\(\int \frac {1}{25} e^{-2+\frac {-50 e^2+25 e^{2+x}+x^4}{25 e^2}} (450 e^2 x+225 e^{2+x} x^2+36 x^5) \, dx\) [537]

Optimal result

Integrand size = 53, antiderivative size = 22 \[ \int \frac {1}{25} e^{-2+\frac {-50 e^2+25 e^{2+x}+x^4}{25 e^2}} \left (450 e^2 x+225 e^{2+x} x^2+36 x^5\right ) \, dx=9 e^{-2+e^x+\frac {x^4}{25 e^2}} x^2 \]

[Out]

9*exp(1/25*x^4/exp(1)^2-2+exp(x))*x^2

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(60\) vs. \(2(22)=44\).

Time = 0.05 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.73, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {12, 2326} \[ \int \frac {1}{25} e^{-2+\frac {-50 e^2+25 e^{2+x}+x^4}{25 e^2}} \left (450 e^2 x+225 e^{2+x} x^2+36 x^5\right ) \, dx=\frac {9 e^{-\frac {-x^4-25 e^{x+2}+50 e^2}{25 e^2}} \left (4 x^5+25 e^{x+2} x^2\right )}{4 x^3+25 e^{x+2}} \]

[In]

Int[(E^(-2 + (-50*E^2 + 25*E^(2 + x) + x^4)/(25*E^2))*(450*E^2*x + 225*E^(2 + x)*x^2 + 36*x^5))/25,x]

[Out]

(9*(25*E^(2 + x)*x^2 + 4*x^5))/(E^((50*E^2 - 25*E^(2 + x) - x^4)/(25*E^2))*(25*E^(2 + x) + 4*x^3))

Rule 12

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

Rule 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{25} \int e^{-2+\frac {-50 e^2+25 e^{2+x}+x^4}{25 e^2}} \left (450 e^2 x+225 e^{2+x} x^2+36 x^5\right ) \, dx \\ & = \frac {9 e^{-\frac {50 e^2-25 e^{2+x}-x^4}{25 e^2}} \left (25 e^{2+x} x^2+4 x^5\right )}{25 e^{2+x}+4 x^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {1}{25} e^{-2+\frac {-50 e^2+25 e^{2+x}+x^4}{25 e^2}} \left (450 e^2 x+225 e^{2+x} x^2+36 x^5\right ) \, dx=9 e^{-2+e^x+\frac {x^4}{25 e^2}} x^2 \]

[In]

Integrate[(E^(-2 + (-50*E^2 + 25*E^(2 + x) + x^4)/(25*E^2))*(450*E^2*x + 225*E^(2 + x)*x^2 + 36*x^5))/25,x]

[Out]

9*E^(-2 + E^x + x^4/(25*E^2))*x^2

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23

method	result	size
risch	\(9 x^{2} {\mathrm e}^{-\frac {\left (-25 \,{\mathrm e}^{2+x}+50 \,{\mathrm e}^{2}-x^{4}\right ) {\mathrm e}^{-2}}{25}}\)	\(27\)
norman	\(9 x^{2} {\mathrm e}^{\frac {\left (25 \,{\mathrm e}^{2} {\mathrm e}^{x}-50 \,{\mathrm e}^{2}+x^{4}\right ) {\mathrm e}^{-2}}{25}}\)	\(31\)
parallelrisch	\(9 x^{2} {\mathrm e}^{\frac {\left (25 \,{\mathrm e}^{2} {\mathrm e}^{x}-50 \,{\mathrm e}^{2}+x^{4}\right ) {\mathrm e}^{-2}}{25}}\)	\(31\)

[In]

int(1/25*(225*x^2*exp(1)^2*exp(x)+450*x*exp(1)^2+36*x^5)*exp(1/25*(25*exp(1)^2*exp(x)-50*exp(1)^2+x^4)/exp(1)^
2)/exp(1)^2,x,method=_RETURNVERBOSE)

[Out]

9*x^2*exp(-1/25*(-25*exp(2+x)+50*exp(2)-x^4)*exp(-2))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {1}{25} e^{-2+\frac {-50 e^2+25 e^{2+x}+x^4}{25 e^2}} \left (450 e^2 x+225 e^{2+x} x^2+36 x^5\right ) \, dx=9 \, x^{2} e^{\left (\frac {1}{25} \, {\left (x^{4} - 100 \, e^{2} + 25 \, e^{\left (x + 2\right )}\right )} e^{\left (-2\right )} + 2\right )} \]

[In]

integrate(1/25*(225*x^2*exp(1)^2*exp(x)+450*x*exp(1)^2+36*x^5)*exp(1/25*(25*exp(1)^2*exp(x)-50*exp(1)^2+x^4)/e
xp(1)^2)/exp(1)^2,x, algorithm="fricas")

[Out]

9*x^2*e^(1/25*(x^4 - 100*e^2 + 25*e^(x + 2))*e^(-2) + 2)

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {1}{25} e^{-2+\frac {-50 e^2+25 e^{2+x}+x^4}{25 e^2}} \left (450 e^2 x+225 e^{2+x} x^2+36 x^5\right ) \, dx=9 x^{2} e^{\frac {\frac {x^{4}}{25} + e^{2} e^{x} - 2 e^{2}}{e^{2}}} \]

[In]

integrate(1/25*(225*x**2*exp(1)**2*exp(x)+450*x*exp(1)**2+36*x**5)*exp(1/25*(25*exp(1)**2*exp(x)-50*exp(1)**2+
x**4)/exp(1)**2)/exp(1)**2,x)

[Out]

9*x**2*exp((x**4/25 + exp(2)*exp(x) - 2*exp(2))*exp(-2))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {1}{25} e^{-2+\frac {-50 e^2+25 e^{2+x}+x^4}{25 e^2}} \left (450 e^2 x+225 e^{2+x} x^2+36 x^5\right ) \, dx=9 \, x^{2} e^{\left (\frac {1}{25} \, x^{4} e^{\left (-2\right )} + e^{x} - 2\right )} \]

[In]

integrate(1/25*(225*x^2*exp(1)^2*exp(x)+450*x*exp(1)^2+36*x^5)*exp(1/25*(25*exp(1)^2*exp(x)-50*exp(1)^2+x^4)/e
xp(1)^2)/exp(1)^2,x, algorithm="maxima")

[Out]

9*x^2*e^(1/25*x^4*e^(-2) + e^x - 2)

Giac [F]

\[ \int \frac {1}{25} e^{-2+\frac {-50 e^2+25 e^{2+x}+x^4}{25 e^2}} \left (450 e^2 x+225 e^{2+x} x^2+36 x^5\right ) \, dx=\int { \frac {9}{25} \, {\left (4 \, x^{5} + 25 \, x^{2} e^{\left (x + 2\right )} + 50 \, x e^{2}\right )} e^{\left (\frac {1}{25} \, {\left (x^{4} - 50 \, e^{2} + 25 \, e^{\left (x + 2\right )}\right )} e^{\left (-2\right )} - 2\right )} \,d x } \]

[In]

integrate(1/25*(225*x^2*exp(1)^2*exp(x)+450*x*exp(1)^2+36*x^5)*exp(1/25*(25*exp(1)^2*exp(x)-50*exp(1)^2+x^4)/e
xp(1)^2)/exp(1)^2,x, algorithm="giac")

[Out]

integrate(9/25*(4*x^5 + 25*x^2*e^(x + 2) + 50*x*e^2)*e^(1/25*(x^4 - 50*e^2 + 25*e^(x + 2))*e^(-2) - 2), x)

Mupad [B] (verification not implemented)

Time = 7.83 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {1}{25} e^{-2+\frac {-50 e^2+25 e^{2+x}+x^4}{25 e^2}} \left (450 e^2 x+225 e^{2+x} x^2+36 x^5\right ) \, dx=9\,x^2\,{\mathrm {e}}^{\frac {x^4\,{\mathrm {e}}^{-2}}{25}}\,{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^{-2} \]

[In]

int((exp(exp(-2)*(exp(2)*exp(x) - 2*exp(2) + x^4/25))*exp(-2)*(450*x*exp(2) + 36*x^5 + 225*x^2*exp(2)*exp(x)))
/25,x)

[Out]

9*x^2*exp((x^4*exp(-2))/25)*exp(exp(x))*exp(-2)