3.42.73 \(\int \frac {-5+96 x \log ^3(\frac {16}{e x^3})}{x^2} \, dx\)

Optimal. Leaf size=19 \[ \frac {5}{x}-8 \log ^4\left (\frac {16}{e x^3}\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 22, normalized size of antiderivative = 1.16, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {14, 2302, 30} \begin {gather*} \frac {5}{x}-8 \left (-\log \left (\frac {1}{x^3}\right )+1-\log (16)\right )^4 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-5 + 96*x*Log[16/(E*x^3)]^3)/x^2,x]

[Out]

5/x - 8*(1 - Log[16] - Log[x^(-3)])^4

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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {5}{x^2}+\frac {96 \left (-1+\log (16)+\log \left (\frac {1}{x^3}\right )\right )^3}{x}\right ) \, dx\\ &=\frac {5}{x}+96 \int \frac {\left (-1+\log (16)+\log \left (\frac {1}{x^3}\right )\right )^3}{x} \, dx\\ &=\frac {5}{x}-32 \operatorname {Subst}\left (\int x^3 \, dx,x,-1+\log (16)+\log \left (\frac {1}{x^3}\right )\right )\\ &=\frac {5}{x}-8 \left (1-\log (16)-\log \left (\frac {1}{x^3}\right )\right )^4\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 18, normalized size = 0.95 \begin {gather*} \frac {5}{x}-8 \left (-1+\log (16)+\log \left (\frac {1}{x^3}\right )\right )^4 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5 + 96*x*Log[16/(E*x^3)]^3)/x^2,x]

[Out]

5/x - 8*(-1 + Log[16] + Log[x^(-3)])^4

________________________________________________________________________________________

fricas [A]  time = 1.78, size = 20, normalized size = 1.05 \begin {gather*} -\frac {8 \, x \log \left (\frac {16 \, e^{\left (-1\right )}}{x^{3}}\right )^{4} - 5}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((96*x*log(16/x^3/exp(1))^3-5)/x^2,x, algorithm="fricas")

[Out]

-(8*x*log(16*e^(-1)/x^3)^4 - 5)/x

________________________________________________________________________________________

giac [B]  time = 0.25, size = 40, normalized size = 2.11 \begin {gather*} -8 \, \log \left (\frac {16}{x^{3}}\right )^{4} + 32 \, \log \left (\frac {16}{x^{3}}\right )^{3} - 48 \, \log \left (\frac {16}{x^{3}}\right )^{2} + \frac {5}{x} - 96 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((96*x*log(16/x^3/exp(1))^3-5)/x^2,x, algorithm="giac")

[Out]

-8*log(16/x^3)^4 + 32*log(16/x^3)^3 - 48*log(16/x^3)^2 + 5/x - 96*log(x)

________________________________________________________________________________________

maple [A]  time = 0.03, size = 19, normalized size = 1.00

method	result	size
risch	\(\frac {5}{x}-8 \ln \left (\frac {16 \,{\mathrm e}^{-1}}{x^{3}}\right )^{4}\)	\(19\)
derivativedivides	\(\frac {5}{x}-8 \ln \left (\frac {16 \,{\mathrm e}^{-1}}{x^{3}}\right )^{4}\)	\(21\)
default	\(\frac {5}{x}-8 \ln \left (\frac {16 \,{\mathrm e}^{-1}}{x^{3}}\right )^{4}\)	\(21\)
norman	\(\frac {5-8 x \ln \left (\frac {16 \,{\mathrm e}^{-1}}{x^{3}}\right )^{4}}{x}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((96*x*ln(16/x^3/exp(1))^3-5)/x^2,x,method=_RETURNVERBOSE)

[Out]

5/x-8*ln(16*exp(-1)/x^3)^4

________________________________________________________________________________________

maxima [A]  time = 0.43, size = 18, normalized size = 0.95 \begin {gather*} -8 \, \log \left (\frac {16 \, e^{\left (-1\right )}}{x^{3}}\right )^{4} + \frac {5}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((96*x*log(16/x^3/exp(1))^3-5)/x^2,x, algorithm="maxima")

[Out]

-8*log(16*e^(-1)/x^3)^4 + 5/x

________________________________________________________________________________________

mupad [B]  time = 2.90, size = 18, normalized size = 0.95 \begin {gather*} \frac {5}{x}-8\,{\ln \left (\frac {16\,{\mathrm {e}}^{-1}}{x^3}\right )}^4 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((96*x*log((16*exp(-1))/x^3)^3 - 5)/x^2,x)

[Out]

5/x - 8*log((16*exp(-1))/x^3)^4

________________________________________________________________________________________

sympy [A]  time = 0.11, size = 15, normalized size = 0.79 \begin {gather*} - 8 \log {\left (\frac {16}{e x^{3}} \right )}^{4} + \frac {5}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((96*x*ln(16/x**3/exp(1))**3-5)/x**2,x)

[Out]

-8*log(16*exp(-1)/x**3)**4 + 5/x

________________________________________________________________________________________