3.60.97 \(\int \frac {4 x^2+(-1+2 x^2) \log (4)}{4 x^2} \, dx\)

Optimal. Leaf size=27 \[ -3+x-\frac {x-\log (4)}{4 x}-\frac {1}{2} (2-x) \log (4) \]

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Rubi [A]  time = 0.01, antiderivative size = 16, normalized size of antiderivative = 0.59, number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {12, 14} \begin {gather*} x (1+\log (2))+\frac {\log (4)}{4 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4*x^2 + (-1 + 2*x^2)*Log[4])/(4*x^2),x]

[Out]

x*(1 + Log[2]) + Log[4]/(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 {1}{4} \int \frac {4 x^2+\left (-1+2 x^2\right ) \log (4)}{x^2} \, dx\\ &=\frac {1}{4} \int \left (4 (1+\log (2))-\frac {\log (4)}{x^2}\right ) \, dx\\ &=x (1+\log (2))+\frac {\log (4)}{4 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 15, normalized size = 0.56 \begin {gather*} x+x \log (2)+\frac {\log (4)}{4 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4*x^2 + (-1 + 2*x^2)*Log[4])/(4*x^2),x]

[Out]

x + x*Log[2] + Log[4]/(4*x)

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fricas [A]  time = 0.53, size = 21, normalized size = 0.78 \begin {gather*} \frac {2 \, x^{2} + {\left (2 \, x^{2} + 1\right )} \log \relax (2)}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*x^2 + (2*x^2 + 1)*log(2))/x

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giac [A]  time = 0.15, size = 13, normalized size = 0.48 \begin {gather*} x \log \relax (2) + x + \frac {\log \relax (2)}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*log(2) + x + 1/2*log(2)/x

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maple [A]  time = 0.04, size = 14, normalized size = 0.52

method	result	size
default	\(x +x \ln \relax (2)+\frac {\ln \relax (2)}{2 x}\)	\(14\)
risch	\(x +x \ln \relax (2)+\frac {\ln \relax (2)}{2 x}\)	\(14\)
norman	\(\frac {\left (1+\ln \relax (2)\right ) x^{2}+\frac {\ln \relax (2)}{2}}{x}\)	\(18\)
gosper	\(\frac {2 x^{2} \ln \relax (2)+2 x^{2}+\ln \relax (2)}{2 x}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x+x*ln(2)+1/2*ln(2)/x

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maxima [A]  time = 0.34, size = 14, normalized size = 0.52 \begin {gather*} x {\left (\log \relax (2) + 1\right )} + \frac {\log \relax (2)}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*(log(2) + 1) + 1/2*log(2)/x

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mupad [B]  time = 0.07, size = 16, normalized size = 0.59 \begin {gather*} x\,\left (\frac {\ln \relax (4)}{2}+1\right )+\frac {\ln \relax (2)}{2\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((log(2)*(2*x^2 - 1))/2 + x^2)/x^2,x)

[Out]

x*(log(4)/2 + 1) + log(2)/(2*x)

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sympy [A]  time = 0.08, size = 15, normalized size = 0.56 \begin {gather*} \frac {x \left (2 \log {\relax (2 )} + 2\right )}{2} + \frac {\log {\relax (2 )}}{2 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*(2*log(2) + 2)/2 + log(2)/(2*x)

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