3.23.66 \(\int \frac {3 x^2+\log (3) (-x^2-16 \log ^4(\frac {125}{3}))}{x^2 \log (3)} \, dx\) [2266]

Optimal. Leaf size=23 \[ -3-x+\frac {3 x}{\log (3)}+\frac {16 \log ^4\left (\frac {125}{3}\right )}{x} \]

[Out]

16*ln(3/125)^4/x-3+3*x/ln(3)-x

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Rubi [A]
time = 0.01, antiderivative size = 24, normalized size of antiderivative = 1.04, number of steps used = 3, number of rules used = 2, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {12, 14} \begin {gather*} \frac {16 \log ^4\left (\frac {125}{3}\right )}{x}+\frac {x (3-\log (3))}{\log (3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3*x^2 + Log[3]*(-x^2 - 16*Log[125/3]^4))/(x^2*Log[3]),x]

[Out]

(x*(3 - Log[3]))/Log[3] + (16*Log[125/3]^4)/x

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]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {3 x^2+\log (3) \left (-x^2-16 \log ^4\left (\frac {125}{3}\right )\right )}{x^2} \, dx}{\log (3)}\\ &=\frac {\int \left (3 \left (1-\frac {\log (3)}{3}\right )-\frac {16 \log (3) \log ^4\left (\frac {125}{3}\right )}{x^2}\right ) \, dx}{\log (3)}\\ &=\frac {x (3-\log (3))}{\log (3)}+\frac {16 \log ^4\left (\frac {125}{3}\right )}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.00, size = 22, normalized size = 0.96 \begin {gather*} -x+\frac {3 x}{\log (3)}+\frac {16 \log ^4\left (\frac {125}{3}\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3*x^2 + Log[3]*(-x^2 - 16*Log[125/3]^4))/(x^2*Log[3]),x]

[Out]

-x + (3*x)/Log[3] + (16*Log[125/3]^4)/x

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Maple [A]
time = 0.12, size = 26, normalized size = 1.13

method	result	size
default	\(\frac {3 x -x \ln \left (3\right )+\frac {16 \ln \left (3\right ) \ln \left (\frac {3}{125}\right )^{4}}{x}}{\ln \left (3\right )}\)	\(26\)
gosper	\(\frac {16 \ln \left (3\right ) \ln \left (\frac {3}{125}\right )^{4}-x^{2} \ln \left (3\right )+3 x^{2}}{\ln \left (3\right ) x}\)	\(30\)
norman	\(\frac {-\frac {\left (\ln \left (3\right )-3\right ) x^{2}}{\ln \left (3\right )}+1296 \ln \left (5\right )^{4}-1728 \ln \left (5\right )^{3} \ln \left (3\right )+864 \ln \left (5\right )^{2} \ln \left (3\right )^{2}-192 \ln \left (5\right ) \ln \left (3\right )^{3}+16 \ln \left (3\right )^{4}}{x}\)	\(57\)
risch	\(\frac {3 x}{\ln \left (3\right )}-x +\frac {1296 \ln \left (5\right )^{4}}{x}-\frac {1728 \ln \left (3\right ) \ln \left (5\right )^{3}}{x}+\frac {864 \ln \left (3\right )^{2} \ln \left (5\right )^{2}}{x}-\frac {192 \ln \left (3\right )^{3} \ln \left (5\right )}{x}+\frac {16 \ln \left (3\right )^{4}}{x}\)	\(65\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-16*ln(3/125)^4-x^2)*ln(3)+3*x^2)/x^2/ln(3),x,method=_RETURNVERBOSE)

[Out]

1/ln(3)*(3*x-x*ln(3)+16*ln(3)*ln(3/125)^4/x)

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Maxima [A]
time = 0.26, size = 24, normalized size = 1.04 \begin {gather*} \frac {\frac {16 \, \log \left (3\right ) \log \left (\frac {3}{125}\right )^{4}}{x} - x {\left (\log \left (3\right ) - 3\right )}}{\log \left (3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*log(3/125)^4-x^2)*log(3)+3*x^2)/x^2/log(3),x, algorithm="maxima")

[Out]

(16*log(3)*log(3/125)^4/x - x*(log(3) - 3))/log(3)

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Fricas [A]
time = 0.39, size = 29, normalized size = 1.26 \begin {gather*} \frac {3 \, x^{2} + {\left (16 \, \log \left (\frac {3}{125}\right )^{4} - x^{2}\right )} \log \left (3\right )}{x \log \left (3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*log(3/125)^4-x^2)*log(3)+3*x^2)/x^2/log(3),x, algorithm="fricas")

[Out]

(3*x^2 + (16*log(3/125)^4 - x^2)*log(3))/(x*log(3))

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (19) = 38\).
time = 0.10, size = 61, normalized size = 2.65 \begin {gather*} \frac {x \left (3 - \log {\left (3 \right )}\right ) + \frac {- 1728 \log {\left (3 \right )}^{2} \log {\left (5 \right )}^{3} - 192 \log {\left (3 \right )}^{4} \log {\left (5 \right )} + 16 \log {\left (3 \right )}^{5} + 864 \log {\left (3 \right )}^{3} \log {\left (5 \right )}^{2} + 1296 \log {\left (3 \right )} \log {\left (5 \right )}^{4}}{x}}{\log {\left (3 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*ln(3/125)**4-x**2)*ln(3)+3*x**2)/x**2/ln(3),x)

[Out]

(x*(3 - log(3)) + (-1728*log(3)**2*log(5)**3 - 192*log(3)**4*log(5) + 16*log(3)**5 + 864*log(3)**3*log(5)**2 +
1296*log(3)*log(5)**4)/x)/log(3)

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Giac [A]
time = 0.40, size = 25, normalized size = 1.09 \begin {gather*} \frac {\frac {16 \, \log \left (3\right ) \log \left (\frac {3}{125}\right )^{4}}{x} - x \log \left (3\right ) + 3 \, x}{\log \left (3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*log(3/125)^4-x^2)*log(3)+3*x^2)/x^2/log(3),x, algorithm="giac")

[Out]

(16*log(3)*log(3/125)^4/x - x*log(3) + 3*x)/log(3)

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Mupad [B]
time = 0.05, size = 21, normalized size = 0.91 \begin {gather*} \frac {16\,{\ln \left (\frac {3}{125}\right )}^4}{x}-\frac {x\,\left (\ln \left (3\right )-3\right )}{\ln \left (3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(3)*(16*log(3/125)^4 + x^2) - 3*x^2)/(x^2*log(3)),x)

[Out]

(16*log(3/125)^4)/x - (x*(log(3) - 3))/log(3)

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