3.52.29 \(\int \frac {x^2+(16 e^8+16 e^{2 e^5}-32 e^{4+e^5}) \log (x)+x^2 \log (x) \log (\log (x))}{x^2 \log (x)} \, dx\)

Optimal. Leaf size=24 \[ x \left (-\frac {16 \left (-e^4+e^{e^5}\right )^2}{x^2}+\log (\log (x))\right ) \]

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Rubi [A]  time = 0.30, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {6688, 2298, 14, 2520} \begin {gather*} x \log (\log (x))-\frac {16 \left (e^4-e^{e^5}\right )^2}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2 + (16*E^8 + 16*E^(2*E^5) - 32*E^(4 + E^5))*Log[x] + x^2*Log[x]*Log[Log[x]])/(x^2*Log[x]),x]

[Out]

(-16*(E^4 - E^E^5)^2)/x + x*Log[Log[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 2298

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

Rule 2520

Int[Log[Log[(d_.)*(x_)^(n_.)]^(p_.)*(c_.)], x_Symbol] :> Simp[x*Log[c*Log[d*x^n]^p], x] - Dist[n*p, Int[1/Log[
d*x^n], x], x] /; FreeQ[{c, d, n, p}, x]

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 (\frac {1}{\log (x)}+\frac {16 \left (e^4-e^{e^5}\right )^2+x^2 \log (\log (x))}{x^2}\right ) \, dx\\ &=\int \frac {1}{\log (x)} \, dx+\int \frac {16 \left (e^4-e^{e^5}\right )^2+x^2 \log (\log (x))}{x^2} \, dx\\ &=\text {li}(x)+\int \left (\frac {16 \left (-e^4+e^{e^5}\right )^2}{x^2}+\log (\log (x))\right ) \, dx\\ &=-\frac {16 \left (e^4-e^{e^5}\right )^2}{x}+\text {li}(x)+\int \log (\log (x)) \, dx\\ &=-\frac {16 \left (e^4-e^{e^5}\right )^2}{x}+x \log (\log (x))+\text {li}(x)-\int \frac {1}{\log (x)} \, dx\\ &=-\frac {16 \left (e^4-e^{e^5}\right )^2}{x}+x \log (\log (x))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.17, size = 27, normalized size = 1.12 \begin {gather*} \frac {-16 \left (e^4-e^{e^5}\right )^2+x^2 \log (\log (x))}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2 + (16*E^8 + 16*E^(2*E^5) - 32*E^(4 + E^5))*Log[x] + x^2*Log[x]*Log[Log[x]])/(x^2*Log[x]),x]

[Out]

(-16*(E^4 - E^E^5)^2 + x^2*Log[Log[x]])/x

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fricas [A]  time = 0.57, size = 36, normalized size = 1.50 \begin {gather*} \frac {{\left (x^{2} e^{8} \log \left (\log \relax (x)\right ) - 16 \, e^{16} - 16 \, e^{\left (2 \, e^{5} + 8\right )} + 32 \, e^{\left (e^{5} + 12\right )}\right )} e^{\left (-8\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*log(x)*log(log(x))+(16*exp(exp(5))^2-32*exp(4)*exp(exp(5))+16*exp(4)^2)*log(x)+x^2)/x^2/log(x),
x, algorithm="fricas")

[Out]

(x^2*e^8*log(log(x)) - 16*e^16 - 16*e^(2*e^5 + 8) + 32*e^(e^5 + 12))*e^(-8)/x

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giac [A]  time = 0.31, size = 30, normalized size = 1.25 \begin {gather*} \frac {x^{2} \log \left (\log \relax (x)\right ) - 16 \, e^{8} - 16 \, e^{\left (2 \, e^{5}\right )} + 32 \, e^{\left (e^{5} + 4\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*log(x)*log(log(x))+(16*exp(exp(5))^2-32*exp(4)*exp(exp(5))+16*exp(4)^2)*log(x)+x^2)/x^2/log(x),
x, algorithm="giac")

[Out]

(x^2*log(log(x)) - 16*e^8 - 16*e^(2*e^5) + 32*e^(e^5 + 4))/x

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maple [A]  time = 0.07, size = 33, normalized size = 1.38

method	result	size
default	\(-\frac {16 \,{\mathrm e}^{2 \,{\mathrm e}^{5}}-32 \,{\mathrm e}^{4} {\mathrm e}^{{\mathrm e}^{5}}+16 \,{\mathrm e}^{8}}{x}+x \ln \left (\ln \relax (x )\right )\)	\(33\)
norman	\(\frac {x^{2} \ln \left (\ln \relax (x )\right )-16 \,{\mathrm e}^{8}+32 \,{\mathrm e}^{4} {\mathrm e}^{{\mathrm e}^{5}}-16 \,{\mathrm e}^{2 \,{\mathrm e}^{5}}}{x}\)	\(33\)
risch	\(x \ln \left (\ln \relax (x )\right )-\frac {16 \,{\mathrm e}^{8}}{x}+\frac {32 \,{\mathrm e}^{4+{\mathrm e}^{5}}}{x}-\frac {16 \,{\mathrm e}^{2 \,{\mathrm e}^{5}}}{x}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*ln(x)*ln(ln(x))+(16*exp(exp(5))^2-32*exp(4)*exp(exp(5))+16*exp(4)^2)*ln(x)+x^2)/x^2/ln(x),x,method=_R
ETURNVERBOSE)

[Out]

-(16*exp(exp(5))^2-32*exp(4)*exp(exp(5))+16*exp(4)^2)/x+x*ln(ln(x))

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maxima [A]  time = 0.38, size = 33, normalized size = 1.38 \begin {gather*} x \log \left (\log \relax (x)\right ) - \frac {16 \, e^{8}}{x} - \frac {16 \, e^{\left (2 \, e^{5}\right )}}{x} + \frac {32 \, e^{\left (e^{5} + 4\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*log(x)*log(log(x))+(16*exp(exp(5))^2-32*exp(4)*exp(exp(5))+16*exp(4)^2)*log(x)+x^2)/x^2/log(x),
x, algorithm="maxima")

[Out]

x*log(log(x)) - 16*e^8/x - 16*e^(2*e^5)/x + 32*e^(e^5 + 4)/x

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mupad [B]  time = 3.34, size = 30, normalized size = 1.25 \begin {gather*} x\,\ln \left (\ln \relax (x)\right )-\frac {16\,{\mathrm {e}}^{2\,{\mathrm {e}}^5}-32\,{\mathrm {e}}^{{\mathrm {e}}^5+4}+16\,{\mathrm {e}}^8}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x)*(16*exp(2*exp(5)) + 16*exp(8) - 32*exp(4)*exp(exp(5))) + x^2 + x^2*log(log(x))*log(x))/(x^2*log(x)
),x)

[Out]

x*log(log(x)) - (16*exp(2*exp(5)) - 32*exp(exp(5) + 4) + 16*exp(8))/x

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sympy [A]  time = 0.30, size = 31, normalized size = 1.29 \begin {gather*} x \log {\left (\log {\relax (x )} \right )} - \frac {- 32 e^{4} e^{e^{5}} + 16 e^{8} + 16 e^{2 e^{5}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2*ln(x)*ln(ln(x))+(16*exp(exp(5))**2-32*exp(4)*exp(exp(5))+16*exp(4)**2)*ln(x)+x**2)/x**2/ln(x),
x)

[Out]

x*log(log(x)) - (-32*exp(4)*exp(exp(5)) + 16*exp(8) + 16*exp(2*exp(5)))/x

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