3.17.25 \(\int \frac {12+x^2+x^2 \log (3)}{4 x^2} \, dx\)

Optimal. Leaf size=19 \[ -5+\frac {-3+x}{x}+\frac {1}{4} (x+x \log (3)) \]

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Rubi [A]  time = 0.01, antiderivative size = 15, normalized size of antiderivative = 0.79, number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6, 12, 14} \begin {gather*} \frac {1}{4} x (1+\log (3))-\frac {3}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(12 + x^2 + x^2*Log[3])/(4*x^2),x]

[Out]

-3/x + (x*(1 + Log[3]))/4

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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} &=\int \frac {12+x^2 (1+\log (3))}{4 x^2} \, dx\\ &=\frac {1}{4} \int \frac {12+x^2 (1+\log (3))}{x^2} \, dx\\ &=\frac {1}{4} \int \left (1+\frac {12}{x^2}+\log (3)\right ) \, dx\\ &=-\frac {3}{x}+\frac {1}{4} x (1+\log (3))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 15, normalized size = 0.79 \begin {gather*} \frac {1}{4} \left (-\frac {12}{x}+x+x \log (3)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(12 + x^2 + x^2*Log[3])/(4*x^2),x]

[Out]

(-12/x + x + x*Log[3])/4

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fricas [A]  time = 0.88, size = 16, normalized size = 0.84 \begin {gather*} \frac {x^{2} \log \relax (3) + x^{2} - 12}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(x^2*log(3)+x^2+12)/x^2,x, algorithm="fricas")

[Out]

1/4*(x^2*log(3) + x^2 - 12)/x

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giac [A]  time = 0.29, size = 14, normalized size = 0.74 \begin {gather*} \frac {1}{4} \, x \log \relax (3) + \frac {1}{4} \, x - \frac {3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(x^2*log(3)+x^2+12)/x^2,x, algorithm="giac")

[Out]

1/4*x*log(3) + 1/4*x - 3/x

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maple [A]  time = 0.03, size = 15, normalized size = 0.79

method	result	size
default	\(\frac {x \ln \relax (3)}{4}+\frac {x}{4}-\frac {3}{x}\)	\(15\)
risch	\(\frac {x \ln \relax (3)}{4}+\frac {x}{4}-\frac {3}{x}\)	\(15\)
gosper	\(\frac {x^{2} \ln \relax (3)+x^{2}-12}{4 x}\)	\(17\)
norman	\(\frac {-3+\left (\frac {\ln \relax (3)}{4}+\frac {1}{4}\right ) x^{2}}{x}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(x^2*ln(3)+x^2+12)/x^2,x,method=_RETURNVERBOSE)

[Out]

1/4*x*ln(3)+1/4*x-3/x

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maxima [A]  time = 0.34, size = 13, normalized size = 0.68 \begin {gather*} \frac {1}{4} \, x {\left (\log \relax (3) + 1\right )} - \frac {3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(x^2*log(3)+x^2+12)/x^2,x, algorithm="maxima")

[Out]

1/4*x*(log(3) + 1) - 3/x

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mupad [B]  time = 0.04, size = 14, normalized size = 0.74 \begin {gather*} x\,\left (\frac {\ln \relax (3)}{4}+\frac {1}{4}\right )-\frac {3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*(log(3)/4 + 1/4) - 3/x

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sympy [A]  time = 0.08, size = 10, normalized size = 0.53 \begin {gather*} \frac {x \left (1 + \log {\relax (3 )}\right )}{4} - \frac {3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(x**2*ln(3)+x**2+12)/x**2,x)

[Out]

x*(1 + log(3))/4 - 3/x

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