\(\int \frac {1}{x^2 \log (c x)} \, dx\) [27]

Optimal result

Integrand size = 10, antiderivative size = 9 \[ \int \frac {1}{x^2 \log (c x)} \, dx=c \operatorname {ExpIntegralEi}(-\log (c x)) \]

[Out]

c*Ei(-ln(c*x))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 9, 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 = {2346, 2209} \[ \int \frac {1}{x^2 \log (c x)} \, dx=c \operatorname {ExpIntegralEi}(-\log (c x)) \]

[In]

Int[1/(x^2*Log[c*x]),x]

[Out]

c*ExpIntegralEi[-Log[c*x]]

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2346

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[E^((m + 1)*x)*(a
+ b*x)^p, x], x, Log[c*x]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = c \text {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,\log (c x)\right ) \\ & = c \text {Ei}(-\log (c x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^2 \log (c x)} \, dx=c \operatorname {ExpIntegralEi}(-\log (c x)) \]

[In]

Integrate[1/(x^2*Log[c*x]),x]

[Out]

c*ExpIntegralEi[-Log[c*x]]

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.11

method	result	size
derivativedivides	\(-c \,\operatorname {Ei}_{1}\left (\ln \left (x c \right )\right )\)	\(10\)
default	\(-c \,\operatorname {Ei}_{1}\left (\ln \left (x c \right )\right )\)	\(10\)
risch	\(-c \,\operatorname {Ei}_{1}\left (\ln \left (x c \right )\right )\)	\(10\)

[In]

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

[Out]

-c*Ei(1,ln(x*c))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.11 \[ \int \frac {1}{x^2 \log (c x)} \, dx=c \operatorname {log\_integral}\left (\frac {1}{c x}\right ) \]

[In]

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

[Out]

c*log_integral(1/(c*x))

Sympy [F]

\[ \int \frac {1}{x^2 \log (c x)} \, dx=\int \frac {1}{x^{2} \log {\left (c x \right )}}\, dx \]

[In]

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

[Out]

Integral(1/(x**2*log(c*x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^2 \log (c x)} \, dx=c {\rm Ei}\left (-\log \left (c x\right )\right ) \]

[In]

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

[Out]

c*Ei(-log(c*x))

Giac [F]

\[ \int \frac {1}{x^2 \log (c x)} \, dx=\int { \frac {1}{x^{2} \log \left (c x\right )} \,d x } \]

[In]

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

[Out]

integrate(1/(x^2*log(c*x)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^2 \log (c x)} \, dx=\int \frac {1}{x^2\,\ln \left (c\,x\right )} \,d x \]

[In]

int(1/(x^2*log(c*x)),x)

[Out]

int(1/(x^2*log(c*x)), x)