3.28.9 \(\int \frac {e^{-4 x} (4 \log (x^2)+(-2-4 x) \log ^2(x^2))}{4 x^3-4 x^3 \log (2)+x^3 \log ^2(2)} \, dx\)

Optimal. Leaf size=21 \[ \frac {e^{-4 x} \log ^2\left (x^2\right )}{x^2 (-2+\log (2))^2} \]

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Rubi [A]  time = 0.18, antiderivative size = 23, normalized size of antiderivative = 1.10, number of steps used = 6, number of rules used = 4, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {6, 12, 6741, 2288} \begin {gather*} \frac {e^{-4 x} \log ^2\left (x^2\right )}{x^2 (2-\log (2))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4*Log[x^2] + (-2 - 4*x)*Log[x^2]^2)/(E^(4*x)*(4*x^3 - 4*x^3*Log[2] + x^3*Log[2]^2)),x]

[Out]

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

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 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-4 x} \left (4 \log \left (x^2\right )+(-2-4 x) \log ^2\left (x^2\right )\right )}{x^3 (4-4 \log (2))+x^3 \log ^2(2)} \, dx\\ &=\int \frac {e^{-4 x} \left (4 \log \left (x^2\right )+(-2-4 x) \log ^2\left (x^2\right )\right )}{x^3 \left (4-4 \log (2)+\log ^2(2)\right )} \, dx\\ &=\frac {\int \frac {e^{-4 x} \left (4 \log \left (x^2\right )+(-2-4 x) \log ^2\left (x^2\right )\right )}{x^3} \, dx}{4-4 \log (2)+\log ^2(2)}\\ &=\frac {\int \frac {2 e^{-4 x} \log \left (x^2\right ) \left (2-\log \left (x^2\right )-2 x \log \left (x^2\right )\right )}{x^3} \, dx}{4-4 \log (2)+\log ^2(2)}\\ &=\frac {2 \int \frac {e^{-4 x} \log \left (x^2\right ) \left (2-\log \left (x^2\right )-2 x \log \left (x^2\right )\right )}{x^3} \, dx}{(2-\log (2))^2}\\ &=\frac {e^{-4 x} \log ^2\left (x^2\right )}{x^2 (2-\log (2))^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 21, normalized size = 1.00 \begin {gather*} \frac {e^{-4 x} \log ^2\left (x^2\right )}{x^2 (-2+\log (2))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4*Log[x^2] + (-2 - 4*x)*Log[x^2]^2)/(E^(4*x)*(4*x^3 - 4*x^3*Log[2] + x^3*Log[2]^2)),x]

[Out]

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

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fricas [A]  time = 0.66, size = 34, normalized size = 1.62 \begin {gather*} \frac {e^{\left (-4 \, x\right )} \log \left (x^{2}\right )^{2}}{x^{2} \log \relax (2)^{2} - 4 \, x^{2} \log \relax (2) + 4 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x-2)*log(x^2)^2+4*log(x^2))/(x^3*log(2)^2-4*x^3*log(2)+4*x^3)/exp(x)^4,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.25, size = 34, normalized size = 1.62 \begin {gather*} \frac {e^{\left (-4 \, x\right )} \log \left (x^{2}\right )^{2}}{x^{2} \log \relax (2)^{2} - 4 \, x^{2} \log \relax (2) + 4 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x-2)*log(x^2)^2+4*log(x^2))/(x^3*log(2)^2-4*x^3*log(2)+4*x^3)/exp(x)^4,x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.14, size = 190, normalized size = 9.05

method	result	size
risch	\(\frac {4 \,{\mathrm e}^{-4 x} \ln \relax (x )^{2}}{x^{2} \left (\ln \relax (2)^{2}-4 \ln \relax (2)+4\right )}-\frac {2 i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (\mathrm {csgn}\left (i x \right )^{2}-2 \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )+\mathrm {csgn}\left (i x^{2}\right )^{2}\right ) {\mathrm e}^{-4 x} \ln \relax (x )}{x^{2} \left (\ln \relax (2)^{2}-4 \ln \relax (2)+4\right )}-\frac {\pi ^{2} \mathrm {csgn}\left (i x^{2}\right )^{2} \left (\mathrm {csgn}\left (i x \right )^{4}-4 \mathrm {csgn}\left (i x \right )^{3} \mathrm {csgn}\left (i x^{2}\right )+6 \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )^{2}-4 \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{3}+\mathrm {csgn}\left (i x^{2}\right )^{4}\right ) {\mathrm e}^{-4 x}}{4 x^{2} \left (\ln \relax (2)^{2}-4 \ln \relax (2)+4\right )}\)	\(190\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x-2)*ln(x^2)^2+4*ln(x^2))/(x^3*ln(2)^2-4*x^3*ln(2)+4*x^3)/exp(x)^4,x,method=_RETURNVERBOSE)

[Out]

4/x^2/(ln(2)^2-4*ln(2)+4)*exp(-4*x)*ln(x)^2-2*I*Pi*csgn(I*x^2)*(csgn(I*x)^2-2*csgn(I*x^2)*csgn(I*x)+csgn(I*x^2
)^2)/x^2/(ln(2)^2-4*ln(2)+4)*exp(-4*x)*ln(x)-1/4*Pi^2*csgn(I*x^2)^2*(csgn(I*x)^4-4*csgn(I*x)^3*csgn(I*x^2)+6*c
sgn(I*x)^2*csgn(I*x^2)^2-4*csgn(I*x)*csgn(I*x^2)^3+csgn(I*x^2)^4)/x^2/(ln(2)^2-4*ln(2)+4)*exp(-4*x)

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maxima [A]  time = 0.64, size = 25, normalized size = 1.19 \begin {gather*} \frac {4 \, e^{\left (-4 \, x\right )} \log \relax (x)^{2}}{{\left (\log \relax (2)^{2} - 4 \, \log \relax (2) + 4\right )} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x-2)*log(x^2)^2+4*log(x^2))/(x^3*log(2)^2-4*x^3*log(2)+4*x^3)/exp(x)^4,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.69, size = 20, normalized size = 0.95 \begin {gather*} \frac {{\ln \left (x^2\right )}^2\,{\mathrm {e}}^{-4\,x}}{x^2\,{\left (\ln \relax (2)-2\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-4*x)*(4*log(x^2) - log(x^2)^2*(4*x + 2)))/(x^3*log(2)^2 - 4*x^3*log(2) + 4*x^3),x)

[Out]

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

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sympy [A]  time = 0.32, size = 32, normalized size = 1.52 \begin {gather*} \frac {e^{- 4 x} \log {\left (x^{2} \right )}^{2}}{- 4 x^{2} \log {\relax (2 )} + x^{2} \log {\relax (2 )}^{2} + 4 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x-2)*ln(x**2)**2+4*ln(x**2))/(x**3*ln(2)**2-4*x**3*ln(2)+4*x**3)/exp(x)**4,x)

[Out]

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

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