3.3.57 \(\int \frac {1}{9} e^{-2 x} (e^{2 x} (1024+256 x)+e^{2+x} (1024-320 x^2-64 x^3)+e^4 (256 x-64 x^2-96 x^3-16 x^4)) \, dx\)

Optimal. Leaf size=22 \[ \frac {8}{9} (4+x)^2 \left (4+e^{2-x} x\right )^2 \]

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Rubi [B]  time = 0.46, antiderivative size = 92, normalized size of antiderivative = 4.18, number of steps used = 29, number of rules used = 5, integrand size = 62, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.081, Rules used = {12, 6742, 2196, 2194, 2176} \begin {gather*} \frac {8}{9} e^{4-2 x} x^4+\frac {64}{9} e^{4-2 x} x^3+\frac {64}{9} e^{2-x} x^3+\frac {128}{9} e^{4-2 x} x^2+\frac {512}{9} e^{2-x} x^2+\frac {1024}{9} e^{2-x} x+\frac {128}{9} (x+4)^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(2*x)*(1024 + 256*x) + E^(2 + x)*(1024 - 320*x^2 - 64*x^3) + E^4*(256*x - 64*x^2 - 96*x^3 - 16*x^4))/(9
*E^(2*x)),x]

[Out]

(1024*E^(2 - x)*x)/9 + (128*E^(4 - 2*x)*x^2)/9 + (512*E^(2 - x)*x^2)/9 + (64*E^(4 - 2*x)*x^3)/9 + (64*E^(2 - x
)*x^3)/9 + (8*E^(4 - 2*x)*x^4)/9 + (128*(4 + x)^2)/9

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

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}{9} \int e^{-2 x} \left (e^{2 x} (1024+256 x)+e^{2+x} \left (1024-320 x^2-64 x^3\right )+e^4 \left (256 x-64 x^2-96 x^3-16 x^4\right )\right ) \, dx\\ &=\frac {1}{9} \int \left (256 (4+x)-64 e^{2-x} \left (-16+5 x^2+x^3\right )-16 e^{4-2 x} x \left (-16+4 x+6 x^2+x^3\right )\right ) \, dx\\ &=\frac {128}{9} (4+x)^2-\frac {16}{9} \int e^{4-2 x} x \left (-16+4 x+6 x^2+x^3\right ) \, dx-\frac {64}{9} \int e^{2-x} \left (-16+5 x^2+x^3\right ) \, dx\\ &=\frac {128}{9} (4+x)^2-\frac {16}{9} \int \left (-16 e^{4-2 x} x+4 e^{4-2 x} x^2+6 e^{4-2 x} x^3+e^{4-2 x} x^4\right ) \, dx-\frac {64}{9} \int \left (-16 e^{2-x}+5 e^{2-x} x^2+e^{2-x} x^3\right ) \, dx\\ &=\frac {128}{9} (4+x)^2-\frac {16}{9} \int e^{4-2 x} x^4 \, dx-\frac {64}{9} \int e^{4-2 x} x^2 \, dx-\frac {64}{9} \int e^{2-x} x^3 \, dx-\frac {32}{3} \int e^{4-2 x} x^3 \, dx+\frac {256}{9} \int e^{4-2 x} x \, dx-\frac {320}{9} \int e^{2-x} x^2 \, dx+\frac {1024}{9} \int e^{2-x} \, dx\\ &=-\frac {1024 e^{2-x}}{9}-\frac {128}{9} e^{4-2 x} x+\frac {32}{9} e^{4-2 x} x^2+\frac {320}{9} e^{2-x} x^2+\frac {16}{3} e^{4-2 x} x^3+\frac {64}{9} e^{2-x} x^3+\frac {8}{9} e^{4-2 x} x^4+\frac {128}{9} (4+x)^2-\frac {32}{9} \int e^{4-2 x} x^3 \, dx-\frac {64}{9} \int e^{4-2 x} x \, dx+\frac {128}{9} \int e^{4-2 x} \, dx-16 \int e^{4-2 x} x^2 \, dx-\frac {64}{3} \int e^{2-x} x^2 \, dx-\frac {640}{9} \int e^{2-x} x \, dx\\ &=-\frac {64}{9} e^{4-2 x}-\frac {1024 e^{2-x}}{9}-\frac {32}{3} e^{4-2 x} x+\frac {640}{9} e^{2-x} x+\frac {104}{9} e^{4-2 x} x^2+\frac {512}{9} e^{2-x} x^2+\frac {64}{9} e^{4-2 x} x^3+\frac {64}{9} e^{2-x} x^3+\frac {8}{9} e^{4-2 x} x^4+\frac {128}{9} (4+x)^2-\frac {32}{9} \int e^{4-2 x} \, dx-\frac {16}{3} \int e^{4-2 x} x^2 \, dx-16 \int e^{4-2 x} x \, dx-\frac {128}{3} \int e^{2-x} x \, dx-\frac {640}{9} \int e^{2-x} \, dx\\ &=-\frac {16}{3} e^{4-2 x}-\frac {128 e^{2-x}}{3}-\frac {8}{3} e^{4-2 x} x+\frac {1024}{9} e^{2-x} x+\frac {128}{9} e^{4-2 x} x^2+\frac {512}{9} e^{2-x} x^2+\frac {64}{9} e^{4-2 x} x^3+\frac {64}{9} e^{2-x} x^3+\frac {8}{9} e^{4-2 x} x^4+\frac {128}{9} (4+x)^2-\frac {16}{3} \int e^{4-2 x} x \, dx-8 \int e^{4-2 x} \, dx-\frac {128}{3} \int e^{2-x} \, dx\\ &=-\frac {4}{3} e^{4-2 x}+\frac {1024}{9} e^{2-x} x+\frac {128}{9} e^{4-2 x} x^2+\frac {512}{9} e^{2-x} x^2+\frac {64}{9} e^{4-2 x} x^3+\frac {64}{9} e^{2-x} x^3+\frac {8}{9} e^{4-2 x} x^4+\frac {128}{9} (4+x)^2-\frac {8}{3} \int e^{4-2 x} \, dx\\ &=\frac {1024}{9} e^{2-x} x+\frac {128}{9} e^{4-2 x} x^2+\frac {512}{9} e^{2-x} x^2+\frac {64}{9} e^{4-2 x} x^3+\frac {64}{9} e^{2-x} x^3+\frac {8}{9} e^{4-2 x} x^4+\frac {128}{9} (4+x)^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 43, normalized size = 1.95 \begin {gather*} \frac {8}{9} e^{-2 x} x \left (8 e^{2+x} (4+x)^2+e^4 x (4+x)^2+16 e^{2 x} (8+x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2*x)*(1024 + 256*x) + E^(2 + x)*(1024 - 320*x^2 - 64*x^3) + E^4*(256*x - 64*x^2 - 96*x^3 - 16*x^
4))/(9*E^(2*x)),x]

[Out]

(8*x*(8*E^(2 + x)*(4 + x)^2 + E^4*x*(4 + x)^2 + 16*E^(2*x)*(8 + x)))/(9*E^(2*x))

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fricas [B]  time = 0.73, size = 59, normalized size = 2.68 \begin {gather*} \frac {8}{9} \, {\left ({\left (x^{4} + 8 \, x^{3} + 16 \, x^{2}\right )} e^{8} + 16 \, {\left (x^{2} + 8 \, x\right )} e^{\left (2 \, x + 4\right )} + 8 \, {\left (x^{3} + 8 \, x^{2} + 16 \, x\right )} e^{\left (x + 6\right )}\right )} e^{\left (-2 \, x - 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*((256*x+1024)*exp(x)^2+(-64*x^3-320*x^2+1024)*exp(2)*exp(x)+(-16*x^4-96*x^3-64*x^2+256*x)*exp(2)
^2)/exp(x)^2,x, algorithm="fricas")

[Out]

8/9*((x^4 + 8*x^3 + 16*x^2)*e^8 + 16*(x^2 + 8*x)*e^(2*x + 4) + 8*(x^3 + 8*x^2 + 16*x)*e^(x + 6))*e^(-2*x - 4)

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giac [B]  time = 0.26, size = 73, normalized size = 3.32 \begin {gather*} \frac {8}{9} \, x^{4} e^{\left (-2 \, x + 4\right )} + \frac {64}{9} \, x^{3} e^{\left (-x + 2\right )} + \frac {64}{9} \, x^{3} e^{\left (-2 \, x + 4\right )} + \frac {512}{9} \, x^{2} e^{\left (-x + 2\right )} + \frac {128}{9} \, x^{2} e^{\left (-2 \, x + 4\right )} + \frac {128}{9} \, x^{2} + \frac {1024}{9} \, x e^{\left (-x + 2\right )} + \frac {1024}{9} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*((256*x+1024)*exp(x)^2+(-64*x^3-320*x^2+1024)*exp(2)*exp(x)+(-16*x^4-96*x^3-64*x^2+256*x)*exp(2)
^2)/exp(x)^2,x, algorithm="giac")

[Out]

8/9*x^4*e^(-2*x + 4) + 64/9*x^3*e^(-x + 2) + 64/9*x^3*e^(-2*x + 4) + 512/9*x^2*e^(-x + 2) + 128/9*x^2*e^(-2*x
+ 4) + 128/9*x^2 + 1024/9*x*e^(-x + 2) + 1024/9*x

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maple [B]  time = 0.12, size = 64, normalized size = 2.91

method	result	size
risch	\(\frac {128 x^{2}}{9}+\frac {1024 x}{9}+\frac {\left (64 x^{3} {\mathrm e}^{2}+512 x^{2} {\mathrm e}^{2}+1024 \,{\mathrm e}^{2} x \right ) {\mathrm e}^{-x}}{9}+\frac {\left (8 x^{4} {\mathrm e}^{4}+64 x^{3} {\mathrm e}^{4}+128 x^{2} {\mathrm e}^{4}\right ) {\mathrm e}^{-2 x}}{9}\)	\(64\)
norman	\(\left (\frac {1024 x \,{\mathrm e}^{2 x}}{9}+\frac {128 x^{2} {\mathrm e}^{4}}{9}+\frac {64 x^{3} {\mathrm e}^{4}}{9}+\frac {8 x^{4} {\mathrm e}^{4}}{9}+\frac {128 \,{\mathrm e}^{2 x} x^{2}}{9}+\frac {1024 x \,{\mathrm e}^{2} {\mathrm e}^{x}}{9}+\frac {512 x^{2} {\mathrm e}^{2} {\mathrm e}^{x}}{9}+\frac {64 x^{3} {\mathrm e}^{2} {\mathrm e}^{x}}{9}\right ) {\mathrm e}^{-2 x}\)	\(75\)
default	\(\frac {128 x^{2}}{9}+\frac {1024 x}{9}-\frac {1024 \,{\mathrm e}^{2} {\mathrm e}^{-x}}{9}+\frac {256 \,{\mathrm e}^{4} \left (-\frac {x \,{\mathrm e}^{-2 x}}{2}-\frac {{\mathrm e}^{-2 x}}{4}\right )}{9}-\frac {64 \,{\mathrm e}^{4} \left (-\frac {x^{2} {\mathrm e}^{-2 x}}{2}-\frac {x \,{\mathrm e}^{-2 x}}{2}-\frac {{\mathrm e}^{-2 x}}{4}\right )}{9}-\frac {32 \,{\mathrm e}^{4} \left (-\frac {{\mathrm e}^{-2 x} x^{3}}{2}-\frac {3 x^{2} {\mathrm e}^{-2 x}}{4}-\frac {3 x \,{\mathrm e}^{-2 x}}{4}-\frac {3 \,{\mathrm e}^{-2 x}}{8}\right )}{3}-\frac {16 \,{\mathrm e}^{4} \left (-\frac {x^{4} {\mathrm e}^{-2 x}}{2}-{\mathrm e}^{-2 x} x^{3}-\frac {3 x^{2} {\mathrm e}^{-2 x}}{2}-\frac {3 x \,{\mathrm e}^{-2 x}}{2}-\frac {3 \,{\mathrm e}^{-2 x}}{4}\right )}{9}-\frac {320 \,{\mathrm e}^{2} \left (-x^{2} {\mathrm e}^{-x}-2 x \,{\mathrm e}^{-x}-2 \,{\mathrm e}^{-x}\right )}{9}-\frac {64 \,{\mathrm e}^{2} \left (-x^{3} {\mathrm e}^{-x}-3 x^{2} {\mathrm e}^{-x}-6 x \,{\mathrm e}^{-x}-6 \,{\mathrm e}^{-x}\right )}{9}\)	\(215\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/9*((256*x+1024)*exp(x)^2+(-64*x^3-320*x^2+1024)*exp(2)*exp(x)+(-16*x^4-96*x^3-64*x^2+256*x)*exp(2)^2)/ex
p(x)^2,x,method=_RETURNVERBOSE)

[Out]

128/9*x^2+1024/9*x+1/9*(64*x^3*exp(2)+512*x^2*exp(2)+1024*exp(2)*x)*exp(-x)+1/9*(8*x^4*exp(4)+64*x^3*exp(4)+12
8*x^2*exp(4))*exp(-2*x)

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maxima [B]  time = 0.48, size = 170, normalized size = 7.73 \begin {gather*} \frac {128}{9} \, x^{2} + \frac {64}{9} \, {\left (x^{3} e^{2} + 3 \, x^{2} e^{2} + 6 \, x e^{2} + 6 \, e^{2}\right )} e^{\left (-x\right )} + \frac {320}{9} \, {\left (x^{2} e^{2} + 2 \, x e^{2} + 2 \, e^{2}\right )} e^{\left (-x\right )} + \frac {4}{9} \, {\left (2 \, x^{4} e^{4} + 4 \, x^{3} e^{4} + 6 \, x^{2} e^{4} + 6 \, x e^{4} + 3 \, e^{4}\right )} e^{\left (-2 \, x\right )} + \frac {4}{3} \, {\left (4 \, x^{3} e^{4} + 6 \, x^{2} e^{4} + 6 \, x e^{4} + 3 \, e^{4}\right )} e^{\left (-2 \, x\right )} + \frac {16}{9} \, {\left (2 \, x^{2} e^{4} + 2 \, x e^{4} + e^{4}\right )} e^{\left (-2 \, x\right )} - \frac {64}{9} \, {\left (2 \, x e^{4} + e^{4}\right )} e^{\left (-2 \, x\right )} + \frac {1024}{9} \, x - \frac {1024}{9} \, e^{\left (-x + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*((256*x+1024)*exp(x)^2+(-64*x^3-320*x^2+1024)*exp(2)*exp(x)+(-16*x^4-96*x^3-64*x^2+256*x)*exp(2)
^2)/exp(x)^2,x, algorithm="maxima")

[Out]

128/9*x^2 + 64/9*(x^3*e^2 + 3*x^2*e^2 + 6*x*e^2 + 6*e^2)*e^(-x) + 320/9*(x^2*e^2 + 2*x*e^2 + 2*e^2)*e^(-x) + 4
/9*(2*x^4*e^4 + 4*x^3*e^4 + 6*x^2*e^4 + 6*x*e^4 + 3*e^4)*e^(-2*x) + 4/3*(4*x^3*e^4 + 6*x^2*e^4 + 6*x*e^4 + 3*e
^4)*e^(-2*x) + 16/9*(2*x^2*e^4 + 2*x*e^4 + e^4)*e^(-2*x) - 64/9*(2*x*e^4 + e^4)*e^(-2*x) + 1024/9*x - 1024/9*e
^(-x + 2)

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mupad [B]  time = 0.44, size = 61, normalized size = 2.77 \begin {gather*} \frac {1024\,x}{9}+{\mathrm {e}}^{-x}\,\left (\frac {64\,{\mathrm {e}}^2\,x^3}{9}+\frac {512\,{\mathrm {e}}^2\,x^2}{9}+\frac {1024\,{\mathrm {e}}^2\,x}{9}\right )+{\mathrm {e}}^{-2\,x}\,\left (\frac {8\,{\mathrm {e}}^4\,x^4}{9}+\frac {64\,{\mathrm {e}}^4\,x^3}{9}+\frac {128\,{\mathrm {e}}^4\,x^2}{9}\right )+\frac {128\,x^2}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(-2*x)*((exp(4)*(64*x^2 - 256*x + 96*x^3 + 16*x^4))/9 - (exp(2*x)*(256*x + 1024))/9 + (exp(2)*exp(x)*(
320*x^2 + 64*x^3 - 1024))/9),x)

[Out]

(1024*x)/9 + exp(-x)*((1024*x*exp(2))/9 + (512*x^2*exp(2))/9 + (64*x^3*exp(2))/9) + exp(-2*x)*((128*x^2*exp(4)
)/9 + (64*x^3*exp(4))/9 + (8*x^4*exp(4))/9) + (128*x^2)/9

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sympy [B]  time = 0.20, size = 71, normalized size = 3.23 \begin {gather*} \frac {128 x^{2}}{9} + \frac {1024 x}{9} + \frac {\left (576 x^{3} e^{2} + 4608 x^{2} e^{2} + 9216 x e^{2}\right ) e^{- x}}{81} + \frac {\left (72 x^{4} e^{4} + 576 x^{3} e^{4} + 1152 x^{2} e^{4}\right ) e^{- 2 x}}{81} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*((256*x+1024)*exp(x)**2+(-64*x**3-320*x**2+1024)*exp(2)*exp(x)+(-16*x**4-96*x**3-64*x**2+256*x)*
exp(2)**2)/exp(x)**2,x)

[Out]

128*x**2/9 + 1024*x/9 + (576*x**3*exp(2) + 4608*x**2*exp(2) + 9216*x*exp(2))*exp(-x)/81 + (72*x**4*exp(4) + 57
6*x**3*exp(4) + 1152*x**2*exp(4))*exp(-2*x)/81

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