3.53.68 \(\int \frac {e^x (4 x-4 x^2)}{625 e^{3 x}-1875 e^{2 x} x+1875 e^x x^2-625 x^3} \, dx\)

Optimal. Leaf size=18 \[ -6+\frac {2}{25 \left (5-\frac {5 e^x}{x}\right )^2} \]

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Rubi [A]  time = 0.39, antiderivative size = 16, normalized size of antiderivative = 0.89, number of steps used = 5, number of rules used = 5, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {1593, 6688, 12, 6712, 32} \begin {gather*} \frac {2}{625 \left (1-\frac {e^x}{x}\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(4*x - 4*x^2))/(625*E^(3*x) - 1875*E^(2*x)*x + 1875*E^x*x^2 - 625*x^3),x]

[Out]

2/(625*(1 - E^x/x)^2)

Rule 12

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

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 1593

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

Rule 6688

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

Rule 6712

Int[(u_)*(v_)^(r_.)*((a_.)*(v_)^(p_.) + (b_.)*(w_)^(q_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(p*w*D[v, x
] - q*v*D[w, x])]}, -Dist[c*q, Subst[Int[(a + b*x^q)^m, x], x, v^(m*p + r + 1)*w], x] /; FreeQ[c, x]] /; FreeQ
[{a, b, m, p, q, r}, x] && EqQ[p + q*(m*p + r + 1), 0] && IntegerQ[q] && IntegerQ[m]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^x (4-4 x) x}{625 e^{3 x}-1875 e^{2 x} x+1875 e^x x^2-625 x^3} \, dx\\ &=\int \frac {4 e^x (1-x) x}{625 \left (e^x-x\right )^3} \, dx\\ &=\frac {4}{625} \int \frac {e^x (1-x) x}{\left (e^x-x\right )^3} \, dx\\ &=-\left (\frac {4}{625} \operatorname {Subst}\left (\int \frac {1}{(-1+x)^3} \, dx,x,\frac {e^x}{x}\right )\right )\\ &=\frac {2}{625 \left (1-\frac {e^x}{x}\right )^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 16, normalized size = 0.89 \begin {gather*} \frac {2 x^2}{625 \left (e^x-x\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(4*x - 4*x^2))/(625*E^(3*x) - 1875*E^(2*x)*x + 1875*E^x*x^2 - 625*x^3),x]

[Out]

(2*x^2)/(625*(E^x - x)^2)

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fricas [A]  time = 0.60, size = 20, normalized size = 1.11 \begin {gather*} \frac {2 \, x^{2}}{625 \, {\left (x^{2} - 2 \, x e^{x} + e^{\left (2 \, x\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x^2+4*x)*exp(x)/(625*exp(x)^3-1875*x*exp(x)^2+1875*exp(x)*x^2-625*x^3),x, algorithm="fricas")

[Out]

2/625*x^2/(x^2 - 2*x*e^x + e^(2*x))

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giac [A]  time = 0.18, size = 20, normalized size = 1.11 \begin {gather*} \frac {2 \, x^{2}}{625 \, {\left (x^{2} - 2 \, x e^{x} + e^{\left (2 \, x\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x^2+4*x)*exp(x)/(625*exp(x)^3-1875*x*exp(x)^2+1875*exp(x)*x^2-625*x^3),x, algorithm="giac")

[Out]

2/625*x^2/(x^2 - 2*x*e^x + e^(2*x))

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maple [A]  time = 0.05, size = 14, normalized size = 0.78

method	result	size
risch	\(\frac {2 x^{2}}{625 \left (x -{\mathrm e}^{x}\right )^{2}}\)	\(14\)
norman	\(\frac {-\frac {2 \,{\mathrm e}^{2 x}}{625}+\frac {4 \,{\mathrm e}^{x} x}{625}}{\left (x -{\mathrm e}^{x}\right )^{2}}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-4*x^2+4*x)*exp(x)/(625*exp(x)^3-1875*x*exp(x)^2+1875*exp(x)*x^2-625*x^3),x,method=_RETURNVERBOSE)

[Out]

2/625*x^2/(x-exp(x))^2

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maxima [A]  time = 0.40, size = 20, normalized size = 1.11 \begin {gather*} \frac {2 \, x^{2}}{625 \, {\left (x^{2} - 2 \, x e^{x} + e^{\left (2 \, x\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x^2+4*x)*exp(x)/(625*exp(x)^3-1875*x*exp(x)^2+1875*exp(x)*x^2-625*x^3),x, algorithm="maxima")

[Out]

2/625*x^2/(x^2 - 2*x*e^x + e^(2*x))

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mupad [B]  time = 0.11, size = 13, normalized size = 0.72 \begin {gather*} \frac {2\,x^2}{625\,{\left (x-{\mathrm {e}}^x\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(4*x - 4*x^2))/(625*exp(3*x) - 1875*x*exp(2*x) + 1875*x^2*exp(x) - 625*x^3),x)

[Out]

(2*x^2)/(625*(x - exp(x))^2)

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sympy [A]  time = 0.11, size = 22, normalized size = 1.22 \begin {gather*} \frac {2 x^{2}}{625 x^{2} - 1250 x e^{x} + 625 e^{2 x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x**2+4*x)*exp(x)/(625*exp(x)**3-1875*x*exp(x)**2+1875*exp(x)*x**2-625*x**3),x)

[Out]

2*x**2/(625*x**2 - 1250*x*exp(x) + 625*exp(2*x))

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