\(\int \frac {1}{36} e^{-2 x} x^2 (-\frac {9 e^{2 x}}{x^2}+100 x-50 x^2+\frac {3 e^x (-80+40 x)}{x}) \, dx\) [8359]

Optimal result

Integrand size = 44, antiderivative size = 22 \[ \int \frac {1}{36} e^{-2 x} x^2 \left (-\frac {9 e^{2 x}}{x^2}+100 x-50 x^2+\frac {3 e^x (-80+40 x)}{x}\right ) \, dx=-\frac {x}{4}+\left (2-\frac {5}{6} e^{-x} x^2\right )^2 \]

[Out]

(2-5/2/exp(ln(x)+ln(3/x^2)+x)*x)^2-1/4*x

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.36, number of steps used = 21, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {12, 6820, 2227, 2207, 2225} \[ \int \frac {1}{36} e^{-2 x} x^2 \left (-\frac {9 e^{2 x}}{x^2}+100 x-50 x^2+\frac {3 e^x (-80+40 x)}{x}\right ) \, dx=\frac {25}{36} e^{-2 x} x^4-\frac {10}{3} e^{-x} x^2-\frac {x}{4} \]

[In]

Int[(x^2*((-9*E^(2*x))/x^2 + 100*x - 50*x^2 + (3*E^x*(-80 + 40*x))/x))/(36*E^(2*x)),x]

[Out]

-1/4*x - (10*x^2)/(3*E^x) + (25*x^4)/(36*E^(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 2207

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^m*
((b*F^(g*(e + f*x)))^n/(f*g*n*Log[F])), x] - Dist[d*(m/(f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !TrueQ[$UseGamma]

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2227

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !TrueQ[$UseGamma]

Rule 6820

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{36} \int e^{-2 x} x^2 \left (-\frac {9 e^{2 x}}{x^2}+100 x-50 x^2+\frac {3 e^x (-80+40 x)}{x}\right ) \, dx \\ & = \frac {1}{36} \int \left (-9+120 e^{-x} (-2+x) x-50 e^{-2 x} (-2+x) x^3\right ) \, dx \\ & = -\frac {x}{4}-\frac {25}{18} \int e^{-2 x} (-2+x) x^3 \, dx+\frac {10}{3} \int e^{-x} (-2+x) x \, dx \\ & = -\frac {x}{4}-\frac {25}{18} \int \left (-2 e^{-2 x} x^3+e^{-2 x} x^4\right ) \, dx+\frac {10}{3} \int \left (-2 e^{-x} x+e^{-x} x^2\right ) \, dx \\ & = -\frac {x}{4}-\frac {25}{18} \int e^{-2 x} x^4 \, dx+\frac {25}{9} \int e^{-2 x} x^3 \, dx+\frac {10}{3} \int e^{-x} x^2 \, dx-\frac {20}{3} \int e^{-x} x \, dx \\ & = -\frac {x}{4}+\frac {20 e^{-x} x}{3}-\frac {10}{3} e^{-x} x^2-\frac {25}{18} e^{-2 x} x^3+\frac {25}{36} e^{-2 x} x^4-\frac {25}{9} \int e^{-2 x} x^3 \, dx+\frac {25}{6} \int e^{-2 x} x^2 \, dx-\frac {20}{3} \int e^{-x} \, dx+\frac {20}{3} \int e^{-x} x \, dx \\ & = \frac {20 e^{-x}}{3}-\frac {x}{4}-\frac {25}{12} e^{-2 x} x^2-\frac {10}{3} e^{-x} x^2+\frac {25}{36} e^{-2 x} x^4+\frac {25}{6} \int e^{-2 x} x \, dx-\frac {25}{6} \int e^{-2 x} x^2 \, dx+\frac {20}{3} \int e^{-x} \, dx \\ & = -\frac {x}{4}-\frac {25}{12} e^{-2 x} x-\frac {10}{3} e^{-x} x^2+\frac {25}{36} e^{-2 x} x^4+\frac {25}{12} \int e^{-2 x} \, dx-\frac {25}{6} \int e^{-2 x} x \, dx \\ & = -\frac {25}{24} e^{-2 x}-\frac {x}{4}-\frac {10}{3} e^{-x} x^2+\frac {25}{36} e^{-2 x} x^4-\frac {25}{12} \int e^{-2 x} \, dx \\ & = -\frac {x}{4}-\frac {10}{3} e^{-x} x^2+\frac {25}{36} e^{-2 x} x^4 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27 \[ \int \frac {1}{36} e^{-2 x} x^2 \left (-\frac {9 e^{2 x}}{x^2}+100 x-50 x^2+\frac {3 e^x (-80+40 x)}{x}\right ) \, dx=\frac {1}{36} \left (-9 x-120 e^{-x} x^2+25 e^{-2 x} x^4\right ) \]

[In]

Integrate[(x^2*((-9*E^(2*x))/x^2 + 100*x - 50*x^2 + (3*E^x*(-80 + 40*x))/x))/(36*E^(2*x)),x]

[Out]

(-9*x - (120*x^2)/E^x + (25*x^4)/E^(2*x))/36

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.77

method	result	size
default	\(-\frac {x}{4}-\frac {10 x^{2} {\mathrm e}^{-x}}{3}+\frac {25 x^{4} {\mathrm e}^{-2 x}}{36}\)	\(39\)
parts	\(-\frac {x}{4}-\frac {10 x^{2} {\mathrm e}^{-x}}{3}+\frac {25 x^{4} {\mathrm e}^{-2 x}}{36}\)	\(39\)
parallelrisch	\(\frac {\left (-\frac {18 \,{\mathrm e}^{2 x}}{x}+50 x^{2}-80 \,{\mathrm e}^{\ln \left (x \right )+\ln \left (\frac {3}{x^{2}}\right )+x} x \right ) {\mathrm e}^{-2 x} x^{2}}{72}\)	\(52\)
risch	\(-\frac {x}{4}+\frac {10 x^{2} {\mathrm e}^{-x} {\mathrm e}^{-\frac {i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}}{2}} {\mathrm e}^{-\frac {i \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )}{2}}}{3}+\frac {25 x^{4} {\mathrm e}^{-2 x}}{36}\)	\(56\)

[In]

int(1/4*(-exp(ln(x)+ln(3/x^2)+x)^2+(40*x-80)*exp(ln(x)+ln(3/x^2)+x)-50*x^2+100*x)/exp(ln(x)+ln(3/x^2)+x)^2,x,m
ethod=_RETURNVERBOSE)

[Out]

-1/4*x-10/exp(ln(x)+ln(3/x^2)+x)*x+25/4*x^2/exp(ln(x)+ln(3/x^2)+x)^2

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.68 \[ \int \frac {1}{36} e^{-2 x} x^2 \left (-\frac {9 e^{2 x}}{x^2}+100 x-50 x^2+\frac {3 e^x (-80+40 x)}{x}\right ) \, dx=\frac {1}{4} \, {\left (25 \, x^{2} - x e^{\left (2 \, x + \log \left (3\right ) + \log \left (\frac {3}{x^{2}}\right )\right )} - 40 \, x e^{\left (x + \frac {1}{2} \, \log \left (3\right ) + \frac {1}{2} \, \log \left (\frac {3}{x^{2}}\right )\right )}\right )} e^{\left (-2 \, x - \log \left (3\right ) - \log \left (\frac {3}{x^{2}}\right )\right )} \]

[In]

integrate(1/4*(-exp(log(x)+log(3/x^2)+x)^2+(40*x-80)*exp(log(x)+log(3/x^2)+x)-50*x^2+100*x)/exp(log(x)+log(3/x
^2)+x)^2,x, algorithm="fricas")

[Out]

1/4*(25*x^2 - x*e^(2*x + log(3) + log(3/x^2)) - 40*x*e^(x + 1/2*log(3) + 1/2*log(3/x^2)))*e^(-2*x - log(3) - l
og(3/x^2))

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {1}{36} e^{-2 x} x^2 \left (-\frac {9 e^{2 x}}{x^2}+100 x-50 x^2+\frac {3 e^x (-80+40 x)}{x}\right ) \, dx=\frac {25 x^{4} e^{- 2 x}}{36} - \frac {10 x^{2} e^{- x}}{3} - \frac {x}{4} \]

[In]

integrate(1/4*(-exp(ln(x)+ln(3/x**2)+x)**2+(40*x-80)*exp(ln(x)+ln(3/x**2)+x)-50*x**2+100*x)/exp(ln(x)+ln(3/x**
2)+x)**2,x)

[Out]

25*x**4*exp(-2*x)/36 - 10*x**2*exp(-x)/3 - x/4

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 74 vs. \(2 (30) = 60\).

Time = 0.21 (sec) , antiderivative size = 74, normalized size of antiderivative = 3.36 \[ \int \frac {1}{36} e^{-2 x} x^2 \left (-\frac {9 e^{2 x}}{x^2}+100 x-50 x^2+\frac {3 e^x (-80+40 x)}{x}\right ) \, dx=-\frac {10}{3} \, {\left (x^{2} + 2 \, x + 2\right )} e^{\left (-x\right )} + \frac {20}{3} \, {\left (x + 1\right )} e^{\left (-x\right )} + \frac {25}{72} \, {\left (2 \, x^{4} + 4 \, x^{3} + 6 \, x^{2} + 6 \, x + 3\right )} e^{\left (-2 \, x\right )} - \frac {25}{72} \, {\left (4 \, x^{3} + 6 \, x^{2} + 6 \, x + 3\right )} e^{\left (-2 \, x\right )} - \frac {1}{4} \, x \]

[In]

integrate(1/4*(-exp(log(x)+log(3/x^2)+x)^2+(40*x-80)*exp(log(x)+log(3/x^2)+x)-50*x^2+100*x)/exp(log(x)+log(3/x
^2)+x)^2,x, algorithm="maxima")

[Out]

-10/3*(x^2 + 2*x + 2)*e^(-x) + 20/3*(x + 1)*e^(-x) + 25/72*(2*x^4 + 4*x^3 + 6*x^2 + 6*x + 3)*e^(-2*x) - 25/72*
(4*x^3 + 6*x^2 + 6*x + 3)*e^(-2*x) - 1/4*x

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {1}{36} e^{-2 x} x^2 \left (-\frac {9 e^{2 x}}{x^2}+100 x-50 x^2+\frac {3 e^x (-80+40 x)}{x}\right ) \, dx=\frac {25}{36} \, x^{4} e^{\left (-2 \, x\right )} - \frac {10}{3} \, x^{2} e^{\left (-x\right )} - \frac {1}{4} \, x \]

[In]

integrate(1/4*(-exp(log(x)+log(3/x^2)+x)^2+(40*x-80)*exp(log(x)+log(3/x^2)+x)-50*x^2+100*x)/exp(log(x)+log(3/x
^2)+x)^2,x, algorithm="giac")

[Out]

25/36*x^4*e^(-2*x) - 10/3*x^2*e^(-x) - 1/4*x

Mupad [B] (verification not implemented)

Time = 14.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {1}{36} e^{-2 x} x^2 \left (-\frac {9 e^{2 x}}{x^2}+100 x-50 x^2+\frac {3 e^x (-80+40 x)}{x}\right ) \, dx=-\frac {x\,\left (120\,x\,{\mathrm {e}}^{-x}-25\,x^3\,{\mathrm {e}}^{-2\,x}+9\right )}{36} \]

[In]

int(exp(- 2*x - 2*log(3/x^2) - 2*log(x))*(25*x - exp(2*x + 2*log(3/x^2) + 2*log(x))/4 + (exp(x + log(3/x^2) +
log(x))*(40*x - 80))/4 - (25*x^2)/2),x)

[Out]

-(x*(120*x*exp(-x) - 25*x^3*exp(-2*x) + 9))/36