3.45.45 \(\int \frac {-1-2 x-4 x^2+(-1-4 x) \log (x)}{2 e^{4 x} x^4+4 e^{4 x} x^3 \log (x)+2 e^{4 x} x^2 \log ^2(x)} \, dx\)

Optimal. Leaf size=20 \[ -9+\frac {e^{-4 x}}{2 x (x+\log (x))} \]

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Rubi [A]  time = 0.22, antiderivative size = 26, normalized size of antiderivative = 1.30, number of steps used = 3, number of rules used = 3, integrand size = 58, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.052, Rules used = {6688, 12, 2288} \begin {gather*} \frac {e^{-4 x} \left (x^2+x \log (x)\right )}{2 x^2 (x+\log (x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 - 2*x - 4*x^2 + (-1 - 4*x)*Log[x])/(2*E^(4*x)*x^4 + 4*E^(4*x)*x^3*Log[x] + 2*E^(4*x)*x^2*Log[x]^2),x]

[Out]

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

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 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-4 x} \left (-1-2 x-4 x^2-\log (x)-4 x \log (x)\right )}{2 x^2 (x+\log (x))^2} \, dx\\ &=\frac {1}{2} \int \frac {e^{-4 x} \left (-1-2 x-4 x^2-\log (x)-4 x \log (x)\right )}{x^2 (x+\log (x))^2} \, dx\\ &=\frac {e^{-4 x} \left (x^2+x \log (x)\right )}{2 x^2 (x+\log (x))^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 19, normalized size = 0.95 \begin {gather*} \frac {e^{-4 x}}{2 \left (x^2+x \log (x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 - 2*x - 4*x^2 + (-1 - 4*x)*Log[x])/(2*E^(4*x)*x^4 + 4*E^(4*x)*x^3*Log[x] + 2*E^(4*x)*x^2*Log[x]^
2),x]

[Out]

1/(2*E^(4*x)*(x^2 + x*Log[x]))

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fricas [A]  time = 0.61, size = 21, normalized size = 1.05 \begin {gather*} \frac {1}{2 \, {\left (x^{2} e^{\left (4 \, x\right )} + x e^{\left (4 \, x\right )} \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x-1)*log(x)-4*x^2-2*x-1)/(2*x^2*exp(2*x)^2*log(x)^2+4*x^3*exp(2*x)^2*log(x)+2*x^4*exp(2*x)^2),x
, algorithm="fricas")

[Out]

1/2/(x^2*e^(4*x) + x*e^(4*x)*log(x))

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giac [A]  time = 0.19, size = 16, normalized size = 0.80 \begin {gather*} \frac {e^{\left (-4 \, x\right )}}{2 \, {\left (x^{2} + x \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x-1)*log(x)-4*x^2-2*x-1)/(2*x^2*exp(2*x)^2*log(x)^2+4*x^3*exp(2*x)^2*log(x)+2*x^4*exp(2*x)^2),x
, algorithm="giac")

[Out]

1/2*e^(-4*x)/(x^2 + x*log(x))

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maple [A]  time = 0.02, size = 16, normalized size = 0.80

method	result	size
risch	\(\frac {{\mathrm e}^{-4 x}}{2 x \left (x +\ln \relax (x )\right )}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x-1)*ln(x)-4*x^2-2*x-1)/(2*x^2*exp(2*x)^2*ln(x)^2+4*x^3*exp(2*x)^2*ln(x)+2*x^4*exp(2*x)^2),x,method=_
RETURNVERBOSE)

[Out]

1/2/x*exp(-4*x)/(x+ln(x))

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maxima [A]  time = 0.40, size = 16, normalized size = 0.80 \begin {gather*} \frac {e^{\left (-4 \, x\right )}}{2 \, {\left (x^{2} + x \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x-1)*log(x)-4*x^2-2*x-1)/(2*x^2*exp(2*x)^2*log(x)^2+4*x^3*exp(2*x)^2*log(x)+2*x^4*exp(2*x)^2),x
, algorithm="maxima")

[Out]

1/2*e^(-4*x)/(x^2 + x*log(x))

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mupad [B]  time = 3.32, size = 21, normalized size = 1.05 \begin {gather*} \frac {1}{2\,x^2\,{\mathrm {e}}^{4\,x}+2\,x\,{\mathrm {e}}^{4\,x}\,\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x + log(x)*(4*x + 1) + 4*x^2 + 1)/(2*x^4*exp(4*x) + 4*x^3*exp(4*x)*log(x) + 2*x^2*exp(4*x)*log(x)^2),x
)

[Out]

1/(2*x^2*exp(4*x) + 2*x*exp(4*x)*log(x))

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sympy [A]  time = 0.28, size = 17, normalized size = 0.85 \begin {gather*} \frac {e^{- 4 x}}{2 x^{2} + 2 x \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x-1)*ln(x)-4*x**2-2*x-1)/(2*x**2*exp(2*x)**2*ln(x)**2+4*x**3*exp(2*x)**2*ln(x)+2*x**4*exp(2*x)*
*2),x)

[Out]

exp(-4*x)/(2*x**2 + 2*x*log(x))

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