3.65.29 \(\int \frac {e^{-2 x} ((-5+6 x-8 x^2) \log ^3(x)+e^{\frac {9}{\log ^2(x)}} (18 x+(-2 x+2 x^2) \log ^3(x)))}{4 \log ^3(x)} \, dx\)

Optimal. Leaf size=27 \[ \frac {1}{4} e^{-2 x} \left (3+x-\left (-4+e^{\frac {9}{\log ^2(x)}}\right ) x^2\right ) \]

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Rubi [B]  time = 0.44, antiderivative size = 71, normalized size of antiderivative = 2.63, number of steps used = 10, number of rules used = 5, integrand size = 56, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {12, 6742, 2194, 2176, 2288} \begin {gather*} e^{-2 x} x^2+\frac {1}{4} e^{-2 x} x+\frac {3 e^{-2 x}}{4}-\frac {x e^{\frac {9}{\log ^2(x)}-2 x} \left (x \log ^3(x)+9\right )}{4 \left (\frac {9}{x \log ^3(x)}+1\right ) \log ^3(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-5 + 6*x - 8*x^2)*Log[x]^3 + E^(9/Log[x]^2)*(18*x + (-2*x + 2*x^2)*Log[x]^3))/(4*E^(2*x)*Log[x]^3),x]

[Out]

3/(4*E^(2*x)) + x/(4*E^(2*x)) + x^2/E^(2*x) - (E^(-2*x + 9/Log[x]^2)*x*(9 + x*Log[x]^3))/(4*(1 + 9/(x*Log[x]^3
))*Log[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 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, 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 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}{4} \int \frac {e^{-2 x} \left (\left (-5+6 x-8 x^2\right ) \log ^3(x)+e^{\frac {9}{\log ^2(x)}} \left (18 x+\left (-2 x+2 x^2\right ) \log ^3(x)\right )\right )}{\log ^3(x)} \, dx\\ &=\frac {1}{4} \int \left (-5 e^{-2 x}+6 e^{-2 x} x-8 e^{-2 x} x^2+\frac {2 e^{-2 x+\frac {9}{\log ^2(x)}} x \left (9-\log ^3(x)+x \log ^3(x)\right )}{\log ^3(x)}\right ) \, dx\\ &=\frac {1}{2} \int \frac {e^{-2 x+\frac {9}{\log ^2(x)}} x \left (9-\log ^3(x)+x \log ^3(x)\right )}{\log ^3(x)} \, dx-\frac {5}{4} \int e^{-2 x} \, dx+\frac {3}{2} \int e^{-2 x} x \, dx-2 \int e^{-2 x} x^2 \, dx\\ &=\frac {5 e^{-2 x}}{8}-\frac {3}{4} e^{-2 x} x+e^{-2 x} x^2-\frac {e^{-2 x+\frac {9}{\log ^2(x)}} x \left (9+x \log ^3(x)\right )}{4 \left (1+\frac {9}{x \log ^3(x)}\right ) \log ^3(x)}+\frac {3}{4} \int e^{-2 x} \, dx-2 \int e^{-2 x} x \, dx\\ &=\frac {e^{-2 x}}{4}+\frac {1}{4} e^{-2 x} x+e^{-2 x} x^2-\frac {e^{-2 x+\frac {9}{\log ^2(x)}} x \left (9+x \log ^3(x)\right )}{4 \left (1+\frac {9}{x \log ^3(x)}\right ) \log ^3(x)}-\int e^{-2 x} \, dx\\ &=\frac {3 e^{-2 x}}{4}+\frac {1}{4} e^{-2 x} x+e^{-2 x} x^2-\frac {e^{-2 x+\frac {9}{\log ^2(x)}} x \left (9+x \log ^3(x)\right )}{4 \left (1+\frac {9}{x \log ^3(x)}\right ) \log ^3(x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.38, size = 27, normalized size = 1.00 \begin {gather*} \frac {1}{4} e^{-2 x} \left (3+x-\left (-4+e^{\frac {9}{\log ^2(x)}}\right ) x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-5 + 6*x - 8*x^2)*Log[x]^3 + E^(9/Log[x]^2)*(18*x + (-2*x + 2*x^2)*Log[x]^3))/(4*E^(2*x)*Log[x]^3)
,x]

[Out]

(3 + x - (-4 + E^(9/Log[x]^2))*x^2)/(4*E^(2*x))

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fricas [A]  time = 0.74, size = 31, normalized size = 1.15 \begin {gather*} -\frac {1}{4} \, x^{2} e^{\left (-2 \, x + \frac {9}{\log \relax (x)^{2}}\right )} + \frac {1}{4} \, {\left (4 \, x^{2} + x + 3\right )} e^{\left (-2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(((2*x^2-2*x)*log(x)^3+18*x)*exp(9/log(x)^2)+(-8*x^2+6*x-5)*log(x)^3)/exp(x)^2/log(x)^3,x, algor
ithm="fricas")

[Out]

-1/4*x^2*e^(-2*x + 9/log(x)^2) + 1/4*(4*x^2 + x + 3)*e^(-2*x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(((2*x^2-2*x)*log(x)^3+18*x)*exp(9/log(x)^2)+(-8*x^2+6*x-5)*log(x)^3)/exp(x)^2/log(x)^3,x, algor
ithm="giac")

[Out]

undef

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maple [A]  time = 0.04, size = 37, normalized size = 1.37

method	result	size
risch	\(\frac {\left (4 x^{2}+x +3\right ) {\mathrm e}^{-2 x}}{4}-\frac {x^{2} {\mathrm e}^{-\frac {2 x \ln \relax (x )^{2}-9}{\ln \relax (x )^{2}}}}{4}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(((2*x^2-2*x)*ln(x)^3+18*x)*exp(9/ln(x)^2)+(-8*x^2+6*x-5)*ln(x)^3)/exp(x)^2/ln(x)^3,x,method=_RETURNVE
RBOSE)

[Out]

1/4*(4*x^2+x+3)*exp(-2*x)-1/4*x^2*exp(-(2*x*ln(x)^2-9)/ln(x)^2)

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maxima [B]  time = 0.44, size = 50, normalized size = 1.85 \begin {gather*} -\frac {1}{4} \, x^{2} e^{\left (-2 \, x + \frac {9}{\log \relax (x)^{2}}\right )} + \frac {1}{2} \, {\left (2 \, x^{2} + 2 \, x + 1\right )} e^{\left (-2 \, x\right )} - \frac {3}{8} \, {\left (2 \, x + 1\right )} e^{\left (-2 \, x\right )} + \frac {5}{8} \, e^{\left (-2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(((2*x^2-2*x)*log(x)^3+18*x)*exp(9/log(x)^2)+(-8*x^2+6*x-5)*log(x)^3)/exp(x)^2/log(x)^3,x, algor
ithm="maxima")

[Out]

-1/4*x^2*e^(-2*x + 9/log(x)^2) + 1/2*(2*x^2 + 2*x + 1)*e^(-2*x) - 3/8*(2*x + 1)*e^(-2*x) + 5/8*e^(-2*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 {{\ln \relax (x)}^3\,\left (8\,x^2-6\,x+5\right )}{4}-\frac {{\mathrm {e}}^{\frac {9}{{\ln \relax (x)}^2}}\,\left (18\,x-{\ln \relax (x)}^3\,\left (2\,x-2\,x^2\right )\right )}{4}\right )}{{\ln \relax (x)}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-2*x)*((log(x)^3*(8*x^2 - 6*x + 5))/4 - (exp(9/log(x)^2)*(18*x - log(x)^3*(2*x - 2*x^2)))/4))/log(x)
^3,x)

[Out]

int(-(exp(-2*x)*((log(x)^3*(8*x^2 - 6*x + 5))/4 - (exp(9/log(x)^2)*(18*x - log(x)^3*(2*x - 2*x^2)))/4))/log(x)
^3, x)

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sympy [A]  time = 13.20, size = 26, normalized size = 0.96 \begin {gather*} \frac {\left (- x^{2} e^{\frac {9}{\log {\relax (x )}^{2}}} + 4 x^{2} + x + 3\right ) e^{- 2 x}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(((2*x**2-2*x)*ln(x)**3+18*x)*exp(9/ln(x)**2)+(-8*x**2+6*x-5)*ln(x)**3)/exp(x)**2/ln(x)**3,x)

[Out]

(-x**2*exp(9/log(x)**2) + 4*x**2 + x + 3)*exp(-2*x)/4

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