3.70.1 \(\int \frac {(4+2 x) \log (x)+(-4-x) \log ^2(x)}{96 x^3} \, dx\)

Optimal. Leaf size=14 \[ \frac {(2+x) \log ^2(x)}{96 x^2} \]

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Rubi [B]  time = 0.17, antiderivative size = 61, normalized size of antiderivative = 4.36, number of steps used = 13, number of rules used = 8, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {12, 14, 37, 2334, 43, 2353, 2305, 2304} \begin {gather*} \frac {\log ^2(x)}{48 x^2}-\frac {(x+2)^2 \log (x)}{192 x^2}+\frac {\log (x)}{48 x^2}+\frac {\log ^2(x)}{96 x}+\frac {\log (x)}{48 x}+\frac {\log (x)}{192} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((4 + 2*x)*Log[x] + (-4 - x)*Log[x]^2)/(96*x^3),x]

[Out]

Log[x]/192 + Log[x]/(48*x^2) + Log[x]/(48*x) - ((2 + x)^2*Log[x])/(192*x^2) + Log[x]^2/(48*x^2) + Log[x]^2/(96
*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 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

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]

Rule 2334

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = I
ntHide[x^m*(d + e*x^r)^q, x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]
] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[q, 0] && IntegerQ[m] &&  !(EqQ[q, 1] && EqQ[m, -1])

Rule 2353

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[
{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && IntegerQ[r
]))

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{96} \int \frac {(4+2 x) \log (x)+(-4-x) \log ^2(x)}{x^3} \, dx\\ &=\frac {1}{96} \int \left (\frac {2 (2+x) \log (x)}{x^3}-\frac {(4+x) \log ^2(x)}{x^3}\right ) \, dx\\ &=-\left (\frac {1}{96} \int \frac {(4+x) \log ^2(x)}{x^3} \, dx\right )+\frac {1}{48} \int \frac {(2+x) \log (x)}{x^3} \, dx\\ &=-\frac {(2+x)^2 \log (x)}{192 x^2}-\frac {1}{96} \int \left (\frac {4 \log ^2(x)}{x^3}+\frac {\log ^2(x)}{x^2}\right ) \, dx-\frac {1}{48} \int -\frac {(2+x)^2}{4 x^3} \, dx\\ &=-\frac {(2+x)^2 \log (x)}{192 x^2}+\frac {1}{192} \int \frac {(2+x)^2}{x^3} \, dx-\frac {1}{96} \int \frac {\log ^2(x)}{x^2} \, dx-\frac {1}{24} \int \frac {\log ^2(x)}{x^3} \, dx\\ &=-\frac {(2+x)^2 \log (x)}{192 x^2}+\frac {\log ^2(x)}{48 x^2}+\frac {\log ^2(x)}{96 x}+\frac {1}{192} \int \left (\frac {4}{x^3}+\frac {4}{x^2}+\frac {1}{x}\right ) \, dx-\frac {1}{48} \int \frac {\log (x)}{x^2} \, dx-\frac {1}{24} \int \frac {\log (x)}{x^3} \, dx\\ &=\frac {\log (x)}{192}+\frac {\log (x)}{48 x^2}+\frac {\log (x)}{48 x}-\frac {(2+x)^2 \log (x)}{192 x^2}+\frac {\log ^2(x)}{48 x^2}+\frac {\log ^2(x)}{96 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 22, normalized size = 1.57 \begin {gather*} \frac {1}{96} \left (\frac {2 \log ^2(x)}{x^2}+\frac {\log ^2(x)}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((4 + 2*x)*Log[x] + (-4 - x)*Log[x]^2)/(96*x^3),x]

[Out]

((2*Log[x]^2)/x^2 + Log[x]^2/x)/96

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fricas [A]  time = 0.64, size = 12, normalized size = 0.86 \begin {gather*} \frac {{\left (x + 2\right )} \log \relax (x)^{2}}{96 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/96*((-x-4)*log(x)^2+(2*x+4)*log(x))/x^3,x, algorithm="fricas")

[Out]

1/96*(x + 2)*log(x)^2/x^2

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giac [A]  time = 0.23, size = 12, normalized size = 0.86 \begin {gather*} \frac {{\left (x + 2\right )} \log \relax (x)^{2}}{96 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/96*((-x-4)*log(x)^2+(2*x+4)*log(x))/x^3,x, algorithm="giac")

[Out]

1/96*(x + 2)*log(x)^2/x^2

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maple [A]  time = 0.02, size = 13, normalized size = 0.93

method	result	size
risch	\(\frac {\ln \relax (x )^{2} \left (2+x \right )}{96 x^{2}}\)	\(13\)
norman	\(\frac {\frac {\ln \relax (x )^{2}}{48}+\frac {x \ln \relax (x )^{2}}{96}}{x^{2}}\)	\(19\)
default	\(\frac {\ln \relax (x )^{2}}{96 x}+\frac {\ln \relax (x )^{2}}{48 x^{2}}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/96*((-x-4)*ln(x)^2+(2*x+4)*ln(x))/x^3,x,method=_RETURNVERBOSE)

[Out]

1/96/x^2*ln(x)^2*(2+x)

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maxima [B]  time = 0.37, size = 57, normalized size = 4.07 \begin {gather*} \frac {\log \relax (x)^{2} + 2 \, \log \relax (x) + 2}{96 \, x} - \frac {\log \relax (x)}{48 \, x} + \frac {2 \, \log \relax (x)^{2} + 2 \, \log \relax (x) + 1}{96 \, x^{2}} - \frac {1}{48 \, x} - \frac {\log \relax (x)}{48 \, x^{2}} - \frac {1}{96 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/96*((-x-4)*log(x)^2+(2*x+4)*log(x))/x^3,x, algorithm="maxima")

[Out]

1/96*(log(x)^2 + 2*log(x) + 2)/x - 1/48*log(x)/x + 1/96*(2*log(x)^2 + 2*log(x) + 1)/x^2 - 1/48/x - 1/48*log(x)
/x^2 - 1/96/x^2

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mupad [B]  time = 4.13, size = 12, normalized size = 0.86 \begin {gather*} \frac {{\ln \relax (x)}^2\,\left (x+2\right )}{96\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((log(x)*(2*x + 4))/96 - (log(x)^2*(x + 4))/96)/x^3,x)

[Out]

(log(x)^2*(x + 2))/(96*x^2)

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sympy [A]  time = 0.13, size = 12, normalized size = 0.86 \begin {gather*} \frac {\left (x + 2\right ) \log {\relax (x )}^{2}}{96 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/96*((-x-4)*ln(x)**2+(2*x+4)*ln(x))/x**3,x)

[Out]

(x + 2)*log(x)**2/(96*x**2)

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