\(\int \frac {1}{x^2 \log ^2(c x)} \, dx\) [34]

Optimal result

Integrand size = 10, antiderivative size = 22 \[ \int \frac {1}{x^2 \log ^2(c x)} \, dx=-c \operatorname {ExpIntegralEi}(-\log (c x))-\frac {1}{x \log (c x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {2343, 2346, 2209} \[ \int \frac {1}{x^2 \log ^2(c x)} \, dx=-c \operatorname {ExpIntegralEi}(-\log (c x))-\frac {1}{x \log (c x)} \]

[In]

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

[Out]

-(c*ExpIntegralEi[-Log[c*x]]) - 1/(x*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 2343

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Log
[c*x^n])^(p + 1)/(b*d*n*(p + 1))), x] - Dist[(m + 1)/(b*n*(p + 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] && LtQ[p, -1]

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}& = -\frac {1}{x \log (c x)}-\int \frac {1}{x^2 \log (c x)} \, dx \\ & = -\frac {1}{x \log (c x)}-c \text {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,\log (c x)\right ) \\ & = -c \text {Ei}(-\log (c x))-\frac {1}{x \log (c x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^2 \log ^2(c x)} \, dx=-c \operatorname {ExpIntegralEi}(-\log (c x))-\frac {1}{x \log (c x)} \]

[In]

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

[Out]

-(c*ExpIntegralEi[-Log[c*x]]) - 1/(x*Log[c*x])

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95

method	result	size
risch	\(-\frac {1}{x \ln \left (x c \right )}+c \,\operatorname {Ei}_{1}\left (\ln \left (x c \right )\right )\)	\(21\)
derivativedivides	\(c \left (-\frac {1}{x c \ln \left (x c \right )}+\operatorname {Ei}_{1}\left (\ln \left (x c \right )\right )\right )\)	\(24\)
default	\(c \left (-\frac {1}{x c \ln \left (x c \right )}+\operatorname {Ei}_{1}\left (\ln \left (x c \right )\right )\right )\)	\(24\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27 \[ \int \frac {1}{x^2 \log ^2(c x)} \, dx=-\frac {c x \log \left (c x\right ) \operatorname {log\_integral}\left (\frac {1}{c x}\right ) + 1}{x \log \left (c x\right )} \]

[In]

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

[Out]

-(c*x*log(c*x)*log_integral(1/(c*x)) + 1)/(x*log(c*x))

Sympy [F]

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

[In]

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

[Out]

-Integral(1/(x**2*log(c*x)), x) - 1/(x*log(c*x))

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.41 \[ \int \frac {1}{x^2 \log ^2(c x)} \, dx=-c \Gamma \left (-1, \log \left (c x\right )\right ) \]

[In]

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

[Out]

-c*gamma(-1, log(c*x))

Giac [F]

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

[In]

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

[Out]

integrate(1/(x^2*log(c*x)^2), x)

Mupad [F(-1)]

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

[In]

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

[Out]

int(1/(x^2*log(c*x)^2), x)