3.62.43 \(\int \frac {-248+20 x-20 x \log (x)+x \log ^2(x)-100 \log ^2(\frac {9}{\log (5)})-10 \log ^4(\frac {9}{\log (5)})}{x \log ^2(x)} \, dx\)

Optimal. Leaf size=31 \[ x-\frac {2 \left (1+5 \left (2 x-\left (5+\log ^2\left (\frac {9}{\log (5)}\right )\right )^2\right )\right )}{\log (x)} \]

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Rubi [A]  time = 0.36, antiderivative size = 39, normalized size of antiderivative = 1.26, number of steps used = 10, number of rules used = 7, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {6741, 6742, 2298, 2353, 2297, 2302, 30} \begin {gather*} x+\frac {2 \left (124+5 \log ^4\left (\frac {9}{\log (5)}\right )+50 \log ^2\left (\frac {9}{\log (5)}\right )\right )}{\log (x)}-\frac {20 x}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-248 + 20*x - 20*x*Log[x] + x*Log[x]^2 - 100*Log[9/Log[5]]^2 - 10*Log[9/Log[5]]^4)/(x*Log[x]^2),x]

[Out]

x - (20*x)/Log[x] + (2*(124 + 50*Log[9/Log[5]]^2 + 5*Log[9/Log[5]]^4))/Log[x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2353

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[
{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && IntegerQ[r
]))

Rule 6741

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

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 \frac {20 x-20 x \log (x)+x \log ^2(x)-248 \left (1+\frac {5}{124} \log ^2\left (\frac {9}{\log (5)}\right ) \left (10+\log ^2\left (\frac {9}{\log (5)}\right )\right )\right )}{x \log ^2(x)} \, dx\\ &=\int \left (1-\frac {20}{\log (x)}+\frac {2 \left (-124+10 x-50 \log ^2\left (\frac {9}{\log (5)}\right )-5 \log ^4\left (\frac {9}{\log (5)}\right )\right )}{x \log ^2(x)}\right ) \, dx\\ &=x+2 \int \frac {-124+10 x-50 \log ^2\left (\frac {9}{\log (5)}\right )-5 \log ^4\left (\frac {9}{\log (5)}\right )}{x \log ^2(x)} \, dx-20 \int \frac {1}{\log (x)} \, dx\\ &=x-20 \text {li}(x)+2 \int \left (\frac {10}{\log ^2(x)}+\frac {-124-50 \log ^2\left (\frac {9}{\log (5)}\right )-5 \log ^4\left (\frac {9}{\log (5)}\right )}{x \log ^2(x)}\right ) \, dx\\ &=x-20 \text {li}(x)+20 \int \frac {1}{\log ^2(x)} \, dx-\left (2 \left (124+50 \log ^2\left (\frac {9}{\log (5)}\right )+5 \log ^4\left (\frac {9}{\log (5)}\right )\right )\right ) \int \frac {1}{x \log ^2(x)} \, dx\\ &=x-\frac {20 x}{\log (x)}-20 \text {li}(x)+20 \int \frac {1}{\log (x)} \, dx-\left (2 \left (124+50 \log ^2\left (\frac {9}{\log (5)}\right )+5 \log ^4\left (\frac {9}{\log (5)}\right )\right )\right ) \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (x)\right )\\ &=x-\frac {20 x}{\log (x)}+\frac {2 \left (124+50 \log ^2\left (\frac {9}{\log (5)}\right )+5 \log ^4\left (\frac {9}{\log (5)}\right )\right )}{\log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 45, normalized size = 1.45 \begin {gather*} x+\frac {248}{\log (x)}-\frac {20 x}{\log (x)}+\frac {100 \log ^2\left (\frac {9}{\log (5)}\right )}{\log (x)}+\frac {10 \log ^4\left (\frac {9}{\log (5)}\right )}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-248 + 20*x - 20*x*Log[x] + x*Log[x]^2 - 100*Log[9/Log[5]]^2 - 10*Log[9/Log[5]]^4)/(x*Log[x]^2),x]

[Out]

x + 248/Log[x] - (20*x)/Log[x] + (100*Log[9/Log[5]]^2)/Log[x] + (10*Log[9/Log[5]]^4)/Log[x]

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fricas [A]  time = 0.91, size = 36, normalized size = 1.16 \begin {gather*} \frac {10 \, \log \left (\frac {9}{\log \relax (5)}\right )^{4} + x \log \relax (x) + 100 \, \log \left (\frac {9}{\log \relax (5)}\right )^{2} - 20 \, x + 248}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*log(x)^2-20*x*log(x)-10*log(9/log(5))^4-100*log(9/log(5))^2+20*x-248)/x/log(x)^2,x, algorithm="fr
icas")

[Out]

(10*log(9/log(5))^4 + x*log(x) + 100*log(9/log(5))^2 - 20*x + 248)/log(x)

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giac [B]  time = 0.16, size = 75, normalized size = 2.42 \begin {gather*} x + \frac {2 \, {\left (80 \, \log \relax (3)^{4} - 160 \, \log \relax (3)^{3} \log \left (\log \relax (5)\right ) + 120 \, \log \relax (3)^{2} \log \left (\log \relax (5)\right )^{2} - 40 \, \log \relax (3) \log \left (\log \relax (5)\right )^{3} + 5 \, \log \left (\log \relax (5)\right )^{4} + 200 \, \log \relax (3)^{2} - 200 \, \log \relax (3) \log \left (\log \relax (5)\right ) + 50 \, \log \left (\log \relax (5)\right )^{2} - 10 \, x + 124\right )}}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*log(x)^2-20*x*log(x)-10*log(9/log(5))^4-100*log(9/log(5))^2+20*x-248)/x/log(x)^2,x, algorithm="gi
ac")

[Out]

x + 2*(80*log(3)^4 - 160*log(3)^3*log(log(5)) + 120*log(3)^2*log(log(5))^2 - 40*log(3)*log(log(5))^3 + 5*log(l
og(5))^4 + 200*log(3)^2 - 200*log(3)*log(log(5)) + 50*log(log(5))^2 - 10*x + 124)/log(x)

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maple [B]  time = 0.07, size = 76, normalized size = 2.45

method	result	size
risch	\(x +\frac {10 \ln \left (\ln \relax (5)\right )^{4}-80 \ln \relax (3) \ln \left (\ln \relax (5)\right )^{3}+240 \ln \relax (3)^{2} \ln \left (\ln \relax (5)\right )^{2}-320 \ln \relax (3)^{3} \ln \left (\ln \relax (5)\right )+160 \ln \relax (3)^{4}+100 \ln \left (\ln \relax (5)\right )^{2}-400 \ln \relax (3) \ln \left (\ln \relax (5)\right )+400 \ln \relax (3)^{2}-20 x +248}{\ln \relax (x )}\)	\(76\)
norman	\(\frac {x \ln \relax (x )-20 x +248+10 \ln \left (\ln \relax (5)\right )^{4}-80 \ln \relax (3) \ln \left (\ln \relax (5)\right )^{3}+240 \ln \relax (3)^{2} \ln \left (\ln \relax (5)\right )^{2}-320 \ln \relax (3)^{3} \ln \left (\ln \relax (5)\right )+160 \ln \relax (3)^{4}+100 \ln \left (\ln \relax (5)\right )^{2}-400 \ln \relax (3) \ln \left (\ln \relax (5)\right )+400 \ln \relax (3)^{2}}{\ln \relax (x )}\)	\(77\)
default	\(x +\frac {160 \ln \relax (3)^{4}}{\ln \relax (x )}-\frac {320 \ln \relax (3)^{3} \ln \left (\ln \relax (5)\right )}{\ln \relax (x )}+\frac {240 \ln \relax (3)^{2} \ln \left (\ln \relax (5)\right )^{2}}{\ln \relax (x )}-\frac {80 \ln \relax (3) \ln \left (\ln \relax (5)\right )^{3}}{\ln \relax (x )}+\frac {10 \ln \left (\ln \relax (5)\right )^{4}}{\ln \relax (x )}+\frac {400 \ln \relax (3)^{2}}{\ln \relax (x )}-\frac {400 \ln \relax (3) \ln \left (\ln \relax (5)\right )}{\ln \relax (x )}+\frac {100 \ln \left (\ln \relax (5)\right )^{2}}{\ln \relax (x )}-\frac {20 x}{\ln \relax (x )}+\frac {248}{\ln \relax (x )}\)	\(110\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*ln(x)^2-20*x*ln(x)-10*ln(9/ln(5))^4-100*ln(9/ln(5))^2+20*x-248)/x/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

x+2*(5*ln(ln(5))^4-40*ln(3)*ln(ln(5))^3+120*ln(3)^2*ln(ln(5))^2-160*ln(3)^3*ln(ln(5))+80*ln(3)^4+50*ln(ln(5))^
2-200*ln(3)*ln(ln(5))+200*ln(3)^2-10*x+124)/ln(x)

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maxima [C]  time = 0.37, size = 51, normalized size = 1.65 \begin {gather*} \frac {10 \, \log \left (\frac {9}{\log \relax (5)}\right )^{4}}{\log \relax (x)} + x + \frac {100 \, \log \left (\frac {9}{\log \relax (5)}\right )^{2}}{\log \relax (x)} + \frac {248}{\log \relax (x)} - 20 \, {\rm Ei}\left (\log \relax (x)\right ) + 20 \, \Gamma \left (-1, -\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*log(x)^2-20*x*log(x)-10*log(9/log(5))^4-100*log(9/log(5))^2+20*x-248)/x/log(x)^2,x, algorithm="ma
xima")

[Out]

10*log(9/log(5))^4/log(x) + x + 100*log(9/log(5))^2/log(x) + 248/log(x) - 20*Ei(log(x)) + 20*gamma(-1, -log(x)
)

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mupad [B]  time = 4.31, size = 34, normalized size = 1.10 \begin {gather*} x+\frac {100\,{\ln \left (\frac {9}{\ln \relax (5)}\right )}^2-20\,x+10\,{\ln \left (\frac {9}{\ln \relax (5)}\right )}^4+248}{\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(20*x*log(x) - x*log(x)^2 - 20*x + 100*log(9/log(5))^2 + 10*log(9/log(5))^4 + 248)/(x*log(x)^2),x)

[Out]

x + (100*log(9/log(5))^2 - 20*x + 10*log(9/log(5))^4 + 248)/log(x)

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sympy [B]  time = 0.12, size = 87, normalized size = 2.81 \begin {gather*} x + \frac {- 20 x - 400 \log {\relax (3 )} \log {\left (\log {\relax (5 )} \right )} - 320 \log {\relax (3 )}^{3} \log {\left (\log {\relax (5 )} \right )} - 80 \log {\relax (3 )} \log {\left (\log {\relax (5 )} \right )}^{3} + 10 \log {\left (\log {\relax (5 )} \right )}^{4} + 100 \log {\left (\log {\relax (5 )} \right )}^{2} + 240 \log {\relax (3 )}^{2} \log {\left (\log {\relax (5 )} \right )}^{2} + 160 \log {\relax (3 )}^{4} + 248 + 400 \log {\relax (3 )}^{2}}{\log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*ln(x)**2-20*x*ln(x)-10*ln(9/ln(5))**4-100*ln(9/ln(5))**2+20*x-248)/x/ln(x)**2,x)

[Out]

x + (-20*x - 400*log(3)*log(log(5)) - 320*log(3)**3*log(log(5)) - 80*log(3)*log(log(5))**3 + 10*log(log(5))**4
+ 100*log(log(5))**2 + 240*log(3)**2*log(log(5))**2 + 160*log(3)**4 + 248 + 400*log(3)**2)/log(x)

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