\(\int \frac {1}{7} e^{\frac {1}{4} (-25+4 x+20 x^2-4 x^4)} (1+10 x-4 x^3) \, dx\) [1744]

Optimal result

Integrand size = 35, antiderivative size = 21 \[ \int \frac {1}{7} e^{\frac {1}{4} \left (-25+4 x+20 x^2-4 x^4\right )} \left (1+10 x-4 x^3\right ) \, dx=\frac {1}{7} e^{x-\left (\frac {5}{2}-x^2\right )^2} \]

[Out]

1/7*exp(-(5/2-x^2)^2+x)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {12, 6838} \[ \int \frac {1}{7} e^{\frac {1}{4} \left (-25+4 x+20 x^2-4 x^4\right )} \left (1+10 x-4 x^3\right ) \, dx=\frac {1}{7} e^{\frac {1}{4} \left (-4 x^4+20 x^2+4 x-25\right )} \]

[In]

Int[(E^((-25 + 4*x + 20*x^2 - 4*x^4)/4)*(1 + 10*x - 4*x^3))/7,x]

[Out]

E^((-25 + 4*x + 20*x^2 - 4*x^4)/4)/7

Rule 12

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

Rule 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{7} \int e^{\frac {1}{4} \left (-25+4 x+20 x^2-4 x^4\right )} \left (1+10 x-4 x^3\right ) \, dx \\ & = \frac {1}{7} e^{\frac {1}{4} \left (-25+4 x+20 x^2-4 x^4\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {1}{7} e^{\frac {1}{4} \left (-25+4 x+20 x^2-4 x^4\right )} \left (1+10 x-4 x^3\right ) \, dx=\frac {1}{7} e^{-\frac {25}{4}+x+5 x^2-x^4} \]

[In]

Integrate[(E^((-25 + 4*x + 20*x^2 - 4*x^4)/4)*(1 + 10*x - 4*x^3))/7,x]

[Out]

E^(-25/4 + x + 5*x^2 - x^4)/7

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81

method	result	size
gosper	\(\frac {{\mathrm e}^{-x^{4}+5 x^{2}+x -\frac {25}{4}}}{7}\)	\(17\)
derivativedivides	\(\frac {{\mathrm e}^{-x^{4}+5 x^{2}+x -\frac {25}{4}}}{7}\)	\(17\)
default	\(\frac {{\mathrm e}^{-x^{4}+5 x^{2}+x -\frac {25}{4}}}{7}\)	\(17\)
norman	\(\frac {{\mathrm e}^{-x^{4}+5 x^{2}+x -\frac {25}{4}}}{7}\)	\(17\)
risch	\(\frac {{\mathrm e}^{-x^{4}+5 x^{2}+x -\frac {25}{4}}}{7}\)	\(17\)
parallelrisch	\(\frac {{\mathrm e}^{-x^{4}+5 x^{2}+x -\frac {25}{4}}}{7}\)	\(17\)

[In]

int(1/7*(-4*x^3+10*x+1)*exp(-x^4+5*x^2+x-25/4),x,method=_RETURNVERBOSE)

[Out]

1/7*exp(-x^4+5*x^2+x-25/4)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \frac {1}{7} e^{\frac {1}{4} \left (-25+4 x+20 x^2-4 x^4\right )} \left (1+10 x-4 x^3\right ) \, dx=\frac {1}{7} \, e^{\left (-x^{4} + 5 \, x^{2} + x - \frac {25}{4}\right )} \]

[In]

integrate(1/7*(-4*x^3+10*x+1)*exp(-x^4+5*x^2+x-25/4),x, algorithm="fricas")

[Out]

1/7*e^(-x^4 + 5*x^2 + x - 25/4)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {1}{7} e^{\frac {1}{4} \left (-25+4 x+20 x^2-4 x^4\right )} \left (1+10 x-4 x^3\right ) \, dx=\frac {e^{- x^{4} + 5 x^{2} + x - \frac {25}{4}}}{7} \]

[In]

integrate(1/7*(-4*x**3+10*x+1)*exp(-x**4+5*x**2+x-25/4),x)

[Out]

exp(-x**4 + 5*x**2 + x - 25/4)/7

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \frac {1}{7} e^{\frac {1}{4} \left (-25+4 x+20 x^2-4 x^4\right )} \left (1+10 x-4 x^3\right ) \, dx=\frac {1}{7} \, e^{\left (-x^{4} + 5 \, x^{2} + x - \frac {25}{4}\right )} \]

[In]

integrate(1/7*(-4*x^3+10*x+1)*exp(-x^4+5*x^2+x-25/4),x, algorithm="maxima")

[Out]

1/7*e^(-x^4 + 5*x^2 + x - 25/4)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \frac {1}{7} e^{\frac {1}{4} \left (-25+4 x+20 x^2-4 x^4\right )} \left (1+10 x-4 x^3\right ) \, dx=\frac {1}{7} \, e^{\left (-x^{4} + 5 \, x^{2} + x - \frac {25}{4}\right )} \]

[In]

integrate(1/7*(-4*x^3+10*x+1)*exp(-x^4+5*x^2+x-25/4),x, algorithm="giac")

[Out]

1/7*e^(-x^4 + 5*x^2 + x - 25/4)

Mupad [B] (verification not implemented)

Time = 8.88 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {1}{7} e^{\frac {1}{4} \left (-25+4 x+20 x^2-4 x^4\right )} \left (1+10 x-4 x^3\right ) \, dx=\frac {{\mathrm {e}}^{-\frac {25}{4}}\,{\mathrm {e}}^{-x^4}\,{\mathrm {e}}^{5\,x^2}\,{\mathrm {e}}^x}{7} \]

[In]

int((exp(x + 5*x^2 - x^4 - 25/4)*(10*x - 4*x^3 + 1))/7,x)

[Out]

(exp(-25/4)*exp(-x^4)*exp(5*x^2)*exp(x))/7