3.1.7 \(\int \frac {e^{x+\frac {1}{\log (x)}} (-1+(1+x) \log ^2(x))}{\log ^2(x)} \, dx\) [7]

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

[Out]

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

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Rubi [F]
time = 0.13, 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 = {} \begin {gather*} \int \frac {e^{x+\frac {1}{\log (x)}} \left (-1+(1+x) \log ^2(x)\right )}{\log ^2(x)} \, dx \end {gather*}

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 = 0.04, size = 10, normalized size = 1.00 \begin {gather*} e^{x+\frac {1}{\log (x)}} x \end {gather*}

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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Maple [A]
time = 0.01, size = 10, normalized size = 1.00

method	result	size
norman	\({\mathrm e}^{x +\frac {1}{\ln \left (x \right )}} x\)	\(10\)
risch	\({\mathrm e}^{\frac {x \ln \left (x \right )+1}{\ln \left (x \right )}} x\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 4.07, size = 9, normalized size = 0.90 \begin {gather*} x e^{\left (x + \frac {1}{\log \left (x\right )}\right )} \end {gather*}

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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Fricas [A]
time = 1.10, size = 14, normalized size = 1.40 \begin {gather*} x e^{\left (\frac {x \log \left (x\right ) + 1}{\log \left (x\right )}\right )} \end {gather*}

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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Sympy [A]
time = 1.14, size = 8, normalized size = 0.80 \begin {gather*} x e^{x + \frac {1}{\log {\left (x \right )}}} \end {gather*}

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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Giac [A]
time = 0.51, size = 14, normalized size = 1.40 \begin {gather*} x e^{\left (\frac {x \log \left (x\right ) + 1}{\log \left (x\right )}\right )} \end {gather*}

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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Mupad [B]
time = 0.31, size = 9, normalized size = 0.90 \begin {gather*} x\,{\mathrm {e}}^{\frac {1}{\ln \left (x\right )}}\,{\mathrm {e}}^x \end {gather*}

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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