3.44.51 \(\int \frac {e^{\frac {1}{12} (x^2-\log (x))} (-1+2 x^2)}{12 x} \, dx\) [4351]

Optimal. Leaf size=14 \[ e^{\frac {1}{12} \left (x^2-\log (x)\right )} \]

[Out]

1/exp(1/12*ln(x)-1/12*x^2)

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Rubi [A]
time = 0.10, antiderivative size = 15, normalized size of antiderivative = 1.07, number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {12, 2306, 2326} \begin {gather*} \frac {e^{\frac {x^2}{12}}}{\sqrt [12]{x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((x^2 - Log[x])/12)*(-1 + 2*x^2))/(12*x),x]

[Out]

E^(x^2/12)/x^(1/12)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2306

Int[(u_.)*(F_)^((a_.)*(Log[z_]*(b_.) + (v_.))), x_Symbol] :> Int[u*F^(a*v)*z^(a*b*Log[F]), x] /; FreeQ[{F, a,
b}, 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 {gather*} \begin {aligned} \text {integral} &=\frac {1}{12} \int \frac {e^{\frac {1}{12} \left (x^2-\log (x)\right )} \left (-1+2 x^2\right )}{x} \, dx\\ &=\frac {1}{12} \int \frac {e^{\frac {x^2}{12}} \left (-1+2 x^2\right )}{x^{13/12}} \, dx\\ &=\frac {e^{\frac {x^2}{12}}}{\sqrt [12]{x}}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 1.12, size = 15, normalized size = 1.07 \begin {gather*} \frac {e^{\frac {x^2}{12}}}{\sqrt [12]{x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((x^2 - Log[x])/12)*(-1 + 2*x^2))/(12*x),x]

[Out]

E^(x^2/12)/x^(1/12)

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Maple [A]
time = 0.60, size = 11, normalized size = 0.79

method	result	size
risch	\(\frac {{\mathrm e}^{\frac {x^{2}}{12}}}{x^{\frac {1}{12}}}\)	\(11\)
gosper	\({\mathrm e}^{\ln \left (\frac {1}{x^{\frac {1}{12}}}\right )+\frac {x^{2}}{12}}\)	\(14\)
norman	\({\mathrm e}^{\ln \left (\frac {1}{x^{\frac {1}{12}}}\right )+\frac {x^{2}}{12}}\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/12*(2*x^2-1)/x/exp(1/12*ln(x)-1/12*x^2),x,method=_RETURNVERBOSE)

[Out]

1/x^(1/12)*exp(1/12*x^2)

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Maxima [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.33, size = 45, normalized size = 3.21 \begin {gather*} -\frac {\left (\frac {1}{12}\right )^{\frac {1}{24}} x^{\frac {23}{12}} \Gamma \left (\frac {23}{24}, -\frac {1}{12} \, x^{2}\right )}{\left (-x^{2}\right )^{\frac {23}{24}}} + \frac {\left (\frac {1}{12}\right )^{\frac {1}{24}} \left (-x^{2}\right )^{\frac {1}{24}} \Gamma \left (-\frac {1}{24}, -\frac {1}{12} \, x^{2}\right )}{24 \, x^{\frac {1}{12}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/12*(2*x^2-1)/x/exp(1/12*log(x)-1/12*x^2),x, algorithm="maxima")

[Out]

-(1/12)^(1/24)*x^(23/12)*gamma(23/24, -1/12*x^2)/(-x^2)^(23/24) + 1/24*(1/12)^(1/24)*(-x^2)^(1/24)*gamma(-1/24
, -1/12*x^2)/x^(1/12)

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Fricas [A]
time = 0.41, size = 11, normalized size = 0.79 \begin {gather*} e^{\left (\frac {1}{12} \, x^{2} - \frac {1}{12} \, \log \left (x\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/12*(2*x^2-1)/x/exp(1/12*log(x)-1/12*x^2),x, algorithm="fricas")

[Out]

e^(1/12*x^2 - 1/12*log(x))

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/12*(2*x**2-1)/x/exp(1/12*ln(x)-1/12*x**2),x)

[Out]

Timed out

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Giac [A]
time = 0.41, size = 11, normalized size = 0.79 \begin {gather*} e^{\left (\frac {1}{12} \, x^{2} - \frac {1}{12} \, \log \left (x\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/12*(2*x^2-1)/x/exp(1/12*log(x)-1/12*x^2),x, algorithm="giac")

[Out]

e^(1/12*x^2 - 1/12*log(x))

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Mupad [B]
time = 3.18, size = 10, normalized size = 0.71 \begin {gather*} \frac {{\mathrm {e}}^{\frac {x^2}{12}}}{x^{1/12}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x^2/12 - log(x)/12)*(x^2/6 - 1/12))/x,x)

[Out]

exp(x^2/12)/x^(1/12)

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