3.93.8 \(\int \frac {1}{256} (-256+48 x^2-24 e x^2+3 e^2 x^2+e^{-5+x} (-32 x-16 x^2+e^2 (-2 x-x^2)+e (16 x+8 x^2))) \, dx\)

Optimal. Leaf size=27 \[ -x+\frac {1}{256} (4-e)^2 x^2 \left (-e^{-5+x}+x\right ) \]

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Rubi [A]  time = 0.13, antiderivative size = 47, normalized size of antiderivative = 1.74, number of steps used = 18, number of rules used = 5, integrand size = 64, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.078, Rules used = {6, 12, 2196, 2176, 2194} \begin {gather*} \frac {1}{256} (4-e)^2 x^3-\frac {1}{16} e^{x-5} x^2+\frac {1}{256} (8-e) e^{x-4} x^2-x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-256 + 48*x^2 - 24*E*x^2 + 3*E^2*x^2 + E^(-5 + x)*(-32*x - 16*x^2 + E^2*(-2*x - x^2) + E*(16*x + 8*x^2)))
/256,x]

[Out]

-x - (E^(-5 + x)*x^2)/16 + ((8 - E)*E^(-4 + x)*x^2)/256 + ((4 - E)^2*x^3)/256

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

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 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {1}{256} \left (-256+(48-24 e) x^2+3 e^2 x^2+e^{-5+x} \left (-32 x-16 x^2+e^2 \left (-2 x-x^2\right )+e \left (16 x+8 x^2\right )\right )\right ) \, dx\\ &=\int \frac {1}{256} \left (-256+\left (48-24 e+3 e^2\right ) x^2+e^{-5+x} \left (-32 x-16 x^2+e^2 \left (-2 x-x^2\right )+e \left (16 x+8 x^2\right )\right )\right ) \, dx\\ &=\frac {1}{256} \int \left (-256+\left (48-24 e+3 e^2\right ) x^2+e^{-5+x} \left (-32 x-16 x^2+e^2 \left (-2 x-x^2\right )+e \left (16 x+8 x^2\right )\right )\right ) \, dx\\ &=-x+\frac {1}{256} (4-e)^2 x^3+\frac {1}{256} \int e^{-5+x} \left (-32 x-16 x^2+e^2 \left (-2 x-x^2\right )+e \left (16 x+8 x^2\right )\right ) \, dx\\ &=-x+\frac {1}{256} (4-e)^2 x^3+\frac {1}{256} \int \left (-32 e^{-5+x} x-16 e^{-5+x} x^2+8 \left (1-\frac {e}{8}\right ) e^{-4+x} x (2+x)\right ) \, dx\\ &=-x+\frac {1}{256} (4-e)^2 x^3-\frac {1}{16} \int e^{-5+x} x^2 \, dx-\frac {1}{8} \int e^{-5+x} x \, dx+\frac {1}{256} (8-e) \int e^{-4+x} x (2+x) \, dx\\ &=-x-\frac {1}{8} e^{-5+x} x-\frac {1}{16} e^{-5+x} x^2+\frac {1}{256} (4-e)^2 x^3+\frac {1}{8} \int e^{-5+x} \, dx+\frac {1}{8} \int e^{-5+x} x \, dx+\frac {1}{256} (8-e) \int \left (2 e^{-4+x} x+e^{-4+x} x^2\right ) \, dx\\ &=\frac {e^{-5+x}}{8}-x-\frac {1}{16} e^{-5+x} x^2+\frac {1}{256} (4-e)^2 x^3-\frac {1}{8} \int e^{-5+x} \, dx+\frac {1}{256} (8-e) \int e^{-4+x} x^2 \, dx+\frac {1}{128} (8-e) \int e^{-4+x} x \, dx\\ &=-x+\frac {1}{128} (8-e) e^{-4+x} x-\frac {1}{16} e^{-5+x} x^2+\frac {1}{256} (8-e) e^{-4+x} x^2+\frac {1}{256} (4-e)^2 x^3+\frac {1}{128} (-8+e) \int e^{-4+x} \, dx+\frac {1}{128} (-8+e) \int e^{-4+x} x \, dx\\ &=-\frac {1}{128} (8-e) e^{-4+x}-x-\frac {1}{16} e^{-5+x} x^2+\frac {1}{256} (8-e) e^{-4+x} x^2+\frac {1}{256} (4-e)^2 x^3+\frac {1}{128} (8-e) \int e^{-4+x} \, dx\\ &=-x-\frac {1}{16} e^{-5+x} x^2+\frac {1}{256} (8-e) e^{-4+x} x^2+\frac {1}{256} (4-e)^2 x^3\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.03, size = 65, normalized size = 2.41 \begin {gather*} \frac {-256 e^5 x-16 e^x x^2+8 e^{1+x} x^2-e^{2+x} x^2+16 e^5 x^3-8 e^6 x^3+e^7 x^3}{256 e^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-256 + 48*x^2 - 24*E*x^2 + 3*E^2*x^2 + E^(-5 + x)*(-32*x - 16*x^2 + E^2*(-2*x - x^2) + E*(16*x + 8*
x^2)))/256,x]

[Out]

(-256*E^5*x - 16*E^x*x^2 + 8*E^(1 + x)*x^2 - E^(2 + x)*x^2 + 16*E^5*x^3 - 8*E^6*x^3 + E^7*x^3)/(256*E^5)

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fricas [B]  time = 0.60, size = 48, normalized size = 1.78 \begin {gather*} \frac {1}{256} \, x^{3} e^{2} - \frac {1}{32} \, x^{3} e + \frac {1}{16} \, x^{3} - \frac {1}{256} \, {\left (x^{2} e^{2} - 8 \, x^{2} e + 16 \, x^{2}\right )} e^{\left (x - 5\right )} - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/256*((-x^2-2*x)*exp(1)^2+(8*x^2+16*x)*exp(1)-16*x^2-32*x)*exp(x-5)+3/256*x^2*exp(1)^2-3/32*x^2*exp
(1)+3/16*x^2-1,x, algorithm="fricas")

[Out]

1/256*x^3*e^2 - 1/32*x^3*e + 1/16*x^3 - 1/256*(x^2*e^2 - 8*x^2*e + 16*x^2)*e^(x - 5) - x

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giac [B]  time = 0.12, size = 50, normalized size = 1.85 \begin {gather*} \frac {1}{256} \, x^{3} e^{2} - \frac {1}{32} \, x^{3} e + \frac {1}{16} \, x^{3} - \frac {1}{256} \, x^{2} e^{\left (x - 3\right )} + \frac {1}{32} \, x^{2} e^{\left (x - 4\right )} - \frac {1}{16} \, x^{2} e^{\left (x - 5\right )} - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/256*((-x^2-2*x)*exp(1)^2+(8*x^2+16*x)*exp(1)-16*x^2-32*x)*exp(x-5)+3/256*x^2*exp(1)^2-3/32*x^2*exp
(1)+3/16*x^2-1,x, algorithm="giac")

[Out]

1/256*x^3*e^2 - 1/32*x^3*e + 1/16*x^3 - 1/256*x^2*e^(x - 3) + 1/32*x^2*e^(x - 4) - 1/16*x^2*e^(x - 5) - x

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maple [A]  time = 0.07, size = 41, normalized size = 1.52

method	result	size
norman	\(\left (\frac {{\mathrm e}^{2}}{256}-\frac {{\mathrm e}}{32}+\frac {1}{16}\right ) x^{3}+\left (-\frac {{\mathrm e}^{2}}{256}+\frac {{\mathrm e}}{32}-\frac {1}{16}\right ) x^{2} {\mathrm e}^{x -5}-x\)	\(41\)
risch	\(-\frac {\left ({\mathrm e}^{2}-8 \,{\mathrm e}+16\right ) x^{2} {\mathrm e}^{x -5}}{256}+\frac {x^{3} {\mathrm e}^{2}}{256}-\frac {x^{3} {\mathrm e}}{32}+\frac {x^{3}}{16}-x\)	\(41\)
default	\(-x +\frac {x^{3}}{16}-\frac {x^{3} {\mathrm e}}{32}+\frac {x^{3} {\mathrm e}^{2}}{256}-\frac {5 \,{\mathrm e}^{x -5} \left (x -5\right )}{8}-\frac {25 \,{\mathrm e}^{x -5}}{16}-\frac {{\mathrm e}^{x -5} \left (x -5\right )^{2}}{16}+\frac {35 \,{\mathrm e} \,{\mathrm e}^{x -5}}{32}-\frac {35 \,{\mathrm e}^{2} {\mathrm e}^{x -5}}{256}+\frac {3 \,{\mathrm e} \left ({\mathrm e}^{x -5} \left (x -5\right )-{\mathrm e}^{x -5}\right )}{8}+\frac {{\mathrm e} \left ({\mathrm e}^{x -5} \left (x -5\right )^{2}-2 \,{\mathrm e}^{x -5} \left (x -5\right )+2 \,{\mathrm e}^{x -5}\right )}{32}-\frac {3 \,{\mathrm e}^{2} \left ({\mathrm e}^{x -5} \left (x -5\right )-{\mathrm e}^{x -5}\right )}{64}-\frac {{\mathrm e}^{2} \left ({\mathrm e}^{x -5} \left (x -5\right )^{2}-2 \,{\mathrm e}^{x -5} \left (x -5\right )+2 \,{\mathrm e}^{x -5}\right )}{256}\)	\(172\)
derivativedivides	\(-x +5+\frac {x^{3}}{16}-\frac {x^{3} {\mathrm e}}{32}+\frac {x^{3} {\mathrm e}^{2}}{256}-\frac {5 \,{\mathrm e}^{x -5} \left (x -5\right )}{8}-\frac {25 \,{\mathrm e}^{x -5}}{16}-\frac {{\mathrm e}^{x -5} \left (x -5\right )^{2}}{16}+\frac {35 \,{\mathrm e} \,{\mathrm e}^{x -5}}{32}-\frac {35 \,{\mathrm e}^{2} {\mathrm e}^{x -5}}{256}+\frac {3 \,{\mathrm e} \left ({\mathrm e}^{x -5} \left (x -5\right )-{\mathrm e}^{x -5}\right )}{8}+\frac {{\mathrm e} \left ({\mathrm e}^{x -5} \left (x -5\right )^{2}-2 \,{\mathrm e}^{x -5} \left (x -5\right )+2 \,{\mathrm e}^{x -5}\right )}{32}-\frac {3 \,{\mathrm e}^{2} \left ({\mathrm e}^{x -5} \left (x -5\right )-{\mathrm e}^{x -5}\right )}{64}-\frac {{\mathrm e}^{2} \left ({\mathrm e}^{x -5} \left (x -5\right )^{2}-2 \,{\mathrm e}^{x -5} \left (x -5\right )+2 \,{\mathrm e}^{x -5}\right )}{256}\)	\(173\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/256*((-x^2-2*x)*exp(1)^2+(8*x^2+16*x)*exp(1)-16*x^2-32*x)*exp(x-5)+3/256*x^2*exp(1)^2-3/32*x^2*exp(1)+3/
16*x^2-1,x,method=_RETURNVERBOSE)

[Out]

(1/256*exp(1)^2-1/32*exp(1)+1/16)*x^3+(-1/256*exp(1)^2+1/32*exp(1)-1/16)*x^2*exp(x-5)-x

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maxima [A]  time = 0.36, size = 40, normalized size = 1.48 \begin {gather*} \frac {1}{256} \, x^{3} e^{2} - \frac {1}{32} \, x^{3} e - \frac {1}{256} \, x^{2} {\left (e^{2} - 8 \, e + 16\right )} e^{\left (x - 5\right )} + \frac {1}{16} \, x^{3} - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/256*((-x^2-2*x)*exp(1)^2+(8*x^2+16*x)*exp(1)-16*x^2-32*x)*exp(x-5)+3/256*x^2*exp(1)^2-3/32*x^2*exp
(1)+3/16*x^2-1,x, algorithm="maxima")

[Out]

1/256*x^3*e^2 - 1/32*x^3*e - 1/256*x^2*(e^2 - 8*e + 16)*e^(x - 5) + 1/16*x^3 - x

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mupad [B]  time = 8.08, size = 30, normalized size = 1.11 \begin {gather*} \frac {x^3\,{\left (\mathrm {e}-4\right )}^2}{256}-x-\frac {x^2\,{\mathrm {e}}^{x-5}\,{\left (\mathrm {e}-4\right )}^2}{256} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x^2*exp(2))/256 - (3*x^2*exp(1))/32 - (exp(x - 5)*(32*x - exp(1)*(16*x + 8*x^2) + exp(2)*(2*x + x^2) +
16*x^2))/256 + (3*x^2)/16 - 1,x)

[Out]

(x^3*(exp(1) - 4)^2)/256 - x - (x^2*exp(x - 5)*(exp(1) - 4)^2)/256

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sympy [B]  time = 0.19, size = 44, normalized size = 1.63 \begin {gather*} x^{3} \left (- \frac {e}{32} + \frac {e^{2}}{256} + \frac {1}{16}\right ) - x + \frac {\left (- 16 x^{2} - x^{2} e^{2} + 8 e x^{2}\right ) e^{x - 5}}{256} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/256*((-x**2-2*x)*exp(1)**2+(8*x**2+16*x)*exp(1)-16*x**2-32*x)*exp(x-5)+3/256*x**2*exp(1)**2-3/32*x
**2*exp(1)+3/16*x**2-1,x)

[Out]

x**3*(-E/32 + exp(2)/256 + 1/16) - x + (-16*x**2 - x**2*exp(2) + 8*E*x**2)*exp(x - 5)/256

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