3.87.82 \(\int \frac {-118-125 x-123 x^2}{25 x^2} \, dx\)

Optimal. Leaf size=29 \[ -5 x+\frac {2 \left (2+x+\frac {1}{25} (3+x)^2\right )}{x}+5 (4-\log (x)) \]

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Rubi [A]  time = 0.01, antiderivative size = 17, normalized size of antiderivative = 0.59, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {12, 14} \begin {gather*} -\frac {123 x}{25}+\frac {118}{25 x}-5 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-118 - 125*x - 123*x^2)/(25*x^2),x]

[Out]

118/(25*x) - (123*x)/25 - 5*Log[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}{25} \int \frac {-118-125 x-123 x^2}{x^2} \, dx\\ &=\frac {1}{25} \int \left (-123-\frac {118}{x^2}-\frac {125}{x}\right ) \, dx\\ &=\frac {118}{25 x}-\frac {123 x}{25}-5 \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 17, normalized size = 0.59 \begin {gather*} \frac {1}{25} \left (\frac {118}{x}-123 x-125 \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-118 - 125*x - 123*x^2)/(25*x^2),x]

[Out]

(118/x - 123*x - 125*Log[x])/25

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fricas [A]  time = 0.81, size = 17, normalized size = 0.59 \begin {gather*} -\frac {123 \, x^{2} + 125 \, x \log \relax (x) - 118}{25 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(-123*x^2-125*x-118)/x^2,x, algorithm="fricas")

[Out]

-1/25*(123*x^2 + 125*x*log(x) - 118)/x

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giac [A]  time = 0.14, size = 14, normalized size = 0.48 \begin {gather*} -\frac {123}{25} \, x + \frac {118}{25 \, x} - 5 \, \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(-123*x^2-125*x-118)/x^2,x, algorithm="giac")

[Out]

-123/25*x + 118/25/x - 5*log(abs(x))

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maple [A]  time = 0.02, size = 14, normalized size = 0.48




method result size



default \(-\frac {123 x}{25}-5 \ln \relax (x )+\frac {118}{25 x}\) \(14\)
risch \(-\frac {123 x}{25}-5 \ln \relax (x )+\frac {118}{25 x}\) \(14\)
norman \(\frac {\frac {118}{25}-\frac {123 x^{2}}{25}}{x}-5 \ln \relax (x )\) \(17\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/25*(-123*x^2-125*x-118)/x^2,x,method=_RETURNVERBOSE)

[Out]

-123/25*x-5*ln(x)+118/25/x

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maxima [A]  time = 0.35, size = 13, normalized size = 0.45 \begin {gather*} -\frac {123}{25} \, x + \frac {118}{25 \, x} - 5 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(-123*x^2-125*x-118)/x^2,x, algorithm="maxima")

[Out]

-123/25*x + 118/25/x - 5*log(x)

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mupad [B]  time = 5.89, size = 13, normalized size = 0.45 \begin {gather*} \frac {118}{25\,x}-5\,\ln \relax (x)-\frac {123\,x}{25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*x + (123*x^2)/25 + 118/25)/x^2,x)

[Out]

118/(25*x) - 5*log(x) - (123*x)/25

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sympy [A]  time = 0.07, size = 14, normalized size = 0.48 \begin {gather*} - \frac {123 x}{25} - 5 \log {\relax (x )} + \frac {118}{25 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(-123*x**2-125*x-118)/x**2,x)

[Out]

-123*x/25 - 5*log(x) + 118/(25*x)

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