\(\int \frac {1}{875} (-500-240 x+e^{\frac {2}{25} (4 x^2-4 x^3+x^4)} (80 x-120 x^2+40 x^3)+e^{\frac {1}{25} (4 x^2-4 x^3+x^4)} (50+16 x^2-24 x^3+8 x^4)) \, dx\) [6762]

Optimal result

Integrand size = 82, antiderivative size = 36 \[ \int \frac {1}{875} \left (-500-240 x+e^{\frac {2}{25} \left (4 x^2-4 x^3+x^4\right )} \left (80 x-120 x^2+40 x^3\right )+e^{\frac {1}{25} \left (4 x^2-4 x^3+x^4\right )} \left (50+16 x^2-24 x^3+8 x^4\right )\right ) \, dx=\frac {1}{7} \left (-4+\frac {\left (e^{\frac {1}{25} (-2+x)^2 x^2}+\frac {x}{5}\right )^2}{x}-x\right ) x \]

[Out]

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

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(89\) vs. \(2(36)=72\).

Time = 0.12 (sec) , antiderivative size = 89, normalized size of antiderivative = 2.47, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {12, 1608, 6838, 2326} \[ \int \frac {1}{875} \left (-500-240 x+e^{\frac {2}{25} \left (4 x^2-4 x^3+x^4\right )} \left (80 x-120 x^2+40 x^3\right )+e^{\frac {1}{25} \left (4 x^2-4 x^3+x^4\right )} \left (50+16 x^2-24 x^3+8 x^4\right )\right ) \, dx=-\frac {24 x^2}{175}+\frac {1}{7} e^{\frac {2}{25} \left (x^4-4 x^3+4 x^2\right )}+\frac {2 e^{\frac {1}{25} \left (x^4-4 x^3+4 x^2\right )} \left (x^4-3 x^3+2 x^2\right )}{35 \left (x^3-3 x^2+2 x\right )}-\frac {4 x}{7} \]

[In]

Int[(-500 - 240*x + E^((2*(4*x^2 - 4*x^3 + x^4))/25)*(80*x - 120*x^2 + 40*x^3) + E^((4*x^2 - 4*x^3 + x^4)/25)*
(50 + 16*x^2 - 24*x^3 + 8*x^4))/875,x]

[Out]

E^((2*(4*x^2 - 4*x^3 + x^4))/25)/7 - (4*x)/7 - (24*x^2)/175 + (2*E^((4*x^2 - 4*x^3 + x^4)/25)*(2*x^2 - 3*x^3 +
x^4))/(35*(2*x - 3*x^2 + 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 1608

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

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]

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}{875} \int \left (-500-240 x+e^{\frac {2}{25} \left (4 x^2-4 x^3+x^4\right )} \left (80 x-120 x^2+40 x^3\right )+e^{\frac {1}{25} \left (4 x^2-4 x^3+x^4\right )} \left (50+16 x^2-24 x^3+8 x^4\right )\right ) \, dx \\ & = -\frac {4 x}{7}-\frac {24 x^2}{175}+\frac {1}{875} \int e^{\frac {2}{25} \left (4 x^2-4 x^3+x^4\right )} \left (80 x-120 x^2+40 x^3\right ) \, dx+\frac {1}{875} \int e^{\frac {1}{25} \left (4 x^2-4 x^3+x^4\right )} \left (50+16 x^2-24 x^3+8 x^4\right ) \, dx \\ & = -\frac {4 x}{7}-\frac {24 x^2}{175}+\frac {2 e^{\frac {1}{25} \left (4 x^2-4 x^3+x^4\right )} \left (2 x^2-3 x^3+x^4\right )}{35 \left (2 x-3 x^2+x^3\right )}+\frac {1}{875} \int e^{\frac {2}{25} \left (4 x^2-4 x^3+x^4\right )} x \left (80-120 x+40 x^2\right ) \, dx \\ & = \frac {1}{7} e^{\frac {2}{25} \left (4 x^2-4 x^3+x^4\right )}-\frac {4 x}{7}-\frac {24 x^2}{175}+\frac {2 e^{\frac {1}{25} \left (4 x^2-4 x^3+x^4\right )} \left (2 x^2-3 x^3+x^4\right )}{35 \left (2 x-3 x^2+x^3\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.95 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.28 \[ \int \frac {1}{875} \left (-500-240 x+e^{\frac {2}{25} \left (4 x^2-4 x^3+x^4\right )} \left (80 x-120 x^2+40 x^3\right )+e^{\frac {1}{25} \left (4 x^2-4 x^3+x^4\right )} \left (50+16 x^2-24 x^3+8 x^4\right )\right ) \, dx=\frac {1}{175} \left (25 e^{\frac {2}{25} (-2+x)^2 x^2}+10 e^{\frac {1}{25} (-2+x)^2 x^2} x-4 x (25+6 x)\right ) \]

[In]

Integrate[(-500 - 240*x + E^((2*(4*x^2 - 4*x^3 + x^4))/25)*(80*x - 120*x^2 + 40*x^3) + E^((4*x^2 - 4*x^3 + x^4
)/25)*(50 + 16*x^2 - 24*x^3 + 8*x^4))/875,x]

[Out]

(25*E^((2*(-2 + x)^2*x^2)/25) + 10*E^(((-2 + x)^2*x^2)/25)*x - 4*x*(25 + 6*x))/175

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.03

method	result	size
risch	\(-\frac {4 x}{7}+\frac {{\mathrm e}^{\frac {2 x^{2} \left (-2+x \right )^{2}}{25}}}{7}+\frac {2 \,{\mathrm e}^{\frac {x^{2} \left (-2+x \right )^{2}}{25}} x}{35}-\frac {24 x^{2}}{175}\)	\(37\)
parallelrisch	\(-\frac {24 x^{2}}{175}+\frac {2 x \,{\mathrm e}^{\frac {x^{2} \left (x^{2}-4 x +4\right )}{25}}}{35}+\frac {{\mathrm e}^{\frac {2 x^{2} \left (x^{2}-4 x +4\right )}{25}}}{7}-\frac {4 x}{7}\)	\(45\)
default	\(-\frac {4 x}{7}+\frac {{\mathrm e}^{\frac {2}{25} x^{4}-\frac {8}{25} x^{3}+\frac {8}{25} x^{2}}}{7}+\frac {2 \,{\mathrm e}^{\frac {1}{25} x^{4}-\frac {4}{25} x^{3}+\frac {4}{25} x^{2}} x}{35}-\frac {24 x^{2}}{175}\)	\(51\)
norman	\(-\frac {4 x}{7}+\frac {{\mathrm e}^{\frac {2}{25} x^{4}-\frac {8}{25} x^{3}+\frac {8}{25} x^{2}}}{7}+\frac {2 \,{\mathrm e}^{\frac {1}{25} x^{4}-\frac {4}{25} x^{3}+\frac {4}{25} x^{2}} x}{35}-\frac {24 x^{2}}{175}\)	\(51\)
parts	\(-\frac {4 x}{7}+\frac {{\mathrm e}^{\frac {2}{25} x^{4}-\frac {8}{25} x^{3}+\frac {8}{25} x^{2}}}{7}+\frac {2 \,{\mathrm e}^{\frac {1}{25} x^{4}-\frac {4}{25} x^{3}+\frac {4}{25} x^{2}} x}{35}-\frac {24 x^{2}}{175}\)	\(51\)

[In]

int(1/875*(40*x^3-120*x^2+80*x)*exp(1/25*x^4-4/25*x^3+4/25*x^2)^2+1/875*(8*x^4-24*x^3+16*x^2+50)*exp(1/25*x^4-
4/25*x^3+4/25*x^2)-48/175*x-4/7,x,method=_RETURNVERBOSE)

[Out]

-4/7*x+1/7*exp(2/25*x^2*(-2+x)^2)+2/35*exp(1/25*x^2*(-2+x)^2)*x-24/175*x^2

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.33 \[ \int \frac {1}{875} \left (-500-240 x+e^{\frac {2}{25} \left (4 x^2-4 x^3+x^4\right )} \left (80 x-120 x^2+40 x^3\right )+e^{\frac {1}{25} \left (4 x^2-4 x^3+x^4\right )} \left (50+16 x^2-24 x^3+8 x^4\right )\right ) \, dx=-\frac {24}{175} \, x^{2} + \frac {2}{35} \, x e^{\left (\frac {1}{25} \, x^{4} - \frac {4}{25} \, x^{3} + \frac {4}{25} \, x^{2}\right )} - \frac {4}{7} \, x + \frac {1}{7} \, e^{\left (\frac {2}{25} \, x^{4} - \frac {8}{25} \, x^{3} + \frac {8}{25} \, x^{2}\right )} \]

[In]

integrate(1/875*(40*x^3-120*x^2+80*x)*exp(1/25*x^4-4/25*x^3+4/25*x^2)^2+1/875*(8*x^4-24*x^3+16*x^2+50)*exp(1/2
5*x^4-4/25*x^3+4/25*x^2)-48/175*x-4/7,x, algorithm="fricas")

[Out]

-24/175*x^2 + 2/35*x*e^(1/25*x^4 - 4/25*x^3 + 4/25*x^2) - 4/7*x + 1/7*e^(2/25*x^4 - 8/25*x^3 + 8/25*x^2)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.67 \[ \int \frac {1}{875} \left (-500-240 x+e^{\frac {2}{25} \left (4 x^2-4 x^3+x^4\right )} \left (80 x-120 x^2+40 x^3\right )+e^{\frac {1}{25} \left (4 x^2-4 x^3+x^4\right )} \left (50+16 x^2-24 x^3+8 x^4\right )\right ) \, dx=- \frac {24 x^{2}}{175} + \frac {2 x e^{\frac {x^{4}}{25} - \frac {4 x^{3}}{25} + \frac {4 x^{2}}{25}}}{35} - \frac {4 x}{7} + \frac {e^{\frac {2 x^{4}}{25} - \frac {8 x^{3}}{25} + \frac {8 x^{2}}{25}}}{7} \]

[In]

integrate(1/875*(40*x**3-120*x**2+80*x)*exp(1/25*x**4-4/25*x**3+4/25*x**2)**2+1/875*(8*x**4-24*x**3+16*x**2+50
)*exp(1/25*x**4-4/25*x**3+4/25*x**2)-48/175*x-4/7,x)

[Out]

-24*x**2/175 + 2*x*exp(x**4/25 - 4*x**3/25 + 4*x**2/25)/35 - 4*x/7 + exp(2*x**4/25 - 8*x**3/25 + 8*x**2/25)/7

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.33 \[ \int \frac {1}{875} \left (-500-240 x+e^{\frac {2}{25} \left (4 x^2-4 x^3+x^4\right )} \left (80 x-120 x^2+40 x^3\right )+e^{\frac {1}{25} \left (4 x^2-4 x^3+x^4\right )} \left (50+16 x^2-24 x^3+8 x^4\right )\right ) \, dx=-\frac {24}{175} \, x^{2} + \frac {2}{35} \, x e^{\left (\frac {1}{25} \, x^{4} - \frac {4}{25} \, x^{3} + \frac {4}{25} \, x^{2}\right )} - \frac {4}{7} \, x + \frac {1}{7} \, e^{\left (\frac {2}{25} \, x^{4} - \frac {8}{25} \, x^{3} + \frac {8}{25} \, x^{2}\right )} \]

[In]

integrate(1/875*(40*x^3-120*x^2+80*x)*exp(1/25*x^4-4/25*x^3+4/25*x^2)^2+1/875*(8*x^4-24*x^3+16*x^2+50)*exp(1/2
5*x^4-4/25*x^3+4/25*x^2)-48/175*x-4/7,x, algorithm="maxima")

[Out]

-24/175*x^2 + 2/35*x*e^(1/25*x^4 - 4/25*x^3 + 4/25*x^2) - 4/7*x + 1/7*e^(2/25*x^4 - 8/25*x^3 + 8/25*x^2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.33 \[ \int \frac {1}{875} \left (-500-240 x+e^{\frac {2}{25} \left (4 x^2-4 x^3+x^4\right )} \left (80 x-120 x^2+40 x^3\right )+e^{\frac {1}{25} \left (4 x^2-4 x^3+x^4\right )} \left (50+16 x^2-24 x^3+8 x^4\right )\right ) \, dx=-\frac {24}{175} \, x^{2} + \frac {2}{35} \, x e^{\left (\frac {1}{25} \, x^{4} - \frac {4}{25} \, x^{3} + \frac {4}{25} \, x^{2}\right )} - \frac {4}{7} \, x + \frac {1}{7} \, e^{\left (\frac {2}{25} \, x^{4} - \frac {8}{25} \, x^{3} + \frac {8}{25} \, x^{2}\right )} \]

[In]

integrate(1/875*(40*x^3-120*x^2+80*x)*exp(1/25*x^4-4/25*x^3+4/25*x^2)^2+1/875*(8*x^4-24*x^3+16*x^2+50)*exp(1/2
5*x^4-4/25*x^3+4/25*x^2)-48/175*x-4/7,x, algorithm="giac")

[Out]

-24/175*x^2 + 2/35*x*e^(1/25*x^4 - 4/25*x^3 + 4/25*x^2) - 4/7*x + 1/7*e^(2/25*x^4 - 8/25*x^3 + 8/25*x^2)

Mupad [B] (verification not implemented)

Time = 11.90 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.33 \[ \int \frac {1}{875} \left (-500-240 x+e^{\frac {2}{25} \left (4 x^2-4 x^3+x^4\right )} \left (80 x-120 x^2+40 x^3\right )+e^{\frac {1}{25} \left (4 x^2-4 x^3+x^4\right )} \left (50+16 x^2-24 x^3+8 x^4\right )\right ) \, dx=\frac {{\mathrm {e}}^{\frac {2\,x^4}{25}-\frac {8\,x^3}{25}+\frac {8\,x^2}{25}}}{7}-\frac {4\,x}{7}+\frac {2\,x\,{\mathrm {e}}^{\frac {x^4}{25}-\frac {4\,x^3}{25}+\frac {4\,x^2}{25}}}{35}-\frac {24\,x^2}{175} \]

[In]

int((exp((8*x^2)/25 - (8*x^3)/25 + (2*x^4)/25)*(80*x - 120*x^2 + 40*x^3))/875 - (48*x)/175 + (exp((4*x^2)/25 -
(4*x^3)/25 + x^4/25)*(16*x^2 - 24*x^3 + 8*x^4 + 50))/875 - 4/7,x)

[Out]

exp((8*x^2)/25 - (8*x^3)/25 + (2*x^4)/25)/7 - (4*x)/7 + (2*x*exp((4*x^2)/25 - (4*x^3)/25 + x^4/25))/35 - (24*x
^2)/175