3.19.86 \(\int \frac {-x+25 x^2-8 x^3-\log (3)}{4 x^2} \, dx\)

Optimal. Leaf size=27 \[ x \left (6-x+\frac {4+x+\frac {\log (3)}{x}-\log (x)}{4 x}\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 26, normalized size of antiderivative = 0.96, number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {12, 14} \begin {gather*} -x^2+\frac {25 x}{4}-\frac {\log (x)}{4}+\frac {\log (3)}{4 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-x + 25*x^2 - 8*x^3 - Log[3])/(4*x^2),x]

[Out]

(25*x)/4 - x^2 + Log[3]/(4*x) - Log[x]/4

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 {-x+25 x^2-8 x^3-\log (3)}{x^2} \, dx\\ &=\frac {1}{4} \int \left (25-\frac {1}{x}-8 x-\frac {\log (3)}{x^2}\right ) \, dx\\ &=\frac {25 x}{4}-x^2+\frac {\log (3)}{4 x}-\frac {\log (x)}{4}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 23, normalized size = 0.85 \begin {gather*} \frac {1}{4} \left (25 x-4 x^2+\frac {\log (3)}{x}-\log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-x + 25*x^2 - 8*x^3 - Log[3])/(4*x^2),x]

[Out]

(25*x - 4*x^2 + Log[3]/x - Log[x])/4

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fricas [A]  time = 1.00, size = 24, normalized size = 0.89 \begin {gather*} -\frac {4 \, x^{3} - 25 \, x^{2} + x \log \relax (x) - \log \relax (3)}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-log(3)-8*x^3+25*x^2-x)/x^2,x, algorithm="fricas")

[Out]

-1/4*(4*x^3 - 25*x^2 + x*log(x) - log(3))/x

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giac [A]  time = 0.26, size = 21, normalized size = 0.78 \begin {gather*} -x^{2} + \frac {25}{4} \, x + \frac {\log \relax (3)}{4 \, x} - \frac {1}{4} \, \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-log(3)-8*x^3+25*x^2-x)/x^2,x, algorithm="giac")

[Out]

-x^2 + 25/4*x + 1/4*log(3)/x - 1/4*log(abs(x))

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maple [A]  time = 0.04, size = 21, normalized size = 0.78

method	result	size
default	\(\frac {25 x}{4}-x^{2}-\frac {\ln \relax (x )}{4}+\frac {\ln \relax (3)}{4 x}\)	\(21\)
risch	\(\frac {25 x}{4}-x^{2}-\frac {\ln \relax (x )}{4}+\frac {\ln \relax (3)}{4 x}\)	\(21\)
norman	\(\frac {\frac {25 x^{2}}{4}-x^{3}+\frac {\ln \relax (3)}{4}}{x}-\frac {\ln \relax (x )}{4}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(-ln(3)-8*x^3+25*x^2-x)/x^2,x,method=_RETURNVERBOSE)

[Out]

25/4*x-x^2-1/4*ln(x)+1/4*ln(3)/x

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maxima [A]  time = 0.54, size = 20, normalized size = 0.74 \begin {gather*} -x^{2} + \frac {25}{4} \, x + \frac {\log \relax (3)}{4 \, x} - \frac {1}{4} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-log(3)-8*x^3+25*x^2-x)/x^2,x, algorithm="maxima")

[Out]

-x^2 + 25/4*x + 1/4*log(3)/x - 1/4*log(x)

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mupad [B]  time = 1.12, size = 20, normalized size = 0.74 \begin {gather*} \frac {25\,x}{4}-\frac {\ln \relax (x)}{4}+\frac {\ln \relax (3)}{4\,x}-x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x/4 + log(3)/4 - (25*x^2)/4 + 2*x^3)/x^2,x)

[Out]

(25*x)/4 - log(x)/4 + log(3)/(4*x) - x^2

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sympy [A]  time = 0.09, size = 19, normalized size = 0.70 \begin {gather*} - x^{2} + \frac {25 x}{4} - \frac {\log {\relax (x )}}{4} + \frac {\log {\relax (3 )}}{4 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-ln(3)-8*x**3+25*x**2-x)/x**2,x)

[Out]

-x**2 + 25*x/4 - log(x)/4 + log(3)/(4*x)

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