3.93.96 \(\int \frac {-520-6 x-190 \log (x)-5 \log ^2(x)}{5 x^2} \, dx\)

Optimal. Leaf size=17 \[ \frac {(36+\log (x)) \left (4-\frac {6 x}{5}+\log (x)\right )}{x} \]

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Rubi [A]  time = 0.05, antiderivative size = 27, normalized size of antiderivative = 1.59, number of steps used = 8, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {12, 14, 43, 2304, 2305} \begin {gather*} \frac {144}{x}+\frac {\log ^2(x)}{x}+\frac {40 \log (x)}{x}-\frac {6 \log (x)}{5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-520 - 6*x - 190*Log[x] - 5*Log[x]^2)/(5*x^2),x]

[Out]

144/x - (6*Log[x])/5 + (40*Log[x])/x + Log[x]^2/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]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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
]

Rule 2305

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Lo
g[c*x^n])^p)/(d*(m + 1)), x] - Dist[(b*n*p)/(m + 1), Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{
a, b, c, d, m, n}, x] && NeQ[m, -1] && GtQ[p, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \frac {-520-6 x-190 \log (x)-5 \log ^2(x)}{x^2} \, dx\\ &=\frac {1}{5} \int \left (-\frac {2 (260+3 x)}{x^2}-\frac {190 \log (x)}{x^2}-\frac {5 \log ^2(x)}{x^2}\right ) \, dx\\ &=-\left (\frac {2}{5} \int \frac {260+3 x}{x^2} \, dx\right )-38 \int \frac {\log (x)}{x^2} \, dx-\int \frac {\log ^2(x)}{x^2} \, dx\\ &=\frac {38}{x}+\frac {38 \log (x)}{x}+\frac {\log ^2(x)}{x}-\frac {2}{5} \int \left (\frac {260}{x^2}+\frac {3}{x}\right ) \, dx-2 \int \frac {\log (x)}{x^2} \, dx\\ &=\frac {144}{x}-\frac {6 \log (x)}{5}+\frac {40 \log (x)}{x}+\frac {\log ^2(x)}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 27, normalized size = 1.59 \begin {gather*} \frac {144}{x}-\frac {6 \log (x)}{5}+\frac {40 \log (x)}{x}+\frac {\log ^2(x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-520 - 6*x - 190*Log[x] - 5*Log[x]^2)/(5*x^2),x]

[Out]

144/x - (6*Log[x])/5 + (40*Log[x])/x + Log[x]^2/x

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fricas [A]  time = 1.10, size = 22, normalized size = 1.29 \begin {gather*} -\frac {2 \, {\left (3 \, x - 100\right )} \log \relax (x) - 5 \, \log \relax (x)^{2} - 720}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-5*log(x)^2-190*log(x)-6*x-520)/x^2,x, algorithm="fricas")

[Out]

-1/5*(2*(3*x - 100)*log(x) - 5*log(x)^2 - 720)/x

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giac [A]  time = 0.21, size = 25, normalized size = 1.47 \begin {gather*} \frac {\log \relax (x)^{2}}{x} + \frac {40 \, \log \relax (x)}{x} + \frac {144}{x} - \frac {6}{5} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-5*log(x)^2-190*log(x)-6*x-520)/x^2,x, algorithm="giac")

[Out]

log(x)^2/x + 40*log(x)/x + 144/x - 6/5*log(x)

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maple [A]  time = 0.02, size = 20, normalized size = 1.18

method	result	size
norman	\(\frac {144+\ln \relax (x )^{2}-\frac {6 x \ln \relax (x )}{5}+40 \ln \relax (x )}{x}\)	\(20\)
default	\(\frac {\ln \relax (x )^{2}}{x}+\frac {40 \ln \relax (x )}{x}+\frac {144}{x}-\frac {6 \ln \relax (x )}{5}\)	\(26\)
risch	\(\frac {\ln \relax (x )^{2}}{x}+\frac {40 \ln \relax (x )}{x}-\frac {6 \left (x \ln \relax (x )-120\right )}{5 x}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(-5*ln(x)^2-190*ln(x)-6*x-520)/x^2,x,method=_RETURNVERBOSE)

[Out]

(144+ln(x)^2-6/5*x*ln(x)+40*ln(x))/x

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maxima [A]  time = 0.35, size = 31, normalized size = 1.82 \begin {gather*} \frac {\log \relax (x)^{2} + 2 \, \log \relax (x) + 2}{x} + \frac {38 \, \log \relax (x)}{x} + \frac {142}{x} - \frac {6}{5} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-5*log(x)^2-190*log(x)-6*x-520)/x^2,x, algorithm="maxima")

[Out]

(log(x)^2 + 2*log(x) + 2)/x + 38*log(x)/x + 142/x - 6/5*log(x)

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mupad [B]  time = 5.35, size = 19, normalized size = 1.12 \begin {gather*} \frac {{\ln \relax (x)}^2+40\,\ln \relax (x)+144}{x}-\frac {6\,\ln \relax (x)}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((6*x)/5 + 38*log(x) + log(x)^2 + 104)/x^2,x)

[Out]

(40*log(x) + log(x)^2 + 144)/x - (6*log(x))/5

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sympy [A]  time = 0.14, size = 22, normalized size = 1.29 \begin {gather*} - \frac {6 \log {\relax (x )}}{5} + \frac {\log {\relax (x )}^{2}}{x} + \frac {40 \log {\relax (x )}}{x} + \frac {144}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-5*ln(x)**2-190*ln(x)-6*x-520)/x**2,x)

[Out]

-6*log(x)/5 + log(x)**2/x + 40*log(x)/x + 144/x

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