3.69.13 \(\int \frac {-100 x^2+40 x^3-4 x^4+(200 x^2-120 x^3+16 x^4) \log (x)+(-1-76 x^2+24 x^3) \log ^2(x)-\log (3 e^{\frac {100 x^2-40 x^3+4 x^4+(-40 x^2+8 x^3) \log (x)+4 x^2 \log ^2(x)}{\log (x)}}) \log ^2(x)+8 x^2 \log ^3(x)}{x^2 \log ^2(x)} \, dx\)

Optimal. Leaf size=27 \[ \frac {1+\log \left (3 e^{\frac {4 x^2 (-5+x+\log (x))^2}{\log (x)}}\right )}{x} \]

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Rubi [A]  time = 0.67, antiderivative size = 51, normalized size of antiderivative = 1.89, number of steps used = 47, number of rules used = 11, integrand size = 124, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {6688, 2320, 2330, 2298, 2309, 2178, 2356, 2295, 30, 2555, 12} \begin {gather*} \frac {\log \left (3 e^{\frac {4 (5-x)^2 x^2}{\log (x)}} x^{4 x^2-\frac {8 (5-x) x^2}{\log (x)}}\right )}{x}+\frac {1}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-100*x^2 + 40*x^3 - 4*x^4 + (200*x^2 - 120*x^3 + 16*x^4)*Log[x] + (-1 - 76*x^2 + 24*x^3)*Log[x]^2 - Log[3
*E^((100*x^2 - 40*x^3 + 4*x^4 + (-40*x^2 + 8*x^3)*Log[x] + 4*x^2*Log[x]^2)/Log[x])]*Log[x]^2 + 8*x^2*Log[x]^3)
/(x^2*Log[x]^2),x]

[Out]

x^(-1) + Log[3*E^((4*(5 - x)^2*x^2)/Log[x])*x^(4*x^2 - (8*(5 - x)*x^2)/Log[x])]/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 30

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

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2295

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

Rule 2298

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

Rule 2309

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[E^((m + 1)*x)*(a
+ b*x)^p, x], x, Log[c*x]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[m]

Rule 2320

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[(x*(d + e*x)^q*(a
+ b*Log[c*x^n])^(p + 1))/(b*n*(p + 1)), x] + (-Dist[(q + 1)/(b*n*(p + 1)), Int[(d + e*x)^q*(a + b*Log[c*x^n])^
(p + 1), x], x] + Dist[(d*q)/(b*n*(p + 1)), Int[(d + e*x)^(q - 1)*(a + b*Log[c*x^n])^(p + 1), x], x]) /; FreeQ
[{a, b, c, d, e, n}, x] && LtQ[p, -1] && GtQ[q, 0]

Rule 2330

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

Rule 2356

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(Polyx_), x_Symbol] :> Int[ExpandIntegrand[Polyx*(a + b*Log[c*
x^n])^p, x], x] /; FreeQ[{a, b, c, n, p}, x] && PolynomialQ[Polyx, x]

Rule 2555

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[w*Simplify
[D[u, x]/u], x], x] /; InverseFunctionFreeQ[w, x]] /; ProductQ[u]

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 (-76-\frac {1}{x^2}+24 x-\frac {4 (-5+x)^2}{\log ^2(x)}+\frac {8 \left (25-15 x+2 x^2\right )}{\log (x)}+8 \log (x)-\frac {\log \left (3 \exp \left (\frac {4 (-5+x) x^2 (-5+x+2 \log (x))}{\log (x)}\right ) x^{4 x^2}\right )}{x^2}\right ) \, dx\\ &=\frac {1}{x}-76 x+12 x^2-4 \int \frac {(-5+x)^2}{\log ^2(x)} \, dx+8 \int \frac {25-15 x+2 x^2}{\log (x)} \, dx+8 \int \log (x) \, dx-\int \frac {\log \left (3 \exp \left (\frac {4 (-5+x) x^2 (-5+x+2 \log (x))}{\log (x)}\right ) x^{4 x^2}\right )}{x^2} \, dx\\ &=\frac {1}{x}-84 x+12 x^2+\frac {4 (5-x)^2 x}{\log (x)}+8 x \log (x)+\frac {\log \left (3 e^{\frac {4 (5-x)^2 x^2}{\log (x)}} x^{4 x^2-\frac {8 (5-x) x^2}{\log (x)}}\right )}{x}+8 \int \left (\frac {25}{\log (x)}-\frac {15 x}{\log (x)}+\frac {2 x^2}{\log (x)}\right ) \, dx-12 \int \frac {(-5+x)^2}{\log (x)} \, dx-40 \int \frac {-5+x}{\log (x)} \, dx+\int 4 \left (19-6 x+\frac {(-5+x)^2}{\log ^2(x)}+\frac {-50+30 x-4 x^2}{\log (x)}-2 \log (x)\right ) \, dx\\ &=\frac {1}{x}-84 x+12 x^2+\frac {4 (5-x)^2 x}{\log (x)}+8 x \log (x)+\frac {\log \left (3 e^{\frac {4 (5-x)^2 x^2}{\log (x)}} x^{4 x^2-\frac {8 (5-x) x^2}{\log (x)}}\right )}{x}+4 \int \left (19-6 x+\frac {(-5+x)^2}{\log ^2(x)}+\frac {-50+30 x-4 x^2}{\log (x)}-2 \log (x)\right ) \, dx-12 \int \left (\frac {25}{\log (x)}-\frac {10 x}{\log (x)}+\frac {x^2}{\log (x)}\right ) \, dx+16 \int \frac {x^2}{\log (x)} \, dx-40 \int \left (-\frac {5}{\log (x)}+\frac {x}{\log (x)}\right ) \, dx-120 \int \frac {x}{\log (x)} \, dx+200 \int \frac {1}{\log (x)} \, dx\\ &=\frac {1}{x}-8 x+\frac {4 (5-x)^2 x}{\log (x)}+8 x \log (x)+\frac {\log \left (3 e^{\frac {4 (5-x)^2 x^2}{\log (x)}} x^{4 x^2-\frac {8 (5-x) x^2}{\log (x)}}\right )}{x}+200 \text {li}(x)+4 \int \frac {(-5+x)^2}{\log ^2(x)} \, dx+4 \int \frac {-50+30 x-4 x^2}{\log (x)} \, dx-8 \int \log (x) \, dx-12 \int \frac {x^2}{\log (x)} \, dx+16 \operatorname {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (x)\right )-40 \int \frac {x}{\log (x)} \, dx+120 \int \frac {x}{\log (x)} \, dx-120 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )+200 \int \frac {1}{\log (x)} \, dx-300 \int \frac {1}{\log (x)} \, dx\\ &=\frac {1}{x}-120 \text {Ei}(2 \log (x))+16 \text {Ei}(3 \log (x))+\frac {\log \left (3 e^{\frac {4 (5-x)^2 x^2}{\log (x)}} x^{4 x^2-\frac {8 (5-x) x^2}{\log (x)}}\right )}{x}+100 \text {li}(x)+4 \int \left (-\frac {50}{\log (x)}+\frac {30 x}{\log (x)}-\frac {4 x^2}{\log (x)}\right ) \, dx+12 \int \frac {(-5+x)^2}{\log (x)} \, dx-12 \operatorname {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (x)\right )+40 \int \frac {-5+x}{\log (x)} \, dx-40 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )+120 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )\\ &=\frac {1}{x}-40 \text {Ei}(2 \log (x))+4 \text {Ei}(3 \log (x))+\frac {\log \left (3 e^{\frac {4 (5-x)^2 x^2}{\log (x)}} x^{4 x^2-\frac {8 (5-x) x^2}{\log (x)}}\right )}{x}+100 \text {li}(x)+12 \int \left (\frac {25}{\log (x)}-\frac {10 x}{\log (x)}+\frac {x^2}{\log (x)}\right ) \, dx-16 \int \frac {x^2}{\log (x)} \, dx+40 \int \left (-\frac {5}{\log (x)}+\frac {x}{\log (x)}\right ) \, dx+120 \int \frac {x}{\log (x)} \, dx-200 \int \frac {1}{\log (x)} \, dx\\ &=\frac {1}{x}-40 \text {Ei}(2 \log (x))+4 \text {Ei}(3 \log (x))+\frac {\log \left (3 e^{\frac {4 (5-x)^2 x^2}{\log (x)}} x^{4 x^2-\frac {8 (5-x) x^2}{\log (x)}}\right )}{x}-100 \text {li}(x)+12 \int \frac {x^2}{\log (x)} \, dx-16 \operatorname {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (x)\right )+40 \int \frac {x}{\log (x)} \, dx-120 \int \frac {x}{\log (x)} \, dx+120 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )-200 \int \frac {1}{\log (x)} \, dx+300 \int \frac {1}{\log (x)} \, dx\\ &=\frac {1}{x}+80 \text {Ei}(2 \log (x))-12 \text {Ei}(3 \log (x))+\frac {\log \left (3 e^{\frac {4 (5-x)^2 x^2}{\log (x)}} x^{4 x^2-\frac {8 (5-x) x^2}{\log (x)}}\right )}{x}+12 \operatorname {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (x)\right )+40 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )-120 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )\\ &=\frac {1}{x}+\frac {\log \left (3 e^{\frac {4 (5-x)^2 x^2}{\log (x)}} x^{4 x^2-\frac {8 (5-x) x^2}{\log (x)}}\right )}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 37, normalized size = 1.37 \begin {gather*} \frac {1+\log \left (3 e^{\frac {4 (-5+x) x^2 (-5+x+2 \log (x))}{\log (x)}} x^{4 x^2}\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-100*x^2 + 40*x^3 - 4*x^4 + (200*x^2 - 120*x^3 + 16*x^4)*Log[x] + (-1 - 76*x^2 + 24*x^3)*Log[x]^2 -
Log[3*E^((100*x^2 - 40*x^3 + 4*x^4 + (-40*x^2 + 8*x^3)*Log[x] + 4*x^2*Log[x]^2)/Log[x])]*Log[x]^2 + 8*x^2*Log
[x]^3)/(x^2*Log[x]^2),x]

[Out]

(1 + Log[3*E^((4*(-5 + x)*x^2*(-5 + x + 2*Log[x]))/Log[x])*x^(4*x^2)])/x

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fricas [A]  time = 0.50, size = 50, normalized size = 1.85 \begin {gather*} \frac {4 \, x^{4} + 4 \, x^{2} \log \relax (x)^{2} - 40 \, x^{3} + 100 \, x^{2} + {\left (8 \, x^{3} - 40 \, x^{2} + \log \relax (3) + 1\right )} \log \relax (x)}{x \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x)^2*log(3*exp((4*x^2*log(x)^2+(8*x^3-40*x^2)*log(x)+4*x^4-40*x^3+100*x^2)/log(x)))+8*x^2*log(
x)^3+(24*x^3-76*x^2-1)*log(x)^2+(16*x^4-120*x^3+200*x^2)*log(x)-4*x^4+40*x^3-100*x^2)/x^2/log(x)^2,x, algorith
m="fricas")

[Out]

(4*x^4 + 4*x^2*log(x)^2 - 40*x^3 + 100*x^2 + (8*x^3 - 40*x^2 + log(3) + 1)*log(x))/(x*log(x))

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giac [A]  time = 0.62, size = 40, normalized size = 1.48 \begin {gather*} 8 \, x^{2} + 4 \, x \log \relax (x) - 40 \, x + \frac {\log \relax (3) + 1}{x} + \frac {4 \, {\left (x^{3} - 10 \, x^{2} + 25 \, x\right )}}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x)^2*log(3*exp((4*x^2*log(x)^2+(8*x^3-40*x^2)*log(x)+4*x^4-40*x^3+100*x^2)/log(x)))+8*x^2*log(
x)^3+(24*x^3-76*x^2-1)*log(x)^2+(16*x^4-120*x^3+200*x^2)*log(x)-4*x^4+40*x^3-100*x^2)/x^2/log(x)^2,x, algorith
m="giac")

[Out]

8*x^2 + 4*x*log(x) - 40*x + (log(3) + 1)/x + 4*(x^3 - 10*x^2 + 25*x)/log(x)

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maple [A]  time = 0.15, size = 33, normalized size = 1.22

method	result	size
risch	\(\frac {\ln \left ({\mathrm e}^{\frac {4 \left (\ln \relax (x )+x -5\right )^{2} x^{2}}{\ln \relax (x )}}\right )}{x}+\frac {\ln \relax (3)}{x}+\frac {1}{x}\)	\(33\)
default	\(\frac {4 x^{3}}{\ln \relax (x )}-\frac {40 x^{2}}{\ln \relax (x )}+\frac {100 x}{\ln \relax (x )}+8 x^{2}-40 x +\frac {1}{x}+4 x \ln \relax (x )+\frac {\ln \left (3 \,{\mathrm e}^{\frac {4 x^{2} \ln \relax (x )^{2}+\left (8 x^{3}-40 x^{2}\right ) \ln \relax (x )+4 x^{4}-40 x^{3}+100 x^{2}}{\ln \relax (x )}}\right )-\frac {4 x^{2} \ln \relax (x )^{2}+\left (8 x^{3}-40 x^{2}\right ) \ln \relax (x )+4 x^{4}-40 x^{3}+100 x^{2}}{\ln \relax (x )}}{x}\)	\(141\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-ln(x)^2*ln(3*exp((4*x^2*ln(x)^2+(8*x^3-40*x^2)*ln(x)+4*x^4-40*x^3+100*x^2)/ln(x)))+8*x^2*ln(x)^3+(24*x^3
-76*x^2-1)*ln(x)^2+(16*x^4-120*x^3+200*x^2)*ln(x)-4*x^4+40*x^3-100*x^2)/x^2/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

1/x*ln(exp(4/ln(x)*(ln(x)+x-5)^2*x^2))+ln(3)/x+1/x

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 12 \, x^{2} + 8 \, x \log \relax (x) - 84 \, x + \frac {1}{x} - \frac {4 \, x^{4} + 8 \, x^{2} \log \relax (x)^{2} - 40 \, x^{3} + 100 \, x^{2} + {\left (4 \, x^{3} - 44 \, x^{2} - \log \relax (3)\right )} \log \relax (x) - 4 \, \log \relax (x) \log \left (x^{\left (x^{2}\right )}\right ) - 4 \, \log \relax (x) \log \left (e^{\left (\frac {x^{4}}{\log \relax (x)}\right )}\right ) + 40 \, \log \relax (x) \log \left (e^{\left (\frac {x^{3}}{\log \relax (x)}\right )}\right ) - 100 \, \log \relax (x) \log \left (e^{\left (\frac {x^{2}}{\log \relax (x)}\right )}\right )}{x \log \relax (x)} + 16 \, {\rm Ei}\left (3 \, \log \relax (x)\right ) - 120 \, {\rm Ei}\left (2 \, \log \relax (x)\right ) + 200 \, {\rm Ei}\left (\log \relax (x)\right ) - 100 \, \Gamma \left (-1, -\log \relax (x)\right ) + 80 \, \Gamma \left (-1, -2 \, \log \relax (x)\right ) - 12 \, \Gamma \left (-1, -3 \, \log \relax (x)\right ) - \int \frac {4 \, {\left (x^{2} - 10 \, x + 25\right )}}{\log \relax (x)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x)^2*log(3*exp((4*x^2*log(x)^2+(8*x^3-40*x^2)*log(x)+4*x^4-40*x^3+100*x^2)/log(x)))+8*x^2*log(
x)^3+(24*x^3-76*x^2-1)*log(x)^2+(16*x^4-120*x^3+200*x^2)*log(x)-4*x^4+40*x^3-100*x^2)/x^2/log(x)^2,x, algorith
m="maxima")

[Out]

12*x^2 + 8*x*log(x) - 84*x + 1/x - (4*x^4 + 8*x^2*log(x)^2 - 40*x^3 + 100*x^2 + (4*x^3 - 44*x^2 - log(3))*log(
x) - 4*log(x)*log(x^(x^2)) - 4*log(x)*log(e^(x^4/log(x))) + 40*log(x)*log(e^(x^3/log(x))) - 100*log(x)*log(e^(
x^2/log(x))))/(x*log(x)) + 16*Ei(3*log(x)) - 120*Ei(2*log(x)) + 200*Ei(log(x)) - 100*gamma(-1, -log(x)) + 80*g
amma(-1, -2*log(x)) - 12*gamma(-1, -3*log(x)) - integrate(4*(x^2 - 10*x + 25)/log(x), x)

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mupad [B]  time = 4.83, size = 122, normalized size = 4.52 \begin {gather*} 60\,x+\frac {4\,x\,{\left (x-5\right )}^2-4\,x\,\ln \relax (x)\,\left (3\,x^2-20\,x+25\right )}{\ln \relax (x)}+4\,x\,\ln \relax (x)-72\,x^2+12\,x^3+\frac {\ln \left (3\,x^{4\,x^2}\,{\mathrm {e}}^{8\,x^3}\,{\mathrm {e}}^{-40\,x^2}\,{\mathrm {e}}^{\frac {4\,x^4}{\ln \relax (x)}}\,{\mathrm {e}}^{-\frac {40\,x^3}{\ln \relax (x)}}\,{\mathrm {e}}^{\frac {100\,x^2}{\ln \relax (x)}}\right )-\frac {4\,x^2\,{\left (x+\ln \relax (x)-5\right )}^2}{\ln \relax (x)}+1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(3*exp((4*x^2*log(x)^2 - log(x)*(40*x^2 - 8*x^3) + 100*x^2 - 40*x^3 + 4*x^4)/log(x)))*log(x)^2 - log(
x)*(200*x^2 - 120*x^3 + 16*x^4) + log(x)^2*(76*x^2 - 24*x^3 + 1) - 8*x^2*log(x)^3 + 100*x^2 - 40*x^3 + 4*x^4)/
(x^2*log(x)^2),x)

[Out]

60*x + (4*x*(x - 5)^2 - 4*x*log(x)*(3*x^2 - 20*x + 25))/log(x) + 4*x*log(x) - 72*x^2 + 12*x^3 + (log(3*x^(4*x^
2)*exp(8*x^3)*exp(-40*x^2)*exp((4*x^4)/log(x))*exp(-(40*x^3)/log(x))*exp((100*x^2)/log(x))) - (4*x^2*(x + log(
x) - 5)^2)/log(x) + 1)/x

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sympy [A]  time = 0.79, size = 37, normalized size = 1.37 \begin {gather*} 8 x^{2} + 4 x \log {\relax (x )} - 40 x + \frac {4 x^{3} - 40 x^{2} + 100 x}{\log {\relax (x )}} + \frac {1 + \log {\relax (3 )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-ln(x)**2*ln(3*exp((4*x**2*ln(x)**2+(8*x**3-40*x**2)*ln(x)+4*x**4-40*x**3+100*x**2)/ln(x)))+8*x**2*
ln(x)**3+(24*x**3-76*x**2-1)*ln(x)**2+(16*x**4-120*x**3+200*x**2)*ln(x)-4*x**4+40*x**3-100*x**2)/x**2/ln(x)**2
,x)

[Out]

8*x**2 + 4*x*log(x) - 40*x + (4*x**3 - 40*x**2 + 100*x)/log(x) + (1 + log(3))/x

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