3.41.32 \(\int \frac {1600 x^4+1250 x^5+300 x^6+25 x^7+e^{\frac {6}{25 x^3}} (-1152-864 x-216 x^2-18 x^3)}{1600 x^4+1200 x^5+300 x^6+25 x^7} \, dx\)

Optimal. Leaf size=24 \[ 1+e^{\frac {6}{25 x^3}}+x+\frac {x^2}{4 (4+x)^2} \]

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Rubi [A]  time = 0.16, antiderivative size = 25, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 3, integrand size = 70, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {6688, 2209, 1850} \begin {gather*} e^{\frac {6}{25 x^3}}+x-\frac {2}{x+4}+\frac {4}{(x+4)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1600*x^4 + 1250*x^5 + 300*x^6 + 25*x^7 + E^(6/(25*x^3))*(-1152 - 864*x - 216*x^2 - 18*x^3))/(1600*x^4 + 1
200*x^5 + 300*x^6 + 25*x^7),x]

[Out]

E^(6/(25*x^3)) + x + 4/(4 + x)^2 - 2/(4 + x)

Rule 1850

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

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 \left (-\frac {18 e^{\frac {6}{25 x^3}}}{25 x^4}+\frac {64+50 x+12 x^2+x^3}{(4+x)^3}\right ) \, dx\\ &=-\left (\frac {18}{25} \int \frac {e^{\frac {6}{25 x^3}}}{x^4} \, dx\right )+\int \frac {64+50 x+12 x^2+x^3}{(4+x)^3} \, dx\\ &=e^{\frac {6}{25 x^3}}+\int \left (1-\frac {8}{(4+x)^3}+\frac {2}{(4+x)^2}\right ) \, dx\\ &=e^{\frac {6}{25 x^3}}+x+\frac {4}{(4+x)^2}-\frac {2}{4+x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 25, normalized size = 1.04 \begin {gather*} e^{\frac {6}{25 x^3}}+x+\frac {4}{(4+x)^2}-\frac {2}{4+x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1600*x^4 + 1250*x^5 + 300*x^6 + 25*x^7 + E^(6/(25*x^3))*(-1152 - 864*x - 216*x^2 - 18*x^3))/(1600*x
^4 + 1200*x^5 + 300*x^6 + 25*x^7),x]

[Out]

E^(6/(25*x^3)) + x + 4/(4 + x)^2 - 2/(4 + x)

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fricas [B]  time = 0.67, size = 39, normalized size = 1.62 \begin {gather*} \frac {x^{3} + 8 \, x^{2} + {\left (x^{2} + 8 \, x + 16\right )} e^{\left (\frac {6}{25 \, x^{3}}\right )} + 14 \, x - 4}{x^{2} + 8 \, x + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-18*x^3-216*x^2-864*x-1152)*exp(6/25/x^3)+25*x^7+300*x^6+1250*x^5+1600*x^4)/(25*x^7+300*x^6+1200*x
^5+1600*x^4),x, algorithm="fricas")

[Out]

(x^3 + 8*x^2 + (x^2 + 8*x + 16)*e^(6/25/x^3) + 14*x - 4)/(x^2 + 8*x + 16)

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giac [A]  time = 0.15, size = 23, normalized size = 0.96 \begin {gather*} x - \frac {2 \, {\left (x + 2\right )}}{x^{2} + 8 \, x + 16} + e^{\left (\frac {6}{25 \, x^{3}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-18*x^3-216*x^2-864*x-1152)*exp(6/25/x^3)+25*x^7+300*x^6+1250*x^5+1600*x^4)/(25*x^7+300*x^6+1200*x
^5+1600*x^4),x, algorithm="giac")

[Out]

x - 2*(x + 2)/(x^2 + 8*x + 16) + e^(6/25/x^3)

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maple [A]  time = 0.29, size = 25, normalized size = 1.04




method result size



risch \(x +\frac {-2 x -4}{x^{2}+8 x +16}+{\mathrm e}^{\frac {6}{25 x^{3}}}\) \(25\)
norman \(\frac {x^{6}-132 x^{3}-50 x^{4}+{\mathrm e}^{\frac {6}{25 x^{3}}} x^{5}+16 \,{\mathrm e}^{\frac {6}{25 x^{3}}} x^{3}+8 \,{\mathrm e}^{\frac {6}{25 x^{3}}} x^{4}}{x^{3} \left (4+x \right )^{2}}\) \(56\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-18*x^3-216*x^2-864*x-1152)*exp(6/25/x^3)+25*x^7+300*x^6+1250*x^5+1600*x^4)/(25*x^7+300*x^6+1200*x^5+160
0*x^4),x,method=_RETURNVERBOSE)

[Out]

x+(-2*x-4)/(x^2+8*x+16)+exp(6/25/x^3)

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maxima [B]  time = 0.40, size = 67, normalized size = 2.79 \begin {gather*} x - \frac {16 \, {\left (3 \, x + 10\right )}}{x^{2} + 8 \, x + 16} + \frac {96 \, {\left (x + 3\right )}}{x^{2} + 8 \, x + 16} - \frac {50 \, {\left (x + 2\right )}}{x^{2} + 8 \, x + 16} - \frac {32}{x^{2} + 8 \, x + 16} + e^{\left (\frac {6}{25 \, x^{3}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-18*x^3-216*x^2-864*x-1152)*exp(6/25/x^3)+25*x^7+300*x^6+1250*x^5+1600*x^4)/(25*x^7+300*x^6+1200*x
^5+1600*x^4),x, algorithm="maxima")

[Out]

x - 16*(3*x + 10)/(x^2 + 8*x + 16) + 96*(x + 3)/(x^2 + 8*x + 16) - 50*(x + 2)/(x^2 + 8*x + 16) - 32/(x^2 + 8*x
 + 16) + e^(6/25/x^3)

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mupad [B]  time = 3.08, size = 20, normalized size = 0.83 \begin {gather*} x+{\mathrm {e}}^{\frac {6}{25\,x^3}}-\frac {2\,x+4}{{\left (x+4\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1600*x^4 - exp(6/(25*x^3))*(864*x + 216*x^2 + 18*x^3 + 1152) + 1250*x^5 + 300*x^6 + 25*x^7)/(1600*x^4 + 1
200*x^5 + 300*x^6 + 25*x^7),x)

[Out]

x + exp(6/(25*x^3)) - (2*x + 4)/(x + 4)^2

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sympy [A]  time = 0.19, size = 24, normalized size = 1.00 \begin {gather*} x + \frac {- 2 x - 4}{x^{2} + 8 x + 16} + e^{\frac {6}{25 x^{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-18*x**3-216*x**2-864*x-1152)*exp(6/25/x**3)+25*x**7+300*x**6+1250*x**5+1600*x**4)/(25*x**7+300*x*
*6+1200*x**5+1600*x**4),x)

[Out]

x + (-2*x - 4)/(x**2 + 8*x + 16) + exp(6/(25*x**3))

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