3.46.30 \(\int \frac {e^{\frac {(-125+25 x^3) \log ^2(4)+4 x^6 \log (x)}{25 x^2 \log ^2(4)}} (4 x^6+(250+25 x^3) \log ^2(4)+16 x^6 \log (x))}{25 x^3 \log ^2(4)} \, dx\)

Optimal. Leaf size=29 \[ e^{x-\left (-\frac {4 x^4}{25 \log ^2(4)}+\frac {5}{x^2 \log (x)}\right ) \log (x)} \]

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Rubi [A]  time = 0.51, antiderivative size = 28, normalized size of antiderivative = 0.97, number of steps used = 2, number of rules used = 2, integrand size = 69, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {12, 6706} \begin {gather*} e^{-\frac {5-x^3}{x^2}} x^{\frac {4 x^4}{25 \log ^2(4)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(((-125 + 25*x^3)*Log[4]^2 + 4*x^6*Log[x])/(25*x^2*Log[4]^2))*(4*x^6 + (250 + 25*x^3)*Log[4]^2 + 16*x^6
*Log[x]))/(25*x^3*Log[4]^2),x]

[Out]

x^((4*x^4)/(25*Log[4]^2))/E^((5 - x^3)/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 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {\exp \left (\frac {\left (-125+25 x^3\right ) \log ^2(4)+4 x^6 \log (x)}{25 x^2 \log ^2(4)}\right ) \left (4 x^6+\left (250+25 x^3\right ) \log ^2(4)+16 x^6 \log (x)\right )}{x^3} \, dx}{25 \log ^2(4)}\\ &=e^{-\frac {5-x^3}{x^2}} x^{\frac {4 x^4}{25 \log ^2(4)}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 23, normalized size = 0.79 \begin {gather*} e^{-\frac {5}{x^2}+x} x^{\frac {4 x^4}{25 \log ^2(4)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(((-125 + 25*x^3)*Log[4]^2 + 4*x^6*Log[x])/(25*x^2*Log[4]^2))*(4*x^6 + (250 + 25*x^3)*Log[4]^2 +
16*x^6*Log[x]))/(25*x^3*Log[4]^2),x]

[Out]

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

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fricas [A]  time = 0.63, size = 28, normalized size = 0.97 \begin {gather*} e^{\left (\frac {x^{6} \log \relax (x) + 25 \, {\left (x^{3} - 5\right )} \log \relax (2)^{2}}{25 \, x^{2} \log \relax (2)^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/100*(16*x^6*log(x)+4*(25*x^3+250)*log(2)^2+4*x^6)*exp(1/100*(4*x^6*log(x)+4*(25*x^3-125)*log(2)^2)
/x^2/log(2)^2)/x^3/log(2)^2,x, algorithm="fricas")

[Out]

e^(1/25*(x^6*log(x) + 25*(x^3 - 5)*log(2)^2)/(x^2*log(2)^2))

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giac [A]  time = 0.15, size = 19, normalized size = 0.66 \begin {gather*} e^{\left (\frac {x^{4} \log \relax (x)}{25 \, \log \relax (2)^{2}} + x - \frac {5}{x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/100*(16*x^6*log(x)+4*(25*x^3+250)*log(2)^2+4*x^6)*exp(1/100*(4*x^6*log(x)+4*(25*x^3-125)*log(2)^2)
/x^2/log(2)^2)/x^3/log(2)^2,x, algorithm="giac")

[Out]

e^(1/25*x^4*log(x)/log(2)^2 + x - 5/x^2)

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maple [A]  time = 0.03, size = 23, normalized size = 0.79

method	result	size
risch	\(x^{\frac {x^{4}}{25 \ln \relax (2)^{2}}} {\mathrm e}^{\frac {x^{3}-5}{x^{2}}}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/100*(16*x^6*ln(x)+4*(25*x^3+250)*ln(2)^2+4*x^6)*exp(1/100*(4*x^6*ln(x)+4*(25*x^3-125)*ln(2)^2)/x^2/ln(2)
^2)/x^3/ln(2)^2,x,method=_RETURNVERBOSE)

[Out]

x^(1/25*x^4/ln(2)^2)*exp((x^3-5)/x^2)

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maxima [A]  time = 0.67, size = 19, normalized size = 0.66 \begin {gather*} e^{\left (\frac {x^{4} \log \relax (x)}{25 \, \log \relax (2)^{2}} + x - \frac {5}{x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/100*(16*x^6*log(x)+4*(25*x^3+250)*log(2)^2+4*x^6)*exp(1/100*(4*x^6*log(x)+4*(25*x^3-125)*log(2)^2)
/x^2/log(2)^2)/x^3/log(2)^2,x, algorithm="maxima")

[Out]

e^(1/25*x^4*log(x)/log(2)^2 + x - 5/x^2)

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mupad [B]  time = 3.37, size = 21, normalized size = 0.72 \begin {gather*} {\mathrm {e}}^{\frac {x^4\,\ln \relax (x)}{25\,{\ln \relax (2)}^2}}\,{\mathrm {e}}^{-\frac {5}{x^2}}\,{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(((x^6*log(x))/25 + (log(2)^2*(25*x^3 - 125))/25)/(x^2*log(2)^2))*(16*x^6*log(x) + 4*log(2)^2*(25*x^3
+ 250) + 4*x^6))/(100*x^3*log(2)^2),x)

[Out]

exp((x^4*log(x))/(25*log(2)^2))*exp(-5/x^2)*exp(x)

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sympy [A]  time = 0.40, size = 31, normalized size = 1.07 \begin {gather*} e^{\frac {\frac {x^{6} \log {\relax (x )}}{25} + \frac {\left (100 x^{3} - 500\right ) \log {\relax (2 )}^{2}}{100}}{x^{2} \log {\relax (2 )}^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/100*(16*x**6*ln(x)+4*(25*x**3+250)*ln(2)**2+4*x**6)*exp(1/100*(4*x**6*ln(x)+4*(25*x**3-125)*ln(2)*
*2)/x**2/ln(2)**2)/x**3/ln(2)**2,x)

[Out]

exp((x**6*log(x)/25 + (100*x**3 - 500)*log(2)**2/100)/(x**2*log(2)**2))

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