3.93.9 \(\int \frac {e^{-2 e^4} (-10 e^{2 x} x^2+e^{2 x} (15 x^2+10 x^3) \log (x))}{\log ^3(x)} \, dx\)

Optimal. Leaf size=20 \[ \frac {5 e^{-2 e^4+2 x} x^3}{\log ^2(x)} \]

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Rubi [A]  time = 0.13, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {12, 6688, 2202} \begin {gather*} \frac {5 e^{2 x-2 e^4} x^3}{\log ^2(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-10*E^(2*x)*x^2 + E^(2*x)*(15*x^2 + 10*x^3)*Log[x])/(E^(2*E^4)*Log[x]^3),x]

[Out]

(5*E^(-2*E^4 + 2*x)*x^3)/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 2202

Int[Log[(d_.)*(x_)]^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_)))*(x_)^(m_.)*((e_) + Log[(d_.)*(x_)]*(h_.)*((f_.) +
(g_.)*(x_))), x_Symbol] :> Simp[(e*x^(m + 1)*F^(c*(a + b*x))*Log[d*x]^(n + 1))/(n + 1), x] /; FreeQ[{F, a, b,
c, d, e, f, g, h, m, n}, x] && EqQ[e*(m + 1) - f*h*(n + 1), 0] && EqQ[g*h*(n + 1) - b*c*e*Log[F], 0] && NeQ[n,
-1]

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} &=e^{-2 e^4} \int \frac {-10 e^{2 x} x^2+e^{2 x} \left (15 x^2+10 x^3\right ) \log (x)}{\log ^3(x)} \, dx\\ &=e^{-2 e^4} \int \frac {5 e^{2 x} x^2 (-2+(3+2 x) \log (x))}{\log ^3(x)} \, dx\\ &=\left (5 e^{-2 e^4}\right ) \int \frac {e^{2 x} x^2 (-2+(3+2 x) \log (x))}{\log ^3(x)} \, dx\\ &=\frac {5 e^{-2 e^4+2 x} x^3}{\log ^2(x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 20, normalized size = 1.00 \begin {gather*} \frac {5 e^{-2 e^4+2 x} x^3}{\log ^2(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10*E^(2*x)*x^2 + E^(2*x)*(15*x^2 + 10*x^3)*Log[x])/(E^(2*E^4)*Log[x]^3),x]

[Out]

(5*E^(-2*E^4 + 2*x)*x^3)/Log[x]^2

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fricas [A]  time = 0.70, size = 18, normalized size = 0.90 \begin {gather*} \frac {5 \, x^{3} e^{\left (2 \, x - 2 \, e^{4}\right )}}{\log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^3+15*x^2)*exp(x)^2*log(x)-10*exp(x)^2*x^2)/exp(exp(4))^2/log(x)^3,x, algorithm="fricas")

[Out]

5*x^3*e^(2*x - 2*e^4)/log(x)^2

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giac [A]  time = 0.17, size = 18, normalized size = 0.90 \begin {gather*} \frac {5 \, x^{3} e^{\left (2 \, x - 2 \, e^{4}\right )}}{\log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^3+15*x^2)*exp(x)^2*log(x)-10*exp(x)^2*x^2)/exp(exp(4))^2/log(x)^3,x, algorithm="giac")

[Out]

5*x^3*e^(2*x - 2*e^4)/log(x)^2

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maple [A]  time = 0.03, size = 19, normalized size = 0.95

method	result	size
risch	\(\frac {5 x^{3} {\mathrm e}^{2 x -2 \,{\mathrm e}^{4}}}{\ln \relax (x )^{2}}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((10*x^3+15*x^2)*exp(x)^2*ln(x)-10*exp(x)^2*x^2)/exp(exp(4))^2/ln(x)^3,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.41, size = 18, normalized size = 0.90 \begin {gather*} \frac {5 \, x^{3} e^{\left (2 \, x - 2 \, e^{4}\right )}}{\log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^3+15*x^2)*exp(x)^2*log(x)-10*exp(x)^2*x^2)/exp(exp(4))^2/log(x)^3,x, algorithm="maxima")

[Out]

5*x^3*e^(2*x - 2*e^4)/log(x)^2

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mupad [B]  time = 7.43, size = 18, normalized size = 0.90 \begin {gather*} \frac {5\,x^3\,{\mathrm {e}}^{-2\,{\mathrm {e}}^4}\,{\mathrm {e}}^{2\,x}}{{\ln \relax (x)}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-2*exp(4))*(10*x^2*exp(2*x) - exp(2*x)*log(x)*(15*x^2 + 10*x^3)))/log(x)^3,x)

[Out]

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

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sympy [A]  time = 0.45, size = 20, normalized size = 1.00 \begin {gather*} \frac {5 x^{3} e^{2 x}}{e^{2 e^{4}} \log {\relax (x )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x**3+15*x**2)*exp(x)**2*ln(x)-10*exp(x)**2*x**2)/exp(exp(4))**2/ln(x)**3,x)

[Out]

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

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