\(\int \frac {-4 x^2+\log ^2(2)}{4 x^2} \, dx\) [8456]

Optimal result

Integrand size = 17, antiderivative size = 19 \[ \int \frac {-4 x^2+\log ^2(2)}{4 x^2} \, dx=-\frac {19}{12}-\frac {\left (x+\frac {\log (2)}{2}\right )^2}{x} \]

[Out]

-19/12-(x+1/2*ln(2))^2/x

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {12, 14} \[ \int \frac {-4 x^2+\log ^2(2)}{4 x^2} \, dx=-x-\frac {\log ^2(2)}{4 x} \]

[In]

Int[(-4*x^2 + Log[2]^2)/(4*x^2),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {-4 x^2+\log ^2(2)}{4 x^2} \, dx=-x-\frac {\log ^2(2)}{4 x} \]

[In]

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

[Out]

-x - Log[2]^2/(4*x)

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74

method	result	size
default	\(-x -\frac {\ln \left (2\right )^{2}}{4 x}\)	\(14\)
risch	\(-x -\frac {\ln \left (2\right )^{2}}{4 x}\)	\(14\)
gosper	\(-\frac {\ln \left (2\right )^{2}+4 x^{2}}{4 x}\)	\(16\)
parallelrisch	\(-\frac {\ln \left (2\right )^{2}+4 x^{2}}{4 x}\)	\(16\)

[In]

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

[Out]

-x-1/4*ln(2)^2/x

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {-4 x^2+\log ^2(2)}{4 x^2} \, dx=-\frac {4 \, x^{2} + \log \left (2\right )^{2}}{4 \, x} \]

[In]

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

[Out]

-1/4*(4*x^2 + log(2)^2)/x

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.53 \[ \int \frac {-4 x^2+\log ^2(2)}{4 x^2} \, dx=- x - \frac {\log {\left (2 \right )}^{2}}{4 x} \]

[In]

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

[Out]

-x - log(2)**2/(4*x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \frac {-4 x^2+\log ^2(2)}{4 x^2} \, dx=-x - \frac {\log \left (2\right )^{2}}{4 \, x} \]

[In]

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

[Out]

-x - 1/4*log(2)^2/x

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \frac {-4 x^2+\log ^2(2)}{4 x^2} \, dx=-x - \frac {\log \left (2\right )^{2}}{4 \, x} \]

[In]

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

[Out]

-x - 1/4*log(2)^2/x

Mupad [B] (verification not implemented)

Time = 12.85 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \frac {-4 x^2+\log ^2(2)}{4 x^2} \, dx=-x-\frac {{\ln \left (2\right )}^2}{4\,x} \]

[In]

int((log(2)^2/4 - x^2)/x^2,x)

[Out]

- x - log(2)^2/(4*x)