3.1.6 \(\int \frac {e^{1+\frac {1}{\log (x)}} (-1+\log ^2(x))}{\log ^2(x)} \, dx\) [6]

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

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {2326} \begin {gather*} x e^{\frac {1}{\log (x)}+1} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(1 + Log[x]^(-1))*x

Rule 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

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

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Mathematica [A]
time = 0.00, size = 10, normalized size = 1.00 \begin {gather*} e^{1+\frac {1}{\log (x)}} x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(1 + Log[x]^(-1))*x

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Mathics [A]
time = 2.52, size = 13, normalized size = 1.30 \begin {gather*} x E^{\frac {1+\text {Log}\left [x\right ]}{\text {Log}\left [x\right ]}} \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[(Log[x]^2 - 1)*Exp[1 + 1/Log[x]]/Log[x]^2,x]')

[Out]

x E ^ ((1 + Log[x]) / Log[x])

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

method	result	size
norman	\({\mathrm e}^{1+\frac {1}{\ln \left (x \right )}} x\)	\(10\)
risch	\({\mathrm e}^{\frac {1+\ln \left (x \right )}{\ln \left (x \right )}} x\)	\(13\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.32, size = 9, normalized size = 0.90 \begin {gather*} x e^{\left (\frac {1}{\log \left (x\right )} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*e^(1/log(x) + 1)

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Fricas [A]
time = 0.30, size = 12, normalized size = 1.20 \begin {gather*} x e^{\left (\frac {\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(1+1/log(x))*(-1+log(x)^2)/log(x)^2,x, algorithm="fricas")

[Out]

x*e^((log(x) + 1)/log(x))

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Sympy [A]
time = 0.78, size = 8, normalized size = 0.80 \begin {gather*} x e^{1 + \frac {1}{\log {\left (x \right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*exp(1 + 1/log(x))

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Giac [A]
time = 0.01, size = 8, normalized size = 0.80 \begin {gather*} x \mathrm {e} \mathrm {e}^{\frac 1{\ln x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(1/log(x) + 1)*x

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Mupad [B]
time = 0.26, size = 9, normalized size = 0.90 \begin {gather*} x\,\mathrm {e}\,{\mathrm {e}}^{\frac {1}{\ln \left (x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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