3.88.18 \(\int \frac {-5+(10+14 x^2+5 x^3-2 x^4) \log (x)+(-4-5 x+2 x^2) \log (x) \log (\frac {\log (x)}{x^2})}{-5 x^3 \log (x)+5 x \log (x) \log (\frac {\log (x)}{x^2})} \, dx\)

Optimal. Leaf size=35 \[ -x+\frac {1}{5} \left (x^2+\log (x)\right )-\log \left (\frac {1}{5} x \left (-x^2+\log \left (\frac {\log (x)}{x^2}\right )\right )\right ) \]

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Rubi [A]  time = 0.53, antiderivative size = 33, normalized size of antiderivative = 0.94, number of steps used = 7, number of rules used = 5, integrand size = 65, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {6741, 12, 6742, 14, 6684} \begin {gather*} \frac {x^2}{5}-\log \left (x^2-\log \left (\frac {\log (x)}{x^2}\right )\right )-x-\frac {4 \log (x)}{5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-5 + (10 + 14*x^2 + 5*x^3 - 2*x^4)*Log[x] + (-4 - 5*x + 2*x^2)*Log[x]*Log[Log[x]/x^2])/(-5*x^3*Log[x] + 5
*x*Log[x]*Log[Log[x]/x^2]),x]

[Out]

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

Rule 12

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 {5-\left (10+14 x^2+5 x^3-2 x^4\right ) \log (x)-\left (-4-5 x+2 x^2\right ) \log (x) \log \left (\frac {\log (x)}{x^2}\right )}{5 x \log (x) \left (x^2-\log \left (\frac {\log (x)}{x^2}\right )\right )} \, dx\\ &=\frac {1}{5} \int \frac {5-\left (10+14 x^2+5 x^3-2 x^4\right ) \log (x)-\left (-4-5 x+2 x^2\right ) \log (x) \log \left (\frac {\log (x)}{x^2}\right )}{x \log (x) \left (x^2-\log \left (\frac {\log (x)}{x^2}\right )\right )} \, dx\\ &=\frac {1}{5} \int \left (\frac {-4-5 x+2 x^2}{x}-\frac {5 \left (-1+2 \log (x)+2 x^2 \log (x)\right )}{x \log (x) \left (x^2-\log \left (\frac {\log (x)}{x^2}\right )\right )}\right ) \, dx\\ &=\frac {1}{5} \int \frac {-4-5 x+2 x^2}{x} \, dx-\int \frac {-1+2 \log (x)+2 x^2 \log (x)}{x \log (x) \left (x^2-\log \left (\frac {\log (x)}{x^2}\right )\right )} \, dx\\ &=-\log \left (x^2-\log \left (\frac {\log (x)}{x^2}\right )\right )+\frac {1}{5} \int \left (-5-\frac {4}{x}+2 x\right ) \, dx\\ &=-x+\frac {x^2}{5}-\frac {4 \log (x)}{5}-\log \left (x^2-\log \left (\frac {\log (x)}{x^2}\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 33, normalized size = 0.94 \begin {gather*} -x+\frac {x^2}{5}-\frac {4 \log (x)}{5}-\log \left (x^2-\log \left (\frac {\log (x)}{x^2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5 + (10 + 14*x^2 + 5*x^3 - 2*x^4)*Log[x] + (-4 - 5*x + 2*x^2)*Log[x]*Log[Log[x]/x^2])/(-5*x^3*Log[
x] + 5*x*Log[x]*Log[Log[x]/x^2]),x]

[Out]

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

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fricas [A]  time = 0.61, size = 29, normalized size = 0.83 \begin {gather*} \frac {1}{5} \, x^{2} - x - \log \left (-x^{2} + \log \left (\frac {\log \relax (x)}{x^{2}}\right )\right ) - \frac {4}{5} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2-5*x-4)*log(x)*log(log(x)/x^2)+(-2*x^4+5*x^3+14*x^2+10)*log(x)-5)/(5*x*log(x)*log(log(x)/x^2)
-5*x^3*log(x)),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.19, size = 29, normalized size = 0.83 \begin {gather*} \frac {1}{5} \, x^{2} - x - \log \left (x^{2} + 2 \, \log \relax (x) - \log \left (\log \relax (x)\right )\right ) - \frac {4}{5} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2-5*x-4)*log(x)*log(log(x)/x^2)+(-2*x^4+5*x^3+14*x^2+10)*log(x)-5)/(5*x*log(x)*log(log(x)/x^2)
-5*x^3*log(x)),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.09, size = 158, normalized size = 4.51

method	result	size
risch	\(\frac {x^{2}}{5}-x -\frac {4 \ln \relax (x )}{5}-\ln \left (\ln \left (\ln \relax (x )\right )+\frac {i \left (\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+\pi \,\mathrm {csgn}\left (i \ln \relax (x )\right ) \mathrm {csgn}\left (\frac {i \ln \relax (x )}{x^{2}}\right )^{2}-\pi \,\mathrm {csgn}\left (i \ln \relax (x )\right ) \mathrm {csgn}\left (\frac {i \ln \relax (x )}{x^{2}}\right ) \mathrm {csgn}\left (\frac {i}{x^{2}}\right )-\pi \mathrm {csgn}\left (\frac {i \ln \relax (x )}{x^{2}}\right )^{3}+\pi \mathrm {csgn}\left (\frac {i \ln \relax (x )}{x^{2}}\right )^{2} \mathrm {csgn}\left (\frac {i}{x^{2}}\right )+2 i x^{2}+4 i \ln \relax (x )\right )}{2}\right )\)	\(158\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^2-5*x-4)*ln(x)*ln(ln(x)/x^2)+(-2*x^4+5*x^3+14*x^2+10)*ln(x)-5)/(5*x*ln(x)*ln(ln(x)/x^2)-5*x^3*ln(x))
,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.40, size = 29, normalized size = 0.83 \begin {gather*} \frac {1}{5} \, x^{2} - x - \log \left (-x^{2} - 2 \, \log \relax (x) + \log \left (\log \relax (x)\right )\right ) - \frac {4}{5} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2-5*x-4)*log(x)*log(log(x)/x^2)+(-2*x^4+5*x^3+14*x^2+10)*log(x)-5)/(5*x*log(x)*log(log(x)/x^2)
-5*x^3*log(x)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.49, size = 29, normalized size = 0.83 \begin {gather*} \frac {x^2}{5}-\ln \left (\ln \left (\frac {\ln \relax (x)}{x^2}\right )-x^2\right )-\frac {4\,\ln \relax (x)}{5}-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(log(x)/x^2)*log(x)*(5*x - 2*x^2 + 4) - log(x)*(14*x^2 + 5*x^3 - 2*x^4 + 10) + 5)/(5*x^3*log(x) - 5*x*
log(log(x)/x^2)*log(x)),x)

[Out]

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

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sympy [A]  time = 0.35, size = 26, normalized size = 0.74 \begin {gather*} \frac {x^{2}}{5} - x - \frac {4 \log {\relax (x )}}{5} - \log {\left (- x^{2} + \log {\left (\frac {\log {\relax (x )}}{x^{2}} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**2-5*x-4)*ln(x)*ln(ln(x)/x**2)+(-2*x**4+5*x**3+14*x**2+10)*ln(x)-5)/(5*x*ln(x)*ln(ln(x)/x**2)-
5*x**3*ln(x)),x)

[Out]

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

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