\(\int \frac {1}{20} e^{\frac {1}{20} (-4 e^{e^{5+x^2}} x^2-5 \log (x))} (15+e^{e^{5+x^2}} (-8 x^2-8 e^{5+x^2} x^4)) \, dx\) [7973]

Optimal result

Integrand size = 59, antiderivative size = 26 \[ \int \frac {1}{20} e^{\frac {1}{20} \left (-4 e^{e^{5+x^2}} x^2-5 \log (x)\right )} \left (15+e^{e^{5+x^2}} \left (-8 x^2-8 e^{5+x^2} x^4\right )\right ) \, dx=-2+e^{-\frac {1}{5} e^{e^{5+x^2}} x^2} x^{3/4} \]

[Out]

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

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(79\) vs. \(2(26)=52\).

Time = 0.13 (sec) , antiderivative size = 79, normalized size of antiderivative = 3.04, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {12, 2306, 2326} \[ \int \frac {1}{20} e^{\frac {1}{20} \left (-4 e^{e^{5+x^2}} x^2-5 \log (x)\right )} \left (15+e^{e^{5+x^2}} \left (-8 x^2-8 e^{5+x^2} x^4\right )\right ) \, dx=\frac {e^{e^{x^2+5}-\frac {1}{5} e^{e^{x^2+5}} x^2} \left (x^2+e^{x^2+5} x^4\right )}{\sqrt [4]{x} \left (e^{e^{x^2+5}} x+e^{x^2+e^{x^2+5}+5} x^3\right )} \]

[In]

Int[(E^((-4*E^E^(5 + x^2)*x^2 - 5*Log[x])/20)*(15 + E^E^(5 + x^2)*(-8*x^2 - 8*E^(5 + x^2)*x^4)))/20,x]

[Out]

(E^(E^(5 + x^2) - (E^E^(5 + x^2)*x^2)/5)*(x^2 + E^(5 + x^2)*x^4))/(x^(1/4)*(E^E^(5 + x^2)*x + E^(5 + E^(5 + x^
2) + 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 2306

Int[(u_.)*(F_)^((a_.)*(Log[z_]*(b_.) + (v_.))), x_Symbol] :> Int[u*F^(a*v)*z^(a*b*Log[F]), x] /; FreeQ[{F, a,
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}{20} \int e^{\frac {1}{20} \left (-4 e^{e^{5+x^2}} x^2-5 \log (x)\right )} \left (15+e^{e^{5+x^2}} \left (-8 x^2-8 e^{5+x^2} x^4\right )\right ) \, dx \\ & = \frac {1}{20} \int \frac {e^{-\frac {1}{5} e^{e^{5+x^2}} x^2} \left (15+e^{e^{5+x^2}} \left (-8 x^2-8 e^{5+x^2} x^4\right )\right )}{\sqrt [4]{x}} \, dx \\ & = \frac {e^{e^{5+x^2}-\frac {1}{5} e^{e^{5+x^2}} x^2} \left (x^2+e^{5+x^2} x^4\right )}{\sqrt [4]{x} \left (e^{e^{5+x^2}} x+e^{5+e^{5+x^2}+x^2} x^3\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {1}{20} e^{\frac {1}{20} \left (-4 e^{e^{5+x^2}} x^2-5 \log (x)\right )} \left (15+e^{e^{5+x^2}} \left (-8 x^2-8 e^{5+x^2} x^4\right )\right ) \, dx=e^{-\frac {1}{5} e^{e^{5+x^2}} x^2} x^{3/4} \]

[In]

Integrate[(E^((-4*E^E^(5 + x^2)*x^2 - 5*Log[x])/20)*(15 + E^E^(5 + x^2)*(-8*x^2 - 8*E^(5 + x^2)*x^4)))/20,x]

[Out]

x^(3/4)/E^((E^E^(5 + x^2)*x^2)/5)

Maple [A] (verified)

Time = 0.82 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.69

method	result	size
risch	\(x^{\frac {3}{4}} {\mathrm e}^{-\frac {x^{2} {\mathrm e}^{{\mathrm e}^{x^{2}+5}}}{5}}\)	\(18\)
parallelrisch	\(x \,{\mathrm e}^{-\frac {x^{2} {\mathrm e}^{{\mathrm e}^{x^{2}+5}}}{5}+\ln \left (\frac {1}{x^{\frac {1}{4}}}\right )}\)	\(23\)

[In]

int(1/20*((-8*x^4*exp(x^2+5)-8*x^2)*exp(exp(x^2+5))+15)/exp(1/5*x^2*exp(exp(x^2+5))+1/4*ln(x)),x,method=_RETUR
NVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {1}{20} e^{\frac {1}{20} \left (-4 e^{e^{5+x^2}} x^2-5 \log (x)\right )} \left (15+e^{e^{5+x^2}} \left (-8 x^2-8 e^{5+x^2} x^4\right )\right ) \, dx=x e^{\left (-\frac {1}{5} \, x^{2} e^{\left (e^{\left (x^{2} + 5\right )}\right )} - \frac {1}{4} \, \log \left (x\right )\right )} \]

[In]

integrate(1/20*((-8*x^4*exp(x^2+5)-8*x^2)*exp(exp(x^2+5))+15)/exp(1/5*x^2*exp(exp(x^2+5))+1/4*log(x)),x, algor
ithm="fricas")

[Out]

x*e^(-1/5*x^2*e^(e^(x^2 + 5)) - 1/4*log(x))

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{20} e^{\frac {1}{20} \left (-4 e^{e^{5+x^2}} x^2-5 \log (x)\right )} \left (15+e^{e^{5+x^2}} \left (-8 x^2-8 e^{5+x^2} x^4\right )\right ) \, dx=\text {Timed out} \]

[In]

integrate(1/20*((-8*x**4*exp(x**2+5)-8*x**2)*exp(exp(x**2+5))+15)/exp(1/5*x**2*exp(exp(x**2+5))+1/4*ln(x)),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {1}{20} e^{\frac {1}{20} \left (-4 e^{e^{5+x^2}} x^2-5 \log (x)\right )} \left (15+e^{e^{5+x^2}} \left (-8 x^2-8 e^{5+x^2} x^4\right )\right ) \, dx=\int { -\frac {1}{20} \, {\left (8 \, {\left (x^{4} e^{\left (x^{2} + 5\right )} + x^{2}\right )} e^{\left (e^{\left (x^{2} + 5\right )}\right )} - 15\right )} e^{\left (-\frac {1}{5} \, x^{2} e^{\left (e^{\left (x^{2} + 5\right )}\right )} - \frac {1}{4} \, \log \left (x\right )\right )} \,d x } \]

[In]

integrate(1/20*((-8*x^4*exp(x^2+5)-8*x^2)*exp(exp(x^2+5))+15)/exp(1/5*x^2*exp(exp(x^2+5))+1/4*log(x)),x, algor
ithm="maxima")

[Out]

(x^(15/4)*e^(x^2 + 5) + x^(7/4))*e^(-1/5*x^2*e^(e^(x^2 + 5)))/(x^3*e^(x^2 + 5) + x) - 1/20*integrate(-5*(3*x^6
*e^(2*x^2 + 10) - (3*x^2*e^(2*x^2 + 10) + (8*x^2*e^5 + 11*e^5)*e^(x^2))*x^4 + 6*x^4*e^(x^2 + 5) + ((8*x^4*e^5
+ 5*x^2*e^5)*e^(x^2) - 3)*x^2 + 3*x^2)*e^(-1/5*x^2*e^(e^(x^2 + 5)))/((x^6*e^(2*x^2 + 10) + 2*x^4*e^(x^2 + 5) +
x^2)*x^(1/4)), x)

Giac [F]

\[ \int \frac {1}{20} e^{\frac {1}{20} \left (-4 e^{e^{5+x^2}} x^2-5 \log (x)\right )} \left (15+e^{e^{5+x^2}} \left (-8 x^2-8 e^{5+x^2} x^4\right )\right ) \, dx=\int { -\frac {1}{20} \, {\left (8 \, {\left (x^{4} e^{\left (x^{2} + 5\right )} + x^{2}\right )} e^{\left (e^{\left (x^{2} + 5\right )}\right )} - 15\right )} e^{\left (-\frac {1}{5} \, x^{2} e^{\left (e^{\left (x^{2} + 5\right )}\right )} - \frac {1}{4} \, \log \left (x\right )\right )} \,d x } \]

[In]

integrate(1/20*((-8*x^4*exp(x^2+5)-8*x^2)*exp(exp(x^2+5))+15)/exp(1/5*x^2*exp(exp(x^2+5))+1/4*log(x)),x, algor
ithm="giac")

[Out]

integrate(-1/20*(8*(x^4*e^(x^2 + 5) + x^2)*e^(e^(x^2 + 5)) - 15)*e^(-1/5*x^2*e^(e^(x^2 + 5)) - 1/4*log(x)), x)

Mupad [B] (verification not implemented)

Time = 12.03 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.69 \[ \int \frac {1}{20} e^{\frac {1}{20} \left (-4 e^{e^{5+x^2}} x^2-5 \log (x)\right )} \left (15+e^{e^{5+x^2}} \left (-8 x^2-8 e^{5+x^2} x^4\right )\right ) \, dx=x^{3/4}\,{\mathrm {e}}^{-\frac {x^2\,{\mathrm {e}}^{{\mathrm {e}}^{x^2}\,{\mathrm {e}}^5}}{5}} \]

[In]

int(-exp(- log(x)/4 - (x^2*exp(exp(x^2 + 5)))/5)*((exp(exp(x^2 + 5))*(8*x^4*exp(x^2 + 5) + 8*x^2))/20 - 3/4),x
)

[Out]

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