3.55.9 \(\int \frac {e^{-x} (8-1016 x-257 x^2+112 x^3-8 x^4+e^{1+x} (16-2048 x+512 x^2-32 x^3))}{2 x^2-256 x^3+8224 x^4-2048 x^5+128 x^6} \, dx\)

Optimal. Leaf size=27 \[ \frac {e+\frac {e^{-x}}{2}}{\frac {x}{-8+x}+8 x^2} \]

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Rubi [C]  time = 5.72, antiderivative size = 621, normalized size of antiderivative = 23.00, number of steps used = 100, number of rules used = 10, integrand size = 75, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {6741, 12, 6742, 2177, 2178, 2270, 2268, 1646, 1586, 30} \begin {gather*} -\frac {4}{127} \left (8128+511 \sqrt {254}\right ) e^{\frac {1}{4} \left (\sqrt {254}-16\right )} \text {Ei}\left (\frac {1}{4} \left (-4 x-\sqrt {254}+16\right )\right )+2 \left (127+8 \sqrt {254}\right ) e^{\frac {1}{4} \left (\sqrt {254}-16\right )} \text {Ei}\left (\frac {1}{4} \left (-4 x-\sqrt {254}+16\right )\right )-\frac {255 \left (16-\sqrt {254}\right ) e^{\frac {1}{4} \left (\sqrt {254}-16\right )} \text {Ei}\left (\frac {1}{4} \left (-4 x-\sqrt {254}+16\right )\right )}{2032}-\frac {e^{\frac {1}{4} \left (\sqrt {254}-16\right )} \text {Ei}\left (\frac {1}{4} \left (-4 x-\sqrt {254}+16\right )\right )}{1016 \sqrt {254}}-\frac {500}{127} \sqrt {\frac {2}{127}} e^{\frac {1}{4} \left (\sqrt {254}-16\right )} \text {Ei}\left (\frac {1}{4} \left (-4 x-\sqrt {254}+16\right )\right )+\frac {509}{127} e^{\frac {1}{4} \left (\sqrt {254}-16\right )} \text {Ei}\left (\frac {1}{4} \left (-4 x-\sqrt {254}+16\right )\right )-\frac {255 \left (16+\sqrt {254}\right ) e^{-4-\frac {\sqrt {\frac {127}{2}}}{2}} \text {Ei}\left (\frac {1}{4} \left (-4 x+\sqrt {254}+16\right )\right )}{2032}+2 \left (127-8 \sqrt {254}\right ) e^{-4-\frac {\sqrt {\frac {127}{2}}}{2}} \text {Ei}\left (\frac {1}{4} \left (-4 x+\sqrt {254}+16\right )\right )-\frac {4}{127} \left (8128-511 \sqrt {254}\right ) e^{-4-\frac {\sqrt {\frac {127}{2}}}{2}} \text {Ei}\left (\frac {1}{4} \left (-4 x+\sqrt {254}+16\right )\right )+\frac {e^{-4-\frac {\sqrt {\frac {127}{2}}}{2}} \text {Ei}\left (\frac {1}{4} \left (-4 x+\sqrt {254}+16\right )\right )}{1016 \sqrt {254}}+\frac {500}{127} \sqrt {\frac {2}{127}} e^{-4-\frac {\sqrt {\frac {127}{2}}}{2}} \text {Ei}\left (\frac {1}{4} \left (-4 x+\sqrt {254}+16\right )\right )+\frac {509}{127} e^{-4-\frac {\sqrt {\frac {127}{2}}}{2}} \text {Ei}\left (\frac {1}{4} \left (-4 x+\sqrt {254}+16\right )\right )-\frac {e (511-64 x)}{8 x^2-64 x+1}+\frac {255 \left (16-\sqrt {254}\right ) e^{-x}}{508 \left (-4 x-\sqrt {254}+16\right )}-\frac {2036 e^{-x}}{127 \left (-4 x-\sqrt {254}+16\right )}+\frac {255 \left (16+\sqrt {254}\right ) e^{-x}}{508 \left (-4 x+\sqrt {254}+16\right )}-\frac {2036 e^{-x}}{127 \left (-4 x+\sqrt {254}+16\right )}-\frac {4 e^{-x}}{x}-\frac {8 e}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(8 - 1016*x - 257*x^2 + 112*x^3 - 8*x^4 + E^(1 + x)*(16 - 2048*x + 512*x^2 - 32*x^3))/(E^x*(2*x^2 - 256*x^
3 + 8224*x^4 - 2048*x^5 + 128*x^6)),x]

[Out]

-2036/(127*E^x*(16 - Sqrt[254] - 4*x)) + (255*(16 - Sqrt[254]))/(508*E^x*(16 - Sqrt[254] - 4*x)) - 2036/(127*E
^x*(16 + Sqrt[254] - 4*x)) + (255*(16 + Sqrt[254]))/(508*E^x*(16 + Sqrt[254] - 4*x)) - (8*E)/x - 4/(E^x*x) - (
E*(511 - 64*x))/(1 - 64*x + 8*x^2) + (509*E^((-16 + Sqrt[254])/4)*ExpIntegralEi[(16 - Sqrt[254] - 4*x)/4])/127
- (500*Sqrt[2/127]*E^((-16 + Sqrt[254])/4)*ExpIntegralEi[(16 - Sqrt[254] - 4*x)/4])/127 - (E^((-16 + Sqrt[254
])/4)*ExpIntegralEi[(16 - Sqrt[254] - 4*x)/4])/(1016*Sqrt[254]) - (255*(16 - Sqrt[254])*E^((-16 + Sqrt[254])/4
)*ExpIntegralEi[(16 - Sqrt[254] - 4*x)/4])/2032 + 2*(127 + 8*Sqrt[254])*E^((-16 + Sqrt[254])/4)*ExpIntegralEi[
(16 - Sqrt[254] - 4*x)/4] - (4*(8128 + 511*Sqrt[254])*E^((-16 + Sqrt[254])/4)*ExpIntegralEi[(16 - Sqrt[254] -
4*x)/4])/127 + (509*E^(-4 - Sqrt[127/2]/2)*ExpIntegralEi[(16 + Sqrt[254] - 4*x)/4])/127 + (500*Sqrt[2/127]*E^(
-4 - Sqrt[127/2]/2)*ExpIntegralEi[(16 + Sqrt[254] - 4*x)/4])/127 + (E^(-4 - Sqrt[127/2]/2)*ExpIntegralEi[(16 +
Sqrt[254] - 4*x)/4])/(1016*Sqrt[254]) - (4*(8128 - 511*Sqrt[254])*E^(-4 - Sqrt[127/2]/2)*ExpIntegralEi[(16 +
Sqrt[254] - 4*x)/4])/127 + 2*(127 - 8*Sqrt[254])*E^(-4 - Sqrt[127/2]/2)*ExpIntegralEi[(16 + Sqrt[254] - 4*x)/4
] - (255*(16 + Sqrt[254])*E^(-4 - Sqrt[127/2]/2)*ExpIntegralEi[(16 + Sqrt[254] - 4*x)/4])/2032

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rule 1646

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = Polynomi
alQuotient[(d + e*x)^m*Pq, a + b*x + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + b*x + c*x^2,
x], x, 0], g = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + b*x + c*x^2, x], x, 1]}, Simp[((b*f - 2*a*g + (2
*c*f - b*g)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(d
+ e*x)^m*(a + b*x + c*x^2)^(p + 1)*ExpandToSum[((p + 1)*(b^2 - 4*a*c)*Q)/(d + e*x)^m - ((2*p + 3)*(2*c*f - b*
g))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2
- b*d*e + a*e^2, 0] && LtQ[p, -1] && ILtQ[m, 0]

Rule 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), 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] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2268

Int[(F_)^((g_.)*((d_.) + (e_.)*(x_))^(n_.))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegr
and[F^(g*(d + e*x)^n), 1/(a + b*x + c*x^2), x], x] /; FreeQ[{F, a, b, c, d, e, g, n}, x]

Rule 2270

Int[((F_)^((g_.)*((d_.) + (e_.)*(x_))^(n_.))*(u_)^(m_.))/((a_.) + (b_.)*(x_) + (c_)*(x_)^2), x_Symbol] :> Int[
ExpandIntegrand[F^(g*(d + e*x)^n), u^m/(a + b*x + c*x^2), x], x] /; FreeQ[{F, a, b, c, d, e, g, n}, x] && Poly
nomialQ[u, x] && IntegerQ[m]

Rule 6741

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-x} \left (8-1016 x-257 x^2+112 x^3-8 x^4+e^{1+x} \left (16-2048 x+512 x^2-32 x^3\right )\right )}{2 x^2 \left (1-64 x+8 x^2\right )^2} \, dx\\ &=\frac {1}{2} \int \frac {e^{-x} \left (8-1016 x-257 x^2+112 x^3-8 x^4+e^{1+x} \left (16-2048 x+512 x^2-32 x^3\right )\right )}{x^2 \left (1-64 x+8 x^2\right )^2} \, dx\\ &=\frac {1}{2} \int \left (-\frac {257 e^{-x}}{\left (1-64 x+8 x^2\right )^2}+\frac {8 e^{-x}}{x^2 \left (1-64 x+8 x^2\right )^2}-\frac {1016 e^{-x}}{x \left (1-64 x+8 x^2\right )^2}+\frac {112 e^{-x} x}{\left (1-64 x+8 x^2\right )^2}-\frac {8 e^{-x} x^2}{\left (1-64 x+8 x^2\right )^2}-\frac {16 e \left (-1+128 x-32 x^2+2 x^3\right )}{x^2 \left (1-64 x+8 x^2\right )^2}\right ) \, dx\\ &=4 \int \frac {e^{-x}}{x^2 \left (1-64 x+8 x^2\right )^2} \, dx-4 \int \frac {e^{-x} x^2}{\left (1-64 x+8 x^2\right )^2} \, dx+56 \int \frac {e^{-x} x}{\left (1-64 x+8 x^2\right )^2} \, dx-\frac {257}{2} \int \frac {e^{-x}}{\left (1-64 x+8 x^2\right )^2} \, dx-508 \int \frac {e^{-x}}{x \left (1-64 x+8 x^2\right )^2} \, dx-(8 e) \int \frac {-1+128 x-32 x^2+2 x^3}{x^2 \left (1-64 x+8 x^2\right )^2} \, dx\\ &=-\frac {e (511-64 x)}{1-64 x+8 x^2}-4 \int \left (\frac {e^{-x} (-1+64 x)}{8 \left (1-64 x+8 x^2\right )^2}+\frac {e^{-x}}{8 \left (1-64 x+8 x^2\right )}\right ) \, dx+4 \int \left (\frac {e^{-x}}{x^2}+\frac {128 e^{-x}}{x}-\frac {8 e^{-x} (-511+64 x)}{\left (1-64 x+8 x^2\right )^2}-\frac {8 e^{-x} (-1023+128 x)}{1-64 x+8 x^2}\right ) \, dx+56 \int \left (\frac {\left (64+4 \sqrt {254}\right ) e^{-x}}{254 \left (64+4 \sqrt {254}-16 x\right )^2}+\frac {4 \sqrt {\frac {2}{127}} e^{-x}}{127 \left (64+4 \sqrt {254}-16 x\right )}+\frac {\left (64-4 \sqrt {254}\right ) e^{-x}}{254 \left (-64+4 \sqrt {254}+16 x\right )^2}+\frac {4 \sqrt {\frac {2}{127}} e^{-x}}{127 \left (-64+4 \sqrt {254}+16 x\right )}\right ) \, dx-\frac {257}{2} \int \left (\frac {8 e^{-x}}{127 \left (64+4 \sqrt {254}-16 x\right )^2}+\frac {\sqrt {\frac {2}{127}} e^{-x}}{127 \left (64+4 \sqrt {254}-16 x\right )}+\frac {8 e^{-x}}{127 \left (-64+4 \sqrt {254}+16 x\right )^2}+\frac {\sqrt {\frac {2}{127}} e^{-x}}{127 \left (-64+4 \sqrt {254}+16 x\right )}\right ) \, dx-508 \int \left (\frac {e^{-x}}{x}-\frac {8 e^{-x} (-8+x)}{\left (1-64 x+8 x^2\right )^2}-\frac {8 e^{-x} (-8+x)}{1-64 x+8 x^2}\right ) \, dx+\frac {1}{508} e \int \frac {4064-260096 x+32512 x^2}{x^2 \left (1-64 x+8 x^2\right )} \, dx\\ &=-\frac {e (511-64 x)}{1-64 x+8 x^2}-\frac {1}{2} \int \frac {e^{-x} (-1+64 x)}{\left (1-64 x+8 x^2\right )^2} \, dx-\frac {1}{2} \int \frac {e^{-x}}{1-64 x+8 x^2} \, dx+4 \int \frac {e^{-x}}{x^2} \, dx-\frac {1028}{127} \int \frac {e^{-x}}{\left (64+4 \sqrt {254}-16 x\right )^2} \, dx-\frac {1028}{127} \int \frac {e^{-x}}{\left (-64+4 \sqrt {254}+16 x\right )^2} \, dx-32 \int \frac {e^{-x} (-511+64 x)}{\left (1-64 x+8 x^2\right )^2} \, dx-32 \int \frac {e^{-x} (-1023+128 x)}{1-64 x+8 x^2} \, dx-508 \int \frac {e^{-x}}{x} \, dx+512 \int \frac {e^{-x}}{x} \, dx+4064 \int \frac {e^{-x} (-8+x)}{\left (1-64 x+8 x^2\right )^2} \, dx+4064 \int \frac {e^{-x} (-8+x)}{1-64 x+8 x^2} \, dx+\frac {1}{127} \left (224 \sqrt {\frac {2}{127}}\right ) \int \frac {e^{-x}}{64+4 \sqrt {254}-16 x} \, dx+\frac {1}{127} \left (224 \sqrt {\frac {2}{127}}\right ) \int \frac {e^{-x}}{-64+4 \sqrt {254}+16 x} \, dx-\frac {257 \int \frac {e^{-x}}{64+4 \sqrt {254}-16 x} \, dx}{127 \sqrt {254}}-\frac {257 \int \frac {e^{-x}}{-64+4 \sqrt {254}+16 x} \, dx}{127 \sqrt {254}}+\frac {1}{127} \left (112 \left (16-\sqrt {254}\right )\right ) \int \frac {e^{-x}}{\left (-64+4 \sqrt {254}+16 x\right )^2} \, dx+\frac {1}{127} \left (112 \left (16+\sqrt {254}\right )\right ) \int \frac {e^{-x}}{\left (64+4 \sqrt {254}-16 x\right )^2} \, dx+\frac {1}{508} e \int \frac {4064}{x^2} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.76, size = 36, normalized size = 1.33 \begin {gather*} \frac {e^{-x} \left (1+2 e^{1+x}\right ) (-8+x)}{2 x \left (1-64 x+8 x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(8 - 1016*x - 257*x^2 + 112*x^3 - 8*x^4 + E^(1 + x)*(16 - 2048*x + 512*x^2 - 32*x^3))/(E^x*(2*x^2 -
256*x^3 + 8224*x^4 - 2048*x^5 + 128*x^6)),x]

[Out]

((1 + 2*E^(1 + x))*(-8 + x))/(2*E^x*x*(1 - 64*x + 8*x^2))

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fricas [A]  time = 0.49, size = 38, normalized size = 1.41 \begin {gather*} \frac {{\left ({\left (x - 8\right )} e + 2 \, {\left (x - 8\right )} e^{\left (x + 2\right )}\right )} e^{\left (-x - 1\right )}}{2 \, {\left (8 \, x^{3} - 64 \, x^{2} + x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-32*x^3+512*x^2-2048*x+16)*exp(1)*exp(x)-8*x^4+112*x^3-257*x^2-1016*x+8)/(128*x^6-2048*x^5+8224*x^
4-256*x^3+2*x^2)/exp(x),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.16, size = 38, normalized size = 1.41 \begin {gather*} \frac {2 \, x e + x e^{\left (-x\right )} - 16 \, e - 8 \, e^{\left (-x\right )}}{2 \, {\left (8 \, x^{3} - 64 \, x^{2} + x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-32*x^3+512*x^2-2048*x+16)*exp(1)*exp(x)-8*x^4+112*x^3-257*x^2-1016*x+8)/(128*x^6-2048*x^5+8224*x^
4-256*x^3+2*x^2)/exp(x),x, algorithm="giac")

[Out]

1/2*(2*x*e + x*e^(-x) - 16*e - 8*e^(-x))/(8*x^3 - 64*x^2 + x)

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maple [A]  time = 0.14, size = 50, normalized size = 1.85

method	result	size
norman	\(\frac {\left (-4+64 x^{2} {\mathrm e} \,{\mathrm e}^{x}-8 x^{3} {\mathrm e} \,{\mathrm e}^{x}+\frac {x}{2}-8 \,{\mathrm e} \,{\mathrm e}^{x}\right ) {\mathrm e}^{-x}}{x \left (8 x^{2}-64 x +1\right )}\)	\(50\)
risch	\(\frac {x \,{\mathrm e}-8 \,{\mathrm e}}{\left (8 x^{2}-64 x +1\right ) x}+\frac {\left (-8+x \right ) {\mathrm e}^{-x}}{2 \left (8 x^{2}-64 x +1\right ) x}\)	\(53\)
default	\(\frac {1028 \,{\mathrm e} x}{127 \left (x^{2}-8 x +\frac {1}{8}\right )}-\frac {8176 \,{\mathrm e}}{127 \left (x^{2}-8 x +\frac {1}{8}\right )}-\frac {24 \,{\mathrm e} \sqrt {254}\, \arctanh \left (\frac {\left (16 x -64\right ) \sqrt {254}}{1016}\right )}{16129}-\frac {8 \,{\mathrm e}}{x}-\frac {4 \,{\mathrm e}^{-x} \left (2036 x^{2}-16272 x +127\right )}{127 \left (8 x^{2}-64 x +1\right ) x}+256 \,{\mathrm e} \left (-\frac {16 x -64}{4064 \left (8 x^{2}-64 x +1\right )}+\frac {\sqrt {254}\, \arctanh \left (\frac {\left (16 x -64\right ) \sqrt {254}}{1016}\right )}{129032}\right )-16 \,{\mathrm e} \left (-\frac {64 x -2}{4064 \left (8 x^{2}-64 x +1\right )}+\frac {\sqrt {254}\, \arctanh \left (\frac {\left (16 x -64\right ) \sqrt {254}}{1016}\right )}{32258}\right )+\frac {2 \,{\mathrm e}^{-x} \left (32 x -255\right )}{8 x^{2}-64 x +1}+\frac {257 \,{\mathrm e}^{-x} \left (x -4\right )}{508 \left (8 x^{2}-64 x +1\right )}-\frac {7 \,{\mathrm e}^{-x} \left (32 x -1\right )}{254 \left (8 x^{2}-64 x +1\right )}+\frac {{\mathrm e}^{-x} \left (255 x -4\right )}{4064 x^{2}-32512 x +508}\)	\(257\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-32*x^3+512*x^2-2048*x+16)*exp(1)*exp(x)-8*x^4+112*x^3-257*x^2-1016*x+8)/(128*x^6-2048*x^5+8224*x^4-256*
x^3+2*x^2)/exp(x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.41, size = 34, normalized size = 1.26 \begin {gather*} \frac {2 \, x e + {\left (x - 8\right )} e^{\left (-x\right )} - 16 \, e}{2 \, {\left (8 \, x^{3} - 64 \, x^{2} + x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-32*x^3+512*x^2-2048*x+16)*exp(1)*exp(x)-8*x^4+112*x^3-257*x^2-1016*x+8)/(128*x^6-2048*x^5+8224*x^
4-256*x^3+2*x^2)/exp(x),x, algorithm="maxima")

[Out]

1/2*(2*x*e + (x - 8)*e^(-x) - 16*e)/(8*x^3 - 64*x^2 + x)

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mupad [B]  time = 0.22, size = 40, normalized size = 1.48 \begin {gather*} -\frac {4\,{\mathrm {e}}^{-x}+8\,\mathrm {e}-x\,\left (\frac {{\mathrm {e}}^{-x}}{2}+\mathrm {e}\right )}{x\,\left (8\,x^2-64\,x+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-x)*(1016*x + 257*x^2 - 112*x^3 + 8*x^4 + exp(1)*exp(x)*(2048*x - 512*x^2 + 32*x^3 - 16) - 8))/(2*x^
2 - 256*x^3 + 8224*x^4 - 2048*x^5 + 128*x^6),x)

[Out]

-(4*exp(-x) + 8*exp(1) - x*(exp(-x)/2 + exp(1)))/(x*(8*x^2 - 64*x + 1))

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sympy [B]  time = 0.31, size = 41, normalized size = 1.52 \begin {gather*} \frac {\left (x - 8\right ) e^{- x}}{16 x^{3} - 128 x^{2} + 2 x} - \frac {- e x + 8 e}{8 x^{3} - 64 x^{2} + x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-32*x**3+512*x**2-2048*x+16)*exp(1)*exp(x)-8*x**4+112*x**3-257*x**2-1016*x+8)/(128*x**6-2048*x**5+
8224*x**4-256*x**3+2*x**2)/exp(x),x)

[Out]

(x - 8)*exp(-x)/(16*x**3 - 128*x**2 + 2*x) - (-E*x + 8*E)/(8*x**3 - 64*x**2 + x)

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