3.96.34 \(\int \frac {e^{-2 x} (32 e^x x^4+(-x^9+e^x (-160 x^4+32 x^5) \log (x)) \log (\frac {3}{\log (x)})+(5 x^9-x^{10}) \log (x) \log ^2(\frac {3}{\log (x)}))}{128 \log (x)} \, dx\)

Optimal. Leaf size=23 \[ \left (-2+\frac {1}{16} e^{-x} x^5 \log \left (\frac {3}{\log (x)}\right )\right )^2 \]

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Rubi [A]  time = 1.33, antiderivative size = 41, normalized size of antiderivative = 1.78, number of steps used = 6, number of rules used = 4, integrand size = 76, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {12, 6688, 6742, 2288} \begin {gather*} \frac {1}{256} e^{-2 x} x^{10} \log ^2\left (\frac {3}{\log (x)}\right )-\frac {1}{4} e^{-x} x^5 \log \left (\frac {3}{\log (x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(32*E^x*x^4 + (-x^9 + E^x*(-160*x^4 + 32*x^5)*Log[x])*Log[3/Log[x]] + (5*x^9 - x^10)*Log[x]*Log[3/Log[x]]^
2)/(128*E^(2*x)*Log[x]),x]

[Out]

-1/4*(x^5*Log[3/Log[x]])/E^x + (x^10*Log[3/Log[x]]^2)/(256*E^(2*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 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} &=\frac {1}{128} \int \frac {e^{-2 x} \left (32 e^x x^4+\left (-x^9+e^x \left (-160 x^4+32 x^5\right ) \log (x)\right ) \log \left (\frac {3}{\log (x)}\right )+\left (5 x^9-x^{10}\right ) \log (x) \log ^2\left (\frac {3}{\log (x)}\right )\right )}{\log (x)} \, dx\\ &=\frac {1}{128} \int \frac {e^{-2 x} x^4 \left (32 e^x-x^5 \log \left (\frac {3}{\log (x)}\right )\right ) \left (1+(-5+x) \log (x) \log \left (\frac {3}{\log (x)}\right )\right )}{\log (x)} \, dx\\ &=\frac {1}{128} \int \left (\frac {32 e^{-x} x^4 \left (1-5 \log (x) \log \left (\frac {3}{\log (x)}\right )+x \log (x) \log \left (\frac {3}{\log (x)}\right )\right )}{\log (x)}-\frac {e^{-2 x} x^9 \log \left (\frac {3}{\log (x)}\right ) \left (1-5 \log (x) \log \left (\frac {3}{\log (x)}\right )+x \log (x) \log \left (\frac {3}{\log (x)}\right )\right )}{\log (x)}\right ) \, dx\\ &=-\left (\frac {1}{128} \int \frac {e^{-2 x} x^9 \log \left (\frac {3}{\log (x)}\right ) \left (1-5 \log (x) \log \left (\frac {3}{\log (x)}\right )+x \log (x) \log \left (\frac {3}{\log (x)}\right )\right )}{\log (x)} \, dx\right )+\frac {1}{4} \int \frac {e^{-x} x^4 \left (1-5 \log (x) \log \left (\frac {3}{\log (x)}\right )+x \log (x) \log \left (\frac {3}{\log (x)}\right )\right )}{\log (x)} \, dx\\ &=-\frac {1}{4} e^{-x} x^5 \log \left (\frac {3}{\log (x)}\right )+\frac {1}{256} e^{-2 x} x^{10} \log ^2\left (\frac {3}{\log (x)}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.19, size = 36, normalized size = 1.57 \begin {gather*} \frac {1}{256} e^{-2 x} x^5 \log \left (\frac {3}{\log (x)}\right ) \left (-64 e^x+x^5 \log \left (\frac {3}{\log (x)}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(32*E^x*x^4 + (-x^9 + E^x*(-160*x^4 + 32*x^5)*Log[x])*Log[3/Log[x]] + (5*x^9 - x^10)*Log[x]*Log[3/Lo
g[x]]^2)/(128*E^(2*x)*Log[x]),x]

[Out]

(x^5*Log[3/Log[x]]*(-64*E^x + x^5*Log[3/Log[x]]))/(256*E^(2*x))

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fricas [A]  time = 0.52, size = 34, normalized size = 1.48 \begin {gather*} \frac {1}{256} \, {\left (x^{10} \log \left (\frac {3}{\log \relax (x)}\right )^{2} - 64 \, x^{5} e^{x} \log \left (\frac {3}{\log \relax (x)}\right )\right )} e^{\left (-2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/128*((-x^10+5*x^9)*log(x)*log(3/log(x))^2+((32*x^5-160*x^4)*exp(x)*log(x)-x^9)*log(3/log(x))+32*ex
p(x)*x^4)/exp(x)^2/log(x),x, algorithm="fricas")

[Out]

1/256*(x^10*log(3/log(x))^2 - 64*x^5*e^x*log(3/log(x)))*e^(-2*x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (32 \, x^{4} e^{x} - {\left (x^{10} - 5 \, x^{9}\right )} \log \relax (x) \log \left (\frac {3}{\log \relax (x)}\right )^{2} - {\left (x^{9} - 32 \, {\left (x^{5} - 5 \, x^{4}\right )} e^{x} \log \relax (x)\right )} \log \left (\frac {3}{\log \relax (x)}\right )\right )} e^{\left (-2 \, x\right )}}{128 \, \log \relax (x)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/128*((-x^10+5*x^9)*log(x)*log(3/log(x))^2+((32*x^5-160*x^4)*exp(x)*log(x)-x^9)*log(3/log(x))+32*ex
p(x)*x^4)/exp(x)^2/log(x),x, algorithm="giac")

[Out]

integrate(1/128*(32*x^4*e^x - (x^10 - 5*x^9)*log(x)*log(3/log(x))^2 - (x^9 - 32*(x^5 - 5*x^4)*e^x*log(x))*log(
3/log(x)))*e^(-2*x)/log(x), x)

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maple [B]  time = 0.23, size = 65, normalized size = 2.83

method	result	size
risch	\(\frac {x^{10} {\mathrm e}^{-2 x} \ln \left (\ln \relax (x )\right )^{2}}{256}-\frac {x^{5} \left (2 x^{5} \ln \relax (3)-64 \,{\mathrm e}^{x}\right ) {\mathrm e}^{-2 x} \ln \left (\ln \relax (x )\right )}{256}+\frac {x^{5} \left (4 x^{5} \ln \relax (3)^{2}-256 \ln \relax (3) {\mathrm e}^{x}\right ) {\mathrm e}^{-2 x}}{1024}\)	\(65\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/128*((-x^10+5*x^9)*ln(x)*ln(3/ln(x))^2+((32*x^5-160*x^4)*exp(x)*ln(x)-x^9)*ln(3/ln(x))+32*exp(x)*x^4)/ex
p(x)^2/ln(x),x,method=_RETURNVERBOSE)

[Out]

1/256*x^10*exp(-2*x)*ln(ln(x))^2-1/256*x^5*(2*x^5*ln(3)-64*exp(x))*exp(-2*x)*ln(ln(x))+1/1024*x^5*(4*x^5*ln(3)
^2-256*ln(3)*exp(x))*exp(-2*x)

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maxima [B]  time = 0.48, size = 64, normalized size = 2.78 \begin {gather*} \frac {1}{256} \, x^{10} e^{\left (-2 \, x\right )} \log \relax (3)^{2} + \frac {1}{256} \, x^{10} e^{\left (-2 \, x\right )} \log \left (\log \relax (x)\right )^{2} - \frac {1}{4} \, x^{5} e^{\left (-x\right )} \log \relax (3) - \frac {1}{128} \, {\left (x^{10} e^{\left (-2 \, x\right )} \log \relax (3) - 32 \, x^{5} e^{\left (-x\right )}\right )} \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/128*((-x^10+5*x^9)*log(x)*log(3/log(x))^2+((32*x^5-160*x^4)*exp(x)*log(x)-x^9)*log(3/log(x))+32*ex
p(x)*x^4)/exp(x)^2/log(x),x, algorithm="maxima")

[Out]

1/256*x^10*e^(-2*x)*log(3)^2 + 1/256*x^10*e^(-2*x)*log(log(x))^2 - 1/4*x^5*e^(-x)*log(3) - 1/128*(x^10*e^(-2*x
)*log(3) - 32*x^5*e^(-x))*log(log(x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {{\mathrm {e}}^{-2\,x}\,\left (\frac {x^4\,{\mathrm {e}}^x}{4}-\frac {\ln \left (\frac {3}{\ln \relax (x)}\right )\,\left (x^9+{\mathrm {e}}^x\,\ln \relax (x)\,\left (160\,x^4-32\,x^5\right )\right )}{128}+\frac {{\ln \left (\frac {3}{\ln \relax (x)}\right )}^2\,\ln \relax (x)\,\left (5\,x^9-x^{10}\right )}{128}\right )}{\ln \relax (x)} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-2*x)*((x^4*exp(x))/4 - (log(3/log(x))*(x^9 + exp(x)*log(x)*(160*x^4 - 32*x^5)))/128 + (log(3/log(x))
^2*log(x)*(5*x^9 - x^10))/128))/log(x),x)

[Out]

int((exp(-2*x)*((x^4*exp(x))/4 - (log(3/log(x))*(x^9 + exp(x)*log(x)*(160*x^4 - 32*x^5)))/128 + (log(3/log(x))
^2*log(x)*(5*x^9 - x^10))/128))/log(x), x)

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sympy [A]  time = 82.52, size = 32, normalized size = 1.39 \begin {gather*} \frac {x^{10} e^{- 2 x} \log {\left (\frac {3}{\log {\relax (x )}} \right )}^{2}}{256} - \frac {x^{5} e^{- x} \log {\left (\frac {3}{\log {\relax (x )}} \right )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/128*((-x**10+5*x**9)*ln(x)*ln(3/ln(x))**2+((32*x**5-160*x**4)*exp(x)*ln(x)-x**9)*ln(3/ln(x))+32*ex
p(x)*x**4)/exp(x)**2/ln(x),x)

[Out]

x**10*exp(-2*x)*log(3/log(x))**2/256 - x**5*exp(-x)*log(3/log(x))/4

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