3.61.19 \(\int \frac {e^{\frac {1}{9} (36 x-96 x^2+100 x^3-48 x^4+9 x^5)} (12-64 x+100 x^2-64 x^3+15 x^4)}{3-6 e^{\frac {1}{9} (36 x-96 x^2+100 x^3-48 x^4+9 x^5)}+3 e^{\frac {2}{9} (36 x-96 x^2+100 x^3-48 x^4+9 x^5)}} \, dx\)

Optimal. Leaf size=21 \[ \frac {1}{1-e^{x \left (2+\left (-\frac {8}{3}+x\right ) x\right )^2}} \]

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Rubi [A]  time = 3.53, antiderivative size = 38, normalized size of antiderivative = 1.81, number of steps used = 4, number of rules used = 4, integrand size = 119, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {6688, 12, 6711, 32} \begin {gather*} -\frac {1}{1-e^{-x^5-\frac {100 x^3}{9}+\frac {16}{3} \left (x^2+2\right ) x^2-4 x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((36*x - 96*x^2 + 100*x^3 - 48*x^4 + 9*x^5)/9)*(12 - 64*x + 100*x^2 - 64*x^3 + 15*x^4))/(3 - 6*E^((36*x
- 96*x^2 + 100*x^3 - 48*x^4 + 9*x^5)/9) + 3*E^((2*(36*x - 96*x^2 + 100*x^3 - 48*x^4 + 9*x^5))/9)),x]

[Out]

-(1 - E^(-4*x - (100*x^3)/9 - x^5 + (16*x^2*(2 + x^2))/3))^(-1)

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 6688

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

Rule 6711

Int[(u_)*((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*p, Subst[Int[(b + a*x^p)^m, x], x, v*w^(m*q + 1)], x] /; FreeQ[c, x]] /; FreeQ[{a, b, m, p, q}
, x] && EqQ[p + q*(m*p + 1), 0] && IntegerQ[p] && IntegerQ[m]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {1}{9} x \left (6+8 x+3 x^2\right )^2} \left (12-64 x+100 x^2-64 x^3+15 x^4\right )}{3 \left (e^{\frac {16}{3} x^2 \left (2+x^2\right )}-e^{4 x+\frac {100 x^3}{9}+x^5}\right )^2} \, dx\\ &=\frac {1}{3} \int \frac {e^{\frac {1}{9} x \left (6+8 x+3 x^2\right )^2} \left (12-64 x+100 x^2-64 x^3+15 x^4\right )}{\left (e^{\frac {16}{3} x^2 \left (2+x^2\right )}-e^{4 x+\frac {100 x^3}{9}+x^5}\right )^2} \, dx\\ &=-\operatorname {Subst}\left (\int \frac {1}{(-1+x)^2} \, dx,x,e^{-4 x-\frac {100 x^3}{9}-x^5+\frac {16}{3} x^2 \left (2+x^2\right )}\right )\\ &=-\frac {1}{1-e^{-4 x-\frac {100 x^3}{9}-x^5+\frac {16}{3} x^2 \left (2+x^2\right )}}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 1.43, size = 53, normalized size = 2.52 \begin {gather*} -\frac {e^{4 x+\frac {100 x^3}{9}+x^5}}{-e^{\frac {16}{3} x^2 \left (2+x^2\right )}+e^{4 x+\frac {100 x^3}{9}+x^5}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((36*x - 96*x^2 + 100*x^3 - 48*x^4 + 9*x^5)/9)*(12 - 64*x + 100*x^2 - 64*x^3 + 15*x^4))/(3 - 6*E^
((36*x - 96*x^2 + 100*x^3 - 48*x^4 + 9*x^5)/9) + 3*E^((2*(36*x - 96*x^2 + 100*x^3 - 48*x^4 + 9*x^5))/9)),x]

[Out]

-(E^(4*x + (100*x^3)/9 + x^5)/(-E^((16*x^2*(2 + x^2))/3) + E^(4*x + (100*x^3)/9 + x^5)))

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fricas [A]  time = 0.70, size = 29, normalized size = 1.38 \begin {gather*} -\frac {1}{e^{\left (x^{5} - \frac {16}{3} \, x^{4} + \frac {100}{9} \, x^{3} - \frac {32}{3} \, x^{2} + 4 \, x\right )} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((15*x^4-64*x^3+100*x^2-64*x+12)*exp(x^5-16/3*x^4+100/9*x^3-32/3*x^2+4*x)/(3*exp(x^5-16/3*x^4+100/9*x
^3-32/3*x^2+4*x)^2-6*exp(x^5-16/3*x^4+100/9*x^3-32/3*x^2+4*x)+3),x, algorithm="fricas")

[Out]

-1/(e^(x^5 - 16/3*x^4 + 100/9*x^3 - 32/3*x^2 + 4*x) - 1)

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giac [A]  time = 4.98, size = 29, normalized size = 1.38 \begin {gather*} -\frac {1}{e^{\left (x^{5} - \frac {16}{3} \, x^{4} + \frac {100}{9} \, x^{3} - \frac {32}{3} \, x^{2} + 4 \, x\right )} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((15*x^4-64*x^3+100*x^2-64*x+12)*exp(x^5-16/3*x^4+100/9*x^3-32/3*x^2+4*x)/(3*exp(x^5-16/3*x^4+100/9*x
^3-32/3*x^2+4*x)^2-6*exp(x^5-16/3*x^4+100/9*x^3-32/3*x^2+4*x)+3),x, algorithm="giac")

[Out]

-1/(e^(x^5 - 16/3*x^4 + 100/9*x^3 - 32/3*x^2 + 4*x) - 1)

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maple [A]  time = 0.07, size = 23, normalized size = 1.10

method	result	size
risch	\(-\frac {1}{{\mathrm e}^{\frac {x \left (3 x^{2}-8 x +6\right )^{2}}{9}}-1}\)	\(23\)
norman	\(-\frac {1}{{\mathrm e}^{x^{5}-\frac {16}{3} x^{4}+\frac {100}{9} x^{3}-\frac {32}{3} x^{2}+4 x}-1}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((15*x^4-64*x^3+100*x^2-64*x+12)*exp(x^5-16/3*x^4+100/9*x^3-32/3*x^2+4*x)/(3*exp(x^5-16/3*x^4+100/9*x^3-32/
3*x^2+4*x)^2-6*exp(x^5-16/3*x^4+100/9*x^3-32/3*x^2+4*x)+3),x,method=_RETURNVERBOSE)

[Out]

-1/(exp(1/9*x*(3*x^2-8*x+6)^2)-1)

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maxima [B]  time = 0.45, size = 45, normalized size = 2.14 \begin {gather*} -\frac {e^{\left (x^{5} + \frac {100}{9} \, x^{3} + 4 \, x\right )}}{e^{\left (x^{5} + \frac {100}{9} \, x^{3} + 4 \, x\right )} - e^{\left (\frac {16}{3} \, x^{4} + \frac {32}{3} \, x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((15*x^4-64*x^3+100*x^2-64*x+12)*exp(x^5-16/3*x^4+100/9*x^3-32/3*x^2+4*x)/(3*exp(x^5-16/3*x^4+100/9*x
^3-32/3*x^2+4*x)^2-6*exp(x^5-16/3*x^4+100/9*x^3-32/3*x^2+4*x)+3),x, algorithm="maxima")

[Out]

-e^(x^5 + 100/9*x^3 + 4*x)/(e^(x^5 + 100/9*x^3 + 4*x) - e^(16/3*x^4 + 32/3*x^2))

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mupad [B]  time = 4.35, size = 29, normalized size = 1.38 \begin {gather*} -\frac {1}{{\mathrm {e}}^{x^5-\frac {16\,x^4}{3}+\frac {100\,x^3}{9}-\frac {32\,x^2}{3}+4\,x}-1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(4*x - (32*x^2)/3 + (100*x^3)/9 - (16*x^4)/3 + x^5)*(100*x^2 - 64*x - 64*x^3 + 15*x^4 + 12))/(3*exp(8*
x - (64*x^2)/3 + (200*x^3)/9 - (32*x^4)/3 + 2*x^5) - 6*exp(4*x - (32*x^2)/3 + (100*x^3)/9 - (16*x^4)/3 + x^5)
+ 3),x)

[Out]

-1/(exp(4*x - (32*x^2)/3 + (100*x^3)/9 - (16*x^4)/3 + x^5) - 1)

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sympy [B]  time = 0.20, size = 32, normalized size = 1.52 \begin {gather*} - \frac {1}{e^{x^{5} - \frac {16 x^{4}}{3} + \frac {100 x^{3}}{9} - \frac {32 x^{2}}{3} + 4 x} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((15*x**4-64*x**3+100*x**2-64*x+12)*exp(x**5-16/3*x**4+100/9*x**3-32/3*x**2+4*x)/(3*exp(x**5-16/3*x**
4+100/9*x**3-32/3*x**2+4*x)**2-6*exp(x**5-16/3*x**4+100/9*x**3-32/3*x**2+4*x)+3),x)

[Out]

-1/(exp(x**5 - 16*x**4/3 + 100*x**3/9 - 32*x**2/3 + 4*x) - 1)

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