3.22.9 \(\int \frac {-1+(1-15 x+2 x^2-8 e^{x^2} x^2) \log (x)}{x \log (x)} \, dx\)

Optimal. Leaf size=26 \[ 2-4 e^{x^2}+(-7+x)^2-x-\log \left (\frac {\log (x)}{x}\right ) \]

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Rubi [A]  time = 0.41, antiderivative size = 21, normalized size of antiderivative = 0.81, number of steps used = 7, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {6742, 2209, 6688, 2302, 29} \begin {gather*} x^2-4 e^{x^2}-15 x+\log (x)-\log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 + (1 - 15*x + 2*x^2 - 8*E^x^2*x^2)*Log[x])/(x*Log[x]),x]

[Out]

-4*E^x^2 - 15*x + x^2 + Log[x] - Log[Log[x]]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-8 e^{x^2} x+\frac {-1+\log (x)-15 x \log (x)+2 x^2 \log (x)}{x \log (x)}\right ) \, dx\\ &=-\left (8 \int e^{x^2} x \, dx\right )+\int \frac {-1+\log (x)-15 x \log (x)+2 x^2 \log (x)}{x \log (x)} \, dx\\ &=-4 e^{x^2}+\int \left (-15+\frac {1}{x}+2 x-\frac {1}{x \log (x)}\right ) \, dx\\ &=-4 e^{x^2}-15 x+x^2+\log (x)-\int \frac {1}{x \log (x)} \, dx\\ &=-4 e^{x^2}-15 x+x^2+\log (x)-\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right )\\ &=-4 e^{x^2}-15 x+x^2+\log (x)-\log (\log (x))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 21, normalized size = 0.81 \begin {gather*} -4 e^{x^2}-15 x+x^2+\log (x)-\log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + (1 - 15*x + 2*x^2 - 8*E^x^2*x^2)*Log[x])/(x*Log[x]),x]

[Out]

-4*E^x^2 - 15*x + x^2 + Log[x] - Log[Log[x]]

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fricas [A]  time = 0.74, size = 20, normalized size = 0.77 \begin {gather*} x^{2} - 15 \, x - 4 \, e^{\left (x^{2}\right )} + \log \relax (x) - \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x^2*exp(x^2)+2*x^2-15*x+1)*log(x)-1)/x/log(x),x, algorithm="fricas")

[Out]

x^2 - 15*x - 4*e^(x^2) + log(x) - log(log(x))

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giac [A]  time = 0.24, size = 20, normalized size = 0.77 \begin {gather*} x^{2} - 15 \, x - 4 \, e^{\left (x^{2}\right )} + \log \relax (x) - \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x^2*exp(x^2)+2*x^2-15*x+1)*log(x)-1)/x/log(x),x, algorithm="giac")

[Out]

x^2 - 15*x - 4*e^(x^2) + log(x) - log(log(x))

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maple [A]  time = 0.02, size = 21, normalized size = 0.81

method	result	size
default	\(x^{2}-15 x +\ln \relax (x )-\ln \left (\ln \relax (x )\right )-4 \,{\mathrm e}^{x^{2}}\)	\(21\)
norman	\(x^{2}-15 x +\ln \relax (x )-\ln \left (\ln \relax (x )\right )-4 \,{\mathrm e}^{x^{2}}\)	\(21\)
risch	\(x^{2}-15 x +\ln \relax (x )-\ln \left (\ln \relax (x )\right )-4 \,{\mathrm e}^{x^{2}}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-8*x^2*exp(x^2)+2*x^2-15*x+1)*ln(x)-1)/x/ln(x),x,method=_RETURNVERBOSE)

[Out]

x^2-15*x+ln(x)-ln(ln(x))-4*exp(x^2)

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maxima [A]  time = 0.45, size = 20, normalized size = 0.77 \begin {gather*} x^{2} - 15 \, x - 4 \, e^{\left (x^{2}\right )} + \log \relax (x) - \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x^2*exp(x^2)+2*x^2-15*x+1)*log(x)-1)/x/log(x),x, algorithm="maxima")

[Out]

x^2 - 15*x - 4*e^(x^2) + log(x) - log(log(x))

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mupad [B]  time = 1.21, size = 20, normalized size = 0.77 \begin {gather*} \ln \relax (x)-4\,{\mathrm {e}}^{x^2}-\ln \left (\ln \relax (x)\right )-15\,x+x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x)*(15*x + 8*x^2*exp(x^2) - 2*x^2 - 1) + 1)/(x*log(x)),x)

[Out]

log(x) - 4*exp(x^2) - log(log(x)) - 15*x + x^2

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sympy [A]  time = 0.31, size = 20, normalized size = 0.77 \begin {gather*} x^{2} - 15 x - 4 e^{x^{2}} + \log {\relax (x )} - \log {\left (\log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x**2*exp(x**2)+2*x**2-15*x+1)*ln(x)-1)/x/ln(x),x)

[Out]

x**2 - 15*x - 4*exp(x**2) + log(x) - log(log(x))

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