3.7.88 \(\int \frac {x^3+4 \log (3)+512 x^2 \log (x)}{x^3} \, dx\)

Optimal. Leaf size=18 \[ 2+x-\frac {2 \log (3)}{x^2}+\log (5)+256 \log ^2(x) \]

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Rubi [A]  time = 0.02, antiderivative size = 17, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {14, 2301} \begin {gather*} -\frac {\log (81)}{2 x^2}+x+256 \log ^2(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^3 + 4*Log[3] + 512*x^2*Log[x])/x^3,x]

[Out]

x - Log[81]/(2*x^2) + 256*Log[x]^2

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 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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {x^3+\log (81)}{x^3}+\frac {512 \log (x)}{x}\right ) \, dx\\ &=512 \int \frac {\log (x)}{x} \, dx+\int \frac {x^3+\log (81)}{x^3} \, dx\\ &=256 \log ^2(x)+\int \left (1+\frac {\log (81)}{x^3}\right ) \, dx\\ &=x-\frac {\log (81)}{2 x^2}+256 \log ^2(x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 17, normalized size = 0.94 \begin {gather*} x-\frac {\log (81)}{2 x^2}+256 \log ^2(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^3 + 4*Log[3] + 512*x^2*Log[x])/x^3,x]

[Out]

x - Log[81]/(2*x^2) + 256*Log[x]^2

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fricas [A]  time = 0.98, size = 21, normalized size = 1.17 \begin {gather*} \frac {256 \, x^{2} \log \relax (x)^{2} + x^{3} - 2 \, \log \relax (3)}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((512*x^2*log(x)+4*log(3)+x^3)/x^3,x, algorithm="fricas")

[Out]

(256*x^2*log(x)^2 + x^3 - 2*log(3))/x^2

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giac [A]  time = 0.31, size = 15, normalized size = 0.83 \begin {gather*} 256 \, \log \relax (x)^{2} + x - \frac {2 \, \log \relax (3)}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((512*x^2*log(x)+4*log(3)+x^3)/x^3,x, algorithm="giac")

[Out]

256*log(x)^2 + x - 2*log(3)/x^2

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maple [A]  time = 0.03, size = 16, normalized size = 0.89

method	result	size
default	\(256 \ln \relax (x )^{2}-\frac {2 \ln \relax (3)}{x^{2}}+x\)	\(16\)
norman	\(\frac {x^{3}+256 x^{2} \ln \relax (x )^{2}-2 \ln \relax (3)}{x^{2}}\)	\(22\)
risch	\(256 \ln \relax (x )^{2}-\frac {-x^{3}+2 \ln \relax (3)}{x^{2}}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((512*x^2*ln(x)+4*ln(3)+x^3)/x^3,x,method=_RETURNVERBOSE)

[Out]

256*ln(x)^2-2/x^2*ln(3)+x

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maxima [A]  time = 0.48, size = 15, normalized size = 0.83 \begin {gather*} 256 \, \log \relax (x)^{2} + x - \frac {2 \, \log \relax (3)}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((512*x^2*log(x)+4*log(3)+x^3)/x^3,x, algorithm="maxima")

[Out]

256*log(x)^2 + x - 2*log(3)/x^2

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mupad [B]  time = 0.53, size = 15, normalized size = 0.83 \begin {gather*} x+256\,{\ln \relax (x)}^2-\frac {2\,\ln \relax (3)}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*log(3) + 512*x^2*log(x) + x^3)/x^3,x)

[Out]

x + 256*log(x)^2 - (2*log(3))/x^2

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sympy [A]  time = 0.11, size = 15, normalized size = 0.83 \begin {gather*} x + 256 \log {\relax (x )}^{2} - \frac {2 \log {\relax (3 )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((512*x**2*ln(x)+4*ln(3)+x**3)/x**3,x)

[Out]

x + 256*log(x)**2 - 2*log(3)/x**2

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