3.102.86 \(\int \frac {4 x+e^{-x} \log (5) (-12-8 x+(-4-4 x) \log (x))}{4 x^2+4 x^2 \log (x)+x^2 \log ^2(x)} \, dx\)

Optimal. Leaf size=25 \[ \frac {2 \left (-x+e^{-x} \log (5)\right )}{x+\frac {1}{2} x \log (x)} \]

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Rubi [A]  time = 0.63, antiderivative size = 35, normalized size of antiderivative = 1.40, number of steps used = 6, number of rules used = 5, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.102, Rules used = {6741, 6742, 2302, 30, 2288} \begin {gather*} \frac {4 e^{-x} \log (5) (2 x+x \log (x))}{x^2 (\log (x)+2)^2}-\frac {4}{\log (x)+2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4*x + (Log[5]*(-12 - 8*x + (-4 - 4*x)*Log[x]))/E^x)/(4*x^2 + 4*x^2*Log[x] + x^2*Log[x]^2),x]

[Out]

-4/(2 + Log[x]) + (4*Log[5]*(2*x + x*Log[x]))/(E^x*x^2*(2 + Log[x])^2)

Rule 30

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

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 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]

Rule 6741

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 x+e^{-x} \log (5) (-12-8 x+(-4-4 x) \log (x))}{x^2 (2+\log (x))^2} \, dx\\ &=\int \left (\frac {4}{x (2+\log (x))^2}-\frac {4 e^{-x} \log (5) (3+2 x+\log (x)+x \log (x))}{x^2 (2+\log (x))^2}\right ) \, dx\\ &=4 \int \frac {1}{x (2+\log (x))^2} \, dx-(4 \log (5)) \int \frac {e^{-x} (3+2 x+\log (x)+x \log (x))}{x^2 (2+\log (x))^2} \, dx\\ &=\frac {4 e^{-x} \log (5) (2 x+x \log (x))}{x^2 (2+\log (x))^2}+4 \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,2+\log (x)\right )\\ &=-\frac {4}{2+\log (x)}+\frac {4 e^{-x} \log (5) (2 x+x \log (x))}{x^2 (2+\log (x))^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.20, size = 24, normalized size = 0.96 \begin {gather*} \frac {e^{-x} \left (-4 e^x x+\log (625)\right )}{x (2+\log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4*x + (Log[5]*(-12 - 8*x + (-4 - 4*x)*Log[x]))/E^x)/(4*x^2 + 4*x^2*Log[x] + x^2*Log[x]^2),x]

[Out]

(-4*E^x*x + Log[625])/(E^x*x*(2 + Log[x]))

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fricas [A]  time = 0.78, size = 24, normalized size = 0.96 \begin {gather*} -\frac {4 \, {\left (x - e^{\left (-x + \log \left (\log \relax (5)\right )\right )}\right )}}{x \log \relax (x) + 2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x-4)*log(x)-8*x-12)*exp(log(log(5))-x)+4*x)/(x^2*log(x)^2+4*x^2*log(x)+4*x^2),x, algorithm="fr
icas")

[Out]

-4*(x - e^(-x + log(log(5))))/(x*log(x) + 2*x)

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giac [A]  time = 0.15, size = 23, normalized size = 0.92 \begin {gather*} \frac {4 \, {\left (e^{\left (-x\right )} \log \relax (5) - x\right )}}{x \log \relax (x) + 2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x-4)*log(x)-8*x-12)*exp(log(log(5))-x)+4*x)/(x^2*log(x)^2+4*x^2*log(x)+4*x^2),x, algorithm="gi
ac")

[Out]

4*(e^(-x)*log(5) - x)/(x*log(x) + 2*x)

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maple [A]  time = 0.09, size = 22, normalized size = 0.88

method	result	size
risch	\(-\frac {4 \left (x -\ln \relax (5) {\mathrm e}^{-x}\right )}{x \left (\ln \relax (x )+2\right )}\)	\(22\)
norman	\(\frac {-4 x +4 \,{\mathrm e}^{\ln \left (\ln \relax (5)\right )-x}}{\left (\ln \relax (x )+2\right ) x}\)	\(25\)
default	\(\frac {4 \,{\mathrm e}^{\ln \left (\ln \relax (5)\right )-x}}{x \left (\ln \relax (x )+2\right )}-\frac {4}{\ln \relax (x )+2}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-4*x-4)*ln(x)-8*x-12)*exp(ln(ln(5))-x)+4*x)/(x^2*ln(x)^2+4*x^2*ln(x)+4*x^2),x,method=_RETURNVERBOSE)

[Out]

-4/x*(x-ln(5)*exp(-x))/(ln(x)+2)

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maxima [A]  time = 0.51, size = 27, normalized size = 1.08 \begin {gather*} \frac {4 \, e^{\left (-x\right )} \log \relax (5)}{x \log \relax (x) + 2 \, x} - \frac {4}{\log \relax (x) + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x-4)*log(x)-8*x-12)*exp(log(log(5))-x)+4*x)/(x^2*log(x)^2+4*x^2*log(x)+4*x^2),x, algorithm="ma
xima")

[Out]

4*e^(-x)*log(5)/(x*log(x) + 2*x) - 4/(log(x) + 2)

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mupad [B]  time = 8.11, size = 23, normalized size = 0.92 \begin {gather*} \frac {4\,{\mathrm {e}}^{-x}\,\left (\ln \relax (5)-x\,{\mathrm {e}}^x\right )}{x\,\left (\ln \relax (x)+2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x - exp(log(log(5)) - x)*(8*x + log(x)*(4*x + 4) + 12))/(4*x^2*log(x) + x^2*log(x)^2 + 4*x^2),x)

[Out]

(4*exp(-x)*(log(5) - x*exp(x)))/(x*(log(x) + 2))

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sympy [A]  time = 0.32, size = 22, normalized size = 0.88 \begin {gather*} - \frac {4}{\log {\relax (x )} + 2} + \frac {4 e^{- x} \log {\relax (5 )}}{x \log {\relax (x )} + 2 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x-4)*ln(x)-8*x-12)*exp(ln(ln(5))-x)+4*x)/(x**2*ln(x)**2+4*x**2*ln(x)+4*x**2),x)

[Out]

-4/(log(x) + 2) + 4*exp(-x)*log(5)/(x*log(x) + 2*x)

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