3.24.17 \(\int \frac {x+\frac {1}{5} e^{-3-x+\log ^2(x)} (-x+2 \log (x))}{x} \, dx\)

Optimal. Leaf size=23 \[ -3+e^{e^3}+\frac {1}{5} e^{-3-x+\log ^2(x)}+x \]

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Rubi [A]  time = 0.18, antiderivative size = 17, normalized size of antiderivative = 0.74, number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {14, 6706} \begin {gather*} x+\frac {1}{5} e^{-x+\log ^2(x)-3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x + (E^(-3 - x + Log[x]^2)*(-x + 2*Log[x]))/5)/x,x]

[Out]

E^(-3 - x + Log[x]^2)/5 + x

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (1-\frac {e^{-3-x+\log ^2(x)} (x-2 \log (x))}{5 x}\right ) \, dx\\ &=x-\frac {1}{5} \int \frac {e^{-3-x+\log ^2(x)} (x-2 \log (x))}{x} \, dx\\ &=\frac {1}{5} e^{-3-x+\log ^2(x)}+x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 17, normalized size = 0.74 \begin {gather*} \frac {1}{5} e^{-3-x+\log ^2(x)}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x + (E^(-3 - x + Log[x]^2)*(-x + 2*Log[x]))/5)/x,x]

[Out]

E^(-3 - x + Log[x]^2)/5 + x

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fricas [A]  time = 0.68, size = 16, normalized size = 0.70 \begin {gather*} x + e^{\left (\log \relax (x)^{2} - x - \log \relax (5) - 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*log(x)-x)*exp(log(x)^2-log(5)-3-x)+x)/x,x, algorithm="fricas")

[Out]

x + e^(log(x)^2 - x - log(5) - 3)

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giac [A]  time = 0.21, size = 19, normalized size = 0.83 \begin {gather*} \frac {1}{5} \, {\left (5 \, x e^{3} + e^{\left (\log \relax (x)^{2} - x\right )}\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*log(x)-x)*exp(log(x)^2-log(5)-3-x)+x)/x,x, algorithm="giac")

[Out]

1/5*(5*x*e^3 + e^(log(x)^2 - x))*e^(-3)

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maple [A]  time = 0.03, size = 15, normalized size = 0.65

method	result	size
risch	\(x +\frac {{\mathrm e}^{\ln \relax (x )^{2}-3-x}}{5}\)	\(15\)
default	\(x +{\mathrm e}^{\ln \relax (x )^{2}-\ln \relax (5)-3-x}\)	\(17\)
norman	\(x +{\mathrm e}^{\ln \relax (x )^{2}-\ln \relax (5)-3-x}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*ln(x)-x)*exp(ln(x)^2-ln(5)-3-x)+x)/x,x,method=_RETURNVERBOSE)

[Out]

x+1/5*exp(ln(x)^2-3-x)

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maxima [A]  time = 0.51, size = 14, normalized size = 0.61 \begin {gather*} x + \frac {1}{5} \, e^{\left (\log \relax (x)^{2} - x - 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*log(x)-x)*exp(log(x)^2-log(5)-3-x)+x)/x,x, algorithm="maxima")

[Out]

x + 1/5*e^(log(x)^2 - x - 3)

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mupad [B]  time = 1.44, size = 15, normalized size = 0.65 \begin {gather*} x+\frac {{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-3}\,{\mathrm {e}}^{{\ln \relax (x)}^2}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x - exp(log(x)^2 - log(5) - x - 3)*(x - 2*log(x)))/x,x)

[Out]

x + (exp(-x)*exp(-3)*exp(log(x)^2))/5

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sympy [A]  time = 0.28, size = 12, normalized size = 0.52 \begin {gather*} x + \frac {e^{- x + \log {\relax (x )}^{2} - 3}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*ln(x)-x)*exp(ln(x)**2-ln(5)-3-x)+x)/x,x)

[Out]

x + exp(-x + log(x)**2 - 3)/5

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