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

Optimal. Leaf size=22 \[ \log (8+e) \left (1+x+\frac {-1+x+\log \left (\frac {\log (4)}{x}\right )}{x}\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 31, normalized size of antiderivative = 1.41, number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {14, 2304} \begin {gather*} x \log (8+e)+\frac {\log (8+e) \log \left (\frac {\log (4)}{x}\right )}{x}-\frac {\log (8+e)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(Log[8 + E]/x) + x*Log[8 + E] + (Log[8 + E]*Log[Log[4]/x])/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 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} &=\int \left (\log (8+e)-\frac {\log (8+e) \log \left (\frac {\log (4)}{x}\right )}{x^2}\right ) \, dx\\ &=x \log (8+e)-\log (8+e) \int \frac {\log \left (\frac {\log (4)}{x}\right )}{x^2} \, dx\\ &=-\frac {\log (8+e)}{x}+x \log (8+e)+\frac {\log (8+e) \log \left (\frac {\log (4)}{x}\right )}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 31, normalized size = 1.41 \begin {gather*} -\frac {\log (8+e)}{x}+x \log (8+e)+\frac {\log (8+e) \log \left (\frac {\log (4)}{x}\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(Log[8 + E]/x) + x*Log[8 + E] + (Log[8 + E]*Log[Log[4]/x])/x

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fricas [A]  time = 0.84, size = 30, normalized size = 1.36 \begin {gather*} \frac {{\left (x^{2} - 1\right )} \log \left (e + 8\right ) + \log \left (\frac {2 \, \log \relax (2)}{x}\right ) \log \left (e + 8\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((x^2 - 1)*log(e + 8) + log(2*log(2)/x)*log(e + 8))/x

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giac [B]  time = 0.43, size = 52, normalized size = 2.36 \begin {gather*} \frac {{\left (\log \relax (2)^{2} \log \left (e + 8\right ) + \frac {\log \relax (2)^{2} \log \left (\frac {2 \, \log \relax (2)}{x}\right ) \log \left (e + 8\right )}{x^{2}} - \frac {\log \relax (2)^{2} \log \left (e + 8\right )}{x^{2}}\right )} x}{\log \relax (2)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(2)^2*log(e + 8) + log(2)^2*log(2*log(2)/x)*log(e + 8)/x^2 - log(2)^2*log(e + 8)/x^2)*x/log(2)^2

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maple [A]  time = 0.29, size = 33, normalized size = 1.50

method	result	size
risch	\(\frac {\ln \left ({\mathrm e}+8\right ) \ln \left (\frac {2 \ln \relax (2)}{x}\right )}{x}+\frac {\ln \left ({\mathrm e}+8\right ) \left (x^{2}-1\right )}{x}\)	\(33\)
derivativedivides	\(\frac {\ln \left ({\mathrm e}+8\right ) \ln \left (\frac {2 \ln \relax (2)}{x}\right )}{x}-\frac {\ln \left ({\mathrm e}+8\right )}{x}+x \ln \left ({\mathrm e}+8\right )\)	\(36\)
default	\(\frac {\ln \left ({\mathrm e}+8\right ) \ln \left (\frac {2 \ln \relax (2)}{x}\right )}{x}-\frac {\ln \left ({\mathrm e}+8\right )}{x}+x \ln \left ({\mathrm e}+8\right )\)	\(36\)
norman	\(\frac {x^{2} \ln \left ({\mathrm e}+8\right )+\ln \left ({\mathrm e}+8\right ) \ln \left (\frac {2 \ln \relax (2)}{x}\right )-\ln \left ({\mathrm e}+8\right )}{x}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(exp(1)+8)/x*ln(2*ln(2)/x)+ln(exp(1)+8)*(x^2-1)/x

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maxima [A]  time = 0.36, size = 40, normalized size = 1.82 \begin {gather*} x \log \left (e + 8\right ) + \frac {{\left (\frac {\log \relax (2) \log \left (\frac {2 \, \log \relax (2)}{x}\right )}{x} - \frac {\log \relax (2)}{x}\right )} \log \left (e + 8\right )}{\log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*log(e + 8) + (log(2)*log(2*log(2)/x)/x - log(2)/x)*log(e + 8)/log(2)

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mupad [B]  time = 0.35, size = 22, normalized size = 1.00 \begin {gather*} \frac {\ln \left (\mathrm {e}+8\right )\,\left (\ln \left (\frac {2\,\ln \relax (2)}{x}\right )+x^2-1\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.15, size = 32, normalized size = 1.45 \begin {gather*} x \log {\left (e + 8 \right )} + \frac {\log {\left (\frac {2 \log {\relax (2 )}}{x} \right )} \log {\left (e + 8 \right )}}{x} - \frac {\log {\left (e + 8 \right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*log(E + 8) + log(2*log(2)/x)*log(E + 8)/x - log(E + 8)/x

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