3.101.41 \(\int \frac {16-3 x-16 x^3+3 x^4+(-48+6 x+3 x^4) \log (x)}{5 x^4} \, dx\)

Optimal. Leaf size=27 \[ -1+\frac {(1+3 (5-x)) \left (\frac {1}{x^2}-x\right ) \log (x)}{5 x} \]

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Rubi [A]  time = 0.07, antiderivative size = 32, normalized size of antiderivative = 1.19, number of steps used = 10, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 14, 2357, 2295, 2304} \begin {gather*} \frac {16 \log (x)}{5 x^3}-\frac {3 \log (x)}{5 x^2}+\frac {3}{5} x \log (x)-\frac {16 \log (x)}{5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(16 - 3*x - 16*x^3 + 3*x^4 + (-48 + 6*x + 3*x^4)*Log[x])/(5*x^4),x]

[Out]

(-16*Log[x])/5 + (16*Log[x])/(5*x^3) - (3*Log[x])/(5*x^2) + (3*x*Log[x])/5

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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

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 2357

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \frac {16-3 x-16 x^3+3 x^4+\left (-48+6 x+3 x^4\right ) \log (x)}{x^4} \, dx\\ &=\frac {1}{5} \int \left (\frac {16-3 x-16 x^3+3 x^4}{x^4}+\frac {3 \left (-16+2 x+x^4\right ) \log (x)}{x^4}\right ) \, dx\\ &=\frac {1}{5} \int \frac {16-3 x-16 x^3+3 x^4}{x^4} \, dx+\frac {3}{5} \int \frac {\left (-16+2 x+x^4\right ) \log (x)}{x^4} \, dx\\ &=\frac {1}{5} \int \left (3+\frac {16}{x^4}-\frac {3}{x^3}-\frac {16}{x}\right ) \, dx+\frac {3}{5} \int \left (\log (x)-\frac {16 \log (x)}{x^4}+\frac {2 \log (x)}{x^3}\right ) \, dx\\ &=-\frac {16}{15 x^3}+\frac {3}{10 x^2}+\frac {3 x}{5}-\frac {16 \log (x)}{5}+\frac {3}{5} \int \log (x) \, dx+\frac {6}{5} \int \frac {\log (x)}{x^3} \, dx-\frac {48}{5} \int \frac {\log (x)}{x^4} \, dx\\ &=-\frac {16 \log (x)}{5}+\frac {16 \log (x)}{5 x^3}-\frac {3 \log (x)}{5 x^2}+\frac {3}{5} x \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 32, normalized size = 1.19 \begin {gather*} -\frac {16 \log (x)}{5}+\frac {16 \log (x)}{5 x^3}-\frac {3 \log (x)}{5 x^2}+\frac {3}{5} x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(16 - 3*x - 16*x^3 + 3*x^4 + (-48 + 6*x + 3*x^4)*Log[x])/(5*x^4),x]

[Out]

(-16*Log[x])/5 + (16*Log[x])/(5*x^3) - (3*Log[x])/(5*x^2) + (3*x*Log[x])/5

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fricas [A]  time = 1.41, size = 22, normalized size = 0.81 \begin {gather*} \frac {{\left (3 \, x^{4} - 16 \, x^{3} - 3 \, x + 16\right )} \log \relax (x)}{5 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((3*x^4+6*x-48)*log(x)+3*x^4-16*x^3-3*x+16)/x^4,x, algorithm="fricas")

[Out]

1/5*(3*x^4 - 16*x^3 - 3*x + 16)*log(x)/x^3

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giac [A]  time = 0.13, size = 23, normalized size = 0.85 \begin {gather*} \frac {1}{5} \, {\left (3 \, x - \frac {3 \, x - 16}{x^{3}}\right )} \log \relax (x) - \frac {16}{5} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((3*x^4+6*x-48)*log(x)+3*x^4-16*x^3-3*x+16)/x^4,x, algorithm="giac")

[Out]

1/5*(3*x - (3*x - 16)/x^3)*log(x) - 16/5*log(x)

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maple [A]  time = 0.02, size = 23, normalized size = 0.85

method	result	size
risch	\(\frac {\left (3 x^{4}-3 x +16\right ) \ln \relax (x )}{5 x^{3}}-\frac {16 \ln \relax (x )}{5}\)	\(23\)
default	\(\frac {3 x \ln \relax (x )}{5}-\frac {16 \ln \relax (x )}{5}-\frac {3 \ln \relax (x )}{5 x^{2}}+\frac {16 \ln \relax (x )}{5 x^{3}}\)	\(25\)
norman	\(\frac {-\frac {16 x^{3} \ln \relax (x )}{5}-\frac {3 x \ln \relax (x )}{5}+\frac {3 x^{4} \ln \relax (x )}{5}+\frac {16 \ln \relax (x )}{5}}{x^{3}}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((3*x^4+6*x-48)*ln(x)+3*x^4-16*x^3-3*x+16)/x^4,x,method=_RETURNVERBOSE)

[Out]

1/5*(3*x^4-3*x+16)/x^3*ln(x)-16/5*ln(x)

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maxima [A]  time = 0.36, size = 24, normalized size = 0.89 \begin {gather*} \frac {3}{5} \, x \log \relax (x) - \frac {3 \, \log \relax (x)}{5 \, x^{2}} + \frac {16 \, \log \relax (x)}{5 \, x^{3}} - \frac {16}{5} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((3*x^4+6*x-48)*log(x)+3*x^4-16*x^3-3*x+16)/x^4,x, algorithm="maxima")

[Out]

3/5*x*log(x) - 3/5*log(x)/x^2 + 16/5*log(x)/x^3 - 16/5*log(x)

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mupad [B]  time = 7.59, size = 22, normalized size = 0.81 \begin {gather*} -\frac {\ln \relax (x)\,\left (-3\,x^4+16\,x^3+3\,x-16\right )}{5\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((log(x)*(6*x + 3*x^4 - 48))/5 - (3*x)/5 - (16*x^3)/5 + (3*x^4)/5 + 16/5)/x^4,x)

[Out]

-(log(x)*(3*x + 16*x^3 - 3*x^4 - 16))/(5*x^3)

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sympy [A]  time = 0.20, size = 24, normalized size = 0.89 \begin {gather*} - \frac {16 \log {\relax (x )}}{5} + \frac {\left (3 x^{4} - 3 x + 16\right ) \log {\relax (x )}}{5 x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((3*x**4+6*x-48)*ln(x)+3*x**4-16*x**3-3*x+16)/x**4,x)

[Out]

-16*log(x)/5 + (3*x**4 - 3*x + 16)*log(x)/(5*x**3)

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