3.75.38 \(\int \frac {80+20 x+e^{256} (8+2 x)+(-40-4 e^{256}) \log (x) \log (\log ^2(x))}{3 x^2 \log (x)} \, dx\)

Optimal. Leaf size=22 \[ 3+\frac {\left (10+e^{256}\right ) (4+x) \log \left (\log ^2(x)\right )}{3 x} \]

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Rubi [A]  time = 0.29, antiderivative size = 30, normalized size of antiderivative = 1.36, number of steps used = 14, number of rules used = 9, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.225, Rules used = {12, 6688, 6742, 2353, 2309, 2178, 2302, 29, 2522} \begin {gather*} \frac {4 \left (10+e^{256}\right ) \log \left (\log ^2(x)\right )}{3 x}+\frac {2}{3} \left (10+e^{256}\right ) \log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(80 + 20*x + E^256*(8 + 2*x) + (-40 - 4*E^256)*Log[x]*Log[Log[x]^2])/(3*x^2*Log[x]),x]

[Out]

(2*(10 + E^256)*Log[Log[x]])/3 + (4*(10 + E^256)*Log[Log[x]^2])/(3*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 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

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 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 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 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 2522

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

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, 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}{3} \int \frac {80+20 x+e^{256} (8+2 x)+\left (-40-4 e^{256}\right ) \log (x) \log \left (\log ^2(x)\right )}{x^2 \log (x)} \, dx\\ &=\frac {1}{3} \int \frac {2 \left (10+e^{256}\right ) \left (4+x-2 \log (x) \log \left (\log ^2(x)\right )\right )}{x^2 \log (x)} \, dx\\ &=\frac {1}{3} \left (2 \left (10+e^{256}\right )\right ) \int \frac {4+x-2 \log (x) \log \left (\log ^2(x)\right )}{x^2 \log (x)} \, dx\\ &=\frac {1}{3} \left (2 \left (10+e^{256}\right )\right ) \int \left (\frac {4+x}{x^2 \log (x)}-\frac {2 \log \left (\log ^2(x)\right )}{x^2}\right ) \, dx\\ &=\frac {1}{3} \left (2 \left (10+e^{256}\right )\right ) \int \frac {4+x}{x^2 \log (x)} \, dx-\frac {1}{3} \left (4 \left (10+e^{256}\right )\right ) \int \frac {\log \left (\log ^2(x)\right )}{x^2} \, dx\\ &=\frac {4 \left (10+e^{256}\right ) \log \left (\log ^2(x)\right )}{3 x}+\frac {1}{3} \left (2 \left (10+e^{256}\right )\right ) \int \left (\frac {4}{x^2 \log (x)}+\frac {1}{x \log (x)}\right ) \, dx-\frac {1}{3} \left (8 \left (10+e^{256}\right )\right ) \int \frac {1}{x^2 \log (x)} \, dx\\ &=\frac {4 \left (10+e^{256}\right ) \log \left (\log ^2(x)\right )}{3 x}+\frac {1}{3} \left (2 \left (10+e^{256}\right )\right ) \int \frac {1}{x \log (x)} \, dx+\frac {1}{3} \left (8 \left (10+e^{256}\right )\right ) \int \frac {1}{x^2 \log (x)} \, dx-\frac {1}{3} \left (8 \left (10+e^{256}\right )\right ) \operatorname {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,\log (x)\right )\\ &=-\frac {8}{3} \left (10+e^{256}\right ) \text {Ei}(-\log (x))+\frac {4 \left (10+e^{256}\right ) \log \left (\log ^2(x)\right )}{3 x}+\frac {1}{3} \left (2 \left (10+e^{256}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right )+\frac {1}{3} \left (8 \left (10+e^{256}\right )\right ) \operatorname {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,\log (x)\right )\\ &=\frac {2}{3} \left (10+e^{256}\right ) \log (\log (x))+\frac {4 \left (10+e^{256}\right ) \log \left (\log ^2(x)\right )}{3 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 23, normalized size = 1.05 \begin {gather*} \frac {2}{3} \left (10+e^{256}\right ) \left (\log (\log (x))+\frac {2 \log \left (\log ^2(x)\right )}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(80 + 20*x + E^256*(8 + 2*x) + (-40 - 4*E^256)*Log[x]*Log[Log[x]^2])/(3*x^2*Log[x]),x]

[Out]

(2*(10 + E^256)*(Log[Log[x]] + (2*Log[Log[x]^2])/x))/3

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fricas [A]  time = 0.79, size = 21, normalized size = 0.95 \begin {gather*} \frac {{\left ({\left (x + 4\right )} e^{256} + 10 \, x + 40\right )} \log \left (\log \relax (x)^{2}\right )}{3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-4*exp(256)-40)*log(x)*log(log(x)^2)+(2*x+8)*exp(256)+20*x+80)/x^2/log(x),x, algorithm="fricas
")

[Out]

1/3*((x + 4)*e^256 + 10*x + 40)*log(log(x)^2)/x

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giac [A]  time = 0.23, size = 35, normalized size = 1.59 \begin {gather*} \frac {2 \, {\left (x e^{256} \log \left (\log \relax (x)\right ) + 2 \, e^{256} \log \left (\log \relax (x)^{2}\right ) + 10 \, x \log \left (\log \relax (x)\right ) + 20 \, \log \left (\log \relax (x)^{2}\right )\right )}}{3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-4*exp(256)-40)*log(x)*log(log(x)^2)+(2*x+8)*exp(256)+20*x+80)/x^2/log(x),x, algorithm="giac")

[Out]

2/3*(x*e^256*log(log(x)) + 2*e^256*log(log(x)^2) + 10*x*log(log(x)) + 20*log(log(x)^2))/x

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maple [A]  time = 0.18, size = 31, normalized size = 1.41

method	result	size
norman	\(\frac {\left (\frac {4 \,{\mathrm e}^{256}}{3}+\frac {40}{3}\right ) \ln \left (\ln \relax (x )^{2}\right )+\left (\frac {10}{3}+\frac {{\mathrm e}^{256}}{3}\right ) x \ln \left (\ln \relax (x )^{2}\right )}{x}\)	\(31\)
risch	\(\frac {8 \left (10+{\mathrm e}^{256}\right ) \ln \left (\ln \relax (x )\right )}{3 x}+\frac {-\frac {2 i \pi \,{\mathrm e}^{256} \mathrm {csgn}\left (i \ln \relax (x )\right )^{2} \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )}{3}+\frac {4 i \pi \,{\mathrm e}^{256} \mathrm {csgn}\left (i \ln \relax (x )\right ) \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )^{2}}{3}-\frac {2 i \pi \,{\mathrm e}^{256} \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )^{3}}{3}-\frac {20 i \pi \mathrm {csgn}\left (i \ln \relax (x )\right )^{2} \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )}{3}+\frac {40 i \pi \,\mathrm {csgn}\left (i \ln \relax (x )\right ) \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )^{2}}{3}-\frac {20 i \pi \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )^{3}}{3}+\frac {2 \ln \left (\ln \relax (x )\right ) {\mathrm e}^{256} x}{3}+\frac {20 x \ln \left (\ln \relax (x )\right )}{3}}{x}\)	\(147\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*((-4*exp(256)-40)*ln(x)*ln(ln(x)^2)+(2*x+8)*exp(256)+20*x+80)/x^2/ln(x),x,method=_RETURNVERBOSE)

[Out]

((4/3*exp(256)+40/3)*ln(ln(x)^2)+(10/3+1/3*exp(256))*x*ln(ln(x)^2))/x

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maxima [C]  time = 0.41, size = 53, normalized size = 2.41 \begin {gather*} \frac {4}{3} \, {\left (\frac {\log \left (\log \relax (x)^{2}\right )}{x} - 2 \, {\rm Ei}\left (-\log \relax (x)\right )\right )} e^{256} + \frac {8}{3} \, {\rm Ei}\left (-\log \relax (x)\right ) e^{256} + \frac {2}{3} \, e^{256} \log \left (\log \relax (x)\right ) + \frac {40 \, \log \left (\log \relax (x)^{2}\right )}{3 \, x} + \frac {20}{3} \, \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-4*exp(256)-40)*log(x)*log(log(x)^2)+(2*x+8)*exp(256)+20*x+80)/x^2/log(x),x, algorithm="maxima
")

[Out]

4/3*(log(log(x)^2)/x - 2*Ei(-log(x)))*e^256 + 8/3*Ei(-log(x))*e^256 + 2/3*e^256*log(log(x)) + 40/3*log(log(x)^
2)/x + 20/3*log(log(x))

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mupad [B]  time = 4.67, size = 17, normalized size = 0.77 \begin {gather*} \frac {\ln \left ({\ln \relax (x)}^2\right )\,\left ({\mathrm {e}}^{256}+10\right )\,\left (x+4\right )}{3\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((20*x)/3 + (exp(256)*(2*x + 8))/3 - (log(log(x)^2)*log(x)*(4*exp(256) + 40))/3 + 80/3)/(x^2*log(x)),x)

[Out]

(log(log(x)^2)*(exp(256) + 10)*(x + 4))/(3*x)

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sympy [A]  time = 10.82, size = 29, normalized size = 1.32 \begin {gather*} \frac {2 \left (10 + e^{256}\right ) \log {\left (\log {\relax (x )} \right )}}{3} + \frac {\left (40 + 4 e^{256}\right ) \log {\left (\log {\relax (x )}^{2} \right )}}{3 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-4*exp(256)-40)*ln(x)*ln(ln(x)**2)+(2*x+8)*exp(256)+20*x+80)/x**2/ln(x),x)

[Out]

2*(10 + exp(256))*log(log(x))/3 + (40 + 4*exp(256))*log(log(x)**2)/(3*x)

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