3.35.34 \(\int \frac {1}{2} e^{8-2 x} (64 x-112 x^2+40 x^3-4 x^4+e^{-4+x} (16-24 x+4 x^2)+3^{2 x} (-1+\log (3))+3^x (8-20 x+4 x^2+(8 x-2 x^2) \log (3)+e^{-4+x} (-2+2 \log (3)))) \, dx\)

Optimal. Leaf size=28 \[ \left (1+e^{4-x} \left (\frac {3^x}{2}+4 x-x^2\right )\right )^2 \]

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Rubi [B]  time = 3.56, antiderivative size = 249, normalized size of antiderivative = 8.89, number of steps used = 43, number of rules used = 7, integrand size = 94, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {12, 6688, 6742, 2194, 2176, 2287, 2196} \begin {gather*} e^{8-2 x} x^4-8 e^{8-2 x} x^3+16 e^{8-2 x} x^2-2 e^{4-x} x^2-x^2 e^{8-x (2-\log (3))}+8 e^{4-x} x+\frac {2 x (5-\log (9)) e^{8-x (2-\log (3))}}{2-\log (3)}-\frac {2 x e^{8-x (2-\log (3))}}{2-\log (3)}+e^{4-x (1-\log (3))}+\frac {2 (5-\log (9)) e^{8-x (2-\log (3))}}{(2-\log (3))^2}+\frac {(1-\log (3)) e^{8-x (2-\log (9))}}{2 (2-\log (9))}-\frac {4 e^{8-x (2-\log (3))}}{2-\log (3)}-\frac {2 e^{8-x (2-\log (3))}}{(2-\log (3))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(8 - 2*x)*(64*x - 112*x^2 + 40*x^3 - 4*x^4 + E^(-4 + x)*(16 - 24*x + 4*x^2) + 3^(2*x)*(-1 + Log[3]) + 3
^x*(8 - 20*x + 4*x^2 + (8*x - 2*x^2)*Log[3] + E^(-4 + x)*(-2 + 2*Log[3]))))/2,x]

[Out]

E^(4 - x*(1 - Log[3])) + 8*E^(4 - x)*x + 16*E^(8 - 2*x)*x^2 - 2*E^(4 - x)*x^2 - E^(8 - x*(2 - Log[3]))*x^2 - 8
*E^(8 - 2*x)*x^3 + E^(8 - 2*x)*x^4 - (2*E^(8 - x*(2 - Log[3])))/(2 - Log[3])^2 - (4*E^(8 - x*(2 - Log[3])))/(2
- Log[3]) - (2*E^(8 - x*(2 - Log[3]))*x)/(2 - Log[3]) + (E^(8 - x*(2 - Log[9]))*(1 - Log[3]))/(2*(2 - Log[9])
) + (2*E^(8 - x*(2 - Log[3]))*(5 - Log[9]))/(2 - Log[3])^2 + (2*E^(8 - x*(2 - Log[3]))*x*(5 - Log[9]))/(2 - Lo
g[3])

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 2287

Int[(u_.)*(F_)^(v_)*(G_)^(w_), x_Symbol] :> With[{z = v*Log[F] + w*Log[G]}, Int[u*NormalizeIntegrand[E^z, x],
x] /; BinomialQ[z, x] || (PolynomialQ[z, x] && LeQ[Exponent[z, x], 2])] /; FreeQ[{F, G}, x]

Rule 6688

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

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}{2} \int e^{8-2 x} \left (64 x-112 x^2+40 x^3-4 x^4+e^{-4+x} \left (16-24 x+4 x^2\right )+3^{2 x} (-1+\log (3))+3^x \left (8-20 x+4 x^2+\left (8 x-2 x^2\right ) \log (3)+e^{-4+x} (-2+2 \log (3))\right )\right ) \, dx\\ &=\frac {1}{2} \int e^{4-2 x} \left (2 e^x+e^4 \left (3^x+8 x-2 x^2\right )\right ) \left (8-12 x+2 x^2-3^x (1-\log (3))\right ) \, dx\\ &=\frac {1}{2} \int \left (2 e^{4-x} \left (8-12 x+2 x^2-3^x (1-\log (3))\right )+e^{8-2 x} \left (3^x+8 x-2 x^2\right ) \left (8-12 x+2 x^2-3^x (1-\log (3))\right )\right ) \, dx\\ &=\frac {1}{2} \int e^{8-2 x} \left (3^x+8 x-2 x^2\right ) \left (8-12 x+2 x^2-3^x (1-\log (3))\right ) \, dx+\int e^{4-x} \left (8-12 x+2 x^2-3^x (1-\log (3))\right ) \, dx\\ &=\frac {1}{2} \int \left (-4 e^{8-2 x} x \left (-16+28 x-10 x^2+x^3\right )+9^x e^{8-2 x} (-1+\log (3))+2\ 3^x e^{8-2 x} \left (4+x^2 (2-\log (3))-2 x (5-\log (9))\right )\right ) \, dx+\int \left (8 e^{4-x}-12 e^{4-x} x+2 e^{4-x} x^2+3^x e^{4-x} (-1+\log (3))\right ) \, dx\\ &=2 \int e^{4-x} x^2 \, dx-2 \int e^{8-2 x} x \left (-16+28 x-10 x^2+x^3\right ) \, dx+8 \int e^{4-x} \, dx-12 \int e^{4-x} x \, dx+\frac {1}{2} (-1+\log (3)) \int 9^x e^{8-2 x} \, dx+(-1+\log (3)) \int 3^x e^{4-x} \, dx+\int 3^x e^{8-2 x} \left (4+x^2 (2-\log (3))-2 x (5-\log (9))\right ) \, dx\\ &=-8 e^{4-x}+12 e^{4-x} x-2 e^{4-x} x^2-2 \int \left (-16 e^{8-2 x} x+28 e^{8-2 x} x^2-10 e^{8-2 x} x^3+e^{8-2 x} x^4\right ) \, dx+4 \int e^{4-x} x \, dx-12 \int e^{4-x} \, dx+\frac {1}{2} (-1+\log (3)) \int e^{8-x (2-\log (9))} \, dx+(-1+\log (3)) \int e^{4-x (1-\log (3))} \, dx+\int e^{8-x (2-\log (3))} \left (4+x^2 (2-\log (3))-2 x (5-\log (9))\right ) \, dx\\ &=4 e^{4-x}+e^{4-x (1-\log (3))}+8 e^{4-x} x-2 e^{4-x} x^2+\frac {e^{8-x (2-\log (9))} (1-\log (3))}{2 (2-\log (9))}-2 \int e^{8-2 x} x^4 \, dx+4 \int e^{4-x} \, dx+20 \int e^{8-2 x} x^3 \, dx+32 \int e^{8-2 x} x \, dx-56 \int e^{8-2 x} x^2 \, dx+\int \left (4 e^{8-x (2-\log (3))}-e^{8-x (2-\log (3))} x^2 (-2+\log (3))+2 e^{8-x (2-\log (3))} x (-5+\log (9))\right ) \, dx\\ &=e^{4-x (1-\log (3))}-16 e^{8-2 x} x+8 e^{4-x} x+28 e^{8-2 x} x^2-2 e^{4-x} x^2-10 e^{8-2 x} x^3+e^{8-2 x} x^4+\frac {e^{8-x (2-\log (9))} (1-\log (3))}{2 (2-\log (9))}+4 \int e^{8-x (2-\log (3))} \, dx-4 \int e^{8-2 x} x^3 \, dx+16 \int e^{8-2 x} \, dx+30 \int e^{8-2 x} x^2 \, dx-56 \int e^{8-2 x} x \, dx+(2-\log (3)) \int e^{8-x (2-\log (3))} x^2 \, dx+(2 (-5+\log (9))) \int e^{8-x (2-\log (3))} x \, dx\\ &=-8 e^{8-2 x}+e^{4-x (1-\log (3))}+12 e^{8-2 x} x+8 e^{4-x} x+13 e^{8-2 x} x^2-2 e^{4-x} x^2-e^{8-x (2-\log (3))} x^2-8 e^{8-2 x} x^3+e^{8-2 x} x^4-\frac {4 e^{8-x (2-\log (3))}}{2-\log (3)}+\frac {e^{8-x (2-\log (9))} (1-\log (3))}{2 (2-\log (9))}+\frac {2 e^{8-x (2-\log (3))} x (5-\log (9))}{2-\log (3)}+2 \int e^{8+x (-2+\log (3))} x \, dx-6 \int e^{8-2 x} x^2 \, dx-28 \int e^{8-2 x} \, dx+30 \int e^{8-2 x} x \, dx-\frac {(2 (5-\log (9))) \int e^{8+x (-2+\log (3))} \, dx}{2-\log (3)}\\ &=6 e^{8-2 x}+e^{4-x (1-\log (3))}-3 e^{8-2 x} x+8 e^{4-x} x+16 e^{8-2 x} x^2-2 e^{4-x} x^2-e^{8-x (2-\log (3))} x^2-8 e^{8-2 x} x^3+e^{8-2 x} x^4-\frac {4 e^{8-x (2-\log (3))}}{2-\log (3)}-\frac {2 e^{8-x (2-\log (3))} x}{2-\log (3)}+\frac {e^{8-x (2-\log (9))} (1-\log (3))}{2 (2-\log (9))}+\frac {2 e^{8-x (2-\log (3))} (5-\log (9))}{(2-\log (3))^2}+\frac {2 e^{8-x (2-\log (3))} x (5-\log (9))}{2-\log (3)}-6 \int e^{8-2 x} x \, dx+15 \int e^{8-2 x} \, dx+\frac {2 \int e^{8+x (-2+\log (3))} \, dx}{2-\log (3)}\\ &=-\frac {3}{2} e^{8-2 x}+e^{4-x (1-\log (3))}+8 e^{4-x} x+16 e^{8-2 x} x^2-2 e^{4-x} x^2-e^{8-x (2-\log (3))} x^2-8 e^{8-2 x} x^3+e^{8-2 x} x^4-\frac {2 e^{8-x (2-\log (3))}}{(2-\log (3))^2}-\frac {4 e^{8-x (2-\log (3))}}{2-\log (3)}-\frac {2 e^{8-x (2-\log (3))} x}{2-\log (3)}+\frac {e^{8-x (2-\log (9))} (1-\log (3))}{2 (2-\log (9))}+\frac {2 e^{8-x (2-\log (3))} (5-\log (9))}{(2-\log (3))^2}+\frac {2 e^{8-x (2-\log (3))} x (5-\log (9))}{2-\log (3)}-3 \int e^{8-2 x} \, dx\\ &=e^{4-x (1-\log (3))}+8 e^{4-x} x+16 e^{8-2 x} x^2-2 e^{4-x} x^2-e^{8-x (2-\log (3))} x^2-8 e^{8-2 x} x^3+e^{8-2 x} x^4-\frac {2 e^{8-x (2-\log (3))}}{(2-\log (3))^2}-\frac {4 e^{8-x (2-\log (3))}}{2-\log (3)}-\frac {2 e^{8-x (2-\log (3))} x}{2-\log (3)}+\frac {e^{8-x (2-\log (9))} (1-\log (3))}{2 (2-\log (9))}+\frac {2 e^{8-x (2-\log (3))} (5-\log (9))}{(2-\log (3))^2}+\frac {2 e^{8-x (2-\log (3))} x (5-\log (9))}{2-\log (3)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.76, size = 45, normalized size = 1.61 \begin {gather*} \frac {1}{4} e^{4-2 x} \left (3^x+8 x-2 x^2\right ) \left (4 e^x+e^4 \left (3^x+8 x-2 x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(8 - 2*x)*(64*x - 112*x^2 + 40*x^3 - 4*x^4 + E^(-4 + x)*(16 - 24*x + 4*x^2) + 3^(2*x)*(-1 + Log[3
]) + 3^x*(8 - 20*x + 4*x^2 + (8*x - 2*x^2)*Log[3] + E^(-4 + x)*(-2 + 2*Log[3]))))/2,x]

[Out]

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

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fricas [B]  time = 0.50, size = 60, normalized size = 2.14 \begin {gather*} \frac {1}{4} \, {\left (4 \, x^{4} - 32 \, x^{3} - 4 \, {\left (x^{2} - 4 \, x - e^{\left (x - 4\right )}\right )} 3^{x} + 64 \, x^{2} - 8 \, {\left (x^{2} - 4 \, x\right )} e^{\left (x - 4\right )} + 3^{2 \, x}\right )} e^{\left (-2 \, x + 8\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((log(3)-1)*exp(x*log(3))^2+((2*log(3)-2)*exp(x-4)+(-2*x^2+8*x)*log(3)+4*x^2-20*x+8)*exp(x*log(3
))+(4*x^2-24*x+16)*exp(x-4)-4*x^4+40*x^3-112*x^2+64*x)/exp(x-4)^2,x, algorithm="fricas")

[Out]

1/4*(4*x^4 - 32*x^3 - 4*(x^2 - 4*x - e^(x - 4))*3^x + 64*x^2 - 8*(x^2 - 4*x)*e^(x - 4) + 3^(2*x))*e^(-2*x + 8)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {1}{2} \, {\left (4 \, x^{4} - 40 \, x^{3} - 2 \, {\left (2 \, x^{2} + {\left (\log \relax (3) - 1\right )} e^{\left (x - 4\right )} - {\left (x^{2} - 4 \, x\right )} \log \relax (3) - 10 \, x + 4\right )} 3^{x} + 112 \, x^{2} - 3^{2 \, x} {\left (\log \relax (3) - 1\right )} - 4 \, {\left (x^{2} - 6 \, x + 4\right )} e^{\left (x - 4\right )} - 64 \, x\right )} e^{\left (-2 \, x + 8\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((log(3)-1)*exp(x*log(3))^2+((2*log(3)-2)*exp(x-4)+(-2*x^2+8*x)*log(3)+4*x^2-20*x+8)*exp(x*log(3
))+(4*x^2-24*x+16)*exp(x-4)-4*x^4+40*x^3-112*x^2+64*x)/exp(x-4)^2,x, algorithm="giac")

[Out]

integrate(-1/2*(4*x^4 - 40*x^3 - 2*(2*x^2 + (log(3) - 1)*e^(x - 4) - (x^2 - 4*x)*log(3) - 10*x + 4)*3^x + 112*
x^2 - 3^(2*x)*(log(3) - 1) - 4*(x^2 - 6*x + 4)*e^(x - 4) - 64*x)*e^(-2*x + 8), x)

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maple [B]  time = 0.24, size = 80, normalized size = 2.86

method	result	size
risch	\(\frac {\left (-4 x^{2}+16 x \right ) {\mathrm e}^{-x +4}}{2}+\frac {\left (2 x^{4}-16 x^{3}+32 x^{2}\right ) {\mathrm e}^{-2 x +8}}{2}+\frac {{\mathrm e}^{-2 x +8} 3^{2 x}}{4}-\left (x^{2}-4 x -{\mathrm e}^{x -4}\right ) {\mathrm e}^{-2 x +8} 3^{x}\)	\(80\)
default	\(\frac {{\mathrm e}^{8} \ln \relax (3) {\mathrm e}^{2 x \ln \relax (3)} {\mathrm e}^{-2 x}}{4 \ln \relax (3)-4}-8 \,{\mathrm e}^{-x} {\mathrm e}^{4}+32 \,{\mathrm e}^{8} \left (-\frac {x \,{\mathrm e}^{-2 x}}{2}-\frac {{\mathrm e}^{-2 x}}{4}\right )-56 \,{\mathrm e}^{8} \left (-\frac {x^{2} {\mathrm e}^{-2 x}}{2}-\frac {x \,{\mathrm e}^{-2 x}}{2}-\frac {{\mathrm e}^{-2 x}}{4}\right )+20 \,{\mathrm e}^{8} \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 )-2 \,{\mathrm e}^{8} \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 )+\frac {4 \,{\mathrm e}^{8} {\mathrm e}^{x \ln \relax (3)} {\mathrm e}^{-2 x}}{\ln \relax (3)-2}-\frac {{\mathrm e}^{8} {\mathrm e}^{2 x \ln \relax (3)} {\mathrm e}^{-2 x}}{4 \left (\ln \relax (3)-1\right )}-12 \,{\mathrm e}^{4} \left (-x \,{\mathrm e}^{-x}-{\mathrm e}^{-x}\right )+2 \,{\mathrm e}^{4} \left (-x^{2} {\mathrm e}^{-x}-2 x \,{\mathrm e}^{-x}-2 \,{\mathrm e}^{-x}\right )-\frac {{\mathrm e}^{4} {\mathrm e}^{x \ln \relax (3)} {\mathrm e}^{-x}}{\ln \relax (3)-1}+\frac {\left (\frac {20 \,{\mathrm e}^{8} {\mathrm e}^{x \ln \relax (3)}}{\ln \relax (3)^{2}-4 \ln \relax (3)+4}-\frac {20 \,{\mathrm e}^{8} x \,{\mathrm e}^{x \ln \relax (3)}}{\ln \relax (3)-2}\right ) {\mathrm e}^{-2 x}}{2}+\frac {\left (\frac {4 \,{\mathrm e}^{8} x^{2} {\mathrm e}^{x \ln \relax (3)}}{\ln \relax (3)-2}+\frac {8 \,{\mathrm e}^{8} {\mathrm e}^{x \ln \relax (3)}}{\left (\ln \relax (3)^{2}-4 \ln \relax (3)+4\right ) \left (\ln \relax (3)-2\right )}-\frac {8 \,{\mathrm e}^{8} x \,{\mathrm e}^{x \ln \relax (3)}}{\ln \relax (3)^{2}-4 \ln \relax (3)+4}\right ) {\mathrm e}^{-2 x}}{2}+\frac {{\mathrm e}^{4} \ln \relax (3) {\mathrm e}^{x \ln \relax (3)} {\mathrm e}^{-x}}{\ln \relax (3)-1}+\frac {\left (-\frac {8 \,{\mathrm e}^{8} \ln \relax (3) {\mathrm e}^{x \ln \relax (3)}}{\ln \relax (3)^{2}-4 \ln \relax (3)+4}+\frac {8 \,{\mathrm e}^{8} \ln \relax (3) x \,{\mathrm e}^{x \ln \relax (3)}}{\ln \relax (3)-2}\right ) {\mathrm e}^{-2 x}}{2}+\frac {\left (-\frac {4 \,{\mathrm e}^{8} \ln \relax (3) {\mathrm e}^{x \ln \relax (3)}}{\left (\ln \relax (3)^{2}-4 \ln \relax (3)+4\right ) \left (\ln \relax (3)-2\right )}+\frac {4 \,{\mathrm e}^{8} \ln \relax (3) x \,{\mathrm e}^{x \ln \relax (3)}}{\ln \relax (3)^{2}-4 \ln \relax (3)+4}-\frac {2 \,{\mathrm e}^{8} \ln \relax (3) x^{2} {\mathrm e}^{x \ln \relax (3)}}{\ln \relax (3)-2}\right ) {\mathrm e}^{-2 x}}{2}\)	\(569\)
meijerg	\(\frac {8 \,{\mathrm e}^{2 x \,{\mathrm e}^{8}-4-2 x} \left (1-{\mathrm e}^{-x \,{\mathrm e}^{8} \left (2-{\mathrm e}^{-8}\right )}\right )}{2-{\mathrm e}^{-8}}+\frac {\ln \relax (3) {\mathrm e}^{2 x \,{\mathrm e}^{8}-4-2 x} \left (1-{\mathrm e}^{-x \,{\mathrm e}^{8} \left (2-{\mathrm e}^{-8}\right ) \left (1-\frac {\ln \relax (3) {\mathrm e}^{-8}}{2-{\mathrm e}^{-8}}\right )}\right )}{\left (2-{\mathrm e}^{-8}\right ) \left (1-\frac {\ln \relax (3) {\mathrm e}^{-8}}{2-{\mathrm e}^{-8}}\right )}-\frac {{\mathrm e}^{2 x \,{\mathrm e}^{8}-4-2 x} \left (1-{\mathrm e}^{-x \,{\mathrm e}^{8} \left (2-{\mathrm e}^{-8}\right ) \left (1-\frac {\ln \relax (3) {\mathrm e}^{-8}}{2-{\mathrm e}^{-8}}\right )}\right )}{\left (2-{\mathrm e}^{-8}\right ) \left (1-\frac {\ln \relax (3) {\mathrm e}^{-8}}{2-{\mathrm e}^{-8}}\right )}+\frac {{\mathrm e}^{-2 x +8+2 x \,{\mathrm e}^{8}} \left (2-\frac {\left (3 x^{2} \ln \relax (3)^{2} \left (1-\frac {2 \,{\mathrm e}^{8}}{\ln \relax (3)}\right )^{2}-6 x \ln \relax (3) \left (1-\frac {2 \,{\mathrm e}^{8}}{\ln \relax (3)}\right )+6\right ) {\mathrm e}^{x \ln \relax (3) \left (1-\frac {2 \,{\mathrm e}^{8}}{\ln \relax (3)}\right )}}{3}\right )}{\ln \relax (3)^{2} \left (1-\frac {2 \,{\mathrm e}^{8}}{\ln \relax (3)}\right )^{3}}-\frac {2 \,{\mathrm e}^{-2 x +8+2 x \,{\mathrm e}^{8}} \left (2-\frac {\left (3 x^{2} \ln \relax (3)^{2} \left (1-\frac {2 \,{\mathrm e}^{8}}{\ln \relax (3)}\right )^{2}-6 x \ln \relax (3) \left (1-\frac {2 \,{\mathrm e}^{8}}{\ln \relax (3)}\right )+6\right ) {\mathrm e}^{x \ln \relax (3) \left (1-\frac {2 \,{\mathrm e}^{8}}{\ln \relax (3)}\right )}}{3}\right )}{\ln \relax (3)^{3} \left (1-\frac {2 \,{\mathrm e}^{8}}{\ln \relax (3)}\right )^{3}}+\frac {4 \,{\mathrm e}^{-2 x +8+2 x \,{\mathrm e}^{8}} \left (1-\frac {\left (2-2 x \ln \relax (3) \left (1-\frac {2 \,{\mathrm e}^{8}}{\ln \relax (3)}\right )\right ) {\mathrm e}^{x \ln \relax (3) \left (1-\frac {2 \,{\mathrm e}^{8}}{\ln \relax (3)}\right )}}{2}\right )}{\left (1-\frac {2 \,{\mathrm e}^{8}}{\ln \relax (3)}\right )^{2} \ln \relax (3)}-\frac {10 \,{\mathrm e}^{-2 x +8+2 x \,{\mathrm e}^{8}} \left (1-\frac {\left (2-2 x \ln \relax (3) \left (1-\frac {2 \,{\mathrm e}^{8}}{\ln \relax (3)}\right )\right ) {\mathrm e}^{x \ln \relax (3) \left (1-\frac {2 \,{\mathrm e}^{8}}{\ln \relax (3)}\right )}}{2}\right )}{\ln \relax (3)^{2} \left (1-\frac {2 \,{\mathrm e}^{8}}{\ln \relax (3)}\right )^{2}}+\frac {2 \,{\mathrm e}^{2 x \,{\mathrm e}^{8}-20-2 x} \left (2-\frac {\left (3 x^{2} {\mathrm e}^{16} \left (2-{\mathrm e}^{-8}\right )^{2}+6 x \,{\mathrm e}^{8} \left (2-{\mathrm e}^{-8}\right )+6\right ) {\mathrm e}^{-x \,{\mathrm e}^{8} \left (2-{\mathrm e}^{-8}\right )}}{3}\right )}{\left (2-{\mathrm e}^{-8}\right )^{3}}-\frac {12 \,{\mathrm e}^{2 x \,{\mathrm e}^{8}-12-2 x} \left (1-\frac {\left (2+2 x \,{\mathrm e}^{8} \left (2-{\mathrm e}^{-8}\right )\right ) {\mathrm e}^{-x \,{\mathrm e}^{8} \left (2-{\mathrm e}^{-8}\right )}}{2}\right )}{\left (2-{\mathrm e}^{-8}\right )^{2}}-7 \,{\mathrm e}^{2 x \,{\mathrm e}^{8}-16-2 x} \left (2-\frac {\left (12 x^{2} {\mathrm e}^{16}+12 x \,{\mathrm e}^{8}+6\right ) {\mathrm e}^{-2 x \,{\mathrm e}^{8}}}{3}\right )+8 \,{\mathrm e}^{2 x \,{\mathrm e}^{8}-2 x -8} \left (1-\frac {\left (4 x \,{\mathrm e}^{8}+2\right ) {\mathrm e}^{-2 x \,{\mathrm e}^{8}}}{2}\right )-\frac {{\mathrm e}^{-2 x +8+2 x \,{\mathrm e}^{8}} \left (1-{\mathrm e}^{2 x \ln \relax (3) \left (1-\frac {{\mathrm e}^{8}}{\ln \relax (3)}\right )}\right )}{4 \left (1-\frac {{\mathrm e}^{8}}{\ln \relax (3)}\right )}+\frac {{\mathrm e}^{-2 x +8+2 x \,{\mathrm e}^{8}} \left (1-{\mathrm e}^{2 x \ln \relax (3) \left (1-\frac {{\mathrm e}^{8}}{\ln \relax (3)}\right )}\right )}{4 \ln \relax (3) \left (1-\frac {{\mathrm e}^{8}}{\ln \relax (3)}\right )}-\frac {4 \,{\mathrm e}^{-2 x +8+2 x \,{\mathrm e}^{8}} \left (1-{\mathrm e}^{x \ln \relax (3) \left (1-\frac {2 \,{\mathrm e}^{8}}{\ln \relax (3)}\right )}\right )}{\ln \relax (3) \left (1-\frac {2 \,{\mathrm e}^{8}}{\ln \relax (3)}\right )}-\frac {{\mathrm e}^{2 x \,{\mathrm e}^{8}-32-2 x} \left (24-\frac {\left (80 x^{4} {\mathrm e}^{32}+160 x^{3} {\mathrm e}^{24}+240 x^{2} {\mathrm e}^{16}+240 x \,{\mathrm e}^{8}+120\right ) {\mathrm e}^{-2 x \,{\mathrm e}^{8}}}{5}\right )}{16}+\frac {5 \,{\mathrm e}^{2 x \,{\mathrm e}^{8}-24-2 x} \left (6-\frac {\left (32 x^{3} {\mathrm e}^{24}+48 x^{2} {\mathrm e}^{16}+48 x \,{\mathrm e}^{8}+24\right ) {\mathrm e}^{-2 x \,{\mathrm e}^{8}}}{4}\right )}{4}\)	\(902\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*((ln(3)-1)*exp(x*ln(3))^2+((2*ln(3)-2)*exp(x-4)+(-2*x^2+8*x)*ln(3)+4*x^2-20*x+8)*exp(x*ln(3))+(4*x^2-2
4*x+16)*exp(x-4)-4*x^4+40*x^3-112*x^2+64*x)/exp(x-4)^2,x,method=_RETURNVERBOSE)

[Out]

1/2*(-4*x^2+16*x)*exp(-x+4)+1/2*(2*x^4-16*x^3+32*x^2)*exp(-2*x+8)+1/4*exp(-2*x+8)*(3^x)^2-(x^2-4*x-exp(x-4))*e
xp(-2*x+8)*3^x

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maxima [B]  time = 0.93, size = 431, normalized size = 15.39 \begin {gather*} -2 \, {\left (x^{2} e^{4} + 2 \, x e^{4} + 2 \, e^{4}\right )} e^{\left (-x\right )} + 12 \, {\left (x e^{4} + e^{4}\right )} e^{\left (-x\right )} + \frac {1}{2} \, {\left (2 \, x^{4} e^{8} + 4 \, x^{3} e^{8} + 6 \, x^{2} e^{8} + 6 \, x e^{8} + 3 \, e^{8}\right )} e^{\left (-2 \, x\right )} - \frac {5}{2} \, {\left (4 \, x^{3} e^{8} + 6 \, x^{2} e^{8} + 6 \, x e^{8} + 3 \, e^{8}\right )} e^{\left (-2 \, x\right )} + 14 \, {\left (2 \, x^{2} e^{8} + 2 \, x e^{8} + e^{8}\right )} e^{\left (-2 \, x\right )} - 8 \, {\left (2 \, x e^{8} + e^{8}\right )} e^{\left (-2 \, x\right )} - \frac {{\left ({\left (\log \relax (3)^{2} - 4 \, \log \relax (3) + 4\right )} x^{2} e^{8} - 2 \, x {\left (\log \relax (3) - 2\right )} e^{8} + 2 \, e^{8}\right )} e^{\left (x \log \relax (3) - 2 \, x\right )} \log \relax (3)}{\log \relax (3)^{3} - 6 \, \log \relax (3)^{2} + 12 \, \log \relax (3) - 8} + \frac {4 \, {\left (x {\left (\log \relax (3) - 2\right )} e^{8} - e^{8}\right )} e^{\left (x \log \relax (3) - 2 \, x\right )} \log \relax (3)}{\log \relax (3)^{2} - 4 \, \log \relax (3) + 4} + \frac {2 \, {\left ({\left (\log \relax (3)^{2} - 4 \, \log \relax (3) + 4\right )} x^{2} e^{8} - 2 \, x {\left (\log \relax (3) - 2\right )} e^{8} + 2 \, e^{8}\right )} e^{\left (x \log \relax (3) - 2 \, x\right )}}{\log \relax (3)^{3} - 6 \, \log \relax (3)^{2} + 12 \, \log \relax (3) - 8} - \frac {10 \, {\left (x {\left (\log \relax (3) - 2\right )} e^{8} - e^{8}\right )} e^{\left (x \log \relax (3) - 2 \, x\right )}}{\log \relax (3)^{2} - 4 \, \log \relax (3) + 4} + \frac {6561 \, e^{\left (2 \, {\left (x - 4\right )} {\left (\log \relax (3) - 1\right )}\right )} \log \relax (3)}{4 \, {\left (\log \relax (3) - 1\right )}} + \frac {81 \, e^{\left ({\left (x - 4\right )} {\left (\log \relax (3) - 1\right )}\right )} \log \relax (3)}{\log \relax (3) - 1} - \frac {6561 \, e^{\left (2 \, {\left (x - 4\right )} {\left (\log \relax (3) - 1\right )}\right )}}{4 \, {\left (\log \relax (3) - 1\right )}} - \frac {81 \, e^{\left ({\left (x - 4\right )} {\left (\log \relax (3) - 1\right )}\right )}}{\log \relax (3) - 1} + \frac {324 \, e^{\left ({\left (x - 4\right )} {\left (\log \relax (3) - 2\right )}\right )}}{\log \relax (3) - 2} - 8 \, e^{\left (-x + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((log(3)-1)*exp(x*log(3))^2+((2*log(3)-2)*exp(x-4)+(-2*x^2+8*x)*log(3)+4*x^2-20*x+8)*exp(x*log(3
))+(4*x^2-24*x+16)*exp(x-4)-4*x^4+40*x^3-112*x^2+64*x)/exp(x-4)^2,x, algorithm="maxima")

[Out]

-2*(x^2*e^4 + 2*x*e^4 + 2*e^4)*e^(-x) + 12*(x*e^4 + e^4)*e^(-x) + 1/2*(2*x^4*e^8 + 4*x^3*e^8 + 6*x^2*e^8 + 6*x
*e^8 + 3*e^8)*e^(-2*x) - 5/2*(4*x^3*e^8 + 6*x^2*e^8 + 6*x*e^8 + 3*e^8)*e^(-2*x) + 14*(2*x^2*e^8 + 2*x*e^8 + e^
8)*e^(-2*x) - 8*(2*x*e^8 + e^8)*e^(-2*x) - ((log(3)^2 - 4*log(3) + 4)*x^2*e^8 - 2*x*(log(3) - 2)*e^8 + 2*e^8)*
e^(x*log(3) - 2*x)*log(3)/(log(3)^3 - 6*log(3)^2 + 12*log(3) - 8) + 4*(x*(log(3) - 2)*e^8 - e^8)*e^(x*log(3) -
2*x)*log(3)/(log(3)^2 - 4*log(3) + 4) + 2*((log(3)^2 - 4*log(3) + 4)*x^2*e^8 - 2*x*(log(3) - 2)*e^8 + 2*e^8)*
e^(x*log(3) - 2*x)/(log(3)^3 - 6*log(3)^2 + 12*log(3) - 8) - 10*(x*(log(3) - 2)*e^8 - e^8)*e^(x*log(3) - 2*x)/
(log(3)^2 - 4*log(3) + 4) + 6561/4*e^(2*(x - 4)*(log(3) - 1))*log(3)/(log(3) - 1) + 81*e^((x - 4)*(log(3) - 1)
)*log(3)/(log(3) - 1) - 6561/4*e^(2*(x - 4)*(log(3) - 1))/(log(3) - 1) - 81*e^((x - 4)*(log(3) - 1))/(log(3) -
1) + 324*e^((x - 4)*(log(3) - 2))/(log(3) - 2) - 8*e^(-x + 4)

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mupad [B]  time = 2.83, size = 140, normalized size = 5.00 \begin {gather*} {\mathrm {e}}^{8-2\,x}\,\left (28\,x^2+28\,x+14\right )-{\mathrm {e}}^{8-2\,x}\,\left (16\,x+8\right )+{\mathrm {e}}^{8-2\,x}\,\left (x^4+2\,x^3+3\,x^2+3\,x+\frac {3}{2}\right )-{\mathrm {e}}^{8-2\,x}\,\left (10\,x^3+15\,x^2+15\,x+\frac {15}{2}\right )-2\,x\,{\mathrm {e}}^{4-x}\,\left (x-4\right )+3^x\,{\mathrm {e}}^{4-2\,x}\,\left ({\mathrm {e}}^x+4\,x\,{\mathrm {e}}^4-x^2\,{\mathrm {e}}^4\right )+\frac {3^{2\,x}\,{\mathrm {e}}^{8-2\,x}\,\left (\ln \relax (3)-1\right )}{2\,\left (\ln \relax (9)-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(8 - 2*x)*(32*x + (exp(x - 4)*(4*x^2 - 24*x + 16))/2 + (exp(x*log(3))*(log(3)*(8*x - 2*x^2) - 20*x + ex
p(x - 4)*(2*log(3) - 2) + 4*x^2 + 8))/2 + (exp(2*x*log(3))*(log(3) - 1))/2 - 56*x^2 + 20*x^3 - 2*x^4),x)

[Out]

exp(8 - 2*x)*(28*x + 28*x^2 + 14) - exp(8 - 2*x)*(16*x + 8) + exp(8 - 2*x)*(3*x + 3*x^2 + 2*x^3 + x^4 + 3/2) -
exp(8 - 2*x)*(15*x + 15*x^2 + 10*x^3 + 15/2) - 2*x*exp(4 - x)*(x - 4) + 3^x*exp(4 - 2*x)*(exp(x) + 4*x*exp(4)
- x^2*exp(4)) + (3^(2*x)*exp(8 - 2*x)*(log(3) - 1))/(2*(log(9) - 2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((ln(3)-1)*exp(x*ln(3))**2+((2*ln(3)-2)*exp(x-4)+(-2*x**2+8*x)*ln(3)+4*x**2-20*x+8)*exp(x*ln(3))
+(4*x**2-24*x+16)*exp(x-4)-4*x**4+40*x**3-112*x**2+64*x)/exp(x-4)**2,x)

[Out]

Timed out

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