\(\int \frac {\log ^2(c x)}{x^3} \, dx\) [14]

Optimal result

Integrand size = 10, antiderivative size = 32 \[ \int \frac {\log ^2(c x)}{x^3} \, dx=-\frac {1}{4 x^2}-\frac {\log (c x)}{2 x^2}-\frac {\log ^2(c x)}{2 x^2} \]

[Out]

-1/4/x^2-1/2*ln(c*x)/x^2-1/2*ln(c*x)^2/x^2

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} \[ \int \frac {\log ^2(c x)}{x^3} \, dx=-\frac {\log ^2(c x)}{2 x^2}-\frac {\log (c x)}{2 x^2}-\frac {1}{4 x^2} \]

[In]

Int[Log[c*x]^2/x^3,x]

[Out]

-1/4*1/x^2 - Log[c*x]/(2*x^2) - Log[c*x]^2/(2*x^2)

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(c x)}{2 x^2}+\int \frac {\log (c x)}{x^3} \, dx \\ & = -\frac {1}{4 x^2}-\frac {\log (c x)}{2 x^2}-\frac {\log ^2(c x)}{2 x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {\log ^2(c x)}{x^3} \, dx=-\frac {1}{4 x^2}-\frac {\log (c x)}{2 x^2}-\frac {\log ^2(c x)}{2 x^2} \]

[In]

Integrate[Log[c*x]^2/x^3,x]

[Out]

-1/4*1/x^2 - Log[c*x]/(2*x^2) - Log[c*x]^2/(2*x^2)

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.66

method	result	size
norman	\(\frac {-\frac {1}{4}-\frac {\ln \left (x c \right )^{2}}{2}-\frac {\ln \left (x c \right )}{2}}{x^{2}}\)	\(21\)
parallelrisch	\(\frac {-1-2 \ln \left (x c \right )^{2}-2 \ln \left (x c \right )}{4 x^{2}}\)	\(22\)
risch	\(-\frac {1}{4 x^{2}}-\frac {\ln \left (x c \right )}{2 x^{2}}-\frac {\ln \left (x c \right )^{2}}{2 x^{2}}\)	\(27\)
parts	\(-\frac {\ln \left (x c \right )^{2}}{2 x^{2}}+c^{2} \left (-\frac {\ln \left (x c \right )}{2 x^{2} c^{2}}-\frac {1}{4 x^{2} c^{2}}\right )\)	\(38\)
derivativedivides	\(c^{2} \left (-\frac {\ln \left (x c \right )^{2}}{2 x^{2} c^{2}}-\frac {\ln \left (x c \right )}{2 x^{2} c^{2}}-\frac {1}{4 x^{2} c^{2}}\right )\)	\(40\)
default	\(c^{2} \left (-\frac {\ln \left (x c \right )^{2}}{2 x^{2} c^{2}}-\frac {\ln \left (x c \right )}{2 x^{2} c^{2}}-\frac {1}{4 x^{2} c^{2}}\right )\)	\(40\)

[In]

int(ln(x*c)^2/x^3,x,method=_RETURNVERBOSE)

[Out]

(-1/4-1/2*ln(x*c)^2-1/2*ln(x*c))/x^2

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.66 \[ \int \frac {\log ^2(c x)}{x^3} \, dx=-\frac {2 \, \log \left (c x\right )^{2} + 2 \, \log \left (c x\right ) + 1}{4 \, x^{2}} \]

[In]

integrate(log(c*x)^2/x^3,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {\log ^2(c x)}{x^3} \, dx=- \frac {\log {\left (c x \right )}^{2}}{2 x^{2}} - \frac {\log {\left (c x \right )}}{2 x^{2}} - \frac {1}{4 x^{2}} \]

[In]

integrate(ln(c*x)**2/x**3,x)

[Out]

-log(c*x)**2/(2*x**2) - log(c*x)/(2*x**2) - 1/(4*x**2)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.66 \[ \int \frac {\log ^2(c x)}{x^3} \, dx=-\frac {2 \, \log \left (c x\right )^{2} + 2 \, \log \left (c x\right ) + 1}{4 \, x^{2}} \]

[In]

integrate(log(c*x)^2/x^3,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {\log ^2(c x)}{x^3} \, dx=-\frac {\log \left (c x\right )^{2}}{2 \, x^{2}} - \frac {\log \left (c x\right )}{2 \, x^{2}} - \frac {1}{4 \, x^{2}} \]

[In]

integrate(log(c*x)^2/x^3,x, algorithm="giac")

[Out]

-1/2*log(c*x)^2/x^2 - 1/2*log(c*x)/x^2 - 1/4/x^2

Mupad [B] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.66 \[ \int \frac {\log ^2(c x)}{x^3} \, dx=-\frac {\frac {{\ln \left (c\,x\right )}^2}{2}+\frac {\ln \left (c\,x\right )}{2}+\frac {1}{4}}{x^2} \]

[In]

int(log(c*x)^2/x^3,x)

[Out]

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