3.28.74 \(\int \frac {16-4 x+x \log (4)+(-32+4 x-x \log (4)) \log (x)}{20 x^3} \, dx\)

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

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Rubi [B]  time = 0.09, antiderivative size = 77, normalized size of antiderivative = 3.08, number of steps used = 9, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {6, 12, 14, 37, 2334, 43} \begin {gather*} \frac {2}{5 x^2}-\frac {(16-x (4-\log (4)))^2}{640 x^2}+\frac {(32-x (4-\log (4)))^2 \log (x)}{1280 x^2}-\frac {(4-\log (4))^2 \log (x)}{1280}-\frac {4-\log (4)}{20 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(16 - 4*x + x*Log[4] + (-32 + 4*x - x*Log[4])*Log[x])/(20*x^3),x]

[Out]

2/(5*x^2) - (16 - x*(4 - Log[4]))^2/(640*x^2) - (4 - Log[4])/(20*x) + ((32 - x*(4 - Log[4]))^2*Log[x])/(1280*x
^2) - ((4 - Log[4])^2*Log[x])/1280

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 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])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {16+x (-4+\log (4))+(-32+4 x-x \log (4)) \log (x)}{20 x^3} \, dx\\ &=\frac {1}{20} \int \frac {16+x (-4+\log (4))+(-32+4 x-x \log (4)) \log (x)}{x^3} \, dx\\ &=\frac {1}{20} \int \left (\frac {16-x (4-\log (4))}{x^3}+\frac {(-32+x (4-\log (4))) \log (x)}{x^3}\right ) \, dx\\ &=\frac {1}{20} \int \frac {16-x (4-\log (4))}{x^3} \, dx+\frac {1}{20} \int \frac {(-32+x (4-\log (4))) \log (x)}{x^3} \, dx\\ &=-\frac {(16-x (4-\log (4)))^2}{640 x^2}+\frac {(32-x (4-\log (4)))^2 \log (x)}{1280 x^2}-\frac {1}{20} \int \frac {(32+x (-4+\log (4)))^2}{64 x^3} \, dx\\ &=-\frac {(16-x (4-\log (4)))^2}{640 x^2}+\frac {(32-x (4-\log (4)))^2 \log (x)}{1280 x^2}-\frac {\int \frac {(32+x (-4+\log (4)))^2}{x^3} \, dx}{1280}\\ &=-\frac {(16-x (4-\log (4)))^2}{640 x^2}+\frac {(32-x (4-\log (4)))^2 \log (x)}{1280 x^2}-\frac {\int \left (\frac {1024}{x^3}+\frac {64 (-4+\log (4))}{x^2}+\frac {(-4+\log (4))^2}{x}\right ) \, dx}{1280}\\ &=\frac {2}{5 x^2}-\frac {(16-x (4-\log (4)))^2}{640 x^2}-\frac {4-\log (4)}{20 x}+\frac {(32-x (4-\log (4)))^2 \log (x)}{1280 x^2}-\frac {(4-\log (4))^2 \log (x)}{1280}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 27, normalized size = 1.08 \begin {gather*} \frac {1}{20} \left (\frac {16 \log (x)}{x^2}-\frac {4 \log (x)}{x}+\frac {\log (4) \log (x)}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(16 - 4*x + x*Log[4] + (-32 + 4*x - x*Log[4])*Log[x])/(20*x^3),x]

[Out]

((16*Log[x])/x^2 - (4*Log[x])/x + (Log[4]*Log[x])/x)/20

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fricas [A]  time = 0.69, size = 16, normalized size = 0.64 \begin {gather*} \frac {{\left (x \log \relax (2) - 2 \, x + 8\right )} \log \relax (x)}{10 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/20*((-2*x*log(2)+4*x-32)*log(x)+2*x*log(2)-4*x+16)/x^3,x, algorithm="fricas")

[Out]

1/10*(x*log(2) - 2*x + 8)*log(x)/x^2

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giac [A]  time = 0.33, size = 16, normalized size = 0.64 \begin {gather*} \frac {{\left (x \log \relax (2) - 2 \, x + 8\right )} \log \relax (x)}{10 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/20*((-2*x*log(2)+4*x-32)*log(x)+2*x*log(2)-4*x+16)/x^3,x, algorithm="giac")

[Out]

1/10*(x*log(2) - 2*x + 8)*log(x)/x^2

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maple [A]  time = 0.03, size = 17, normalized size = 0.68

method	result	size
risch	\(\frac {\left (x \ln \relax (2)-2 x +8\right ) \ln \relax (x )}{10 x^{2}}\)	\(17\)
norman	\(\frac {\left (\frac {\ln \relax (2)}{10}-\frac {1}{5}\right ) x \ln \relax (x )+\frac {4 \ln \relax (x )}{5}}{x^{2}}\)	\(20\)
default	\(-\frac {\ln \relax (2) \left (-\frac {\ln \relax (x )}{x}-\frac {1}{x}\right )}{10}-\frac {\ln \relax (x )}{5 x}-\frac {\ln \relax (2)}{10 x}+\frac {4 \ln \relax (x )}{5 x^{2}}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/20*((-2*x*ln(2)+4*x-32)*ln(x)+2*x*ln(2)-4*x+16)/x^3,x,method=_RETURNVERBOSE)

[Out]

1/10*(x*ln(2)-2*x+8)/x^2*ln(x)

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maxima [A]  time = 0.40, size = 36, normalized size = 1.44 \begin {gather*} \frac {1}{10} \, {\left (\frac {\log \relax (x)}{x} + \frac {1}{x}\right )} \log \relax (2) - \frac {\log \relax (2)}{10 \, x} - \frac {\log \relax (x)}{5 \, x} + \frac {4 \, \log \relax (x)}{5 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/20*((-2*x*log(2)+4*x-32)*log(x)+2*x*log(2)-4*x+16)/x^3,x, algorithm="maxima")

[Out]

1/10*(log(x)/x + 1/x)*log(2) - 1/10*log(2)/x - 1/5*log(x)/x + 4/5*log(x)/x^2

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mupad [B]  time = 1.89, size = 16, normalized size = 0.64 \begin {gather*} \frac {\ln \relax (x)\,\left (x\,\ln \relax (2)-2\,x+8\right )}{10\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x/5 - (x*log(2))/10 + (log(x)*(2*x*log(2) - 4*x + 32))/20 - 4/5)/x^3,x)

[Out]

(log(x)*(x*log(2) - 2*x + 8))/(10*x^2)

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sympy [A]  time = 0.15, size = 17, normalized size = 0.68 \begin {gather*} \frac {\left (- 2 x + x \log {\relax (2 )} + 8\right ) \log {\relax (x )}}{10 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/20*((-2*x*ln(2)+4*x-32)*ln(x)+2*x*ln(2)-4*x+16)/x**3,x)

[Out]

(-2*x + x*log(2) + 8)*log(x)/(10*x**2)

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