3.61.32 \(\int \frac {-3+e^{2 x} (-48-96 x)+6 e^{x^2} x}{3125+125 e^{2 x^2}-1250 x+125 x^2+32000 e^{4 x} x^2+e^{x^2} (1250-250 x-4000 e^{2 x} x)+e^{2 x} (-20000 x+4000 x^2)} \, dx\)

Optimal. Leaf size=28 \[ 5 \left (5-\frac {3}{625 \left (5+e^{x^2}-x-16 e^{2 x} x\right )}\right ) \]

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Rubi [A]  time = 0.23, antiderivative size = 24, normalized size of antiderivative = 0.86, number of steps used = 3, number of rules used = 3, integrand size = 87, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {6688, 12, 6686} \begin {gather*} -\frac {3}{125 \left (e^{x^2}-16 e^{2 x} x-x+5\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-3 + E^(2*x)*(-48 - 96*x) + 6*E^x^2*x)/(3125 + 125*E^(2*x^2) - 1250*x + 125*x^2 + 32000*E^(4*x)*x^2 + E^x
^2*(1250 - 250*x - 4000*E^(2*x)*x) + E^(2*x)*(-20000*x + 4000*x^2)),x]

[Out]

-3/(125*(5 + E^x^2 - x - 16*E^(2*x)*x))

Rule 12

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

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

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 {-3+6 e^{x^2} x-48 e^{2 x} (1+2 x)}{125 \left (5+e^{x^2}-x-16 e^{2 x} x\right )^2} \, dx\\ &=\frac {1}{125} \int \frac {-3+6 e^{x^2} x-48 e^{2 x} (1+2 x)}{\left (5+e^{x^2}-x-16 e^{2 x} x\right )^2} \, dx\\ &=-\frac {3}{125 \left (5+e^{x^2}-x-16 e^{2 x} x\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 24, normalized size = 0.86 \begin {gather*} -\frac {3}{125 \left (5+e^{x^2}-x-16 e^{2 x} x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3 + E^(2*x)*(-48 - 96*x) + 6*E^x^2*x)/(3125 + 125*E^(2*x^2) - 1250*x + 125*x^2 + 32000*E^(4*x)*x^2
+ E^x^2*(1250 - 250*x - 4000*E^(2*x)*x) + E^(2*x)*(-20000*x + 4000*x^2)),x]

[Out]

-3/(125*(5 + E^x^2 - x - 16*E^(2*x)*x))

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fricas [A]  time = 1.06, size = 20, normalized size = 0.71 \begin {gather*} \frac {3}{125 \, {\left (16 \, x e^{\left (2 \, x\right )} + x - e^{\left (x^{2}\right )} - 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*exp(x^2)*x+(-96*x-48)*exp(x)^2-3)/(125*exp(x^2)^2+(-4000*x*exp(x)^2-250*x+1250)*exp(x^2)+32000*x^
2*exp(x)^4+(4000*x^2-20000*x)*exp(x)^2+125*x^2-1250*x+3125),x, algorithm="fricas")

[Out]

3/125/(16*x*e^(2*x) + x - e^(x^2) - 5)

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giac [A]  time = 0.28, size = 20, normalized size = 0.71 \begin {gather*} \frac {3}{125 \, {\left (16 \, x e^{\left (2 \, x\right )} + x - e^{\left (x^{2}\right )} - 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*exp(x^2)*x+(-96*x-48)*exp(x)^2-3)/(125*exp(x^2)^2+(-4000*x*exp(x)^2-250*x+1250)*exp(x^2)+32000*x^
2*exp(x)^4+(4000*x^2-20000*x)*exp(x)^2+125*x^2-1250*x+3125),x, algorithm="giac")

[Out]

3/125/(16*x*e^(2*x) + x - e^(x^2) - 5)

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maple [A]  time = 0.04, size = 21, normalized size = 0.75

method	result	size
risch	\(\frac {3}{125 \left (16 x \,{\mathrm e}^{2 x}+x -{\mathrm e}^{x^{2}}-5\right )}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*exp(x^2)*x+(-96*x-48)*exp(x)^2-3)/(125*exp(x^2)^2+(-4000*x*exp(x)^2-250*x+1250)*exp(x^2)+32000*x^2*exp(
x)^4+(4000*x^2-20000*x)*exp(x)^2+125*x^2-1250*x+3125),x,method=_RETURNVERBOSE)

[Out]

3/125/(16*x*exp(2*x)+x-exp(x^2)-5)

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maxima [A]  time = 0.39, size = 20, normalized size = 0.71 \begin {gather*} \frac {3}{125 \, {\left (16 \, x e^{\left (2 \, x\right )} + x - e^{\left (x^{2}\right )} - 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*exp(x^2)*x+(-96*x-48)*exp(x)^2-3)/(125*exp(x^2)^2+(-4000*x*exp(x)^2-250*x+1250)*exp(x^2)+32000*x^
2*exp(x)^4+(4000*x^2-20000*x)*exp(x)^2+125*x^2-1250*x+3125),x, algorithm="maxima")

[Out]

3/125/(16*x*e^(2*x) + x - e^(x^2) - 5)

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mupad [B]  time = 4.55, size = 22, normalized size = 0.79 \begin {gather*} \frac {3}{125\,\left (x-{\mathrm {e}}^{x^2}+16\,x\,{\mathrm {e}}^{2\,x}-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2*x)*(96*x + 48) - 6*x*exp(x^2) + 3)/(125*exp(2*x^2) - 1250*x - exp(2*x)*(20000*x - 4000*x^2) + 3200
0*x^2*exp(4*x) + 125*x^2 - exp(x^2)*(250*x + 4000*x*exp(2*x) - 1250) + 3125),x)

[Out]

3/(125*(x - exp(x^2) + 16*x*exp(2*x) - 5))

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sympy [A]  time = 0.20, size = 22, normalized size = 0.79 \begin {gather*} - \frac {3}{- 2000 x e^{2 x} - 125 x + 125 e^{x^{2}} + 625} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*exp(x**2)*x+(-96*x-48)*exp(x)**2-3)/(125*exp(x**2)**2+(-4000*x*exp(x)**2-250*x+1250)*exp(x**2)+32
000*x**2*exp(x)**4+(4000*x**2-20000*x)*exp(x)**2+125*x**2-1250*x+3125),x)

[Out]

-3/(-2000*x*exp(2*x) - 125*x + 125*exp(x**2) + 625)

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