3.53.84 \(\int \frac {10-2 x^3-5 \log (x)}{x^2} \, dx\)

Optimal. Leaf size=22 \[ -e^4-x^2-\frac {5 (1-\log (x))}{x} \]

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Rubi [A]  time = 0.02, antiderivative size = 18, normalized size of antiderivative = 0.82, number of steps used = 5, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {14, 2304} \begin {gather*} -x^2-\frac {5}{x}+\frac {5 \log (x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(10 - 2*x^3 - 5*Log[x])/x^2,x]

[Out]

-5/x - x^2 + (5*Log[x])/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]]

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {2 \left (-5+x^3\right )}{x^2}-\frac {5 \log (x)}{x^2}\right ) \, dx\\ &=-\left (2 \int \frac {-5+x^3}{x^2} \, dx\right )-5 \int \frac {\log (x)}{x^2} \, dx\\ &=\frac {5}{x}+\frac {5 \log (x)}{x}-2 \int \left (-\frac {5}{x^2}+x\right ) \, dx\\ &=-\frac {5}{x}-x^2+\frac {5 \log (x)}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 18, normalized size = 0.82 \begin {gather*} -\frac {5}{x}-x^2+\frac {5 \log (x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10 - 2*x^3 - 5*Log[x])/x^2,x]

[Out]

-5/x - x^2 + (5*Log[x])/x

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fricas [A]  time = 0.80, size = 14, normalized size = 0.64 \begin {gather*} -\frac {x^{3} - 5 \, \log \relax (x) + 5}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*log(x)-2*x^3+10)/x^2,x, algorithm="fricas")

[Out]

-(x^3 - 5*log(x) + 5)/x

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giac [A]  time = 0.13, size = 18, normalized size = 0.82 \begin {gather*} -x^{2} + \frac {5 \, \log \relax (x)}{x} - \frac {5}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*log(x)-2*x^3+10)/x^2,x, algorithm="giac")

[Out]

-x^2 + 5*log(x)/x - 5/x

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maple [A]  time = 0.02, size = 16, normalized size = 0.73

method	result	size
norman	\(\frac {-5-x^{3}+5 \ln \relax (x )}{x}\)	\(16\)
default	\(-x^{2}+\frac {5 \ln \relax (x )}{x}-\frac {5}{x}\)	\(19\)
risch	\(\frac {5 \ln \relax (x )}{x}-\frac {x^{3}+5}{x}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-5*ln(x)-2*x^3+10)/x^2,x,method=_RETURNVERBOSE)

[Out]

(-5-x^3+5*ln(x))/x

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maxima [A]  time = 0.37, size = 18, normalized size = 0.82 \begin {gather*} -x^{2} + \frac {5 \, \log \relax (x)}{x} - \frac {5}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*log(x)-2*x^3+10)/x^2,x, algorithm="maxima")

[Out]

-x^2 + 5*log(x)/x - 5/x

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mupad [B]  time = 3.43, size = 16, normalized size = 0.73 \begin {gather*} \frac {5\,\ln \relax (x)-5}{x}-x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*log(x) + 2*x^3 - 10)/x^2,x)

[Out]

(5*log(x) - 5)/x - x^2

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sympy [A]  time = 0.09, size = 12, normalized size = 0.55 \begin {gather*} - x^{2} + \frac {5 \log {\relax (x )}}{x} - \frac {5}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*ln(x)-2*x**3+10)/x**2,x)

[Out]

-x**2 + 5*log(x)/x - 5/x

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