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

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

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Rubi [A]  time = 0.02, 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 = 0.01, 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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fricas [A]  time = 0.66, 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 \relax (3)}{x \log \relax (3)} \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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giac [A]  time = 0.24, size = 25, normalized size = 1.09 \begin {gather*} \frac {\frac {16 \, \log \relax (3) \log \left (\frac {3}{125}\right )^{4}}{x} - x \log \relax (3) + 3 \, x}{\log \relax (3)} \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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maple [A]  time = 0.07, size = 26, normalized size = 1.13

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

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 = 0.35, size = 24, normalized size = 1.04 \begin {gather*} \frac {\frac {16 \, \log \relax (3) \log \left (\frac {3}{125}\right )^{4}}{x} - x {\left (\log \relax (3) - 3\right )}}{\log \relax (3)} \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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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 \relax (3)-3\right )}{\ln \relax (3)} \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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sympy [B]  time = 0.25, size = 61, normalized size = 2.65 \begin {gather*} \frac {x \left (3 - \log {\relax (3 )}\right ) + \frac {- 1728 \log {\relax (3 )}^{2} \log {\relax (5 )}^{3} - 192 \log {\relax (3 )}^{4} \log {\relax (5 )} + 16 \log {\relax (3 )}^{5} + 864 \log {\relax (3 )}^{3} \log {\relax (5 )}^{2} + 1296 \log {\relax (3 )} \log {\relax (5 )}^{4}}{x}}{\log {\relax (3 )}} \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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