3.13.86 \(\int \frac {e^2 (-1+2 x-x^2)+e^2 (-x+x^2) \log (x)}{x \log ^3(x)} \, dx\)

Optimal. Leaf size=16 \[ \frac {e^2 (-1+x)^2}{2 \log ^2(x)} \]

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Rubi [B]  time = 0.36, antiderivative size = 71, normalized size of antiderivative = 4.44, number of steps used = 22, number of rules used = 13, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.361, Rules used = {6688, 12, 6742, 2353, 2297, 2298, 2302, 30, 2306, 2309, 2178, 2320, 2330} \begin {gather*} \frac {e^2 x^2}{2 \log ^2(x)}+\frac {e^2 x^2}{\log (x)}-\frac {e^2 x}{\log ^2(x)}+\frac {e^2}{2 \log ^2(x)}+\frac {e^2 (1-x) x}{\log (x)}-\frac {e^2 x}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^2*(-1 + 2*x - x^2) + E^2*(-x + x^2)*Log[x])/(x*Log[x]^3),x]

[Out]

E^2/(2*Log[x]^2) - (E^2*x)/Log[x]^2 + (E^2*x^2)/(2*Log[x]^2) - (E^2*x)/Log[x] + (E^2*(1 - x)*x)/Log[x] + (E^2*
x^2)/Log[x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 2297

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2298

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

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]

Rule 2320

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

Rule 2330

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = Expand
Integrand[(a + b*Log[c*x^n])^p, (d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n, p, q, r}
, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[r]))

Rule 2353

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[
{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && IntegerQ[r
]))

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^2 (1-x) (-1+x-x \log (x))}{x \log ^3(x)} \, dx\\ &=e^2 \int \frac {(1-x) (-1+x-x \log (x))}{x \log ^3(x)} \, dx\\ &=e^2 \int \left (-\frac {(-1+x)^2}{x \log ^3(x)}+\frac {-1+x}{\log ^2(x)}\right ) \, dx\\ &=-\left (e^2 \int \frac {(-1+x)^2}{x \log ^3(x)} \, dx\right )+e^2 \int \frac {-1+x}{\log ^2(x)} \, dx\\ &=\frac {e^2 (1-x) x}{\log (x)}-e^2 \int \left (-\frac {2}{\log ^3(x)}+\frac {1}{x \log ^3(x)}+\frac {x}{\log ^3(x)}\right ) \, dx+e^2 \int \frac {1}{\log (x)} \, dx+\left (2 e^2\right ) \int \frac {-1+x}{\log (x)} \, dx\\ &=\frac {e^2 (1-x) x}{\log (x)}+e^2 \text {li}(x)-e^2 \int \frac {1}{x \log ^3(x)} \, dx-e^2 \int \frac {x}{\log ^3(x)} \, dx+\left (2 e^2\right ) \int \left (-\frac {1}{\log (x)}+\frac {x}{\log (x)}\right ) \, dx+\left (2 e^2\right ) \int \frac {1}{\log ^3(x)} \, dx\\ &=-\frac {e^2 x}{\log ^2(x)}+\frac {e^2 x^2}{2 \log ^2(x)}+\frac {e^2 (1-x) x}{\log (x)}+e^2 \text {li}(x)+e^2 \int \frac {1}{\log ^2(x)} \, dx-e^2 \int \frac {x}{\log ^2(x)} \, dx-e^2 \operatorname {Subst}\left (\int \frac {1}{x^3} \, dx,x,\log (x)\right )-\left (2 e^2\right ) \int \frac {1}{\log (x)} \, dx+\left (2 e^2\right ) \int \frac {x}{\log (x)} \, dx\\ &=\frac {e^2}{2 \log ^2(x)}-\frac {e^2 x}{\log ^2(x)}+\frac {e^2 x^2}{2 \log ^2(x)}-\frac {e^2 x}{\log (x)}+\frac {e^2 (1-x) x}{\log (x)}+\frac {e^2 x^2}{\log (x)}-e^2 \text {li}(x)+e^2 \int \frac {1}{\log (x)} \, dx-\left (2 e^2\right ) \int \frac {x}{\log (x)} \, dx+\left (2 e^2\right ) \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )\\ &=2 e^2 \text {Ei}(2 \log (x))+\frac {e^2}{2 \log ^2(x)}-\frac {e^2 x}{\log ^2(x)}+\frac {e^2 x^2}{2 \log ^2(x)}-\frac {e^2 x}{\log (x)}+\frac {e^2 (1-x) x}{\log (x)}+\frac {e^2 x^2}{\log (x)}-\left (2 e^2\right ) \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )\\ &=\frac {e^2}{2 \log ^2(x)}-\frac {e^2 x}{\log ^2(x)}+\frac {e^2 x^2}{2 \log ^2(x)}-\frac {e^2 x}{\log (x)}+\frac {e^2 (1-x) x}{\log (x)}+\frac {e^2 x^2}{\log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 16, normalized size = 1.00 \begin {gather*} \frac {e^2 (-1+x)^2}{2 \log ^2(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^2*(-1 + 2*x - x^2) + E^2*(-x + x^2)*Log[x])/(x*Log[x]^3),x]

[Out]

(E^2*(-1 + x)^2)/(2*Log[x]^2)

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fricas [A]  time = 0.65, size = 16, normalized size = 1.00 \begin {gather*} \frac {{\left (x^{2} - 2 \, x + 1\right )} e^{2}}{2 \, \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2-x)*exp(1)^2*log(x)+(-x^2+2*x-1)*exp(1)^2)/x/log(x)^3,x, algorithm="fricas")

[Out]

1/2*(x^2 - 2*x + 1)*e^2/log(x)^2

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giac [A]  time = 0.31, size = 20, normalized size = 1.25 \begin {gather*} \frac {x^{2} e^{2} - 2 \, x e^{2} + e^{2}}{2 \, \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2-x)*exp(1)^2*log(x)+(-x^2+2*x-1)*exp(1)^2)/x/log(x)^3,x, algorithm="giac")

[Out]

1/2*(x^2*e^2 - 2*x*e^2 + e^2)/log(x)^2

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maple [A]  time = 0.03, size = 17, normalized size = 1.06

method	result	size
risch	\(\frac {{\mathrm e}^{2} \left (x^{2}-2 x +1\right )}{2 \ln \relax (x )^{2}}\)	\(17\)
norman	\(\frac {\frac {{\mathrm e}^{2}}{2}-{\mathrm e}^{2} x +\frac {x^{2} {\mathrm e}^{2}}{2}}{\ln \relax (x )^{2}}\)	\(29\)
default	\({\mathrm e}^{2} \left (-\frac {x^{2}}{\ln \relax (x )}-2 \expIntegralEi \left (1, -2 \ln \relax (x )\right )\right )-{\mathrm e}^{2} \left (-\frac {x}{\ln \relax (x )}-\expIntegralEi \left (1, -\ln \relax (x )\right )\right )-{\mathrm e}^{2} \left (-\frac {x^{2}}{2 \ln \relax (x )^{2}}-\frac {x^{2}}{\ln \relax (x )}-2 \expIntegralEi \left (1, -2 \ln \relax (x )\right )\right )+2 \,{\mathrm e}^{2} \left (-\frac {x}{2 \ln \relax (x )^{2}}-\frac {x}{2 \ln \relax (x )}-\frac {\expIntegralEi \left (1, -\ln \relax (x )\right )}{2}\right )+\frac {{\mathrm e}^{2}}{2 \ln \relax (x )^{2}}\)	\(119\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^2-x)*exp(1)^2*ln(x)+(-x^2+2*x-1)*exp(1)^2)/x/ln(x)^3,x,method=_RETURNVERBOSE)

[Out]

1/2*exp(2)*(x^2-2*x+1)/ln(x)^2

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maxima [C]  time = 0.39, size = 49, normalized size = 3.06 \begin {gather*} -e^{2} \Gamma \left (-1, -\log \relax (x)\right ) + 2 \, e^{2} \Gamma \left (-1, -2 \, \log \relax (x)\right ) - 2 \, e^{2} \Gamma \left (-2, -\log \relax (x)\right ) + 4 \, e^{2} \Gamma \left (-2, -2 \, \log \relax (x)\right ) + \frac {e^{2}}{2 \, \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2-x)*exp(1)^2*log(x)+(-x^2+2*x-1)*exp(1)^2)/x/log(x)^3,x, algorithm="maxima")

[Out]

-e^2*gamma(-1, -log(x)) + 2*e^2*gamma(-1, -2*log(x)) - 2*e^2*gamma(-2, -log(x)) + 4*e^2*gamma(-2, -2*log(x)) +
1/2*e^2/log(x)^2

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mupad [B]  time = 0.89, size = 13, normalized size = 0.81 \begin {gather*} \frac {{\mathrm {e}}^2\,{\left (x-1\right )}^2}{2\,{\ln \relax (x)}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2)*(x^2 - 2*x + 1) + exp(2)*log(x)*(x - x^2))/(x*log(x)^3),x)

[Out]

(exp(2)*(x - 1)^2)/(2*log(x)^2)

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sympy [A]  time = 0.10, size = 22, normalized size = 1.38 \begin {gather*} \frac {x^{2} e^{2} - 2 x e^{2} + e^{2}}{2 \log {\relax (x )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**2-x)*exp(1)**2*ln(x)+(-x**2+2*x-1)*exp(1)**2)/x/ln(x)**3,x)

[Out]

(x**2*exp(2) - 2*x*exp(2) + exp(2))/(2*log(x)**2)

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