3.15.23 \(\int \frac {(1-2 x^2) \log (5)}{4 x^2} \, dx\)

Optimal. Leaf size=28 \[ \frac {1}{2} \left (-\frac {10-4 x}{20 x}-x-\log (2)+\log (4)\right ) \log (5) \]

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Rubi [A]  time = 0.00, antiderivative size = 17, normalized size of antiderivative = 0.61, number of steps used = 3, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 14} \begin {gather*} -\frac {1}{2} x \log (5)-\frac {\log (5)}{4 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((1 - 2*x^2)*Log[5])/(4*x^2),x]

[Out]

-1/4*Log[5]/x - (x*Log[5])/2

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} \log (5) \int \frac {1-2 x^2}{x^2} \, dx\\ &=\frac {1}{4} \log (5) \int \left (-2+\frac {1}{x^2}\right ) \, dx\\ &=-\frac {\log (5)}{4 x}-\frac {1}{2} x \log (5)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 15, normalized size = 0.54 \begin {gather*} \frac {1}{4} \left (-\frac {1}{x}-2 x\right ) \log (5) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((1 - 2*x^2)*Log[5])/(4*x^2),x]

[Out]

((-x^(-1) - 2*x)*Log[5])/4

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fricas [A]  time = 0.77, size = 14, normalized size = 0.50 \begin {gather*} -\frac {{\left (2 \, x^{2} + 1\right )} \log \relax (5)}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-2*x^2+1)*log(5)/x^2,x, algorithm="fricas")

[Out]

-1/4*(2*x^2 + 1)*log(5)/x

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giac [A]  time = 0.53, size = 11, normalized size = 0.39 \begin {gather*} -\frac {1}{4} \, {\left (2 \, x + \frac {1}{x}\right )} \log \relax (5) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-2*x^2+1)*log(5)/x^2,x, algorithm="giac")

[Out]

-1/4*(2*x + 1/x)*log(5)

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maple [A]  time = 0.02, size = 14, normalized size = 0.50

method	result	size
default	\(\frac {\ln \relax (5) \left (-2 x -\frac {1}{x}\right )}{4}\)	\(14\)
risch	\(-\frac {x \ln \relax (5)}{2}-\frac {\ln \relax (5)}{4 x}\)	\(14\)
gosper	\(-\frac {\ln \relax (5) \left (2 x^{2}+1\right )}{4 x}\)	\(15\)
norman	\(\frac {-\frac {x^{2} \ln \relax (5)}{2}-\frac {\ln \relax (5)}{4}}{x}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(-2*x^2+1)*ln(5)/x^2,x,method=_RETURNVERBOSE)

[Out]

1/4*ln(5)*(-2*x-1/x)

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maxima [A]  time = 0.55, size = 11, normalized size = 0.39 \begin {gather*} -\frac {1}{4} \, {\left (2 \, x + \frac {1}{x}\right )} \log \relax (5) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-2*x^2+1)*log(5)/x^2,x, algorithm="maxima")

[Out]

-1/4*(2*x + 1/x)*log(5)

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mupad [B]  time = 0.04, size = 14, normalized size = 0.50 \begin {gather*} -\frac {\ln \relax (5)\,\left (2\,x^2+1\right )}{4\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(5)*(2*x^2 - 1))/(4*x^2),x)

[Out]

-(log(5)*(2*x^2 + 1))/(4*x)

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sympy [A]  time = 0.08, size = 14, normalized size = 0.50 \begin {gather*} - \frac {x \log {\relax (5 )}}{2} - \frac {\log {\relax (5 )}}{4 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-2*x**2+1)*ln(5)/x**2,x)

[Out]

-x*log(5)/2 - log(5)/(4*x)

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