3.44.2 \(\int \frac {x \log (x^2)+\frac {9 x^2 (-2+2 \log (x^2))}{5 e^5 \log (x^2)}}{x \log (x^2)} \, dx\)

Optimal. Leaf size=18 \[ x+\frac {9 x^2}{5 e^5 \log \left (x^2\right )} \]

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Rubi [A]  time = 0.13, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.098, Rules used = {6688, 2306, 2307, 2298} \begin {gather*} \frac {9 x^2}{5 e^5 \log \left (x^2\right )}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x*Log[x^2] + (9*x^2*(-2 + 2*Log[x^2]))/(5*E^5*Log[x^2]))/(x*Log[x^2]),x]

[Out]

x + (9*x^2)/(5*E^5*Log[x^2])

Rule 2298

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

Rule 2306

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

Rule 2307

Int[(x_)^(m_.)/Log[(c_.)*(x_)^(n_)], x_Symbol] :> Dist[1/n, Subst[Int[1/Log[c*x], x], x, x^n], x] /; FreeQ[{c,
m, n}, x] && EqQ[m, n - 1]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (1-\frac {18 x}{5 e^5 \log ^2\left (x^2\right )}+\frac {18 x}{5 e^5 \log \left (x^2\right )}\right ) \, dx\\ &=x-\frac {18 \int \frac {x}{\log ^2\left (x^2\right )} \, dx}{5 e^5}+\frac {18 \int \frac {x}{\log \left (x^2\right )} \, dx}{5 e^5}\\ &=x+\frac {9 x^2}{5 e^5 \log \left (x^2\right )}+\frac {9 \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,x^2\right )}{5 e^5}-\frac {18 \int \frac {x}{\log \left (x^2\right )} \, dx}{5 e^5}\\ &=x+\frac {9 x^2}{5 e^5 \log \left (x^2\right )}+\frac {9 \text {li}\left (x^2\right )}{5 e^5}-\frac {9 \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,x^2\right )}{5 e^5}\\ &=x+\frac {9 x^2}{5 e^5 \log \left (x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 18, normalized size = 1.00 \begin {gather*} x+\frac {9 x^2}{5 e^5 \log \left (x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x*Log[x^2] + (9*x^2*(-2 + 2*Log[x^2]))/(5*E^5*Log[x^2]))/(x*Log[x^2]),x]

[Out]

x + (9*x^2)/(5*E^5*Log[x^2])

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fricas [A]  time = 0.89, size = 26, normalized size = 1.44 \begin {gather*} \frac {3 \, x^{2} e^{\left (\log \relax (3) - 5\right )} + 5 \, x \log \left (x^{2}\right )}{5 \, \log \left (x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*log(x^2)-2)*exp(-log(5/3*log(x^2)/x^2)+log(3)-5)+x*log(x^2))/x/log(x^2),x, algorithm="fricas")

[Out]

1/5*(3*x^2*e^(log(3) - 5) + 5*x*log(x^2))/log(x^2)

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giac [A]  time = 0.16, size = 20, normalized size = 1.11 \begin {gather*} \frac {9 \, x^{2} e^{\left (-5\right )} \mathrm {sgn}\relax (x)^{2}}{10 \, \log \left (x \mathrm {sgn}\relax (x)\right )} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*log(x^2)-2)*exp(-log(5/3*log(x^2)/x^2)+log(3)-5)+x*log(x^2))/x/log(x^2),x, algorithm="giac")

[Out]

9/10*x^2*e^(-5)*sgn(x)^2/log(x*sgn(x)) + x

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maple [A]  time = 0.07, size = 18, normalized size = 1.00

method	result	size
default	\(x +\frac {9 \,{\mathrm e}^{-5} x^{2}}{5 \ln \left (x^{2}\right )}\)	\(18\)
norman	\(\frac {x \ln \left (x^{2}\right )+\frac {9 x^{2} {\mathrm e}^{-5}}{5}}{\ln \left (x^{2}\right )}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*ln(x^2)-2)*exp(-ln(5/3*ln(x^2)/x^2)+ln(3)-5)+x*ln(x^2))/x/ln(x^2),x,method=_RETURNVERBOSE)

[Out]

x+9/5/exp(5)*x^2/ln(x^2)

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maxima [A]  time = 0.38, size = 13, normalized size = 0.72 \begin {gather*} \frac {9 \, x^{2} e^{\left (-5\right )}}{10 \, \log \relax (x)} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*log(x^2)-2)*exp(-log(5/3*log(x^2)/x^2)+log(3)-5)+x*log(x^2))/x/log(x^2),x, algorithm="maxima")

[Out]

9/10*x^2*e^(-5)/log(x) + x

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mupad [B]  time = 3.10, size = 15, normalized size = 0.83 \begin {gather*} x+\frac {9\,x^2\,{\mathrm {e}}^{-5}}{5\,\ln \left (x^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*log(x^2) + exp(log(3) - log((5*log(x^2))/(3*x^2)) - 5)*(2*log(x^2) - 2))/(x*log(x^2)),x)

[Out]

x + (9*x^2*exp(-5))/(5*log(x^2))

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sympy [A]  time = 0.10, size = 15, normalized size = 0.83 \begin {gather*} \frac {9 x^{2}}{5 e^{5} \log {\left (x^{2} \right )}} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*ln(x**2)-2)*exp(-ln(5/3*ln(x**2)/x**2)+ln(3)-5)+x*ln(x**2))/x/ln(x**2),x)

[Out]

9*x**2*exp(-5)/(5*log(x**2)) + x

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