\(\int \frac {1}{4} e^{-8-4 x} x^3 (5 \log ^3(x)+(5-5 x) \log ^4(x)) \, dx\) [7703]

Optimal result

Integrand size = 31, antiderivative size = 18 \[ \int \frac {1}{4} e^{-8-4 x} x^3 \left (5 \log ^3(x)+(5-5 x) \log ^4(x)\right ) \, dx=\frac {5}{16} e^{-8-4 x} x^4 \log ^4(x) \]

[Out]

5/16*ln(x)^4*exp(ln(x)-x-2)^4

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {12, 6873, 2326} \[ \int \frac {1}{4} e^{-8-4 x} x^3 \left (5 \log ^3(x)+(5-5 x) \log ^4(x)\right ) \, dx=\frac {5}{16} e^{-4 x-8} x^4 \log ^4(x) \]

[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 2326

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 6873

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

Rubi steps \begin{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {1}{4} e^{-8-4 x} x^3 \left (5 \log ^3(x)+(5-5 x) \log ^4(x)\right ) \, dx=\frac {5}{16} e^{-8-4 x} x^4 \log ^4(x) \]

[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

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89

method	result	size
risch	\(\frac {5 \ln \left (x \right )^{4} x^{4} {\mathrm e}^{-4 x -8}}{16}\)	\(16\)
parallelrisch	\(\frac {5 \ln \left (x \right )^{4} x^{4} {\mathrm e}^{-4 x -8}}{16}\)	\(17\)

[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)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {1}{4} e^{-8-4 x} x^3 \left (5 \log ^3(x)+(5-5 x) \log ^4(x)\right ) \, dx=\frac {5}{16} \, e^{\left (-4 \, x + 4 \, \log \left (x\right ) - 8\right )} \log \left (x\right )^{4} \]

[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

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \frac {1}{4} e^{-8-4 x} x^3 \left (5 \log ^3(x)+(5-5 x) \log ^4(x)\right ) \, dx=\frac {5 x^{4} e^{- 4 x - 8} \log {\left (x \right )}^{4}}{16} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {1}{4} e^{-8-4 x} x^3 \left (5 \log ^3(x)+(5-5 x) \log ^4(x)\right ) \, dx=\frac {5}{16} \, x^{4} e^{\left (-4 \, x - 8\right )} \log \left (x\right )^{4} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {1}{4} e^{-8-4 x} x^3 \left (5 \log ^3(x)+(5-5 x) \log ^4(x)\right ) \, dx=\frac {5}{16} \, x^{4} e^{\left (-4 \, x - 8\right )} \log \left (x\right )^{4} \]

[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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{4} e^{-8-4 x} x^3 \left (5 \log ^3(x)+(5-5 x) \log ^4(x)\right ) \, dx=\int \frac {{\mathrm {e}}^{4\,\ln \left (x\right )-4\,x-8}\,\left (5\,{\ln \left (x\right )}^3-{\ln \left (x\right )}^4\,\left (5\,x-5\right )\right )}{4\,x} \,d x \]

[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)