\(\int \frac {e^{2 x^2} (4000-8000 x^2)+e^{x^2} (2050 x-4100 x^3)}{125 e^{3 x^2}+75 e^{2 x^2} x+15 e^{x^2} x^2+x^3} \, dx\) [7283]

Optimal result

Integrand size = 67, antiderivative size = 24 \[ \int \frac {e^{2 x^2} \left (4000-8000 x^2\right )+e^{x^2} \left (2050 x-4100 x^3\right )}{125 e^{3 x^2}+75 e^{2 x^2} x+15 e^{x^2} x^2+x^3} \, dx=\frac {1}{5} \left (16+\frac {5 x}{e^{x^2}+\frac {x}{5}}\right )^2 \]

[Out]

1/5*(16+5*x/(1/5*x+exp(x^2)))^2

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.090, Rules used = {6820, 12, 6874, 6844, 32, 6843} \[ \int \frac {e^{2 x^2} \left (4000-8000 x^2\right )+e^{x^2} \left (2050 x-4100 x^3\right )}{125 e^{3 x^2}+75 e^{2 x^2} x+15 e^{x^2} x^2+x^3} \, dx=\frac {160}{\frac {5 e^{x^2}}{x}+1}+\frac {125}{\left (\frac {5 e^{x^2}}{x}+1\right )^2} \]

[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 6820

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

Rule 6843

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 6844

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]] /; Fre
eQ[{a, b, m, p, q, r}, x] && EqQ[p + q*(m*p + r + 1), 0] && IntegerQ[q] && IntegerQ[m]

Rule 6874

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

Rubi steps \begin{align*} \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 \text {Subst}\left (\int \frac {1}{(1+5 x)^2} \, dx,x,\frac {e^{x^2}}{x}\right )\right )-1250 \text {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{align*}

Mathematica [A] (verified)

Time = 0.61 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {e^{2 x^2} \left (4000-8000 x^2\right )+e^{x^2} \left (2050 x-4100 x^3\right )}{125 e^{3 x^2}+75 e^{2 x^2} x+15 e^{x^2} x^2+x^3} \, dx=\frac {5 x \left (160 e^{x^2}+57 x\right )}{\left (5 e^{x^2}+x\right )^2} \]

[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

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 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	\(\frac {-7125 \,{\mathrm e}^{2 x^{2}}-2050 \,{\mathrm e}^{x^{2}} x}{\left (5 \,{\mathrm e}^{x^{2}}+x \right )^{2}}\)	\(28\)
parallelrisch	\(\frac {7125 x^{2}+20000 \,{\mathrm e}^{x^{2}} x}{25 x^{2}+250 \,{\mathrm e}^{x^{2}} x +625 \,{\mathrm e}^{2 x^{2}}}\)	\(37\)

[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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {e^{2 x^2} \left (4000-8000 x^2\right )+e^{x^2} \left (2050 x-4100 x^3\right )}{125 e^{3 x^2}+75 e^{2 x^2} x+15 e^{x^2} x^2+x^3} \, dx=\frac {5 \, {\left (57 \, x^{2} + 160 \, x e^{\left (x^{2}\right )}\right )}}{x^{2} + 10 \, x e^{\left (x^{2}\right )} + 25 \, e^{\left (2 \, x^{2}\right )}} \]

[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))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 34 vs. \(2 (15) = 30\).

Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {e^{2 x^2} \left (4000-8000 x^2\right )+e^{x^2} \left (2050 x-4100 x^3\right )}{125 e^{3 x^2}+75 e^{2 x^2} x+15 e^{x^2} x^2+x^3} \, dx=\frac {57 x^{2} + 160 x e^{x^{2}}}{\frac {x^{2}}{5} + 2 x e^{x^{2}} + 5 e^{2 x^{2}}} \]

[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]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {e^{2 x^2} \left (4000-8000 x^2\right )+e^{x^2} \left (2050 x-4100 x^3\right )}{125 e^{3 x^2}+75 e^{2 x^2} x+15 e^{x^2} x^2+x^3} \, dx=\frac {5 \, {\left (57 \, x^{2} + 160 \, x e^{\left (x^{2}\right )}\right )}}{x^{2} + 10 \, x e^{\left (x^{2}\right )} + 25 \, e^{\left (2 \, x^{2}\right )}} \]

[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))

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {e^{2 x^2} \left (4000-8000 x^2\right )+e^{x^2} \left (2050 x-4100 x^3\right )}{125 e^{3 x^2}+75 e^{2 x^2} x+15 e^{x^2} x^2+x^3} \, dx=\frac {5 \, {\left (57 \, x^{2} + 160 \, x e^{\left (x^{2}\right )}\right )}}{x^{2} + 10 \, x e^{\left (x^{2}\right )} + 25 \, e^{\left (2 \, x^{2}\right )}} \]

[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))

Mupad [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {e^{2 x^2} \left (4000-8000 x^2\right )+e^{x^2} \left (2050 x-4100 x^3\right )}{125 e^{3 x^2}+75 e^{2 x^2} x+15 e^{x^2} x^2+x^3} \, dx=\frac {5\,x\,\left (57\,x+160\,{\mathrm {e}}^{x^2}\right )}{{\left (x+5\,{\mathrm {e}}^{x^2}\right )}^2} \]

[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