3.34 \(\int \frac {1}{x^2 \log ^2(c x)} \, dx\)

Optimal. Leaf size=22 \[ -c \text {Ei}(-\log (c x))-\frac {1}{x \log (c x)} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {2306, 2309, 2178} \[ -c \text {Ei}(-\log (c x))-\frac {1}{x \log (c x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2178

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

Rule 2306

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 2309

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*} \int \frac {1}{x^2 \log ^2(c x)} \, dx &=-\frac {1}{x \log (c x)}-\int \frac {1}{x^2 \log (c x)} \, dx\\ &=-\frac {1}{x \log (c x)}-c \operatorname {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*}

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Mathematica [A]  time = 0.02, size = 22, normalized size = 1.00 \[ -c \text {Ei}(-\log (c x))-\frac {1}{x \log (c x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.40, size = 28, normalized size = 1.27 \[ -\frac {c x \log \left (c x\right ) \operatorname {log\_integral}\left (\frac {1}{c x}\right ) + 1}{x \log \left (c x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{2} \log \left (c x\right )^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(x^2*log(c*x)^2), x)

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maple [A]  time = 0.03, size = 21, normalized size = 0.95 \[ c \Ei \left (1, \ln \left (c x \right )\right )-\frac {1}{x \ln \left (c x \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.84, size = 9, normalized size = 0.41 \[ -c \Gamma \left (-1, \log \left (c x\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c*gamma(-1, log(c*x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \[ \int \frac {1}{x^2\,{\ln \left (c\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/(x^2*log(c*x)^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {1}{x^{2} \log {\left (c x \right )}}\, dx - \frac {1}{x \log {\left (c x \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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