3.7.30 \(\int \frac {4 x^3+8 x^4-8 x^5+e^{\frac {625}{x^2}} (-3750+15000 x-15000 x^2)}{3 x^3-12 x^4+12 x^5} \, dx\)

Optimal. Leaf size=23 \[ e^{\frac {625}{x^2}}-\frac {2 x}{3}+\frac {2 x}{1-2 x} \]

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Rubi [A]  time = 0.30, antiderivative size = 20, normalized size of antiderivative = 0.87, number of steps used = 8, number of rules used = 6, integrand size = 53, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.113, Rules used = {1594, 27, 12, 6688, 2209, 683} \begin {gather*} e^{\frac {625}{x^2}}-\frac {2 x}{3}+\frac {1}{1-2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4*x^3 + 8*x^4 - 8*x^5 + E^(625/x^2)*(-3750 + 15000*x - 15000*x^2))/(3*x^3 - 12*x^4 + 12*x^5),x]

[Out]

E^(625/x^2) + (1 - 2*x)^(-1) - (2*x)/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 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 683

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e,
0] && IGtQ[p, 0] &&  !(EqQ[m, 3] && NeQ[p, 1])

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 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 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 x^3+8 x^4-8 x^5+e^{\frac {625}{x^2}} \left (-3750+15000 x-15000 x^2\right )}{x^3 \left (3-12 x+12 x^2\right )} \, dx\\ &=\int \frac {4 x^3+8 x^4-8 x^5+e^{\frac {625}{x^2}} \left (-3750+15000 x-15000 x^2\right )}{3 x^3 (-1+2 x)^2} \, dx\\ &=\frac {1}{3} \int \frac {4 x^3+8 x^4-8 x^5+e^{\frac {625}{x^2}} \left (-3750+15000 x-15000 x^2\right )}{x^3 (-1+2 x)^2} \, dx\\ &=\frac {1}{3} \int \left (-\frac {3750 e^{\frac {625}{x^2}}}{x^3}+\frac {4+8 x-8 x^2}{(1-2 x)^2}\right ) \, dx\\ &=\frac {1}{3} \int \frac {4+8 x-8 x^2}{(1-2 x)^2} \, dx-1250 \int \frac {e^{\frac {625}{x^2}}}{x^3} \, dx\\ &=e^{\frac {625}{x^2}}+\frac {1}{3} \int \left (-2+\frac {6}{(-1+2 x)^2}\right ) \, dx\\ &=e^{\frac {625}{x^2}}+\frac {1}{1-2 x}-\frac {2 x}{3}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 20, normalized size = 0.87 \begin {gather*} e^{\frac {625}{x^2}}+\frac {1}{1-2 x}-\frac {2 x}{3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4*x^3 + 8*x^4 - 8*x^5 + E^(625/x^2)*(-3750 + 15000*x - 15000*x^2))/(3*x^3 - 12*x^4 + 12*x^5),x]

[Out]

E^(625/x^2) + (1 - 2*x)^(-1) - (2*x)/3

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fricas [A]  time = 0.69, size = 32, normalized size = 1.39 \begin {gather*} -\frac {4 \, x^{2} - 3 \, {\left (2 \, x - 1\right )} e^{\left (\frac {625}{x^{2}}\right )} - 2 \, x + 3}{3 \, {\left (2 \, x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15000*x^2+15000*x-3750)*exp(625/x^2)-8*x^5+8*x^4+4*x^3)/(12*x^5-12*x^4+3*x^3),x, algorithm="frica
s")

[Out]

-1/3*(4*x^2 - 3*(2*x - 1)*e^(625/x^2) - 2*x + 3)/(2*x - 1)

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giac [A]  time = 0.25, size = 19, normalized size = 0.83 \begin {gather*} -\frac {2}{3} \, x - \frac {1}{2 \, x - 1} + e^{\left (\frac {625}{x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15000*x^2+15000*x-3750)*exp(625/x^2)-8*x^5+8*x^4+4*x^3)/(12*x^5-12*x^4+3*x^3),x, algorithm="giac"
)

[Out]

-2/3*x - 1/(2*x - 1) + e^(625/x^2)

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maple [A]  time = 0.54, size = 18, normalized size = 0.78

method	result	size
risch	\(-\frac {2 x}{3}-\frac {1}{2 \left (x -\frac {1}{2}\right )}+{\mathrm e}^{\frac {625}{x^{2}}}\)	\(18\)
norman	\(\frac {-\frac {4 x^{3}}{3}-\frac {4 x^{4}}{3}-{\mathrm e}^{\frac {625}{x^{2}}} x^{2}+2 \,{\mathrm e}^{\frac {625}{x^{2}}} x^{3}}{x^{2} \left (2 x -1\right )}\)	\(45\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-15000*x^2+15000*x-3750)*exp(625/x^2)-8*x^5+8*x^4+4*x^3)/(12*x^5-12*x^4+3*x^3),x,method=_RETURNVERBOSE)

[Out]

-2/3*x-1/2/(x-1/2)+exp(625/x^2)

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maxima [A]  time = 0.57, size = 19, normalized size = 0.83 \begin {gather*} -\frac {2}{3} \, x - \frac {1}{2 \, x - 1} + e^{\left (\frac {625}{x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15000*x^2+15000*x-3750)*exp(625/x^2)-8*x^5+8*x^4+4*x^3)/(12*x^5-12*x^4+3*x^3),x, algorithm="maxim
a")

[Out]

-2/3*x - 1/(2*x - 1) + e^(625/x^2)

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mupad [B]  time = 0.56, size = 19, normalized size = 0.83 \begin {gather*} {\mathrm {e}}^{\frac {625}{x^2}}-\frac {2\,x}{3}-\frac {1}{2\,x-1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(625/x^2)*(15000*x^2 - 15000*x + 3750) - 4*x^3 - 8*x^4 + 8*x^5)/(3*x^3 - 12*x^4 + 12*x^5),x)

[Out]

exp(625/x^2) - (2*x)/3 - 1/(2*x - 1)

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sympy [A]  time = 0.17, size = 17, normalized size = 0.74 \begin {gather*} - \frac {2 x}{3} + e^{\frac {625}{x^{2}}} - \frac {1}{2 x - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15000*x**2+15000*x-3750)*exp(625/x**2)-8*x**5+8*x**4+4*x**3)/(12*x**5-12*x**4+3*x**3),x)

[Out]

-2*x/3 + exp(625/x**2) - 1/(2*x - 1)

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