3.55.28 \(\int \frac {-2 x+2 x^2+(-2 x+4 x^2) \log (x)+e^{\frac {1}{2} (-8+\log (x))} (2 e^{x^2}+e^{x^2} (1+4 x^2) \log (x))}{2 x} \, dx\)

Optimal. Leaf size=25 \[ -1+\left (e^{-4+x^2} \sqrt {x}-x+x^2\right ) \log (x) \]

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Rubi [A]  time = 0.09, antiderivative size = 27, normalized size of antiderivative = 1.08, number of steps used = 7, number of rules used = 5, integrand size = 62, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.081, Rules used = {12, 14, 2295, 2304, 2288} \begin {gather*} x^2 \log (x)+e^{x^2-4} \sqrt {x} \log (x)-x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*x + 2*x^2 + (-2*x + 4*x^2)*Log[x] + E^((-8 + Log[x])/2)*(2*E^x^2 + E^x^2*(1 + 4*x^2)*Log[x]))/(2*x),x]

[Out]

E^(-4 + x^2)*Sqrt[x]*Log[x] - x*Log[x] + x^2*Log[x]

Rule 12

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \frac {-2 x+2 x^2+\left (-2 x+4 x^2\right ) \log (x)+e^{\frac {1}{2} (-8+\log (x))} \left (2 e^{x^2}+e^{x^2} \left (1+4 x^2\right ) \log (x)\right )}{x} \, dx\\ &=\frac {1}{2} \int \left (2 (-1+x-\log (x)+2 x \log (x))+\frac {e^{-4+x^2} \left (2+\log (x)+4 x^2 \log (x)\right )}{\sqrt {x}}\right ) \, dx\\ &=\frac {1}{2} \int \frac {e^{-4+x^2} \left (2+\log (x)+4 x^2 \log (x)\right )}{\sqrt {x}} \, dx+\int (-1+x-\log (x)+2 x \log (x)) \, dx\\ &=-x+\frac {x^2}{2}+e^{-4+x^2} \sqrt {x} \log (x)+2 \int x \log (x) \, dx-\int \log (x) \, dx\\ &=e^{-4+x^2} \sqrt {x} \log (x)-x \log (x)+x^2 \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 27, normalized size = 1.08 \begin {gather*} e^{-4+x^2} \sqrt {x} \log (x)-x \log (x)+x^2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*x + 2*x^2 + (-2*x + 4*x^2)*Log[x] + E^((-8 + Log[x])/2)*(2*E^x^2 + E^x^2*(1 + 4*x^2)*Log[x]))/(2
*x),x]

[Out]

E^(-4 + x^2)*Sqrt[x]*Log[x] - x*Log[x] + x^2*Log[x]

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fricas [A]  time = 0.70, size = 26, normalized size = 1.04 \begin {gather*} {\left ({\left (x^{2} - x\right )} e^{4} \log \relax (x) + \sqrt {x} e^{\left (x^{2}\right )} \log \relax (x)\right )} e^{\left (-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(((4*x^2+1)*exp(x^2)*log(x)+2*exp(x^2))*exp(1/4*log(x)-2)^2+(4*x^2-2*x)*log(x)+2*x^2-2*x)/x,x, a
lgorithm="fricas")

[Out]

((x^2 - x)*e^4*log(x) + sqrt(x)*e^(x^2)*log(x))*e^(-4)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, x^{2} + {\left ({\left (4 \, x^{2} + 1\right )} e^{\left (x^{2}\right )} \log \relax (x) + 2 \, e^{\left (x^{2}\right )}\right )} e^{\left (\frac {1}{2} \, \log \relax (x) - 4\right )} + 2 \, {\left (2 \, x^{2} - x\right )} \log \relax (x) - 2 \, x}{2 \, x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(((4*x^2+1)*exp(x^2)*log(x)+2*exp(x^2))*exp(1/4*log(x)-2)^2+(4*x^2-2*x)*log(x)+2*x^2-2*x)/x,x, a
lgorithm="giac")

[Out]

integrate(1/2*(2*x^2 + ((4*x^2 + 1)*e^(x^2)*log(x) + 2*e^(x^2))*e^(1/2*log(x) - 4) + 2*(2*x^2 - x)*log(x) - 2*
x)/x, x)

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maple [A]  time = 0.14, size = 27, normalized size = 1.08

method	result	size
default	\(x^{2} \ln \relax (x )-x \ln \relax (x )+\sqrt {x}\, {\mathrm e}^{\left (x -2\right ) \left (2+x \right )} \ln \relax (x )\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(((4*x^2+1)*exp(x^2)*ln(x)+2*exp(x^2))*exp(1/4*ln(x)-2)^2+(4*x^2-2*x)*ln(x)+2*x^2-2*x)/x,x,method=_RET
URNVERBOSE)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 2 \, e^{\left (-4\right )} \int x^{\frac {3}{2}} e^{\left (x^{2}\right )} \log \relax (x)\,{d x} - x \log \relax (x) + \frac {1}{2} \, \int 2 \, x {\left (2 \, \log \relax (x) + 1\right )}\,{d x} + \frac {1}{2} \, \int \frac {{\left (\log \relax (x) + 2\right )} e^{\left (x^{2} - 4\right )}}{\sqrt {x}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(((4*x^2+1)*exp(x^2)*log(x)+2*exp(x^2))*exp(1/4*log(x)-2)^2+(4*x^2-2*x)*log(x)+2*x^2-2*x)/x,x, a
lgorithm="maxima")

[Out]

2*e^(-4)*integrate(x^(3/2)*e^(x^2)*log(x), x) - x*log(x) + 1/2*integrate(2*x*(2*log(x) + 1), x) + 1/2*integrat
e((log(x) + 2)*e^(x^2 - 4)/sqrt(x), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} -\int \frac {x-\frac {{\mathrm {e}}^{\frac {\ln \relax (x)}{2}-4}\,\left (2\,{\mathrm {e}}^{x^2}+{\mathrm {e}}^{x^2}\,\ln \relax (x)\,\left (4\,x^2+1\right )\right )}{2}+\frac {\ln \relax (x)\,\left (2\,x-4\,x^2\right )}{2}-x^2}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x - (exp(log(x)/2 - 4)*(2*exp(x^2) + exp(x^2)*log(x)*(4*x^2 + 1)))/2 + (log(x)*(2*x - 4*x^2))/2 - x^2)/x
,x)

[Out]

-int((x - (exp(log(x)/2 - 4)*(2*exp(x^2) + exp(x^2)*log(x)*(4*x^2 + 1)))/2 + (log(x)*(2*x - 4*x^2))/2 - x^2)/x
, x)

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sympy [F(-1)]  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(1/2*(((4*x**2+1)*exp(x**2)*ln(x)+2*exp(x**2))*exp(1/4*ln(x)-2)**2+(4*x**2-2*x)*ln(x)+2*x**2-2*x)/x,x
)

[Out]

Timed out

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