3.73.83 \(\int \frac {e^{2 x^2} (4000-8000 x^2)+e^{x^2} (2050 x-4100 x^3)}{125 e^{3 x^2}+75 e^{2 x^2} x+15 e^{x^2} x^2+x^3} \, dx\)

Optimal. Leaf size=24 \[ \frac {1}{5} \left (16+\frac {5 x}{e^{x^2}+\frac {x}{5}}\right )^2 \]

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Rubi [A]  time = 0.77, antiderivative size = 33, normalized size of antiderivative = 1.38, number of steps used = 8, number of rules used = 6, integrand size = 67, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.090, Rules used = {6688, 12, 6742, 6712, 32, 6711} \begin {gather*} \frac {160}{\frac {5 e^{x^2}}{x}+1}+\frac {125}{\left (\frac {5 e^{x^2}}{x}+1\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(2*x^2)*(4000 - 8000*x^2) + E^x^2*(2050*x - 4100*x^3))/(125*E^(3*x^2) + 75*E^(2*x^2)*x + 15*E^x^2*x^2 +
x^3),x]

[Out]

125/(1 + (5*E^x^2)/x)^2 + 160/(1 + (5*E^x^2)/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 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]

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]

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {50 e^{x^2} \left (80 e^{x^2}+41 x\right ) \left (1-2 x^2\right )}{\left (5 e^{x^2}+x\right )^3} \, dx\\ &=50 \int \frac {e^{x^2} \left (80 e^{x^2}+41 x\right ) \left (1-2 x^2\right )}{\left (5 e^{x^2}+x\right )^3} \, dx\\ &=50 \int \left (-\frac {25 e^{x^2} x \left (-1+2 x^2\right )}{\left (5 e^{x^2}+x\right )^3}-\frac {16 e^{x^2} \left (-1+2 x^2\right )}{\left (5 e^{x^2}+x\right )^2}\right ) \, dx\\ &=-\left (800 \int \frac {e^{x^2} \left (-1+2 x^2\right )}{\left (5 e^{x^2}+x\right )^2} \, dx\right )-1250 \int \frac {e^{x^2} x \left (-1+2 x^2\right )}{\left (5 e^{x^2}+x\right )^3} \, dx\\ &=-\left (800 \operatorname {Subst}\left (\int \frac {1}{(1+5 x)^2} \, dx,x,\frac {e^{x^2}}{x}\right )\right )-1250 \operatorname {Subst}\left (\int \frac {1}{(1+5 x)^3} \, dx,x,\frac {e^{x^2}}{x}\right )\\ &=\frac {125}{\left (1+\frac {5 e^{x^2}}{x}\right )^2}+\frac {160}{1+\frac {5 e^{x^2}}{x}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.23, size = 25, normalized size = 1.04 \begin {gather*} \frac {5 x \left (160 e^{x^2}+57 x\right )}{\left (5 e^{x^2}+x\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2*x^2)*(4000 - 8000*x^2) + E^x^2*(2050*x - 4100*x^3))/(125*E^(3*x^2) + 75*E^(2*x^2)*x + 15*E^x^2
*x^2 + x^3),x]

[Out]

(5*x*(160*E^x^2 + 57*x))/(5*E^x^2 + x)^2

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fricas [A]  time = 0.71, size = 36, normalized size = 1.50 \begin {gather*} \frac {5 \, {\left (57 \, x^{2} + 160 \, x e^{\left (x^{2}\right )}\right )}}{x^{2} + 10 \, x e^{\left (x^{2}\right )} + 25 \, e^{\left (2 \, x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8000*x^2+4000)*exp(x^2)^2+(-4100*x^3+2050*x)*exp(x^2))/(125*exp(x^2)^3+75*x*exp(x^2)^2+15*x^2*exp
(x^2)+x^3),x, algorithm="fricas")

[Out]

5*(57*x^2 + 160*x*e^(x^2))/(x^2 + 10*x*e^(x^2) + 25*e^(2*x^2))

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giac [A]  time = 0.19, size = 36, normalized size = 1.50 \begin {gather*} \frac {5 \, {\left (57 \, x^{2} + 160 \, x e^{\left (x^{2}\right )}\right )}}{x^{2} + 10 \, x e^{\left (x^{2}\right )} + 25 \, e^{\left (2 \, x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8000*x^2+4000)*exp(x^2)^2+(-4100*x^3+2050*x)*exp(x^2))/(125*exp(x^2)^3+75*x*exp(x^2)^2+15*x^2*exp
(x^2)+x^3),x, algorithm="giac")

[Out]

5*(57*x^2 + 160*x*e^(x^2))/(x^2 + 10*x*e^(x^2) + 25*e^(2*x^2))

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maple [A]  time = 0.05, size = 24, normalized size = 1.00

method	result	size
risch	\(\frac {5 x \left (57 x +160 \,{\mathrm e}^{x^{2}}\right )}{\left (5 \,{\mathrm e}^{x^{2}}+x \right )^{2}}\)	\(24\)
norman	\(\frac {-7125 \,{\mathrm e}^{2 x^{2}}-2050 \,{\mathrm e}^{x^{2}} x}{\left (5 \,{\mathrm e}^{x^{2}}+x \right )^{2}}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-8000*x^2+4000)*exp(x^2)^2+(-4100*x^3+2050*x)*exp(x^2))/(125*exp(x^2)^3+75*x*exp(x^2)^2+15*x^2*exp(x^2)+
x^3),x,method=_RETURNVERBOSE)

[Out]

5*x*(57*x+160*exp(x^2))/(5*exp(x^2)+x)^2

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maxima [A]  time = 0.39, size = 36, normalized size = 1.50 \begin {gather*} \frac {5 \, {\left (57 \, x^{2} + 160 \, x e^{\left (x^{2}\right )}\right )}}{x^{2} + 10 \, x e^{\left (x^{2}\right )} + 25 \, e^{\left (2 \, x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8000*x^2+4000)*exp(x^2)^2+(-4100*x^3+2050*x)*exp(x^2))/(125*exp(x^2)^3+75*x*exp(x^2)^2+15*x^2*exp
(x^2)+x^3),x, algorithm="maxima")

[Out]

5*(57*x^2 + 160*x*e^(x^2))/(x^2 + 10*x*e^(x^2) + 25*e^(2*x^2))

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mupad [B]  time = 0.15, size = 23, normalized size = 0.96 \begin {gather*} \frac {5\,x\,\left (57\,x+160\,{\mathrm {e}}^{x^2}\right )}{{\left (x+5\,{\mathrm {e}}^{x^2}\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x^2)*(2050*x - 4100*x^3) - exp(2*x^2)*(8000*x^2 - 4000))/(125*exp(3*x^2) + 75*x*exp(2*x^2) + 15*x^2*e
xp(x^2) + x^3),x)

[Out]

(5*x*(57*x + 160*exp(x^2)))/(x + 5*exp(x^2))^2

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sympy [B]  time = 0.16, size = 32, normalized size = 1.33 \begin {gather*} \frac {285 x^{2} + 800 x e^{x^{2}}}{x^{2} + 10 x e^{x^{2}} + 25 e^{2 x^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8000*x**2+4000)*exp(x**2)**2+(-4100*x**3+2050*x)*exp(x**2))/(125*exp(x**2)**3+75*x*exp(x**2)**2+1
5*x**2*exp(x**2)+x**3),x)

[Out]

(285*x**2 + 800*x*exp(x**2))/(x**2 + 10*x*exp(x**2) + 25*exp(2*x**2))

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