3.8.27 \(\int \frac {-10+2 x+(20 x+2 x^2-x^3) \log (x)+(-15-3 x+x^2) \log (x) \log (\log ^2(x))}{(5 x^2-x^3) \log (x)+(-5 x+x^2) \log (x) \log (\log ^2(x))} \, dx\)

Optimal. Leaf size=23 \[ \log \left (\frac {2 e^x x^3 \left (-x+\log \left (\log ^2(x)\right )\right )}{-5+x}\right ) \]

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Rubi [A]  time = 1.23, antiderivative size = 24, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 4, integrand size = 71, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {6741, 6742, 893, 6684} \begin {gather*} x+\log \left (x-\log \left (\log ^2(x)\right )\right )-\log (5-x)+3 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-10 + 2*x + (20*x + 2*x^2 - x^3)*Log[x] + (-15 - 3*x + x^2)*Log[x]*Log[Log[x]^2])/((5*x^2 - x^3)*Log[x] +
(-5*x + x^2)*Log[x]*Log[Log[x]^2]),x]

[Out]

x - Log[5 - x] + 3*Log[x] + Log[x - Log[Log[x]^2]]

Rule 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 6684

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

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= 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 \frac {-10+2 x+\left (20 x+2 x^2-x^3\right ) \log (x)+\left (-15-3 x+x^2\right ) \log (x) \log \left (\log ^2(x)\right )}{(5-x) x \log (x) \left (x-\log \left (\log ^2(x)\right )\right )} \, dx\\ &=\int \left (\frac {-15-3 x+x^2}{(-5+x) x}+\frac {-2+x \log (x)}{x \log (x) \left (x-\log \left (\log ^2(x)\right )\right )}\right ) \, dx\\ &=\int \frac {-15-3 x+x^2}{(-5+x) x} \, dx+\int \frac {-2+x \log (x)}{x \log (x) \left (x-\log \left (\log ^2(x)\right )\right )} \, dx\\ &=\log \left (x-\log \left (\log ^2(x)\right )\right )+\int \left (1+\frac {1}{5-x}+\frac {3}{x}\right ) \, dx\\ &=x-\log (5-x)+3 \log (x)+\log \left (x-\log \left (\log ^2(x)\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 24, normalized size = 1.04 \begin {gather*} x-\log (5-x)+3 \log (x)+\log \left (x-\log \left (\log ^2(x)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10 + 2*x + (20*x + 2*x^2 - x^3)*Log[x] + (-15 - 3*x + x^2)*Log[x]*Log[Log[x]^2])/((5*x^2 - x^3)*Lo
g[x] + (-5*x + x^2)*Log[x]*Log[Log[x]^2]),x]

[Out]

x - Log[5 - x] + 3*Log[x] + Log[x - Log[Log[x]^2]]

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fricas [A]  time = 0.68, size = 22, normalized size = 0.96 \begin {gather*} x - \log \left (x - 5\right ) + 3 \, \log \relax (x) + \log \left (-x + \log \left (\log \relax (x)^{2}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2-3*x-15)*log(x)*log(log(x)^2)+(-x^3+2*x^2+20*x)*log(x)+2*x-10)/((x^2-5*x)*log(x)*log(log(x)^2)+
(-x^3+5*x^2)*log(x)),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.60, size = 22, normalized size = 0.96 \begin {gather*} x - \log \left (x - 5\right ) + 3 \, \log \relax (x) + \log \left (-x + \log \left (\log \relax (x)^{2}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2-3*x-15)*log(x)*log(log(x)^2)+(-x^3+2*x^2+20*x)*log(x)+2*x-10)/((x^2-5*x)*log(x)*log(log(x)^2)+
(-x^3+5*x^2)*log(x)),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.18, size = 75, normalized size = 3.26

method	result	size
risch	\(x -\ln \left (x -5\right )+3 \ln \relax (x )+\ln \left (\ln \left (\ln \relax (x )\right )-\frac {i \left (\pi \mathrm {csgn}\left (i \ln \relax (x )\right )^{2} \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )-2 \pi \,\mathrm {csgn}\left (i \ln \relax (x )\right ) \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )^{2}+\pi \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )^{3}-2 i x \right )}{4}\right )\)	\(75\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^2-3*x-15)*ln(x)*ln(ln(x)^2)+(-x^3+2*x^2+20*x)*ln(x)+2*x-10)/((x^2-5*x)*ln(x)*ln(ln(x)^2)+(-x^3+5*x^2)*
ln(x)),x,method=_RETURNVERBOSE)

[Out]

x-ln(x-5)+3*ln(x)+ln(ln(ln(x))-1/4*I*(Pi*csgn(I*ln(x))^2*csgn(I*ln(x)^2)-2*Pi*csgn(I*ln(x))*csgn(I*ln(x)^2)^2+
Pi*csgn(I*ln(x)^2)^3-2*I*x))

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maxima [A]  time = 0.56, size = 20, normalized size = 0.87 \begin {gather*} x - \log \left (x - 5\right ) + 3 \, \log \relax (x) + \log \left (-\frac {1}{2} \, x + \log \left (\log \relax (x)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2-3*x-15)*log(x)*log(log(x)^2)+(-x^3+2*x^2+20*x)*log(x)+2*x-10)/((x^2-5*x)*log(x)*log(log(x)^2)+
(-x^3+5*x^2)*log(x)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.65, size = 22, normalized size = 0.96 \begin {gather*} x-\ln \left (x-5\right )+\ln \left (\ln \left ({\ln \relax (x)}^2\right )-x\right )+3\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x + log(x)*(20*x + 2*x^2 - x^3) - log(log(x)^2)*log(x)*(3*x - x^2 + 15) - 10)/(log(x)*(5*x^2 - x^3) - l
og(log(x)^2)*log(x)*(5*x - x^2)),x)

[Out]

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

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sympy [A]  time = 0.35, size = 20, normalized size = 0.87 \begin {gather*} x + 3 \log {\relax (x )} + \log {\left (- x + \log {\left (\log {\relax (x )}^{2} \right )} \right )} - \log {\left (x - 5 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**2-3*x-15)*ln(x)*ln(ln(x)**2)+(-x**3+2*x**2+20*x)*ln(x)+2*x-10)/((x**2-5*x)*ln(x)*ln(ln(x)**2)+(
-x**3+5*x**2)*ln(x)),x)

[Out]

x + 3*log(x) + log(-x + log(log(x)**2)) - log(x - 5)

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