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

Optimal. Leaf size=26 \[ -2+e^{-\frac {1}{5} e^{e^{5+x^2}} x^2} x^{3/4} \]

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Rubi [B]  time = 0.20, antiderivative size = 79, normalized size of antiderivative = 3.04, number of steps used = 3, number of rules used = 3, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {12, 2274, 2288} \begin {gather*} \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 )} \end {gather*}

Antiderivative was successfully verified.

[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 2274

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 2288

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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 24, normalized size = 0.92 \begin {gather*} e^{-\frac {1}{5} e^{e^{5+x^2}} x^2} x^{3/4} \end {gather*}

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.65, size = 20, normalized size = 0.77 \begin {gather*} x e^{\left (-\frac {1}{5} \, x^{2} e^{\left (e^{\left (x^{2} + 5\right )}\right )} - \frac {1}{4} \, \log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 \relax (x)\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.26, size = 18, normalized size = 0.69

method	result	size
risch	\(x^{\frac {3}{4}} {\mathrm e}^{-\frac {x^{2} {\mathrm e}^{{\mathrm e}^{x^{2}+5}}}{5}}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {{\left (x^{\frac {15}{4}} e^{\left (x^{2} + 5\right )} + x^{\frac {7}{4}}\right )} e^{\left (-\frac {1}{5} \, x^{2} e^{\left (e^{\left (x^{2} + 5\right )}\right )}\right )}}{x^{3} e^{\left (x^{2} + 5\right )} + x} - \frac {1}{20} \, \int -\frac {5 \, {\left (3 \, x^{6} e^{\left (2 \, x^{2} + 10\right )} - {\left (3 \, x^{2} e^{\left (2 \, x^{2} + 10\right )} + {\left (8 \, x^{2} e^{5} + 11 \, e^{5}\right )} e^{\left (x^{2}\right )}\right )} x^{4} + 6 \, x^{4} e^{\left (x^{2} + 5\right )} + {\left ({\left (8 \, x^{4} e^{5} + 5 \, x^{2} e^{5}\right )} e^{\left (x^{2}\right )} - 3\right )} x^{2} + 3 \, x^{2}\right )} e^{\left (-\frac {1}{5} \, x^{2} e^{\left (e^{\left (x^{2} + 5\right )}\right )}\right )}}{{\left (x^{6} e^{\left (2 \, x^{2} + 10\right )} + 2 \, x^{4} e^{\left (x^{2} + 5\right )} + x^{2}\right )} x^{\frac {1}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 5.18, size = 18, normalized size = 0.69 \begin {gather*} x^{3/4}\,{\mathrm {e}}^{-\frac {x^2\,{\mathrm {e}}^{{\mathrm {e}}^{x^2}\,{\mathrm {e}}^5}}{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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