3.31.3 \(\int \frac {-2 x+(-1+2 x-4 x^2) \log (x)+\log (x) \log (5 x)}{x^2 \log (x)} \, dx\)

Optimal. Leaf size=26 \[ 4-4 x-\frac {\log (5 x)}{x}+\log \left (\frac {e^3 x^2}{\log ^2(x)}\right ) \]

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Rubi [A]  time = 0.26, antiderivative size = 22, normalized size of antiderivative = 0.85, number of steps used = 9, number of rules used = 5, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {6742, 14, 2302, 29, 2304} \begin {gather*} -4 x+2 \log (x)-2 \log (\log (x))-\frac {\log (5 x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-4*x + 2*Log[x] - Log[5*x]/x - 2*Log[Log[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 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2302

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

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, 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} &=\int \left (\frac {-2 x-\log (x)+2 x \log (x)-4 x^2 \log (x)}{x^2 \log (x)}+\frac {\log (5 x)}{x^2}\right ) \, dx\\ &=\int \frac {-2 x-\log (x)+2 x \log (x)-4 x^2 \log (x)}{x^2 \log (x)} \, dx+\int \frac {\log (5 x)}{x^2} \, dx\\ &=-\frac {1}{x}-\frac {\log (5 x)}{x}+\int \left (\frac {-1+2 x-4 x^2}{x^2}-\frac {2}{x \log (x)}\right ) \, dx\\ &=-\frac {1}{x}-\frac {\log (5 x)}{x}-2 \int \frac {1}{x \log (x)} \, dx+\int \frac {-1+2 x-4 x^2}{x^2} \, dx\\ &=-\frac {1}{x}-\frac {\log (5 x)}{x}-2 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right )+\int \left (-4-\frac {1}{x^2}+\frac {2}{x}\right ) \, dx\\ &=-4 x+2 \log (x)-\frac {\log (5 x)}{x}-2 \log (\log (x))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 22, normalized size = 0.85 \begin {gather*} -4 x+2 \log (x)-\frac {\log (5 x)}{x}-2 \log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-4*x + 2*Log[x] - Log[5*x]/x - 2*Log[Log[x]]

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fricas [A]  time = 0.65, size = 28, normalized size = 1.08 \begin {gather*} -\frac {4 \, x^{2} - {\left (2 \, x - 1\right )} \log \relax (x) + 2 \, x \log \left (\log \relax (x)\right ) + \log \relax (5)}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.26, size = 27, normalized size = 1.04 \begin {gather*} -4 \, x - \frac {\log \relax (5)}{x} - \frac {\log \relax (x)}{x} + 2 \, \log \relax (x) - 2 \, \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 27, normalized size = 1.04

method	result	size
norman	\(\frac {-4 x^{2}-\ln \left (5 x \right )}{x}+2 \ln \relax (x )-2 \ln \left (\ln \relax (x )\right )\)	\(27\)
default	\(-4 x -\frac {\ln \relax (5)}{x}-\frac {\ln \relax (x )}{x}+2 \ln \relax (x )-2 \ln \left (\ln \relax (x )\right )\)	\(28\)
risch	\(-\frac {\ln \relax (x )}{x}-\frac {-4 x \ln \relax (x )+8 x^{2}+2 \ln \relax (5)}{2 x}-2 \ln \left (\ln \relax (x )\right )\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.43, size = 22, normalized size = 0.85 \begin {gather*} -4 \, x - \frac {\log \left (5 \, x\right )}{x} + 2 \, \log \relax (x) - 2 \, \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.79, size = 23, normalized size = 0.88 \begin {gather*} 2\,\ln \relax (x)-2\,\ln \left (\ln \relax (x)\right )-4\,x-\frac {\ln \relax (5)+\ln \relax (x)}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.27, size = 24, normalized size = 0.92 \begin {gather*} - 4 x + 2 \log {\relax (x )} - 2 \log {\left (\log {\relax (x )} \right )} - \frac {\log {\relax (x )}}{x} - \frac {\log {\relax (5 )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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