\(\int \frac {-5+96 x \log ^3(\frac {16}{e x^3})}{x^2} \, dx\) [4173]

Optimal result

Integrand size = 20, antiderivative size = 19 \[ \int \frac {-5+96 x \log ^3\left (\frac {16}{e x^3}\right )}{x^2} \, dx=\frac {5}{x}-8 \log ^4\left (\frac {16}{e x^3}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {14, 2339, 30} \[ \int \frac {-5+96 x \log ^3\left (\frac {16}{e x^3}\right )}{x^2} \, dx=\frac {5}{x}-8 \left (-\log \left (\frac {1}{x^3}\right )+1-\log (16)\right )^4 \]

[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 2339

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{align*} \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 \text {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{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {-5+96 x \log ^3\left (\frac {16}{e x^3}\right )}{x^2} \, dx=\frac {5}{x}-8 \left (-1+\log (16)+\log \left (\frac {1}{x^3}\right )\right )^4 \]

[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

Maple [A] (verified)

Time = 1.48 (sec) , antiderivative size = 19, normalized size of antiderivative = 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	\(\frac {5}{x}-8 \ln \left (\frac {16 \,{\mathrm e}^{-1}}{x^{3}}\right )^{4}\)	\(21\)
parts	\(\frac {5}{x}-8 \ln \left (\frac {16 \,{\mathrm e}^{-1}}{x^{3}}\right )^{4}\)	\(21\)
norman	\(\frac {5-8 \ln \left (\frac {16 \,{\mathrm e}^{-1}}{x^{3}}\right )^{4} x}{x}\)	\(22\)

[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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {-5+96 x \log ^3\left (\frac {16}{e x^3}\right )}{x^2} \, dx=-\frac {8 \, x \log \left (\frac {16 \, e^{\left (-1\right )}}{x^{3}}\right )^{4} - 5}{x} \]

[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

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {-5+96 x \log ^3\left (\frac {16}{e x^3}\right )}{x^2} \, dx=- 8 \log {\left (\frac {16}{e x^{3}} \right )}^{4} + \frac {5}{x} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {-5+96 x \log ^3\left (\frac {16}{e x^3}\right )}{x^2} \, dx=-8 \, \log \left (\frac {16 \, e^{\left (-1\right )}}{x^{3}}\right )^{4} + \frac {5}{x} \]

[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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 40 vs. \(2 (18) = 36\).

Time = 0.27 (sec) , antiderivative size = 40, normalized size of antiderivative = 2.11 \[ \int \frac {-5+96 x \log ^3\left (\frac {16}{e x^3}\right )}{x^2} \, dx=-8 \, \log \left (\frac {16}{x^{3}}\right )^{4} + 32 \, \log \left (\frac {16}{x^{3}}\right )^{3} - 48 \, \log \left (\frac {16}{x^{3}}\right )^{2} + \frac {5}{x} - 96 \, \log \left (x\right ) \]

[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)

Mupad [B] (verification not implemented)

Time = 9.78 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {-5+96 x \log ^3\left (\frac {16}{e x^3}\right )}{x^2} \, dx=\frac {5}{x}-8\,{\ln \left (\frac {16\,{\mathrm {e}}^{-1}}{x^3}\right )}^4 \]

[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