3.86.54 \(\int \frac {-50+x \log ^3(x)+162 \log ^4(x)}{x \log ^3(x)} \, dx\)

Optimal. Leaf size=17 \[ x+\frac {\left (-5-9 \log ^2(x)\right )^2}{\log ^2(x)} \]

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Rubi [A]  time = 0.12, antiderivative size = 14, normalized size of antiderivative = 0.82, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {6742, 2302, 30, 2301} \begin {gather*} x+81 \log ^2(x)+\frac {25}{\log ^2(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-50 + x*Log[x]^3 + 162*Log[x]^4)/(x*Log[x]^3),x]

[Out]

x + 25/Log[x]^2 + 81*Log[x]^2

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, 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 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 (1-\frac {50}{x \log ^3(x)}+\frac {162 \log (x)}{x}\right ) \, dx\\ &=x-50 \int \frac {1}{x \log ^3(x)} \, dx+162 \int \frac {\log (x)}{x} \, dx\\ &=x+81 \log ^2(x)-50 \operatorname {Subst}\left (\int \frac {1}{x^3} \, dx,x,\log (x)\right )\\ &=x+\frac {25}{\log ^2(x)}+81 \log ^2(x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 14, normalized size = 0.82 \begin {gather*} x+\frac {25}{\log ^2(x)}+81 \log ^2(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-50 + x*Log[x]^3 + 162*Log[x]^4)/(x*Log[x]^3),x]

[Out]

x + 25/Log[x]^2 + 81*Log[x]^2

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fricas [A]  time = 0.68, size = 19, normalized size = 1.12 \begin {gather*} \frac {81 \, \log \relax (x)^{4} + x \log \relax (x)^{2} + 25}{\log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((162*log(x)^4+x*log(x)^3-50)/x/log(x)^3,x, algorithm="fricas")

[Out]

(81*log(x)^4 + x*log(x)^2 + 25)/log(x)^2

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giac [A]  time = 0.15, size = 14, normalized size = 0.82 \begin {gather*} 81 \, \log \relax (x)^{2} + x + \frac {25}{\log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((162*log(x)^4+x*log(x)^3-50)/x/log(x)^3,x, algorithm="giac")

[Out]

81*log(x)^2 + x + 25/log(x)^2

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maple [A]  time = 0.02, size = 15, normalized size = 0.88

method	result	size
default	\(81 \ln \relax (x )^{2}+x +\frac {25}{\ln \relax (x )^{2}}\)	\(15\)
risch	\(81 \ln \relax (x )^{2}+x +\frac {25}{\ln \relax (x )^{2}}\)	\(15\)
norman	\(\frac {25+x \ln \relax (x )^{2}+81 \ln \relax (x )^{4}}{\ln \relax (x )^{2}}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((162*ln(x)^4+x*ln(x)^3-50)/x/ln(x)^3,x,method=_RETURNVERBOSE)

[Out]

81*ln(x)^2+x+25/ln(x)^2

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maxima [A]  time = 0.40, size = 14, normalized size = 0.82 \begin {gather*} 81 \, \log \relax (x)^{2} + x + \frac {25}{\log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((162*log(x)^4+x*log(x)^3-50)/x/log(x)^3,x, algorithm="maxima")

[Out]

81*log(x)^2 + x + 25/log(x)^2

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mupad [B]  time = 5.20, size = 14, normalized size = 0.82 \begin {gather*} x+\frac {25}{{\ln \relax (x)}^2}+81\,{\ln \relax (x)}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*log(x)^3 + 162*log(x)^4 - 50)/(x*log(x)^3),x)

[Out]

x + 25/log(x)^2 + 81*log(x)^2

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sympy [A]  time = 0.10, size = 14, normalized size = 0.82 \begin {gather*} x + 81 \log {\relax (x )}^{2} + \frac {25}{\log {\relax (x )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((162*ln(x)**4+x*ln(x)**3-50)/x/ln(x)**3,x)

[Out]

x + 81*log(x)**2 + 25/log(x)**2

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