\(\int (10 x+e^{2 e^{\frac {4+3 x-x^5}{x}}} (-2 x+e^{\frac {4+3 x-x^5}{x}} (8+8 x^5))) \, dx\) [7902]

Optimal result

Integrand size = 53, antiderivative size = 29 \[ \int \left (10 x+e^{2 e^{\frac {4+3 x-x^5}{x}}} \left (-2 x+e^{\frac {4+3 x-x^5}{x}} \left (8+8 x^5\right )\right )\right ) \, dx=\left (5-e^{2 e^{-x^4+\frac {4+3 x}{x}}}\right ) x^2 \]

[Out]

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

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(62\) vs. \(2(29)=58\).

Time = 0.04 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.14, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.019, Rules used = {2326} \[ \int \left (10 x+e^{2 e^{\frac {4+3 x-x^5}{x}}} \left (-2 x+e^{\frac {4+3 x-x^5}{x}} \left (8+8 x^5\right )\right )\right ) \, dx=5 x^2+\frac {4 e^{2 e^{\frac {-x^5+3 x+4}{x}}} \left (x^5+1\right )}{\frac {3-5 x^4}{x}-\frac {-x^5+3 x+4}{x^2}} \]

[In]

Int[10*x + E^(2*E^((4 + 3*x - x^5)/x))*(-2*x + E^((4 + 3*x - x^5)/x)*(8 + 8*x^5)),x]

[Out]

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

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}& = 5 x^2+\int e^{2 e^{\frac {4+3 x-x^5}{x}}} \left (-2 x+e^{\frac {4+3 x-x^5}{x}} \left (8+8 x^5\right )\right ) \, dx \\ & = 5 x^2+\frac {4 e^{2 e^{\frac {4+3 x-x^5}{x}}} \left (1+x^5\right )}{\frac {3-5 x^4}{x}-\frac {4+3 x-x^5}{x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \left (10 x+e^{2 e^{\frac {4+3 x-x^5}{x}}} \left (-2 x+e^{\frac {4+3 x-x^5}{x}} \left (8+8 x^5\right )\right )\right ) \, dx=-\left (\left (-5+e^{2 e^{3+\frac {4}{x}-x^4}}\right ) x^2\right ) \]

[In]

Integrate[10*x + E^(2*E^((4 + 3*x - x^5)/x))*(-2*x + E^((4 + 3*x - x^5)/x)*(8 + 8*x^5)),x]

[Out]

-((-5 + E^(2*E^(3 + 4/x - x^4)))*x^2)

Maple [A] (verified)

Time = 3.98 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00

method	result	size
default	\(-x^{2} {\mathrm e}^{2 \,{\mathrm e}^{-\frac {x^{5}-3 x -4}{x}}}+5 x^{2}\)	\(29\)
risch	\(-x^{2} {\mathrm e}^{2 \,{\mathrm e}^{-\frac {x^{5}-3 x -4}{x}}}+5 x^{2}\)	\(29\)
parallelrisch	\(\frac {-{\mathrm e}^{2 \,{\mathrm e}^{\frac {-x^{5}+3 x +4}{x}}} x^{4}+5 x^{4}}{x^{2}}\)	\(37\)

[In]

int(((8*x^5+8)*exp((exp(ln(x)+ln(3))-x^5+4)/x)-2*x)*exp(exp((exp(ln(x)+ln(3))-x^5+4)/x))^2+10*x,x,method=_RETU
RNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \left (10 x+e^{2 e^{\frac {4+3 x-x^5}{x}}} \left (-2 x+e^{\frac {4+3 x-x^5}{x}} \left (8+8 x^5\right )\right )\right ) \, dx=-x^{2} e^{\left (2 \, e^{\left (-\frac {x^{5} - 3 \, x - 4}{x}\right )}\right )} + 5 \, x^{2} \]

[In]

integrate(((8*x^5+8)*exp((exp(log(x)+log(3))-x^5+4)/x)-2*x)*exp(exp((exp(log(x)+log(3))-x^5+4)/x))^2+10*x,x, a
lgorithm="fricas")

[Out]

-x^2*e^(2*e^(-(x^5 - 3*x - 4)/x)) + 5*x^2

Sympy [A] (verification not implemented)

Time = 0.77 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \left (10 x+e^{2 e^{\frac {4+3 x-x^5}{x}}} \left (-2 x+e^{\frac {4+3 x-x^5}{x}} \left (8+8 x^5\right )\right )\right ) \, dx=- x^{2} e^{2 e^{\frac {- x^{5} + 3 x + 4}{x}}} + 5 x^{2} \]

[In]

integrate(((8*x**5+8)*exp((exp(ln(x)+ln(3))-x**5+4)/x)-2*x)*exp(exp((exp(ln(x)+ln(3))-x**5+4)/x))**2+10*x,x)

[Out]

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

Maxima [F]

\[ \int \left (10 x+e^{2 e^{\frac {4+3 x-x^5}{x}}} \left (-2 x+e^{\frac {4+3 x-x^5}{x}} \left (8+8 x^5\right )\right )\right ) \, dx=\int { 2 \, {\left (4 \, {\left (x^{5} + 1\right )} e^{\left (-\frac {x^{5} - e^{\left (\log \left (3\right ) + \log \left (x\right )\right )} - 4}{x}\right )} - x\right )} e^{\left (2 \, e^{\left (-\frac {x^{5} - e^{\left (\log \left (3\right ) + \log \left (x\right )\right )} - 4}{x}\right )}\right )} + 10 \, x \,d x } \]

[In]

integrate(((8*x^5+8)*exp((exp(log(x)+log(3))-x^5+4)/x)-2*x)*exp(exp((exp(log(x)+log(3))-x^5+4)/x))^2+10*x,x, a
lgorithm="maxima")

[Out]

5*x^2 - 2*integrate((x*e^(x^4) - 4*(x^5*e^3 + e^3)*e^(4/x))*e^(-x^4 + 2*e^(-x^4 + 4/x + 3)), x)

Giac [F]

\[ \int \left (10 x+e^{2 e^{\frac {4+3 x-x^5}{x}}} \left (-2 x+e^{\frac {4+3 x-x^5}{x}} \left (8+8 x^5\right )\right )\right ) \, dx=\int { 2 \, {\left (4 \, {\left (x^{5} + 1\right )} e^{\left (-\frac {x^{5} - e^{\left (\log \left (3\right ) + \log \left (x\right )\right )} - 4}{x}\right )} - x\right )} e^{\left (2 \, e^{\left (-\frac {x^{5} - e^{\left (\log \left (3\right ) + \log \left (x\right )\right )} - 4}{x}\right )}\right )} + 10 \, x \,d x } \]

[In]

integrate(((8*x^5+8)*exp((exp(log(x)+log(3))-x^5+4)/x)-2*x)*exp(exp((exp(log(x)+log(3))-x^5+4)/x))^2+10*x,x, a
lgorithm="giac")

[Out]

integrate(2*(4*(x^5 + 1)*e^(-(x^5 - e^(log(3) + log(x)) - 4)/x) - x)*e^(2*e^(-(x^5 - e^(log(3) + log(x)) - 4)/
x)) + 10*x, x)

Mupad [B] (verification not implemented)

Time = 12.69 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \left (10 x+e^{2 e^{\frac {4+3 x-x^5}{x}}} \left (-2 x+e^{\frac {4+3 x-x^5}{x}} \left (8+8 x^5\right )\right )\right ) \, dx=-x^2\,\left ({\mathrm {e}}^{2\,{\mathrm {e}}^3\,{\mathrm {e}}^{4/x}\,{\mathrm {e}}^{-x^4}}-5\right ) \]

[In]

int(10*x - exp(2*exp((exp(log(3) + log(x)) - x^5 + 4)/x))*(2*x - exp((exp(log(3) + log(x)) - x^5 + 4)/x)*(8*x^
5 + 8)),x)

[Out]

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