3.6.35 \(\int \frac {-16-16 x+5 x^2+(-20 x+5 x^2) \log (x)+(-16-12 x+10 x^2) \log (x) \log (\frac {1}{9} (-4-5 x) \log (x))}{(36+45 x) \log (x)} \, dx\)

Optimal. Leaf size=25 \[ \frac {1}{9} (-4+x) x \log \left (\left (-x+\frac {1}{9} (-4+4 x)\right ) \log (x)\right ) \]

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Rubi [A]  time = 0.56, antiderivative size = 37, normalized size of antiderivative = 1.48, number of steps used = 22, number of rules used = 8, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {6742, 77, 2330, 2298, 2309, 2178, 2555, 12} \begin {gather*} \frac {1}{9} x^2 \log \left (-\frac {1}{9} (5 x+4) \log (x)\right )-\frac {4}{9} x \log \left (-\frac {1}{9} (5 x+4) \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-16 - 16*x + 5*x^2 + (-20*x + 5*x^2)*Log[x] + (-16 - 12*x + 10*x^2)*Log[x]*Log[((-4 - 5*x)*Log[x])/9])/((
36 + 45*x)*Log[x]),x]

[Out]

(-4*x*Log[-1/9*((4 + 5*x)*Log[x])])/9 + (x^2*Log[-1/9*((4 + 5*x)*Log[x])])/9

Rule 12

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

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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 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 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {(-4+x) (4+5 x+5 x \log (x))}{9 (4+5 x) \log (x)}+\frac {2}{9} (-2+x) \log \left (-\frac {1}{9} (4+5 x) \log (x)\right )\right ) \, dx\\ &=\frac {1}{9} \int \frac {(-4+x) (4+5 x+5 x \log (x))}{(4+5 x) \log (x)} \, dx+\frac {2}{9} \int (-2+x) \log \left (-\frac {1}{9} (4+5 x) \log (x)\right ) \, dx\\ &=-\frac {4}{9} x \log \left (-\frac {1}{9} (4+5 x) \log (x)\right )+\frac {1}{9} x^2 \log \left (-\frac {1}{9} (4+5 x) \log (x)\right )+\frac {1}{9} \int \left (\frac {5 (-4+x) x}{4+5 x}+\frac {-4+x}{\log (x)}\right ) \, dx-\frac {2}{9} \int \frac {(4-x) (-4-5 x-5 x \log (x))}{2 (4+5 x) \log (x)} \, dx\\ &=-\frac {4}{9} x \log \left (-\frac {1}{9} (4+5 x) \log (x)\right )+\frac {1}{9} x^2 \log \left (-\frac {1}{9} (4+5 x) \log (x)\right )+\frac {1}{9} \int \frac {-4+x}{\log (x)} \, dx-\frac {1}{9} \int \frac {(4-x) (-4-5 x-5 x \log (x))}{(4+5 x) \log (x)} \, dx+\frac {5}{9} \int \frac {(-4+x) x}{4+5 x} \, dx\\ &=-\frac {4}{9} x \log \left (-\frac {1}{9} (4+5 x) \log (x)\right )+\frac {1}{9} x^2 \log \left (-\frac {1}{9} (4+5 x) \log (x)\right )-\frac {1}{9} \int \left (\frac {5 (-4+x) x}{4+5 x}+\frac {-4+x}{\log (x)}\right ) \, dx+\frac {1}{9} \int \left (-\frac {4}{\log (x)}+\frac {x}{\log (x)}\right ) \, dx+\frac {5}{9} \int \left (-\frac {24}{25}+\frac {x}{5}+\frac {96}{25 (4+5 x)}\right ) \, dx\\ &=-\frac {8 x}{15}+\frac {x^2}{18}+\frac {32}{75} \log (4+5 x)-\frac {4}{9} x \log \left (-\frac {1}{9} (4+5 x) \log (x)\right )+\frac {1}{9} x^2 \log \left (-\frac {1}{9} (4+5 x) \log (x)\right )-\frac {1}{9} \int \frac {-4+x}{\log (x)} \, dx+\frac {1}{9} \int \frac {x}{\log (x)} \, dx-\frac {4}{9} \int \frac {1}{\log (x)} \, dx-\frac {5}{9} \int \frac {(-4+x) x}{4+5 x} \, dx\\ &=-\frac {8 x}{15}+\frac {x^2}{18}+\frac {32}{75} \log (4+5 x)-\frac {4}{9} x \log \left (-\frac {1}{9} (4+5 x) \log (x)\right )+\frac {1}{9} x^2 \log \left (-\frac {1}{9} (4+5 x) \log (x)\right )-\frac {4 \text {li}(x)}{9}-\frac {1}{9} \int \left (-\frac {4}{\log (x)}+\frac {x}{\log (x)}\right ) \, dx+\frac {1}{9} \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )-\frac {5}{9} \int \left (-\frac {24}{25}+\frac {x}{5}+\frac {96}{25 (4+5 x)}\right ) \, dx\\ &=\frac {1}{9} \text {Ei}(2 \log (x))-\frac {4}{9} x \log \left (-\frac {1}{9} (4+5 x) \log (x)\right )+\frac {1}{9} x^2 \log \left (-\frac {1}{9} (4+5 x) \log (x)\right )-\frac {4 \text {li}(x)}{9}-\frac {1}{9} \int \frac {x}{\log (x)} \, dx+\frac {4}{9} \int \frac {1}{\log (x)} \, dx\\ &=\frac {1}{9} \text {Ei}(2 \log (x))-\frac {4}{9} x \log \left (-\frac {1}{9} (4+5 x) \log (x)\right )+\frac {1}{9} x^2 \log \left (-\frac {1}{9} (4+5 x) \log (x)\right )-\frac {1}{9} \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )\\ &=-\frac {4}{9} x \log \left (-\frac {1}{9} (4+5 x) \log (x)\right )+\frac {1}{9} x^2 \log \left (-\frac {1}{9} (4+5 x) \log (x)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 20, normalized size = 0.80 \begin {gather*} \frac {1}{9} (-4+x) x \log \left (-\frac {1}{9} (4+5 x) \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-16 - 16*x + 5*x^2 + (-20*x + 5*x^2)*Log[x] + (-16 - 12*x + 10*x^2)*Log[x]*Log[((-4 - 5*x)*Log[x])/
9])/((36 + 45*x)*Log[x]),x]

[Out]

((-4 + x)*x*Log[-1/9*((4 + 5*x)*Log[x])])/9

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fricas [A]  time = 0.81, size = 19, normalized size = 0.76 \begin {gather*} \frac {1}{9} \, {\left (x^{2} - 4 \, x\right )} \log \left (-\frac {1}{9} \, {\left (5 \, x + 4\right )} \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^2-12*x-16)*log(x)*log(1/9*(-5*x-4)*log(x))+(5*x^2-20*x)*log(x)+5*x^2-16*x-16)/(45*x+36)/log(x
),x, algorithm="fricas")

[Out]

1/9*(x^2 - 4*x)*log(-1/9*(5*x + 4)*log(x))

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giac [B]  time = 0.68, size = 33, normalized size = 1.32 \begin {gather*} -\frac {2}{9} \, x^{2} \log \relax (3) + \frac {8}{9} \, x \log \relax (3) + \frac {1}{9} \, {\left (x^{2} - 4 \, x\right )} \log \left (-5 \, x \log \relax (x) - 4 \, \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^2-12*x-16)*log(x)*log(1/9*(-5*x-4)*log(x))+(5*x^2-20*x)*log(x)+5*x^2-16*x-16)/(45*x+36)/log(x
),x, algorithm="giac")

[Out]

-2/9*x^2*log(3) + 8/9*x*log(3) + 1/9*(x^2 - 4*x)*log(-5*x*log(x) - 4*log(x))

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maple [A]  time = 0.33, size = 30, normalized size = 1.20

method	result	size
norman	\(-\frac {4 x \ln \left (\frac {\left (-5 x -4\right ) \ln \relax (x )}{9}\right )}{9}+\frac {x^{2} \ln \left (\frac {\left (-5 x -4\right ) \ln \relax (x )}{9}\right )}{9}\)	\(30\)
risch	\(\frac {x^{2} \ln \relax (5)}{9}-\frac {2 x^{2} \ln \relax (3)}{9}-\frac {4 x \ln \left (\ln \relax (x )\right )}{9}+\frac {x^{2} \ln \left (\ln \relax (x )\right )}{9}-\frac {4 x \ln \relax (5)}{9}+\frac {8 x \ln \relax (3)}{9}+\frac {i \pi \,x^{2}}{9}-\frac {2 i \pi x \mathrm {csgn}\left (i \ln \relax (x ) \left (x +\frac {4}{5}\right )\right )^{3}}{9}+\frac {i \pi \,x^{2} \mathrm {csgn}\left (i \ln \relax (x ) \left (x +\frac {4}{5}\right )\right )^{3}}{18}-\frac {2 i \pi x \,\mathrm {csgn}\left (i \ln \relax (x )\right ) \mathrm {csgn}\left (i \ln \relax (x ) \left (x +\frac {4}{5}\right )\right )^{2}}{9}-\frac {i \pi \,x^{2} \mathrm {csgn}\left (i \ln \relax (x )\right ) \mathrm {csgn}\left (i \left (x +\frac {4}{5}\right )\right ) \mathrm {csgn}\left (i \ln \relax (x ) \left (x +\frac {4}{5}\right )\right )}{18}-\frac {2 i \pi x \,\mathrm {csgn}\left (i \left (x +\frac {4}{5}\right )\right ) \mathrm {csgn}\left (i \ln \relax (x ) \left (x +\frac {4}{5}\right )\right )^{2}}{9}+\frac {i \pi \,x^{2} \mathrm {csgn}\left (i \left (x +\frac {4}{5}\right )\right ) \mathrm {csgn}\left (i \ln \relax (x ) \left (x +\frac {4}{5}\right )\right )^{2}}{18}+\frac {i \pi \,x^{2} \mathrm {csgn}\left (i \ln \relax (x )\right ) \mathrm {csgn}\left (i \ln \relax (x ) \left (x +\frac {4}{5}\right )\right )^{2}}{18}+\frac {2 i \pi x \,\mathrm {csgn}\left (i \ln \relax (x )\right ) \mathrm {csgn}\left (i \left (x +\frac {4}{5}\right )\right ) \mathrm {csgn}\left (i \ln \relax (x ) \left (x +\frac {4}{5}\right )\right )}{9}-\frac {i \pi \,x^{2} \mathrm {csgn}\left (i \ln \relax (x ) \left (x +\frac {4}{5}\right )\right )^{2}}{9}+\left (\frac {1}{9} x^{2}-\frac {4}{9} x \right ) \ln \left (x +\frac {4}{5}\right )+\frac {4 i \pi x \mathrm {csgn}\left (i \ln \relax (x ) \left (x +\frac {4}{5}\right )\right )^{2}}{9}-\frac {4 i \pi x}{9}\)	\(284\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((10*x^2-12*x-16)*ln(x)*ln(1/9*(-5*x-4)*ln(x))+(5*x^2-20*x)*ln(x)+5*x^2-16*x-16)/(45*x+36)/ln(x),x,method=
_RETURNVERBOSE)

[Out]

-4/9*x*ln(1/9*(-5*x-4)*ln(x))+1/9*x^2*ln(1/9*(-5*x-4)*ln(x))

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maxima [B]  time = 0.67, size = 40, normalized size = 1.60 \begin {gather*} -\frac {2}{9} \, x^{2} \log \relax (3) + \frac {8}{9} \, x \log \relax (3) + \frac {1}{9} \, {\left (x^{2} - 4 \, x\right )} \log \left (-5 \, x - 4\right ) + \frac {1}{9} \, {\left (x^{2} - 4 \, x\right )} \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^2-12*x-16)*log(x)*log(1/9*(-5*x-4)*log(x))+(5*x^2-20*x)*log(x)+5*x^2-16*x-16)/(45*x+36)/log(x
),x, algorithm="maxima")

[Out]

-2/9*x^2*log(3) + 8/9*x*log(3) + 1/9*(x^2 - 4*x)*log(-5*x - 4) + 1/9*(x^2 - 4*x)*log(log(x))

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mupad [B]  time = 0.94, size = 21, normalized size = 0.84 \begin {gather*} -\ln \left (-\frac {\ln \relax (x)\,\left (5\,x+4\right )}{9}\right )\,\left (\frac {4\,x}{9}-\frac {x^2}{9}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(16*x + log(x)*(20*x - 5*x^2) - 5*x^2 + log(x)*log(-(log(x)*(5*x + 4))/9)*(12*x - 10*x^2 + 16) + 16)/(log
(x)*(45*x + 36)),x)

[Out]

-log(-(log(x)*(5*x + 4))/9)*((4*x)/9 - x^2/9)

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sympy [B]  time = 0.66, size = 46, normalized size = 1.84 \begin {gather*} \left (\frac {x^{2}}{9} - \frac {4 x}{9} - \frac {16}{225}\right ) \log {\left (\left (- \frac {5 x}{9} - \frac {4}{9}\right ) \log {\relax (x )} \right )} + \frac {16 \log {\left (225 x + 180 \right )}}{225} + \frac {16 \log {\left (\log {\relax (x )} \right )}}{225} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x**2-12*x-16)*ln(x)*ln(1/9*(-5*x-4)*ln(x))+(5*x**2-20*x)*ln(x)+5*x**2-16*x-16)/(45*x+36)/ln(x),
x)

[Out]

(x**2/9 - 4*x/9 - 16/225)*log((-5*x/9 - 4/9)*log(x)) + 16*log(225*x + 180)/225 + 16*log(log(x))/225

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