3.45.40 \(\int \frac {e^{\frac {8-3 x+e^2 x}{x}} (e^2 x-x^2)+e^{\frac {8-3 x+e^2 x}{x}} (e^2 (-8-3 x)+8 x+4 x^2) \log (x)}{e^4 x^5-2 e^2 x^6+x^7} \, dx\)

Optimal. Leaf size=27 \[ \frac {e^{-3+e^2+\frac {8}{x}} \log (x)}{\left (e^2-x\right ) x^3} \]

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Rubi [B]  time = 3.37, antiderivative size = 272, normalized size of antiderivative = 10.07, number of steps used = 32, number of rules used = 12, integrand size = 88, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1594, 27, 6688, 6742, 2222, 2210, 2228, 2178, 2212, 2209, 2223, 2554} \begin {gather*} \frac {e^{\frac {8}{x}+e^2-5} \log (x)}{x^3}-\frac {3 e^{\frac {8}{x}+e^2-5} \log (x)}{8 x^2}+\frac {\left (8+3 e^2\right ) e^{\frac {8}{x}+e^2-7} \log (x)}{8 x^2}-e^{\frac {8}{x}+e^2-11} \log (x)-\frac {3}{256} e^{\frac {8}{x}+e^2-5} \log (x)+\frac {e^{\frac {8}{x}+e^2-9} \log (x)}{e^2-x}+\frac {3 e^{\frac {8}{x}+e^2-5} \log (x)}{32 x}-\frac {\left (8+3 e^2\right ) e^{\frac {8}{x}+e^2-7} \log (x)}{32 x}+\frac {\left (4+e^2\right ) e^{\frac {8}{x}+e^2-9} \log (x)}{4 x}+\frac {1}{256} \left (8+3 e^2\right ) e^{\frac {8}{x}+e^2-7} \log (x)+\frac {1}{8} \left (8+e^2\right ) e^{\frac {8}{x}+e^2-11} \log (x)-\frac {1}{32} \left (4+e^2\right ) e^{\frac {8}{x}+e^2-9} \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((8 - 3*x + E^2*x)/x)*(E^2*x - x^2) + E^((8 - 3*x + E^2*x)/x)*(E^2*(-8 - 3*x) + 8*x + 4*x^2)*Log[x])/(E
^4*x^5 - 2*E^2*x^6 + x^7),x]

[Out]

-(E^(-11 + E^2 + 8/x)*Log[x]) - (3*E^(-5 + E^2 + 8/x)*Log[x])/256 - (E^(-9 + E^2 + 8/x)*(4 + E^2)*Log[x])/32 +
(E^(-11 + E^2 + 8/x)*(8 + E^2)*Log[x])/8 + (E^(-7 + E^2 + 8/x)*(8 + 3*E^2)*Log[x])/256 + (E^(-9 + E^2 + 8/x)*
Log[x])/(E^2 - x) + (E^(-5 + E^2 + 8/x)*Log[x])/x^3 - (3*E^(-5 + E^2 + 8/x)*Log[x])/(8*x^2) + (E^(-7 + E^2 + 8
/x)*(8 + 3*E^2)*Log[x])/(8*x^2) + (3*E^(-5 + E^2 + 8/x)*Log[x])/(32*x) + (E^(-9 + E^2 + 8/x)*(4 + E^2)*Log[x])
/(4*x) - (E^(-7 + E^2 + 8/x)*(8 + 3*E^2)*Log[x])/(32*x)

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

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 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 2210

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Simp[(F^a*ExpIntegralEi[
b*(c + d*x)^n*Log[F]])/(f*n), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[d*e - c*f, 0]

Rule 2212

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
- n + 1)*F^(a + b*(c + d*x)^n))/(b*d*n*Log[F]), x] - Dist[(m - n + 1)/(b*n*Log[F]), Int[(c + d*x)^(m - n)*F^(
a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[(2*(m + 1))/n] && LtQ[0, (m + 1)/n, 5] &&
IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n, 0])

Rule 2222

Int[(F_)^((a_.) + (b_.)/((c_.) + (d_.)*(x_)))/((e_.) + (f_.)*(x_)), x_Symbol] :> Dist[d/f, Int[F^(a + b/(c + d
*x))/(c + d*x), x], x] - Dist[(d*e - c*f)/f, Int[F^(a + b/(c + d*x))/((c + d*x)*(e + f*x)), x], x] /; FreeQ[{F
, a, b, c, d, e, f}, x] && NeQ[d*e - c*f, 0]

Rule 2223

Int[(F_)^((a_.) + (b_.)/((c_.) + (d_.)*(x_)))*((e_.) + (f_.)*(x_))^(m_), x_Symbol] :> Simp[((e + f*x)^(m + 1)*
F^(a + b/(c + d*x)))/(f*(m + 1)), x] + Dist[(b*d*Log[F])/(f*(m + 1)), Int[((e + f*x)^(m + 1)*F^(a + b/(c + d*x
)))/(c + d*x)^2, x], x] /; FreeQ[{F, a, b, c, d, e, f}, x] && NeQ[d*e - c*f, 0] && ILtQ[m, -1]

Rule 2228

Int[(F_)^((a_.) + (b_.)/((c_.) + (d_.)*(x_)))/(((e_.) + (f_.)*(x_))*((g_.) + (h_.)*(x_))), x_Symbol] :> -Dist[
d/(f*(d*g - c*h)), Subst[Int[F^(a - (b*h)/(d*g - c*h) + (d*b*x)/(d*g - c*h))/x, x], x, (g + h*x)/(c + d*x)], x
] /; FreeQ[{F, a, b, c, d, e, f}, x] && EqQ[d*e - c*f, 0]

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, 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} &=\int \frac {e^{\frac {8-3 x+e^2 x}{x}} \left (e^2 x-x^2\right )+e^{\frac {8-3 x+e^2 x}{x}} \left (e^2 (-8-3 x)+8 x+4 x^2\right ) \log (x)}{x^5 \left (e^4-2 e^2 x+x^2\right )} \, dx\\ &=\int \frac {e^{\frac {8-3 x+e^2 x}{x}} \left (e^2 x-x^2\right )+e^{\frac {8-3 x+e^2 x}{x}} \left (e^2 (-8-3 x)+8 x+4 x^2\right ) \log (x)}{x^5 \left (-e^2+x\right )^2} \, dx\\ &=\int \frac {e^{-3+e^2+\frac {8}{x}} \left (\left (e^2-x\right ) x+\left (4 x (2+x)-e^2 (8+3 x)\right ) \log (x)\right )}{\left (e^2-x\right )^2 x^5} \, dx\\ &=\int \left (\frac {e^{-3+e^2+\frac {8}{x}}}{\left (e^2-x\right ) x^4}+\frac {e^{-3+e^2+\frac {8}{x}} \left (-8 e^2+\left (8-3 e^2\right ) x+4 x^2\right ) \log (x)}{\left (e^2-x\right )^2 x^5}\right ) \, dx\\ &=\int \frac {e^{-3+e^2+\frac {8}{x}}}{\left (e^2-x\right ) x^4} \, dx+\int \frac {e^{-3+e^2+\frac {8}{x}} \left (-8 e^2+\left (8-3 e^2\right ) x+4 x^2\right ) \log (x)}{\left (e^2-x\right )^2 x^5} \, dx\\ &=-e^{-11+e^2+\frac {8}{x}} \log (x)-\frac {3}{256} e^{-5+e^2+\frac {8}{x}} \log (x)-\frac {1}{32} e^{-9+e^2+\frac {8}{x}} \left (4+e^2\right ) \log (x)+\frac {1}{8} e^{-11+e^2+\frac {8}{x}} \left (8+e^2\right ) \log (x)+\frac {1}{256} e^{-7+e^2+\frac {8}{x}} \left (8+3 e^2\right ) \log (x)+\frac {e^{-9+e^2+\frac {8}{x}} \log (x)}{e^2-x}+\frac {e^{-5+e^2+\frac {8}{x}} \log (x)}{x^3}-\frac {3 e^{-5+e^2+\frac {8}{x}} \log (x)}{8 x^2}+\frac {e^{-7+e^2+\frac {8}{x}} \left (8+3 e^2\right ) \log (x)}{8 x^2}+\frac {3 e^{-5+e^2+\frac {8}{x}} \log (x)}{32 x}+\frac {e^{-9+e^2+\frac {8}{x}} \left (4+e^2\right ) \log (x)}{4 x}-\frac {e^{-7+e^2+\frac {8}{x}} \left (8+3 e^2\right ) \log (x)}{32 x}+\int \left (\frac {e^{-11+e^2+\frac {8}{x}}}{e^2-x}+\frac {e^{-5+e^2+\frac {8}{x}}}{x^4}+\frac {e^{-7+e^2+\frac {8}{x}}}{x^3}+\frac {e^{-9+e^2+\frac {8}{x}}}{x^2}+\frac {e^{-11+e^2+\frac {8}{x}}}{x}\right ) \, dx-\int \frac {e^{-3+e^2+\frac {8}{x}}}{\left (e^2-x\right ) x^4} \, dx\\ &=-e^{-11+e^2+\frac {8}{x}} \log (x)-\frac {3}{256} e^{-5+e^2+\frac {8}{x}} \log (x)-\frac {1}{32} e^{-9+e^2+\frac {8}{x}} \left (4+e^2\right ) \log (x)+\frac {1}{8} e^{-11+e^2+\frac {8}{x}} \left (8+e^2\right ) \log (x)+\frac {1}{256} e^{-7+e^2+\frac {8}{x}} \left (8+3 e^2\right ) \log (x)+\frac {e^{-9+e^2+\frac {8}{x}} \log (x)}{e^2-x}+\frac {e^{-5+e^2+\frac {8}{x}} \log (x)}{x^3}-\frac {3 e^{-5+e^2+\frac {8}{x}} \log (x)}{8 x^2}+\frac {e^{-7+e^2+\frac {8}{x}} \left (8+3 e^2\right ) \log (x)}{8 x^2}+\frac {3 e^{-5+e^2+\frac {8}{x}} \log (x)}{32 x}+\frac {e^{-9+e^2+\frac {8}{x}} \left (4+e^2\right ) \log (x)}{4 x}-\frac {e^{-7+e^2+\frac {8}{x}} \left (8+3 e^2\right ) \log (x)}{32 x}-\int \left (\frac {e^{-11+e^2+\frac {8}{x}}}{e^2-x}+\frac {e^{-5+e^2+\frac {8}{x}}}{x^4}+\frac {e^{-7+e^2+\frac {8}{x}}}{x^3}+\frac {e^{-9+e^2+\frac {8}{x}}}{x^2}+\frac {e^{-11+e^2+\frac {8}{x}}}{x}\right ) \, dx+\int \frac {e^{-11+e^2+\frac {8}{x}}}{e^2-x} \, dx+\int \frac {e^{-5+e^2+\frac {8}{x}}}{x^4} \, dx+\int \frac {e^{-7+e^2+\frac {8}{x}}}{x^3} \, dx+\int \frac {e^{-9+e^2+\frac {8}{x}}}{x^2} \, dx+\int \frac {e^{-11+e^2+\frac {8}{x}}}{x} \, dx\\ &=-\frac {1}{8} e^{-9+e^2+\frac {8}{x}}-\frac {e^{-5+e^2+\frac {8}{x}}}{8 x^2}-\frac {e^{-7+e^2+\frac {8}{x}}}{8 x}-e^{-11+e^2} \text {Ei}\left (\frac {8}{x}\right )-e^{-11+e^2+\frac {8}{x}} \log (x)-\frac {3}{256} e^{-5+e^2+\frac {8}{x}} \log (x)-\frac {1}{32} e^{-9+e^2+\frac {8}{x}} \left (4+e^2\right ) \log (x)+\frac {1}{8} e^{-11+e^2+\frac {8}{x}} \left (8+e^2\right ) \log (x)+\frac {1}{256} e^{-7+e^2+\frac {8}{x}} \left (8+3 e^2\right ) \log (x)+\frac {e^{-9+e^2+\frac {8}{x}} \log (x)}{e^2-x}+\frac {e^{-5+e^2+\frac {8}{x}} \log (x)}{x^3}-\frac {3 e^{-5+e^2+\frac {8}{x}} \log (x)}{8 x^2}+\frac {e^{-7+e^2+\frac {8}{x}} \left (8+3 e^2\right ) \log (x)}{8 x^2}+\frac {3 e^{-5+e^2+\frac {8}{x}} \log (x)}{32 x}+\frac {e^{-9+e^2+\frac {8}{x}} \left (4+e^2\right ) \log (x)}{4 x}-\frac {e^{-7+e^2+\frac {8}{x}} \left (8+3 e^2\right ) \log (x)}{32 x}-\frac {1}{8} \int \frac {e^{-7+e^2+\frac {8}{x}}}{x^2} \, dx-\frac {1}{4} \int \frac {e^{-5+e^2+\frac {8}{x}}}{x^3} \, dx+e^2 \int \frac {e^{-11+e^2+\frac {8}{x}}}{\left (e^2-x\right ) x} \, dx-\int \frac {e^{-11+e^2+\frac {8}{x}}}{e^2-x} \, dx-\int \frac {e^{-5+e^2+\frac {8}{x}}}{x^4} \, dx-\int \frac {e^{-7+e^2+\frac {8}{x}}}{x^3} \, dx-\int \frac {e^{-9+e^2+\frac {8}{x}}}{x^2} \, dx-2 \int \frac {e^{-11+e^2+\frac {8}{x}}}{x} \, dx\\ &=\frac {1}{64} e^{-7+e^2+\frac {8}{x}}+\frac {e^{-5+e^2+\frac {8}{x}}}{32 x}+e^{-11+e^2} \text {Ei}\left (\frac {8}{x}\right )-e^{-11+e^2+\frac {8}{x}} \log (x)-\frac {3}{256} e^{-5+e^2+\frac {8}{x}} \log (x)-\frac {1}{32} e^{-9+e^2+\frac {8}{x}} \left (4+e^2\right ) \log (x)+\frac {1}{8} e^{-11+e^2+\frac {8}{x}} \left (8+e^2\right ) \log (x)+\frac {1}{256} e^{-7+e^2+\frac {8}{x}} \left (8+3 e^2\right ) \log (x)+\frac {e^{-9+e^2+\frac {8}{x}} \log (x)}{e^2-x}+\frac {e^{-5+e^2+\frac {8}{x}} \log (x)}{x^3}-\frac {3 e^{-5+e^2+\frac {8}{x}} \log (x)}{8 x^2}+\frac {e^{-7+e^2+\frac {8}{x}} \left (8+3 e^2\right ) \log (x)}{8 x^2}+\frac {3 e^{-5+e^2+\frac {8}{x}} \log (x)}{32 x}+\frac {e^{-9+e^2+\frac {8}{x}} \left (4+e^2\right ) \log (x)}{4 x}-\frac {e^{-7+e^2+\frac {8}{x}} \left (8+3 e^2\right ) \log (x)}{32 x}+\frac {1}{32} \int \frac {e^{-5+e^2+\frac {8}{x}}}{x^2} \, dx+\frac {1}{8} \int \frac {e^{-7+e^2+\frac {8}{x}}}{x^2} \, dx+\frac {1}{4} \int \frac {e^{-5+e^2+\frac {8}{x}}}{x^3} \, dx-e^2 \int \frac {e^{-11+e^2+\frac {8}{x}}}{\left (e^2-x\right ) x} \, dx+\int \frac {e^{-11+e^2+\frac {8}{x}}}{x} \, dx-\operatorname {Subst}\left (\int \frac {e^{-11+\frac {8}{e^2}+e^2+\frac {8 x}{e^2}}}{x} \, dx,x,\frac {e^2-x}{x}\right )\\ &=-\frac {1}{256} e^{-5+e^2+\frac {8}{x}}-e^{-11+\frac {8}{e^2}+e^2} \text {Ei}\left (-\frac {8}{e^2}+\frac {8}{x}\right )-e^{-11+e^2+\frac {8}{x}} \log (x)-\frac {3}{256} e^{-5+e^2+\frac {8}{x}} \log (x)-\frac {1}{32} e^{-9+e^2+\frac {8}{x}} \left (4+e^2\right ) \log (x)+\frac {1}{8} e^{-11+e^2+\frac {8}{x}} \left (8+e^2\right ) \log (x)+\frac {1}{256} e^{-7+e^2+\frac {8}{x}} \left (8+3 e^2\right ) \log (x)+\frac {e^{-9+e^2+\frac {8}{x}} \log (x)}{e^2-x}+\frac {e^{-5+e^2+\frac {8}{x}} \log (x)}{x^3}-\frac {3 e^{-5+e^2+\frac {8}{x}} \log (x)}{8 x^2}+\frac {e^{-7+e^2+\frac {8}{x}} \left (8+3 e^2\right ) \log (x)}{8 x^2}+\frac {3 e^{-5+e^2+\frac {8}{x}} \log (x)}{32 x}+\frac {e^{-9+e^2+\frac {8}{x}} \left (4+e^2\right ) \log (x)}{4 x}-\frac {e^{-7+e^2+\frac {8}{x}} \left (8+3 e^2\right ) \log (x)}{32 x}-\frac {1}{32} \int \frac {e^{-5+e^2+\frac {8}{x}}}{x^2} \, dx+\operatorname {Subst}\left (\int \frac {e^{-11+\frac {8}{e^2}+e^2+\frac {8 x}{e^2}}}{x} \, dx,x,\frac {e^2-x}{x}\right )\\ &=-e^{-11+e^2+\frac {8}{x}} \log (x)-\frac {3}{256} e^{-5+e^2+\frac {8}{x}} \log (x)-\frac {1}{32} e^{-9+e^2+\frac {8}{x}} \left (4+e^2\right ) \log (x)+\frac {1}{8} e^{-11+e^2+\frac {8}{x}} \left (8+e^2\right ) \log (x)+\frac {1}{256} e^{-7+e^2+\frac {8}{x}} \left (8+3 e^2\right ) \log (x)+\frac {e^{-9+e^2+\frac {8}{x}} \log (x)}{e^2-x}+\frac {e^{-5+e^2+\frac {8}{x}} \log (x)}{x^3}-\frac {3 e^{-5+e^2+\frac {8}{x}} \log (x)}{8 x^2}+\frac {e^{-7+e^2+\frac {8}{x}} \left (8+3 e^2\right ) \log (x)}{8 x^2}+\frac {3 e^{-5+e^2+\frac {8}{x}} \log (x)}{32 x}+\frac {e^{-9+e^2+\frac {8}{x}} \left (4+e^2\right ) \log (x)}{4 x}-\frac {e^{-7+e^2+\frac {8}{x}} \left (8+3 e^2\right ) \log (x)}{32 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.51, size = 27, normalized size = 1.00 \begin {gather*} \frac {e^{-3+e^2+\frac {8}{x}} \log (x)}{\left (e^2-x\right ) x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((8 - 3*x + E^2*x)/x)*(E^2*x - x^2) + E^((8 - 3*x + E^2*x)/x)*(E^2*(-8 - 3*x) + 8*x + 4*x^2)*Log[
x])/(E^4*x^5 - 2*E^2*x^6 + x^7),x]

[Out]

(E^(-3 + E^2 + 8/x)*Log[x])/((E^2 - x)*x^3)

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fricas [A]  time = 0.68, size = 31, normalized size = 1.15 \begin {gather*} -\frac {e^{\left (\frac {x e^{2} - 3 \, x + 8}{x}\right )} \log \relax (x)}{x^{4} - x^{3} e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3*x-8)*exp(2)+4*x^2+8*x)*exp((exp(2)*x-3*x+8)/x)*log(x)+(exp(2)*x-x^2)*exp((exp(2)*x-3*x+8)/x))/
(x^5*exp(2)^2-2*x^6*exp(2)+x^7),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.18, size = 34, normalized size = 1.26 \begin {gather*} -\frac {e^{\left (\frac {x e^{2} - x + 8}{x}\right )} \log \relax (x)}{x^{4} e^{2} - x^{3} e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3*x-8)*exp(2)+4*x^2+8*x)*exp((exp(2)*x-3*x+8)/x)*log(x)+(exp(2)*x-x^2)*exp((exp(2)*x-3*x+8)/x))/
(x^5*exp(2)^2-2*x^6*exp(2)+x^7),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.17, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (-3 x -8\right ) {\mathrm e}^{2}+4 x^{2}+8 x \right ) {\mathrm e}^{\frac {{\mathrm e}^{2} x -3 x +8}{x}} \ln \relax (x )+\left ({\mathrm e}^{2} x -x^{2}\right ) {\mathrm e}^{\frac {{\mathrm e}^{2} x -3 x +8}{x}}}{x^{5} {\mathrm e}^{4}-2 x^{6} {\mathrm e}^{2}+x^{7}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-3*x-8)*exp(2)+4*x^2+8*x)*exp((exp(2)*x-3*x+8)/x)*ln(x)+(exp(2)*x-x^2)*exp((exp(2)*x-3*x+8)/x))/(x^5*ex
p(2)^2-2*x^6*exp(2)+x^7),x)

[Out]

int((((-3*x-8)*exp(2)+4*x^2+8*x)*exp((exp(2)*x-3*x+8)/x)*ln(x)+(exp(2)*x-x^2)*exp((exp(2)*x-3*x+8)/x))/(x^5*ex
p(2)^2-2*x^6*exp(2)+x^7),x)

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maxima [A]  time = 0.42, size = 29, normalized size = 1.07 \begin {gather*} -\frac {e^{\left (\frac {8}{x} + e^{2}\right )} \log \relax (x)}{x^{4} e^{3} - x^{3} e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3*x-8)*exp(2)+4*x^2+8*x)*exp((exp(2)*x-3*x+8)/x)*log(x)+(exp(2)*x-x^2)*exp((exp(2)*x-3*x+8)/x))/
(x^5*exp(2)^2-2*x^6*exp(2)+x^7),x, algorithm="maxima")

[Out]

-e^(8/x + e^2)*log(x)/(x^4*e^3 - x^3*e^5)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^2-3\,x+8}{x}}\,\left (x\,{\mathrm {e}}^2-x^2\right )+{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^2-3\,x+8}{x}}\,\ln \relax (x)\,\left (8\,x+4\,x^2-{\mathrm {e}}^2\,\left (3\,x+8\right )\right )}{x^7-2\,{\mathrm {e}}^2\,x^6+{\mathrm {e}}^4\,x^5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((x*exp(2) - 3*x + 8)/x)*(x*exp(2) - x^2) + exp((x*exp(2) - 3*x + 8)/x)*log(x)*(8*x + 4*x^2 - exp(2)*(
3*x + 8)))/(x^5*exp(4) - 2*x^6*exp(2) + x^7),x)

[Out]

int((exp((x*exp(2) - 3*x + 8)/x)*(x*exp(2) - x^2) + exp((x*exp(2) - 3*x + 8)/x)*log(x)*(8*x + 4*x^2 - exp(2)*(
3*x + 8)))/(x^5*exp(4) - 2*x^6*exp(2) + x^7), x)

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sympy [A]  time = 0.35, size = 27, normalized size = 1.00 \begin {gather*} - \frac {e^{\frac {- 3 x + x e^{2} + 8}{x}} \log {\relax (x )}}{x^{4} - x^{3} e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3*x-8)*exp(2)+4*x**2+8*x)*exp((exp(2)*x-3*x+8)/x)*ln(x)+(exp(2)*x-x**2)*exp((exp(2)*x-3*x+8)/x))
/(x**5*exp(2)**2-2*x**6*exp(2)+x**7),x)

[Out]

-exp((-3*x + x*exp(2) + 8)/x)*log(x)/(x**4 - x**3*exp(2))

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