{2}+4 \ln \left (\frac {1}{x}\right )}{32 x^{4}}\)	\(22\)
derivativedivides	\(-\frac {1}{32 x^{4}}+\frac {\ln \left (\frac {1}{x}\right )}{8 x^{4}}-\frac {\ln \left (\frac {1}{x}\right )^{2}}{4 x^{4}}\)	\(27\)
default	\(-\frac {1}{32 x^{4}}+\frac {\ln \left (\frac {1}{x}\right )}{8 x^{4}}-\frac {\ln \left (\frac {1}{x}\right )^{2}}{4 x^{4}}\)	\(27\)
risch	\(-\frac {1}{32 x^{4}}+\frac {\ln \left (\frac {1}{x}\right )}{8 x^{4}}-\frac {\ln \left (\frac {1}{x}\right )^{2}}{4 x^{4}}\)	\(27\)
parts	\(-\frac {1}{32 x^{4}}+\frac {\ln \left (\frac {1}{x}\right )}{8 x^{4}}-\frac {\ln \left (\frac {1}{x}\right )^{2}}{4 x^{4}}\)	\(27\)

[In]

int(ln(1/x)^2/x^5,x,method=_RETURNVERBOSE)

[Out]

(-1/32-1/4*ln(1/x)^2+1/8*ln(1/x))/x^4

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.66 \[ \int \frac {\log ^2\left (\frac {1}{x}\right )}{x^5} \, dx=-\frac {8 \, \log \left (\frac {1}{x}\right )^{2} - 4 \, \log \left (\frac {1}{x}\right ) + 1}{32 \, x^{4}} \]

[In]

integrate(log(1/x)^2/x^5,x, algorithm="fricas")

[Out]

-1/32*(8*log(1/x)^2 - 4*log(1/x) + 1)/x^4

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84 \[ \int \frac {\log ^2\left (\frac {1}{x}\right )}{x^5} \, dx=- \frac {\log {\left (\frac {1}{x} \right )}^{2}}{4 x^{4}} + \frac {\log {\left (\frac {1}{x} \right )}}{8 x^{4}} - \frac {1}{32 x^{4}} \]

[In]

integrate(ln(1/x)**2/x**5,x)

[Out]

-log(1/x)**2/(4*x**4) + log(1/x)/(8*x**4) - 1/(32*x**4)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.53 \[ \int \frac {\log ^2\left (\frac {1}{x}\right )}{x^5} \, dx=-\frac {8 \, \log \left (x\right )^{2} + 4 \, \log \left (x\right ) + 1}{32 \, x^{4}} \]

[In]

integrate(log(1/x)^2/x^5,x, algorithm="maxima")

[Out]

-1/32*(8*log(x)^2 + 4*log(x) + 1)/x^4

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.69 \[ \int \frac {\log ^2\left (\frac {1}{x}\right )}{x^5} \, dx=-\frac {\log \left (x\right )^{2}}{4 \, x^{4}} - \frac {\log \left (x\right )}{8 \, x^{4}} - \frac {1}{32 \, x^{4}} \]

[In]

integrate(log(1/x)^2/x^5,x, algorithm="giac")

[Out]

-1/4*log(x)^2/x^4 - 1/8*log(x)/x^4 - 1/32/x^4

Mupad [B] (verification not implemented)

Time = 1.62 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.66 \[ \int \frac {\log ^2\left (\frac {1}{x}\right )}{x^5} \, dx=-\frac {\frac {{\ln \left (\frac {1}{x}\right )}^2}{4}-\frac {\ln \left (\frac {1}{x}\right )}{8}+\frac {1}{32}}{x^4} \]

[In]

int(log(1/x)^2/x^5,x)

[Out]

-(log(1/x)^2/4 - log(1/x)/8 + 1/32)/x^4