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

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

[Out]

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

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Rubi [A]  time = 0.03, 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 = {2288} \[ x e^{\frac {1}{\log (x)}+1} \]

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 2288

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 = 0.01, size = 10, normalized size = 1.00 \[ x e^{\frac {1}{\log (x)}+1} \]

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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fricas [A]  time = 0.42, size = 12, normalized size = 1.20 \[ x e^{\left (\frac {\log \relax (x) + 1}{\log \relax (x)}\right )} \]

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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giac [A]  time = 1.10, size = 9, normalized size = 0.90 \[ x e^{\left (\frac {1}{\log \relax (x)} + 1\right )} \]

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="giac")

[Out]

x*e^(1/log(x) + 1)

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maple [A]  time = 0.02, size = 10, normalized size = 1.00 \[ x \,{\mathrm e}^{\frac {1}{\ln \relax (x )}+1} \]

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)

[Out]

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

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maxima [A]  time = 0.65, size = 9, normalized size = 0.90 \[ x e^{\left (\frac {1}{\log \relax (x)} + 1\right )} \]

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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mupad [B]  time = 0.26, size = 9, normalized size = 0.90 \[ x\,\mathrm {e}\,{\mathrm {e}}^{\frac {1}{\ln \relax (x)}} \]

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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sympy [A]  time = 1.59, size = 8, normalized size = 0.80 \[ x e^{1 + \frac {1}{\log {\relax (x )}}} \]

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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