3.45.100 \(\int \frac {-2+(2 x+e^{4 x^2} (1+8 x^2)) \log (x)-\log (x) \log (\frac {1}{12} x \log ^2(x))}{\log (x)} \, dx\)

Optimal. Leaf size=24 \[ x \left (1+e^{4 x^2}+x-\log \left (\frac {1}{12} x \log ^2(x)\right )\right ) \]

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Rubi [A]  time = 0.22, antiderivative size = 27, normalized size of antiderivative = 1.12, number of steps used = 15, number of rules used = 7, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {6742, 2226, 2204, 2212, 6688, 2298, 2549} \begin {gather*} x^2+e^{4 x^2} x+x-x \log \left (\frac {1}{12} x \log ^2(x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2 + (2*x + E^(4*x^2)*(1 + 8*x^2))*Log[x] - Log[x]*Log[(x*Log[x]^2)/12])/Log[x],x]

[Out]

x + E^(4*x^2)*x + x^2 - x*Log[(x*Log[x]^2)/12]

Rule 2204

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erfi[(c + d*x)*Rt[b*Log[F], 2
]])/(2*d*Rt[b*Log[F], 2]), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2212

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
- n + 1)*F^(a + b*(c + d*x)^n))/(b*d*n*Log[F]), x] - Dist[(m - n + 1)/(b*n*Log[F]), Int[(c + d*x)^(m - n)*F^(
a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[(2*(m + 1))/n] && LtQ[0, (m + 1)/n, 5] &&
IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n, 0])

Rule 2226

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*(u_), x_Symbol] :> Int[ExpandLinearProduct[F^(a + b*(c + d*
x)^n), u, c, d, x], x] /; FreeQ[{F, a, b, c, d, n}, x] && PolynomialQ[u, x]

Rule 2298

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

Rule 2549

Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[x*Simplify[D[u, x]/u], x], x] /; ProductQ[
u]

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 (e^{4 x^2} \left (1+8 x^2\right )+\frac {-2+2 x \log (x)-\log (x) \log \left (\frac {1}{12} x \log ^2(x)\right )}{\log (x)}\right ) \, dx\\ &=\int e^{4 x^2} \left (1+8 x^2\right ) \, dx+\int \frac {-2+2 x \log (x)-\log (x) \log \left (\frac {1}{12} x \log ^2(x)\right )}{\log (x)} \, dx\\ &=\int \left (e^{4 x^2}+8 e^{4 x^2} x^2\right ) \, dx+\int \left (\frac {2 (-1+x \log (x))}{\log (x)}-\log \left (\frac {1}{12} x \log ^2(x)\right )\right ) \, dx\\ &=2 \int \frac {-1+x \log (x)}{\log (x)} \, dx+8 \int e^{4 x^2} x^2 \, dx+\int e^{4 x^2} \, dx-\int \log \left (\frac {1}{12} x \log ^2(x)\right ) \, dx\\ &=e^{4 x^2} x+\frac {1}{4} \sqrt {\pi } \text {erfi}(2 x)-x \log \left (\frac {1}{12} x \log ^2(x)\right )+2 \int \left (x-\frac {1}{\log (x)}\right ) \, dx-\int e^{4 x^2} \, dx+\int \left (1+\frac {2}{\log (x)}\right ) \, dx\\ &=x+e^{4 x^2} x+x^2-x \log \left (\frac {1}{12} x \log ^2(x)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 27, normalized size = 1.12 \begin {gather*} x+e^{4 x^2} x+x^2-x \log \left (\frac {1}{12} x \log ^2(x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 + (2*x + E^(4*x^2)*(1 + 8*x^2))*Log[x] - Log[x]*Log[(x*Log[x]^2)/12])/Log[x],x]

[Out]

x + E^(4*x^2)*x + x^2 - x*Log[(x*Log[x]^2)/12]

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fricas [A]  time = 0.68, size = 24, normalized size = 1.00 \begin {gather*} x^{2} + x e^{\left (4 \, x^{2}\right )} - x \log \left (\frac {1}{12} \, x \log \relax (x)^{2}\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x)*log(1/12*x*log(x)^2)+((8*x^2+1)*exp(4*x^2)+2*x)*log(x)-2)/log(x),x, algorithm="fricas")

[Out]

x^2 + x*e^(4*x^2) - x*log(1/12*x*log(x)^2) + x

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giac [A]  time = 0.18, size = 30, normalized size = 1.25 \begin {gather*} x^{2} + x e^{\left (4 \, x^{2}\right )} + x \log \left (12\right ) - x \log \left (\log \relax (x)^{2}\right ) - x \log \relax (x) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x)*log(1/12*x*log(x)^2)+((8*x^2+1)*exp(4*x^2)+2*x)*log(x)-2)/log(x),x, algorithm="giac")

[Out]

x^2 + x*e^(4*x^2) + x*log(12) - x*log(log(x)^2) - x*log(x) + x

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maple [A]  time = 0.18, size = 28, normalized size = 1.17

method	result	size
default	\(x \,{\mathrm e}^{4 x^{2}}+x^{2}+\ln \left (12\right ) x -\ln \left (x \ln \relax (x )^{2}\right ) x +x\)	\(28\)
risch	\(-2 x \ln \left (\ln \relax (x )\right )-x \ln \relax (x )+\frac {i \pi \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )^{3} x}{2}-\frac {i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \ln \relax (x )^{2}\right )^{2} x}{2}-\frac {i \pi \,\mathrm {csgn}\left (i \ln \relax (x )^{2}\right ) \mathrm {csgn}\left (i x \ln \relax (x )^{2}\right )^{2} x}{2}+\frac {i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \ln \relax (x )^{2}\right ) \mathrm {csgn}\left (i x \ln \relax (x )^{2}\right ) x}{2}+\frac {i \pi \mathrm {csgn}\left (i \ln \relax (x )\right )^{2} \mathrm {csgn}\left (i \ln \relax (x )^{2}\right ) x}{2}-i \pi \,\mathrm {csgn}\left (i \ln \relax (x )\right ) \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )^{2} x +\frac {i \pi \mathrm {csgn}\left (i x \ln \relax (x )^{2}\right )^{3} x}{2}+2 x \ln \relax (2)+x \ln \relax (3)+x^{2}+x +x \,{\mathrm e}^{4 x^{2}}\)	\(179\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-ln(x)*ln(1/12*x*ln(x)^2)+((8*x^2+1)*exp(4*x^2)+2*x)*ln(x)-2)/ln(x),x,method=_RETURNVERBOSE)

[Out]

x*exp(x^2)^4+x^2+ln(12)*x-ln(x*ln(x)^2)*x+x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x)*log(1/12*x*log(x)^2)+((8*x^2+1)*exp(4*x^2)+2*x)*log(x)-2)/log(x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.37, size = 24, normalized size = 1.00 \begin {gather*} x-x\,\ln \left (\frac {x\,{\ln \relax (x)}^2}{12}\right )+x\,{\mathrm {e}}^{4\,x^2}+x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log((x*log(x)^2)/12)*log(x) - log(x)*(2*x + exp(4*x^2)*(8*x^2 + 1)) + 2)/log(x),x)

[Out]

x - x*log((x*log(x)^2)/12) + x*exp(4*x^2) + x^2

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sympy [A]  time = 0.44, size = 24, normalized size = 1.00 \begin {gather*} x^{2} + x e^{4 x^{2}} - x \log {\left (\frac {x \log {\relax (x )}^{2}}{12} \right )} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-ln(x)*ln(1/12*x*ln(x)**2)+((8*x**2+1)*exp(4*x**2)+2*x)*ln(x)-2)/ln(x),x)

[Out]

x**2 + x*exp(4*x**2) - x*log(x*log(x)**2/12) + x

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