3.78.3 \(\int \frac {1}{4} e^{-8-4 x} x^3 (5 \log ^3(x)+(5-5 x) \log ^4(x)) \, dx\)

Optimal. Leaf size=18 \[ \frac {5}{16} e^{-8-4 x} x^4 \log ^4(x) \]

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Rubi [A]  time = 0.23, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {12, 6741, 2288} \begin {gather*} \frac {5}{16} e^{-4 x-8} x^4 \log ^4(x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*E^(-8 - 4*x)*x^4*Log[x]^4)/16

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 6741

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int e^{-8-4 x} x^3 \left (5 \log ^3(x)+(5-5 x) \log ^4(x)\right ) \, dx\\ &=\frac {1}{4} \int 5 e^{-8-4 x} x^3 \log ^3(x) (1+\log (x)-x \log (x)) \, dx\\ &=\frac {5}{4} \int e^{-8-4 x} x^3 \log ^3(x) (1+\log (x)-x \log (x)) \, dx\\ &=\frac {5}{16} e^{-8-4 x} x^4 \log ^4(x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 18, normalized size = 1.00 \begin {gather*} \frac {5}{16} e^{-8-4 x} x^4 \log ^4(x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*E^(-8 - 4*x)*x^4*Log[x]^4)/16

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fricas [A]  time = 0.66, size = 16, normalized size = 0.89 \begin {gather*} \frac {5}{16} \, e^{\left (-4 \, x + 4 \, \log \relax (x) - 8\right )} \log \relax (x)^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-5*x+5)*log(x)^4+5*log(x)^3)*exp(log(x)-x-2)^4/x,x, algorithm="fricas")

[Out]

5/16*e^(-4*x + 4*log(x) - 8)*log(x)^4

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giac [A]  time = 0.14, size = 15, normalized size = 0.83 \begin {gather*} \frac {5}{16} \, x^{4} e^{\left (-4 \, x - 8\right )} \log \relax (x)^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-5*x+5)*log(x)^4+5*log(x)^3)*exp(log(x)-x-2)^4/x,x, algorithm="giac")

[Out]

5/16*x^4*e^(-4*x - 8)*log(x)^4

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maple [A]  time = 0.02, size = 16, normalized size = 0.89

method	result	size
risch	\(\frac {5 \ln \relax (x )^{4} x^{4} {\mathrm e}^{-4 x -8}}{16}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*((-5*x+5)*ln(x)^4+5*ln(x)^3)*exp(ln(x)-x-2)^4/x,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.52, size = 15, normalized size = 0.83 \begin {gather*} \frac {5}{16} \, x^{4} e^{\left (-4 \, x - 8\right )} \log \relax (x)^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-5*x+5)*log(x)^4+5*log(x)^3)*exp(log(x)-x-2)^4/x,x, algorithm="maxima")

[Out]

5/16*x^4*e^(-4*x - 8)*log(x)^4

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mupad [F]  time = 0.00, size = -1, normalized size = -0.06 \begin {gather*} \int \frac {{\mathrm {e}}^{4\,\ln \relax (x)-4\,x-8}\,\left (5\,{\ln \relax (x)}^3-{\ln \relax (x)}^4\,\left (5\,x-5\right )\right )}{4\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(4*log(x) - 4*x - 8)*(5*log(x)^3 - log(x)^4*(5*x - 5)))/(4*x),x)

[Out]

int((exp(4*log(x) - 4*x - 8)*(5*log(x)^3 - log(x)^4*(5*x - 5)))/(4*x), x)

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sympy [A]  time = 0.47, size = 19, normalized size = 1.06 \begin {gather*} \frac {5 x^{4} e^{- 4 x - 8} \log {\relax (x )}^{4}}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*x**4*exp(-4*x - 8)*log(x)**4/16

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