3.76.43 \(\int \frac {5-2 x+(-35+18 x+6 x^2) \log (x)+(5-4 x) \log (x) \log (\log (x))}{5 \log (x)} \, dx\)

Optimal. Leaf size=18 \[ \frac {1}{5} x (-5+2 x) (7+x-\log (\log (x))) \]

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Rubi [A]  time = 0.19, antiderivative size = 33, normalized size of antiderivative = 1.83, number of steps used = 17, number of rules used = 8, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.216, Rules used = {12, 6742, 2330, 2298, 2309, 2178, 2520, 2522} \begin {gather*} \frac {2 x^3}{5}+\frac {9 x^2}{5}-\frac {2}{5} x^2 \log (\log (x))-7 x+x \log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5 - 2*x + (-35 + 18*x + 6*x^2)*Log[x] + (5 - 4*x)*Log[x]*Log[Log[x]])/(5*Log[x]),x]

[Out]

-7*x + (9*x^2)/5 + (2*x^3)/5 + x*Log[Log[x]] - (2*x^2*Log[Log[x]])/5

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 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 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 2520

Int[Log[Log[(d_.)*(x_)^(n_.)]^(p_.)*(c_.)], x_Symbol] :> Simp[x*Log[c*Log[d*x^n]^p], x] - Dist[n*p, Int[1/Log[
d*x^n], x], x] /; FreeQ[{c, d, n, p}, x]

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 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 {5-2 x+\left (-35+18 x+6 x^2\right ) \log (x)+(5-4 x) \log (x) \log (\log (x))}{\log (x)} \, dx\\ &=\frac {1}{5} \int \left (\frac {5-2 x-35 \log (x)+18 x \log (x)+6 x^2 \log (x)}{\log (x)}-(-5+4 x) \log (\log (x))\right ) \, dx\\ &=\frac {1}{5} \int \frac {5-2 x-35 \log (x)+18 x \log (x)+6 x^2 \log (x)}{\log (x)} \, dx-\frac {1}{5} \int (-5+4 x) \log (\log (x)) \, dx\\ &=\frac {1}{5} \int \left (-35+18 x+6 x^2+\frac {5-2 x}{\log (x)}\right ) \, dx-\frac {1}{5} \int (-5 \log (\log (x))+4 x \log (\log (x))) \, dx\\ &=-7 x+\frac {9 x^2}{5}+\frac {2 x^3}{5}+\frac {1}{5} \int \frac {5-2 x}{\log (x)} \, dx-\frac {4}{5} \int x \log (\log (x)) \, dx+\int \log (\log (x)) \, dx\\ &=-7 x+\frac {9 x^2}{5}+\frac {2 x^3}{5}+x \log (\log (x))-\frac {2}{5} x^2 \log (\log (x))+\frac {1}{5} \int \left (\frac {5}{\log (x)}-\frac {2 x}{\log (x)}\right ) \, dx+\frac {2}{5} \int \frac {x}{\log (x)} \, dx-\int \frac {1}{\log (x)} \, dx\\ &=-7 x+\frac {9 x^2}{5}+\frac {2 x^3}{5}+x \log (\log (x))-\frac {2}{5} x^2 \log (\log (x))-\text {li}(x)-\frac {2}{5} \int \frac {x}{\log (x)} \, dx+\frac {2}{5} \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )+\int \frac {1}{\log (x)} \, dx\\ &=-7 x+\frac {9 x^2}{5}+\frac {2 x^3}{5}+\frac {2}{5} \text {Ei}(2 \log (x))+x \log (\log (x))-\frac {2}{5} x^2 \log (\log (x))-\frac {2}{5} \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )\\ &=-7 x+\frac {9 x^2}{5}+\frac {2 x^3}{5}+x \log (\log (x))-\frac {2}{5} x^2 \log (\log (x))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 18, normalized size = 1.00 \begin {gather*} \frac {1}{5} x (-5+2 x) (7+x-\log (\log (x))) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5 - 2*x + (-35 + 18*x + 6*x^2)*Log[x] + (5 - 4*x)*Log[x]*Log[Log[x]])/(5*Log[x]),x]

[Out]

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

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fricas [A]  time = 0.85, size = 28, normalized size = 1.56 \begin {gather*} \frac {2}{5} \, x^{3} + \frac {9}{5} \, x^{2} - \frac {1}{5} \, {\left (2 \, x^{2} - 5 \, x\right )} \log \left (\log \relax (x)\right ) - 7 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-4*x+5)*log(x)*log(log(x))+(6*x^2+18*x-35)*log(x)-2*x+5)/log(x),x, algorithm="fricas")

[Out]

2/5*x^3 + 9/5*x^2 - 1/5*(2*x^2 - 5*x)*log(log(x)) - 7*x

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giac [A]  time = 0.19, size = 27, normalized size = 1.50 \begin {gather*} \frac {2}{5} \, x^{3} - \frac {2}{5} \, x^{2} \log \left (\log \relax (x)\right ) + \frac {9}{5} \, x^{2} + x \log \left (\log \relax (x)\right ) - 7 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-4*x+5)*log(x)*log(log(x))+(6*x^2+18*x-35)*log(x)-2*x+5)/log(x),x, algorithm="giac")

[Out]

2/5*x^3 - 2/5*x^2*log(log(x)) + 9/5*x^2 + x*log(log(x)) - 7*x

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maple [A]  time = 0.03, size = 28, normalized size = 1.56

method	result	size
norman	\(x \ln \left (\ln \relax (x )\right )-7 x +\frac {9 x^{2}}{5}+\frac {2 x^{3}}{5}-\frac {2 x^{2} \ln \left (\ln \relax (x )\right )}{5}\)	\(28\)
risch	\(\frac {\left (-2 x^{2}+5 x \right ) \ln \left (\ln \relax (x )\right )}{5}+\frac {2 x^{3}}{5}+\frac {9 x^{2}}{5}-7 x\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((-4*x+5)*ln(x)*ln(ln(x))+(6*x^2+18*x-35)*ln(x)-2*x+5)/ln(x),x,method=_RETURNVERBOSE)

[Out]

x*ln(ln(x))-7*x+9/5*x^2+2/5*x^3-2/5*x^2*ln(ln(x))

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maxima [A]  time = 0.41, size = 27, normalized size = 1.50 \begin {gather*} \frac {2}{5} \, x^{3} - \frac {2}{5} \, x^{2} \log \left (\log \relax (x)\right ) + \frac {9}{5} \, x^{2} + x \log \left (\log \relax (x)\right ) - 7 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-4*x+5)*log(x)*log(log(x))+(6*x^2+18*x-35)*log(x)-2*x+5)/log(x),x, algorithm="maxima")

[Out]

2/5*x^3 - 2/5*x^2*log(log(x)) + 9/5*x^2 + x*log(log(x)) - 7*x

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mupad [B]  time = 4.65, size = 16, normalized size = 0.89 \begin {gather*} \frac {x\,\left (2\,x-5\right )\,\left (x-\ln \left (\ln \relax (x)\right )+7\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((2*x)/5 - (log(x)*(18*x + 6*x^2 - 35))/5 + (log(log(x))*log(x)*(4*x - 5))/5 - 1)/log(x),x)

[Out]

(x*(2*x - 5)*(x - log(log(x)) + 7))/5

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sympy [A]  time = 0.35, size = 29, normalized size = 1.61 \begin {gather*} \frac {2 x^{3}}{5} + \frac {9 x^{2}}{5} - 7 x + \left (- \frac {2 x^{2}}{5} + x\right ) \log {\left (\log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-4*x+5)*ln(x)*ln(ln(x))+(6*x**2+18*x-35)*ln(x)-2*x+5)/ln(x),x)

[Out]

2*x**3/5 + 9*x**2/5 - 7*x + (-2*x**2/5 + x)*log(log(x))

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