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

Optimal. Leaf size=10 \[ x e^{x+\frac {1}{\log (x)}} \]

[Out]

exp(x+1/ln(x))*x

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Rubi [F]  time = 0.19, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \frac {e^{x+\frac {1}{\log (x)}} \left (-1+(1+x) \log ^2(x)\right )}{\log ^2(x)} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

Rubi steps

\begin {align*} \int \frac {e^{x+\frac {1}{\log (x)}} \left (-1+(1+x) \log ^2(x)\right )}{\log ^2(x)} \, dx &=\int \left (e^{x+\frac {1}{\log (x)}}+e^{x+\frac {1}{\log (x)}} x-\frac {e^{x+\frac {1}{\log (x)}}}{\log ^2(x)}\right ) \, dx\\ &=\int e^{x+\frac {1}{\log (x)}} \, dx+\int e^{x+\frac {1}{\log (x)}} x \, dx-\int \frac {e^{x+\frac {1}{\log (x)}}}{\log ^2(x)} \, dx\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 10, normalized size = 1.00 \[ x e^{x+\frac {1}{\log (x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

E^(x + Log[x]^(-1))*x

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fricas [A]  time = 0.42, size = 14, normalized size = 1.40 \[ x e^{\left (\frac {x \log \relax (x) + 1}{\log \relax (x)}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*e^((x*log(x) + 1)/log(x))

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giac [A]  time = 1.05, size = 14, normalized size = 1.40 \[ x e^{\left (\frac {x \log \relax (x) + 1}{\log \relax (x)}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*e^((x*log(x) + 1)/log(x))

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maple [A]  time = 0.03, size = 10, normalized size = 1.00 \[ x \,{\mathrm e}^{x +\frac {1}{\ln \relax (x )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(x+1/ln(x))*x

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maxima [A]  time = 0.90, size = 9, normalized size = 0.90 \[ x e^{\left (x + \frac {1}{\log \relax (x)}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*e^(x + 1/log(x))

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mupad [B]  time = 0.31, size = 9, normalized size = 0.90 \[ x\,{\mathrm {e}}^{\frac {1}{\ln \relax (x)}}\,{\mathrm {e}}^x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*exp(1/log(x))*exp(x)

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sympy [A]  time = 2.34, size = 8, normalized size = 0.80 \[ x e^{x + \frac {1}{\log {\relax (x )}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*exp(x + 1/log(x))

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