3.71.14 \(\int \frac {-25+20 x^2+(5 x-4 x^3) \log (5)+(20 x^2-4 x^3 \log (5)) \log ^2(x)+(-5+x \log (5)) \log (\frac {5-x \log (5)}{\log (5)})+\log (x) (-25-20 x^2+(4 x+4 x^3) \log (5)+(-5+x \log (5)) \log (\frac {5-x \log (5)}{\log (5)}))}{(60 x^2-20 e^3 x^2+20 x^3+(-12 x^3+4 e^3 x^3-4 x^4) \log (5)) \log ^2(x)+\log (x) (25 x-20 x^3+(-5 x^2+4 x^4) \log (5)+(5 x-x^2 \log (5)) \log (\frac {5-x \log (5)}{\log (5)}))} \, dx\)

Optimal. Leaf size=36 \[ \log \left (-3+e^3-x+\frac {x-\frac {5+\log \left (-x+\frac {5}{\log (5)}\right )}{4 x}}{\log (x)}\right ) \]

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Rubi [A]  time = 6.35, antiderivative size = 48, normalized size of antiderivative = 1.33, number of steps used = 7, number of rules used = 5, integrand size = 200, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {6688, 6742, 2365, 43, 6684} \begin {gather*} \log \left (-4 x^2+4 x^2 \log (x)+4 \left (3-e^3\right ) x \log (x)+\log \left (\frac {5}{\log (5)}-x\right )+5\right )-\log (x)-\log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-25 + 20*x^2 + (5*x - 4*x^3)*Log[5] + (20*x^2 - 4*x^3*Log[5])*Log[x]^2 + (-5 + x*Log[5])*Log[(5 - x*Log[5
])/Log[5]] + Log[x]*(-25 - 20*x^2 + (4*x + 4*x^3)*Log[5] + (-5 + x*Log[5])*Log[(5 - x*Log[5])/Log[5]]))/((60*x
^2 - 20*E^3*x^2 + 20*x^3 + (-12*x^3 + 4*E^3*x^3 - 4*x^4)*Log[5])*Log[x]^2 + Log[x]*(25*x - 20*x^3 + (-5*x^2 +
4*x^4)*Log[5] + (5*x - x^2*Log[5])*Log[(5 - x*Log[5])/Log[5]])),x]

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2365

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

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

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} &=\int \frac {-4 x^2 (-5+x \log (5)) \log ^2(x)-(-5+x \log (5)) \left (-5+4 x^2-\log \left (-x+\frac {5}{\log (5)}\right )\right )-\log (x) \left (25+20 x^2-4 x \log (5)-4 x^3 \log (5)+(5-x \log (5)) \log \left (-x+\frac {5}{\log (5)}\right )\right )}{x (5-x \log (5)) \log (x) \left (5-4 x^2+4 x \left (3-e^3+x\right ) \log (x)+\log \left (-x+\frac {5}{\log (5)}\right )\right )} \, dx\\ &=\int \left (\frac {-1-\log (x)}{x \log (x)}+\frac {60 \left (1-\frac {e^3}{3}-\frac {\log (5)}{60}\right )+4 x^2 \log (5)-20 x \left (1-\frac {1}{5} \left (-3+e^3\right ) \log (5)\right )+60 \left (1-\frac {e^3}{3}\right ) \log (x)-8 x^2 \log (5) \log (x)+40 x \left (1+\frac {1}{10} \left (-3+e^3\right ) \log (5)\right ) \log (x)}{(5-x \log (5)) \left (5-4 x^2+12 \left (1-\frac {e^3}{3}\right ) x \log (x)+4 x^2 \log (x)+\log \left (-x+\frac {5}{\log (5)}\right )\right )}\right ) \, dx\\ &=\int \frac {-1-\log (x)}{x \log (x)} \, dx+\int \frac {60 \left (1-\frac {e^3}{3}-\frac {\log (5)}{60}\right )+4 x^2 \log (5)-20 x \left (1-\frac {1}{5} \left (-3+e^3\right ) \log (5)\right )+60 \left (1-\frac {e^3}{3}\right ) \log (x)-8 x^2 \log (5) \log (x)+40 x \left (1+\frac {1}{10} \left (-3+e^3\right ) \log (5)\right ) \log (x)}{(5-x \log (5)) \left (5-4 x^2+12 \left (1-\frac {e^3}{3}\right ) x \log (x)+4 x^2 \log (x)+\log \left (-x+\frac {5}{\log (5)}\right )\right )} \, dx\\ &=\log \left (5-4 x^2+4 \left (3-e^3\right ) x \log (x)+4 x^2 \log (x)+\log \left (-x+\frac {5}{\log (5)}\right )\right )+\operatorname {Subst}\left (\int \frac {-1-x}{x} \, dx,x,\log (x)\right )\\ &=\log \left (5-4 x^2+4 \left (3-e^3\right ) x \log (x)+4 x^2 \log (x)+\log \left (-x+\frac {5}{\log (5)}\right )\right )+\operatorname {Subst}\left (\int \left (-1-\frac {1}{x}\right ) \, dx,x,\log (x)\right )\\ &=-\log (x)-\log (\log (x))+\log \left (5-4 x^2+4 \left (3-e^3\right ) x \log (x)+4 x^2 \log (x)+\log \left (-x+\frac {5}{\log (5)}\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 49, normalized size = 1.36 \begin {gather*} -\log (x)-\log (\log (x))+\log \left (5-4 x^2+12 x \log (x)-4 e^3 x \log (x)+4 x^2 \log (x)+\log \left (-x+\frac {5}{\log (5)}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-25 + 20*x^2 + (5*x - 4*x^3)*Log[5] + (20*x^2 - 4*x^3*Log[5])*Log[x]^2 + (-5 + x*Log[5])*Log[(5 - x
*Log[5])/Log[5]] + Log[x]*(-25 - 20*x^2 + (4*x + 4*x^3)*Log[5] + (-5 + x*Log[5])*Log[(5 - x*Log[5])/Log[5]]))/
((60*x^2 - 20*E^3*x^2 + 20*x^3 + (-12*x^3 + 4*E^3*x^3 - 4*x^4)*Log[5])*Log[x]^2 + Log[x]*(25*x - 20*x^3 + (-5*
x^2 + 4*x^4)*Log[5] + (5*x - x^2*Log[5])*Log[(5 - x*Log[5])/Log[5]])),x]

[Out]

-Log[x] - Log[Log[x]] + Log[5 - 4*x^2 + 12*x*Log[x] - 4*E^3*x*Log[x] + 4*x^2*Log[x] + Log[-x + 5/Log[5]]]

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fricas [A]  time = 0.66, size = 68, normalized size = 1.89 \begin {gather*} \log \left (x - e^{3} + 3\right ) + \log \left (\frac {4 \, x^{2} - 4 \, {\left (x^{2} - x e^{3} + 3 \, x\right )} \log \relax (x) - \log \left (-\frac {x \log \relax (5) - 5}{\log \relax (5)}\right ) - 5}{x^{2} - x e^{3} + 3 \, x}\right ) - \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^3*log(5)+20*x^2)*log(x)^2+((x*log(5)-5)*log((-x*log(5)+5)/log(5))+(4*x^3+4*x)*log(5)-20*x^2-2
5)*log(x)+(x*log(5)-5)*log((-x*log(5)+5)/log(5))+(-4*x^3+5*x)*log(5)+20*x^2-25)/(((4*x^3*exp(3)-4*x^4-12*x^3)*
log(5)-20*x^2*exp(3)+20*x^3+60*x^2)*log(x)^2+((-x^2*log(5)+5*x)*log((-x*log(5)+5)/log(5))+(4*x^4-5*x^2)*log(5)
-20*x^3+25*x)*log(x)),x, algorithm="fricas")

[Out]

log(x - e^3 + 3) + log((4*x^2 - 4*(x^2 - x*e^3 + 3*x)*log(x) - log(-(x*log(5) - 5)/log(5)) - 5)/(x^2 - x*e^3 +
3*x)) - log(log(x))

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giac [A]  time = 0.46, size = 50, normalized size = 1.39 \begin {gather*} \log \left (4 \, x^{2} \log \relax (x) - 4 \, x e^{3} \log \relax (x) - 4 \, x^{2} + 12 \, x \log \relax (x) + \log \left (-x \log \relax (5) + 5\right ) - \log \left (\log \relax (5)\right ) + 5\right ) - \log \relax (x) - \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^3*log(5)+20*x^2)*log(x)^2+((x*log(5)-5)*log((-x*log(5)+5)/log(5))+(4*x^3+4*x)*log(5)-20*x^2-2
5)*log(x)+(x*log(5)-5)*log((-x*log(5)+5)/log(5))+(-4*x^3+5*x)*log(5)+20*x^2-25)/(((4*x^3*exp(3)-4*x^4-12*x^3)*
log(5)-20*x^2*exp(3)+20*x^3+60*x^2)*log(x)^2+((-x^2*log(5)+5*x)*log((-x*log(5)+5)/log(5))+(4*x^4-5*x^2)*log(5)
-20*x^3+25*x)*log(x)),x, algorithm="giac")

[Out]

log(4*x^2*log(x) - 4*x*e^3*log(x) - 4*x^2 + 12*x*log(x) + log(-x*log(5) + 5) - log(log(5)) + 5) - log(x) - log
(log(x))

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maple [A]  time = 0.05, size = 51, normalized size = 1.42

method	result	size
risch	\(-\ln \relax (x )-\ln \left (\ln \relax (x )\right )+\ln \left (-4 x \,{\mathrm e}^{3} \ln \relax (x )+4 x^{2} \ln \relax (x )-4 x^{2}+12 x \ln \relax (x )+\ln \left (\frac {-x \ln \relax (5)+5}{\ln \relax (5)}\right )+5\right )\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x^3*ln(5)+20*x^2)*ln(x)^2+((x*ln(5)-5)*ln((-x*ln(5)+5)/ln(5))+(4*x^3+4*x)*ln(5)-20*x^2-25)*ln(x)+(x*l
n(5)-5)*ln((-x*ln(5)+5)/ln(5))+(-4*x^3+5*x)*ln(5)+20*x^2-25)/(((4*x^3*exp(3)-4*x^4-12*x^3)*ln(5)-20*x^2*exp(3)
+20*x^3+60*x^2)*ln(x)^2+((-x^2*ln(5)+5*x)*ln((-x*ln(5)+5)/ln(5))+(4*x^4-5*x^2)*ln(5)-20*x^3+25*x)*ln(x)),x,met
hod=_RETURNVERBOSE)

[Out]

-ln(x)-ln(ln(x))+ln(-4*x*exp(3)*ln(x)+4*x^2*ln(x)-4*x^2+12*x*ln(x)+ln((-x*ln(5)+5)/ln(5))+5)

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maxima [A]  time = 0.53, size = 46, normalized size = 1.28 \begin {gather*} \log \left (-4 \, x^{2} + 4 \, {\left (x^{2} - x {\left (e^{3} - 3\right )}\right )} \log \relax (x) + \log \left (-x \log \relax (5) + 5\right ) - \log \left (\log \relax (5)\right ) + 5\right ) - \log \relax (x) - \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^3*log(5)+20*x^2)*log(x)^2+((x*log(5)-5)*log((-x*log(5)+5)/log(5))+(4*x^3+4*x)*log(5)-20*x^2-2
5)*log(x)+(x*log(5)-5)*log((-x*log(5)+5)/log(5))+(-4*x^3+5*x)*log(5)+20*x^2-25)/(((4*x^3*exp(3)-4*x^4-12*x^3)*
log(5)-20*x^2*exp(3)+20*x^3+60*x^2)*log(x)^2+((-x^2*log(5)+5*x)*log((-x*log(5)+5)/log(5))+(4*x^4-5*x^2)*log(5)
-20*x^3+25*x)*log(x)),x, algorithm="maxima")

[Out]

log(-4*x^2 + 4*(x^2 - x*(e^3 - 3))*log(x) + log(-x*log(5) + 5) - log(log(5)) + 5) - log(x) - log(log(x))

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mupad [B]  time = 4.84, size = 50, normalized size = 1.39 \begin {gather*} \ln \left (\ln \left (5-x\,\ln \relax (5)\right )-\ln \left (\ln \relax (5)\right )+4\,x^2\,\ln \relax (x)+12\,x\,\ln \relax (x)-4\,x^2-4\,x\,{\mathrm {e}}^3\,\ln \relax (x)+5\right )-\ln \left (\ln \relax (x)\right )-\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(5)*(5*x - 4*x^3) + log(-(x*log(5) - 5)/log(5))*(x*log(5) - 5) + log(x)*(log(5)*(4*x + 4*x^3) + log(-
(x*log(5) - 5)/log(5))*(x*log(5) - 5) - 20*x^2 - 25) + 20*x^2 - log(x)^2*(4*x^3*log(5) - 20*x^2) - 25)/(log(x)
^2*(log(5)*(12*x^3 - 4*x^3*exp(3) + 4*x^4) + 20*x^2*exp(3) - 60*x^2 - 20*x^3) - log(x)*(25*x + log(-(x*log(5)
- 5)/log(5))*(5*x - x^2*log(5)) - log(5)*(5*x^2 - 4*x^4) - 20*x^3)),x)

[Out]

log(log(5 - x*log(5)) - log(log(5)) + 4*x^2*log(x) + 12*x*log(x) - 4*x^2 - 4*x*exp(3)*log(x) + 5) - log(log(x)
) - log(x)

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sympy [A]  time = 1.00, size = 53, normalized size = 1.47 \begin {gather*} - \log {\relax (x )} + \log {\left (4 x^{2} \log {\relax (x )} - 4 x^{2} - 4 x e^{3} \log {\relax (x )} + 12 x \log {\relax (x )} + \log {\left (\frac {- x \log {\relax (5 )} + 5}{\log {\relax (5 )}} \right )} + 5 \right )} - \log {\left (\log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x**3*ln(5)+20*x**2)*ln(x)**2+((x*ln(5)-5)*ln((-x*ln(5)+5)/ln(5))+(4*x**3+4*x)*ln(5)-20*x**2-25)
*ln(x)+(x*ln(5)-5)*ln((-x*ln(5)+5)/ln(5))+(-4*x**3+5*x)*ln(5)+20*x**2-25)/(((4*x**3*exp(3)-4*x**4-12*x**3)*ln(
5)-20*x**2*exp(3)+20*x**3+60*x**2)*ln(x)**2+((-x**2*ln(5)+5*x)*ln((-x*ln(5)+5)/ln(5))+(4*x**4-5*x**2)*ln(5)-20
*x**3+25*x)*ln(x)),x)

[Out]

-log(x) + log(4*x**2*log(x) - 4*x**2 - 4*x*exp(3)*log(x) + 12*x*log(x) + log((-x*log(5) + 5)/log(5)) + 5) - lo
g(log(x))

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