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

Optimal. Leaf size=21 \[ \frac {1}{2} \left (\frac {1}{e^3}-x\right ) \left (-x+\frac {x}{\log (x)}\right ) \]

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Rubi [A]  time = 0.14, antiderivative size = 36, normalized size of antiderivative = 1.71, number of steps used = 15, number of rules used = 7, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {12, 6742, 2320, 2330, 2298, 2309, 2178} \begin {gather*} \frac {x^2}{2}-\frac {x}{2 e^3}+\frac {\left (1-e^3 x\right ) x}{2 e^3 \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*x/E^3 + x^2/2 + (x*(1 - E^3*x))/(2*E^3*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 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 2298

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

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 6742

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

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 23, normalized size = 1.10 \begin {gather*} \frac {x \left (-1+e^3 x\right ) (-1+\log (x))}{2 e^3 \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.85, size = 32, normalized size = 1.52 \begin {gather*} -\frac {{\left (x^{2} e^{3} - {\left (x^{2} e^{3} - x\right )} \log \relax (x) - x\right )} e^{\left (-3\right )}}{2 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(x^2*e^3 - (x^2*e^3 - x)*log(x) - x)*e^(-3)/log(x)

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giac [A]  time = 0.13, size = 31, normalized size = 1.48 \begin {gather*} \frac {1}{2} \, {\left (x^{2} e^{3} - \frac {x^{2} e^{3}}{\log \relax (x)} - x + \frac {x}{\log \relax (x)}\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(x^2*e^3 - x^2*e^3/log(x) - x + x/log(x))*e^(-3)

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maple [A]  time = 0.04, size = 27, normalized size = 1.29

method	result	size
risch	\(\frac {x^{2}}{2}-\frac {x \,{\mathrm e}^{-3}}{2}-\frac {{\mathrm e}^{-3} x \left (x \,{\mathrm e}^{3}-1\right )}{2 \ln \relax (x )}\)	\(27\)
norman	\(\frac {-\frac {x^{2}}{2}+\frac {x \,{\mathrm e}^{-3}}{2}+\frac {x^{2} \ln \relax (x )}{2}-\frac {x \,{\mathrm e}^{-3} \ln \relax (x )}{2}}{\ln \relax (x )}\)	\(35\)
default	\(\frac {{\mathrm e}^{-3} \left (x^{2} {\mathrm e}^{3}+2 \,{\mathrm e}^{3} \expIntegralEi \left (1, -2 \ln \relax (x )\right )-x +{\mathrm e}^{3} \left (-\frac {x^{2}}{\ln \relax (x )}-2 \expIntegralEi \left (1, -2 \ln \relax (x )\right )\right )+\frac {x}{\ln \relax (x )}\right )}{2}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2-1/2*x*exp(-3)-1/2*exp(-3)*x*(x*exp(3)-1)/ln(x)

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maxima [C]  time = 0.39, size = 44, normalized size = 2.10 \begin {gather*} \frac {1}{2} \, {\left (x^{2} e^{3} - 2 \, {\rm Ei}\left (2 \, \log \relax (x)\right ) e^{3} + 2 \, e^{3} \Gamma \left (-1, -2 \, \log \relax (x)\right ) - x + {\rm Ei}\left (\log \relax (x)\right ) - \Gamma \left (-1, -\log \relax (x)\right )\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(x^2*e^3 - 2*Ei(2*log(x))*e^3 + 2*e^3*gamma(-1, -2*log(x)) - x + Ei(log(x)) - gamma(-1, -log(x)))*e^(-3)

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mupad [B]  time = 6.19, size = 19, normalized size = 0.90 \begin {gather*} \frac {x\,{\mathrm {e}}^{-3}\,\left (\ln \relax (x)-1\right )\,\left (x\,{\mathrm {e}}^3-1\right )}{2\,\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*exp(-3)*(log(x) - 1)*(x*exp(3) - 1))/(2*log(x))

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sympy [A]  time = 0.10, size = 27, normalized size = 1.29 \begin {gather*} \frac {x^{2}}{2} - \frac {x}{2 e^{3}} + \frac {- x^{2} e^{3} + x}{2 e^{3} \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**2/2 - x*exp(-3)/2 + (-x**2*exp(3) + x)*exp(-3)/(2*log(x))

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