3.31.52 \(\int \frac {x^{\frac {1}{64} (-64 x^2-x^3)} (e^{-5+x} (64+64 x-64 x^2-x^3)+e^{-5+x} (-128 x^2-3 x^3) \log (x))}{256 e^5} \, dx\)

Optimal. Leaf size=25 \[ \frac {1}{4} e^{-10+x} x^{1-x \left (x+\frac {x^2}{64}\right )} \]

________________________________________________________________________________________

Rubi [A]  time = 0.15, antiderivative size = 23, normalized size of antiderivative = 0.92, number of steps used = 3, number of rules used = 3, integrand size = 65, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.046, Rules used = {12, 6741, 2288} \begin {gather*} \frac {1}{4} e^{x-10} x^{1-\frac {1}{64} x^2 (x+64)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^((-64*x^2 - x^3)/64)*(E^(-5 + x)*(64 + 64*x - 64*x^2 - x^3) + E^(-5 + x)*(-128*x^2 - 3*x^3)*Log[x]))/(2
56*E^5),x]

[Out]

(E^(-10 + x)*x^(1 - (x^2*(64 + x))/64))/4

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 {\int x^{\frac {1}{64} \left (-64 x^2-x^3\right )} \left (e^{-5+x} \left (64+64 x-64 x^2-x^3\right )+e^{-5+x} \left (-128 x^2-3 x^3\right ) \log (x)\right ) \, dx}{256 e^5}\\ &=\frac {\int e^{-5+x} x^{\frac {1}{64} (-64-x) x^2} \left (64+64 x-64 x^2-x^3-128 x^2 \log (x)-3 x^3 \log (x)\right ) \, dx}{256 e^5}\\ &=\frac {1}{4} e^{-10+x} x^{1-\frac {1}{64} x^2 (64+x)}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 23, normalized size = 0.92 \begin {gather*} \frac {1}{4} e^{-10+x} x^{1-\frac {1}{64} x^2 (64+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^((-64*x^2 - x^3)/64)*(E^(-5 + x)*(64 + 64*x - 64*x^2 - x^3) + E^(-5 + x)*(-128*x^2 - 3*x^3)*Log[x
]))/(256*E^5),x]

[Out]

(E^(-10 + x)*x^(1 - (x^2*(64 + x))/64))/4

________________________________________________________________________________________

fricas [A]  time = 0.55, size = 20, normalized size = 0.80 \begin {gather*} \frac {x e^{\left (x - 10\right )}}{4 \, x^{\frac {1}{64} \, x^{3} + x^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/256*((-3*x^3-128*x^2)*exp(x-5)*log(x)+(-x^3-64*x^2+64*x+64)*exp(x-5))/exp(5)/exp(1/64*(x^3+64*x^2)
*log(x)),x, algorithm="fricas")

[Out]

1/4*x*e^(x - 10)/x^(1/64*x^3 + x^2)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left ({\left (3 \, x^{3} + 128 \, x^{2}\right )} e^{\left (x - 5\right )} \log \relax (x) + {\left (x^{3} + 64 \, x^{2} - 64 \, x - 64\right )} e^{\left (x - 5\right )}\right )} e^{\left (-5\right )}}{256 \, x^{\frac {1}{64} \, x^{3} + x^{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/256*((-3*x^3-128*x^2)*exp(x-5)*log(x)+(-x^3-64*x^2+64*x+64)*exp(x-5))/exp(5)/exp(1/64*(x^3+64*x^2)
*log(x)),x, algorithm="giac")

[Out]

integrate(-1/256*((3*x^3 + 128*x^2)*e^(x - 5)*log(x) + (x^3 + 64*x^2 - 64*x - 64)*e^(x - 5))*e^(-5)/x^(1/64*x^
3 + x^2), x)

________________________________________________________________________________________

maple [A]  time = 0.03, size = 20, normalized size = 0.80

method	result	size
risch	\(\frac {x^{-\frac {x^{2} \left (x +64\right )}{64}} x \,{\mathrm e}^{x -10}}{4}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/256*((-3*x^3-128*x^2)*exp(x-5)*ln(x)+(-x^3-64*x^2+64*x+64)*exp(x-5))/exp(5)/exp(1/64*(x^3+64*x^2)*ln(x))
,x,method=_RETURNVERBOSE)

[Out]

1/4/(x^(1/64*x^2*(x+64)))*x*exp(x-10)

________________________________________________________________________________________

maxima [A]  time = 0.97, size = 21, normalized size = 0.84 \begin {gather*} \frac {1}{4} \, x e^{\left (-\frac {1}{64} \, x^{3} \log \relax (x) - x^{2} \log \relax (x) + x - 10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/256*((-3*x^3-128*x^2)*exp(x-5)*log(x)+(-x^3-64*x^2+64*x+64)*exp(x-5))/exp(5)/exp(1/64*(x^3+64*x^2)
*log(x)),x, algorithm="maxima")

[Out]

1/4*x*e^(-1/64*x^3*log(x) - x^2*log(x) + x - 10)

________________________________________________________________________________________

mupad [B]  time = 2.06, size = 20, normalized size = 0.80 \begin {gather*} \frac {{\mathrm {e}}^{-10}\,{\mathrm {e}}^x}{4\,x^{\frac {x^3}{64}+x^2-1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-(log(x)*(64*x^2 + x^3))/64)*exp(-5)*((exp(x - 5)*(64*x - 64*x^2 - x^3 + 64))/256 - (exp(x - 5)*log(x)
*(128*x^2 + 3*x^3))/256),x)

[Out]

(exp(-10)*exp(x))/(4*x^(x^2 + x^3/64 - 1))

________________________________________________________________________________________

sympy [A]  time = 106.78, size = 24, normalized size = 0.96 \begin {gather*} \frac {x e^{- \left (\frac {x^{3}}{64} + x^{2}\right ) \log {\relax (x )}} e^{x - 5}}{4 e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/256*((-3*x**3-128*x**2)*exp(x-5)*ln(x)+(-x**3-64*x**2+64*x+64)*exp(x-5))/exp(5)/exp(1/64*(x**3+64*
x**2)*ln(x)),x)

[Out]

x*exp(-5)*exp(-(x**3/64 + x**2)*log(x))*exp(x - 5)/4

________________________________________________________________________________________