3.52.83 \(\int \frac {e^{-8/x} (e^{4/x} (-2560 x-1280 x^2-160 x^3) \log (x)+e^{4/x} (-5120-2560 x+80 x^3) \log ^2(x)+(64 x+16 x^2) \log ^3(x)+(128+32 x-8 x^2) \log ^4(x))}{1600 x^2+1200 x^3+300 x^4+25 x^5} \, dx\)

Optimal. Leaf size=24 \[ \left (4-\frac {2 e^{-4/x} \log ^2(x)}{5 (4+x)}\right )^2 \]

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Rubi [B]  time = 6.13, antiderivative size = 57, normalized size of antiderivative = 2.38, number of steps used = 6, number of rules used = 7, integrand size = 107, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {6688, 12, 6742, 2228, 2178, 2288, 2554} \begin {gather*} \frac {4 e^{-8/x} \log ^3(x) (x \log (x)+4 \log (x))}{25 (x+4)^3}-\frac {16 e^{-4/x} \log (x) (x \log (x)+4 \log (x))}{5 (x+4)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(4/x)*(-2560*x - 1280*x^2 - 160*x^3)*Log[x] + E^(4/x)*(-5120 - 2560*x + 80*x^3)*Log[x]^2 + (64*x + 16*x
^2)*Log[x]^3 + (128 + 32*x - 8*x^2)*Log[x]^4)/(E^(8/x)*(1600*x^2 + 1200*x^3 + 300*x^4 + 25*x^5)),x]

[Out]

(-16*Log[x]*(4*Log[x] + x*Log[x]))/(5*E^(4/x)*(4 + x)^2) + (4*Log[x]^3*(4*Log[x] + x*Log[x]))/(25*E^(8/x)*(4 +
x)^3)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2228

Int[(F_)^((a_.) + (b_.)/((c_.) + (d_.)*(x_)))/(((e_.) + (f_.)*(x_))*((g_.) + (h_.)*(x_))), x_Symbol] :> -Dist[
d/(f*(d*g - c*h)), Subst[Int[F^(a - (b*h)/(d*g - c*h) + (d*b*x)/(d*g - c*h))/x, x], x, (g + h*x)/(c + d*x)], x
] /; FreeQ[{F, a, b, c, d, e, f}, x] && EqQ[d*e - c*f, 0]

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 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 6688

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

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 {8 e^{-8/x} \log (x) \left (2 x (4+x)-\left (-16-4 x+x^2\right ) \log (x)\right ) \left (-10 e^{4/x} (4+x)+\log ^2(x)\right )}{25 x^2 (4+x)^3} \, dx\\ &=\frac {8}{25} \int \frac {e^{-8/x} \log (x) \left (2 x (4+x)-\left (-16-4 x+x^2\right ) \log (x)\right ) \left (-10 e^{4/x} (4+x)+\log ^2(x)\right )}{x^2 (4+x)^3} \, dx\\ &=\frac {8}{25} \int \left (\frac {10 e^{-4/x} \log (x) \left (-8 x-2 x^2-16 \log (x)-4 x \log (x)+x^2 \log (x)\right )}{x^2 (4+x)^2}-\frac {e^{-8/x} \log ^3(x) \left (-8 x-2 x^2-16 \log (x)-4 x \log (x)+x^2 \log (x)\right )}{x^2 (4+x)^3}\right ) \, dx\\ &=-\left (\frac {8}{25} \int \frac {e^{-8/x} \log ^3(x) \left (-8 x-2 x^2-16 \log (x)-4 x \log (x)+x^2 \log (x)\right )}{x^2 (4+x)^3} \, dx\right )+\frac {16}{5} \int \frac {e^{-4/x} \log (x) \left (-8 x-2 x^2-16 \log (x)-4 x \log (x)+x^2 \log (x)\right )}{x^2 (4+x)^2} \, dx\\ &=-\frac {16 e^{-4/x} \log (x) (4 \log (x)+x \log (x))}{5 (4+x)^2}+\frac {4 e^{-8/x} \log ^3(x) (4 \log (x)+x \log (x))}{25 (4+x)^3}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.24, size = 39, normalized size = 1.62 \begin {gather*} -\frac {4 e^{-8/x} \left (20 e^{4/x} (4+x) \log ^2(x)-\log ^4(x)\right )}{25 (4+x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(4/x)*(-2560*x - 1280*x^2 - 160*x^3)*Log[x] + E^(4/x)*(-5120 - 2560*x + 80*x^3)*Log[x]^2 + (64*x
+ 16*x^2)*Log[x]^3 + (128 + 32*x - 8*x^2)*Log[x]^4)/(E^(8/x)*(1600*x^2 + 1200*x^3 + 300*x^4 + 25*x^5)),x]

[Out]

(-4*(20*E^(4/x)*(4 + x)*Log[x]^2 - Log[x]^4))/(25*E^(8/x)*(4 + x)^2)

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fricas [A]  time = 0.48, size = 40, normalized size = 1.67 \begin {gather*} -\frac {4 \, {\left (20 \, {\left (x + 4\right )} e^{\frac {4}{x}} \log \relax (x)^{2} - \log \relax (x)^{4}\right )} e^{\left (-\frac {8}{x}\right )}}{25 \, {\left (x^{2} + 8 \, x + 16\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x^2+32*x+128)*log(x)^4+(16*x^2+64*x)*log(x)^3+(80*x^3-2560*x-5120)*exp(4/x)*log(x)^2+(-160*x^3-
1280*x^2-2560*x)*exp(4/x)*log(x))/(25*x^5+300*x^4+1200*x^3+1600*x^2)/exp(4/x)^2,x, algorithm="fricas")

[Out]

-4/25*(20*(x + 4)*e^(4/x)*log(x)^2 - log(x)^4)*e^(-8/x)/(x^2 + 8*x + 16)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x^2+32*x+128)*log(x)^4+(16*x^2+64*x)*log(x)^3+(80*x^3-2560*x-5120)*exp(4/x)*log(x)^2+(-160*x^3-
1280*x^2-2560*x)*exp(4/x)*log(x))/(25*x^5+300*x^4+1200*x^3+1600*x^2)/exp(4/x)^2,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.05, size = 36, normalized size = 1.50

method	result	size
risch	\(\frac {4 \,{\mathrm e}^{-\frac {8}{x}} \ln \relax (x )^{4}}{25 \left (4+x \right )^{2}}-\frac {16 \,{\mathrm e}^{-\frac {4}{x}} \ln \relax (x )^{2}}{5 \left (4+x \right )}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-8*x^2+32*x+128)*ln(x)^4+(16*x^2+64*x)*ln(x)^3+(80*x^3-2560*x-5120)*exp(4/x)*ln(x)^2+(-160*x^3-1280*x^2-
2560*x)*exp(4/x)*ln(x))/(25*x^5+300*x^4+1200*x^3+1600*x^2)/exp(4/x)^2,x,method=_RETURNVERBOSE)

[Out]

4/25/(4+x)^2*exp(-8/x)*ln(x)^4-16/5*exp(-4/x)/(4+x)*ln(x)^2

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maxima [A]  time = 0.42, size = 39, normalized size = 1.62 \begin {gather*} \frac {4 \, {\left (e^{\left (-\frac {8}{x}\right )} \log \relax (x)^{4} - 20 \, {\left (x + 4\right )} e^{\left (-\frac {4}{x}\right )} \log \relax (x)^{2}\right )}}{25 \, {\left (x^{2} + 8 \, x + 16\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x^2+32*x+128)*log(x)^4+(16*x^2+64*x)*log(x)^3+(80*x^3-2560*x-5120)*exp(4/x)*log(x)^2+(-160*x^3-
1280*x^2-2560*x)*exp(4/x)*log(x))/(25*x^5+300*x^4+1200*x^3+1600*x^2)/exp(4/x)^2,x, algorithm="maxima")

[Out]

4/25*(e^(-8/x)*log(x)^4 - 20*(x + 4)*e^(-4/x)*log(x)^2)/(x^2 + 8*x + 16)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {{\mathrm {e}}^{-\frac {8}{x}}\,\left (\left (-8\,x^2+32\,x+128\right )\,{\ln \relax (x)}^4+\left (16\,x^2+64\,x\right )\,{\ln \relax (x)}^3-{\mathrm {e}}^{4/x}\,\left (-80\,x^3+2560\,x+5120\right )\,{\ln \relax (x)}^2-{\mathrm {e}}^{4/x}\,\left (160\,x^3+1280\,x^2+2560\,x\right )\,\ln \relax (x)\right )}{25\,x^5+300\,x^4+1200\,x^3+1600\,x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-8/x)*(log(x)^3*(64*x + 16*x^2) + log(x)^4*(32*x - 8*x^2 + 128) - exp(4/x)*log(x)*(2560*x + 1280*x^2
+ 160*x^3) - exp(4/x)*log(x)^2*(2560*x - 80*x^3 + 5120)))/(1600*x^2 + 1200*x^3 + 300*x^4 + 25*x^5),x)

[Out]

int((exp(-8/x)*(log(x)^3*(64*x + 16*x^2) + log(x)^4*(32*x - 8*x^2 + 128) - exp(4/x)*log(x)*(2560*x + 1280*x^2
+ 160*x^3) - exp(4/x)*log(x)^2*(2560*x - 80*x^3 + 5120)))/(1600*x^2 + 1200*x^3 + 300*x^4 + 25*x^5), x)

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sympy [B]  time = 0.46, size = 66, normalized size = 2.75 \begin {gather*} \frac {\left (20 x \log {\relax (x )}^{4} + 80 \log {\relax (x )}^{4}\right ) e^{- \frac {8}{x}} + \left (- 400 x^{2} \log {\relax (x )}^{2} - 3200 x \log {\relax (x )}^{2} - 6400 \log {\relax (x )}^{2}\right ) e^{- \frac {4}{x}}}{125 x^{3} + 1500 x^{2} + 6000 x + 8000} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x**2+32*x+128)*ln(x)**4+(16*x**2+64*x)*ln(x)**3+(80*x**3-2560*x-5120)*exp(4/x)*ln(x)**2+(-160*x
**3-1280*x**2-2560*x)*exp(4/x)*ln(x))/(25*x**5+300*x**4+1200*x**3+1600*x**2)/exp(4/x)**2,x)

[Out]

((20*x*log(x)**4 + 80*log(x)**4)*exp(-8/x) + (-400*x**2*log(x)**2 - 3200*x*log(x)**2 - 6400*log(x)**2)*exp(-4/
x))/(125*x**3 + 1500*x**2 + 6000*x + 8000)

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