3.95.58 \(\int \frac {-5 \log (256)+5 \log (256) \log (x)+(-14+10 x+5 \log (4)) \log ^2(x)}{5 \log ^2(x)} \, dx\)

Optimal. Leaf size=16 \[ x \left (-\frac {14}{5}+x+\log (4)+\frac {\log (256)}{\log (x)}\right ) \]

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Rubi [A]  time = 0.07, antiderivative size = 23, normalized size of antiderivative = 1.44, number of steps used = 6, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {12, 6742, 2297, 2298} \begin {gather*} x^2+\frac {x \log (256)}{\log (x)}-\frac {2}{5} x (7-5 \log (2)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-5*Log[256] + 5*Log[256]*Log[x] + (-14 + 10*x + 5*Log[4])*Log[x]^2)/(5*Log[x]^2),x]

[Out]

x^2 - (2*x*(7 - 5*Log[2]))/5 + (x*Log[256])/Log[x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2297

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2298

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, 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} &=\frac {1}{5} \int \frac {-5 \log (256)+5 \log (256) \log (x)+(-14+10 x+5 \log (4)) \log ^2(x)}{\log ^2(x)} \, dx\\ &=\frac {1}{5} \int \left (10 x-14 \left (1-\frac {5 \log (2)}{7}\right )-\frac {5 \log (256)}{\log ^2(x)}+\frac {5 \log (256)}{\log (x)}\right ) \, dx\\ &=x^2-\frac {2}{5} x (7-5 \log (2))-\log (256) \int \frac {1}{\log ^2(x)} \, dx+\log (256) \int \frac {1}{\log (x)} \, dx\\ &=x^2-\frac {2}{5} x (7-5 \log (2))+\frac {x \log (256)}{\log (x)}+\log (256) \text {li}(x)-\log (256) \int \frac {1}{\log (x)} \, dx\\ &=x^2-\frac {2}{5} x (7-5 \log (2))+\frac {x \log (256)}{\log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 21, normalized size = 1.31 \begin {gather*} -\frac {14 x}{5}+x^2+x \log (4)+\frac {x \log (256)}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5*Log[256] + 5*Log[256]*Log[x] + (-14 + 10*x + 5*Log[4])*Log[x]^2)/(5*Log[x]^2),x]

[Out]

(-14*x)/5 + x^2 + x*Log[4] + (x*Log[256])/Log[x]

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fricas [A]  time = 0.63, size = 29, normalized size = 1.81 \begin {gather*} \frac {40 \, x \log \relax (2) + {\left (5 \, x^{2} + 10 \, x \log \relax (2) - 14 \, x\right )} \log \relax (x)}{5 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((10*log(2)+10*x-14)*log(x)^2+40*log(2)*log(x)-40*log(2))/log(x)^2,x, algorithm="fricas")

[Out]

1/5*(40*x*log(2) + (5*x^2 + 10*x*log(2) - 14*x)*log(x))/log(x)

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giac [A]  time = 0.17, size = 21, normalized size = 1.31 \begin {gather*} x^{2} + 2 \, x \log \relax (2) - \frac {14}{5} \, x + \frac {8 \, x \log \relax (2)}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((10*log(2)+10*x-14)*log(x)^2+40*log(2)*log(x)-40*log(2))/log(x)^2,x, algorithm="giac")

[Out]

x^2 + 2*x*log(2) - 14/5*x + 8*x*log(2)/log(x)

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maple [A]  time = 0.04, size = 22, normalized size = 1.38

method	result	size
risch	\(2 x \ln \relax (2)+x^{2}-\frac {14 x}{5}+\frac {8 x \ln \relax (2)}{\ln \relax (x )}\)	\(22\)
norman	\(\frac {x^{2} \ln \relax (x )+\left (2 \ln \relax (2)-\frac {14}{5}\right ) x \ln \relax (x )+8 x \ln \relax (2)}{\ln \relax (x )}\)	\(28\)
default	\(2 x \ln \relax (2)+x^{2}-\frac {14 x}{5}-8 \ln \relax (2) \expIntegralEi \left (1, -\ln \relax (x )\right )-8 \ln \relax (2) \left (-\frac {x}{\ln \relax (x )}-\expIntegralEi \left (1, -\ln \relax (x )\right )\right )\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((10*ln(2)+10*x-14)*ln(x)^2+40*ln(2)*ln(x)-40*ln(2))/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

2*x*ln(2)+x^2-14/5*x+8*x*ln(2)/ln(x)

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maxima [C]  time = 0.37, size = 29, normalized size = 1.81 \begin {gather*} x^{2} + 2 \, x \log \relax (2) + 8 \, {\rm Ei}\left (\log \relax (x)\right ) \log \relax (2) - 8 \, \Gamma \left (-1, -\log \relax (x)\right ) \log \relax (2) - \frac {14}{5} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((10*log(2)+10*x-14)*log(x)^2+40*log(2)*log(x)-40*log(2))/log(x)^2,x, algorithm="maxima")

[Out]

x^2 + 2*x*log(2) + 8*Ei(log(x))*log(2) - 8*gamma(-1, -log(x))*log(2) - 14/5*x

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mupad [B]  time = 5.45, size = 22, normalized size = 1.38 \begin {gather*} \frac {x\,\left (5\,x+10\,\ln \relax (2)-14\right )}{5}+\frac {8\,x\,\ln \relax (2)}{\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((log(x)^2*(10*x + 10*log(2) - 14))/5 - 8*log(2) + 8*log(2)*log(x))/log(x)^2,x)

[Out]

(x*(5*x + 10*log(2) - 14))/5 + (8*x*log(2))/log(x)

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sympy [A]  time = 0.11, size = 22, normalized size = 1.38 \begin {gather*} x^{2} + x \left (- \frac {14}{5} + 2 \log {\relax (2 )}\right ) + \frac {8 x \log {\relax (2 )}}{\log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((10*ln(2)+10*x-14)*ln(x)**2+40*ln(2)*ln(x)-40*ln(2))/ln(x)**2,x)

[Out]

x**2 + x*(-14/5 + 2*log(2)) + 8*x*log(2)/log(x)

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