3.25.72 \(\int \frac {(-4+8 x) \log (\frac {e^{4 x}}{x^2})}{x} \, dx\)

Optimal. Leaf size=12 \[ \log ^2\left (\frac {e^{4 x}}{x^2}\right ) \]

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Rubi [A]  time = 0.04, antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {43, 6686} \begin {gather*} \log ^2\left (\frac {e^{4 x}}{x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-4 + 8*x)*Log[E^(4*x)/x^2])/x,x]

[Out]

Log[E^(4*x)/x^2]^2

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 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\log ^2\left (\frac {e^{4 x}}{x^2}\right )\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.03, size = 44, normalized size = 3.67 \begin {gather*} \log ^2\left (\frac {1}{x^2}\right )+4 \log \left (\frac {1}{x^2}\right ) \log (x)-4 \left (4 x (x-\log (x))+\log \left (\frac {e^{4 x}}{x^2}\right ) (-2 x+\log (x))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-4 + 8*x)*Log[E^(4*x)/x^2])/x,x]

[Out]

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

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fricas [A]  time = 0.56, size = 11, normalized size = 0.92 \begin {gather*} \log \left (\frac {e^{\left (4 \, x\right )}}{x^{2}}\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x-4)*log(exp(x)^4/x^2)/x,x, algorithm="fricas")

[Out]

log(e^(4*x)/x^2)^2

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giac [A]  time = 0.31, size = 19, normalized size = 1.58 \begin {gather*} 16 \, x^{2} - 8 \, x \log \left (x^{2}\right ) + \log \left (x^{2}\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x-4)*log(exp(x)^4/x^2)/x,x, algorithm="giac")

[Out]

16*x^2 - 8*x*log(x^2) + log(x^2)^2

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maple [A]  time = 0.19, size = 12, normalized size = 1.00

method	result	size
norman	\(\ln \left (\frac {{\mathrm e}^{4 x}}{x^{2}}\right )^{2}\)	\(12\)
default	\(-4 \ln \left (\frac {{\mathrm e}^{4 x}}{x^{2}}\right ) \ln \relax (x )+8 \ln \left (\frac {{\mathrm e}^{4 x}}{x^{2}}\right ) x -4 \ln \relax (x )^{2}+16 x \ln \relax (x )-16 x^{2}\)	\(43\)
risch	\(4 \ln \relax (x )^{2}-16 x^{2}+\left (32 x -16 \ln \relax (x )\right ) \ln \left ({\mathrm e}^{x}\right )-4 i \pi x \,\mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{3 x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{4 x}\right )-4 i \pi x \,\mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (i {\mathrm e}^{4 x}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{4 x}}{x^{2}}\right )-4 i \pi x \,\mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{3 x}\right )+2 i \pi \ln \relax (x ) \mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{3 x}\right )+2 i \pi \ln \relax (x ) \mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{3 x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{4 x}\right )+2 i \pi \ln \relax (x ) \mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (i {\mathrm e}^{4 x}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{4 x}}{x^{2}}\right )+2 i \pi \ln \relax (x ) \mathrm {csgn}\left (i {\mathrm e}^{4 x}\right )^{3}+2 i \pi \ln \relax (x ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{4 x}}{x^{2}}\right )^{3}-4 i \pi x \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}-4 i \pi x \mathrm {csgn}\left (i {\mathrm e}^{3 x}\right )^{3}-4 i \pi x \mathrm {csgn}\left (i {\mathrm e}^{4 x}\right )^{3}-4 i \pi x \mathrm {csgn}\left (\frac {i {\mathrm e}^{4 x}}{x^{2}}\right )^{3}+2 i \pi \ln \relax (x ) \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}+2 i \pi \ln \relax (x ) \mathrm {csgn}\left (i {\mathrm e}^{3 x}\right )^{3}-2 i \pi \ln \relax (x ) \mathrm {csgn}\left (i x^{2}\right )^{3}+4 i \pi x \mathrm {csgn}\left (i x^{2}\right )^{3}-2 i \pi \ln \relax (x ) \mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{4 x}\right )^{2}-2 i \pi \ln \relax (x ) \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{3 x}\right )^{2}-2 i \pi \ln \relax (x ) \mathrm {csgn}\left (i {\mathrm e}^{3 x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{4 x}\right )^{2}-2 i \pi \ln \relax (x ) \mathrm {csgn}\left (i {\mathrm e}^{4 x}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{4 x}}{x^{2}}\right )^{2}-2 i \pi \ln \relax (x ) \mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{4 x}}{x^{2}}\right )^{2}-2 i \pi \ln \relax (x ) \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+4 i \pi \ln \relax (x ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}-4 i \pi x \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )+8 i \pi x \,\mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{2}+4 i \pi x \,\mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{3 x}\right )^{2}+2 i \pi \ln \relax (x ) \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )-4 i \pi \ln \relax (x ) \mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{2}-2 i \pi \ln \relax (x ) \mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{3 x}\right )^{2}+4 i \pi x \,\mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{4 x}\right )^{2}+4 i \pi x \,\mathrm {csgn}\left (i {\mathrm e}^{2 x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{3 x}\right )^{2}+4 i \pi x \,\mathrm {csgn}\left (i {\mathrm e}^{3 x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{4 x}\right )^{2}+4 i \pi x \,\mathrm {csgn}\left (i {\mathrm e}^{4 x}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{4 x}}{x^{2}}\right )^{2}+4 i \pi x \,\mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{4 x}}{x^{2}}\right )^{2}-8 i \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+4 i \pi x \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )\)	\(805\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x-4)*ln(exp(x)^4/x^2)/x,x,method=_RETURNVERBOSE)

[Out]

ln(exp(x)^4/x^2)^2

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maxima [B]  time = 0.46, size = 36, normalized size = 3.00 \begin {gather*} -16 \, x^{2} + 16 \, x \log \relax (x) - 4 \, \log \relax (x)^{2} + 4 \, {\left (2 \, x - \log \relax (x)\right )} \log \left (\frac {e^{\left (4 \, x\right )}}{x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x-4)*log(exp(x)^4/x^2)/x,x, algorithm="maxima")

[Out]

-16*x^2 + 16*x*log(x) - 4*log(x)^2 + 4*(2*x - log(x))*log(e^(4*x)/x^2)

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mupad [B]  time = 1.46, size = 10, normalized size = 0.83 \begin {gather*} {\left (4\,x+\ln \left (\frac {1}{x^2}\right )\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(exp(4*x)/x^2)*(8*x - 4))/x,x)

[Out]

(4*x + log(1/x^2))^2

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sympy [A]  time = 0.13, size = 10, normalized size = 0.83 \begin {gather*} \log {\left (\frac {e^{4 x}}{x^{2}} \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x-4)*ln(exp(x)**4/x**2)/x,x)

[Out]

log(exp(4*x)/x**2)**2

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