3.96.94 \(\int \frac {e^{e^{x^2}} (2-2 e^9+e^3 (-6-20 x)+10 x+e^6 (6+10 x)+e^{x^2} (-10 x^3+20 e^3 x^3-10 e^6 x^3)+e^{x^2} (-2 x^2+6 e^3 x^2-6 e^6 x^2+2 e^9 x^2) \log (x))}{-125 x^4+(-75 x^3+75 e^3 x^3) \log (x)+(-15 x^2+30 e^3 x^2-15 e^6 x^2) \log ^2(x)+(-x+3 e^3 x-3 e^6 x+e^9 x) \log ^3(x)} \, dx\)

Optimal. Leaf size=23 \[ \frac {e^{e^{x^2}}}{\left (-\frac {5 x}{-1+e^3}+\log (x)\right )^2} \]

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Rubi [B]  time = 0.92, antiderivative size = 144, normalized size of antiderivative = 6.26, number of steps used = 1, number of rules used = 1, integrand size = 180, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.006, Rules used = {2288} \begin {gather*} \frac {e^{e^{x^2}-x^2} \left (e^{x^2} \left (-e^9 x^2+3 e^6 x^2-3 e^3 x^2+x^2\right ) \log (x)+5 e^{x^2} \left (e^6 x^3-2 e^3 x^3+x^3\right )\right )}{x \left (125 x^4+75 \left (x^3-e^3 x^3\right ) \log (x)+15 \left (e^6 x^2-2 e^3 x^2+x^2\right ) \log ^2(x)+\left (1-e^3\right )^3 x \log ^3(x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^E^x^2*(2 - 2*E^9 + E^3*(-6 - 20*x) + 10*x + E^6*(6 + 10*x) + E^x^2*(-10*x^3 + 20*E^3*x^3 - 10*E^6*x^3)
+ E^x^2*(-2*x^2 + 6*E^3*x^2 - 6*E^6*x^2 + 2*E^9*x^2)*Log[x]))/(-125*x^4 + (-75*x^3 + 75*E^3*x^3)*Log[x] + (-15
*x^2 + 30*E^3*x^2 - 15*E^6*x^2)*Log[x]^2 + (-x + 3*E^3*x - 3*E^6*x + E^9*x)*Log[x]^3),x]

[Out]

(E^(E^x^2 - x^2)*(5*E^x^2*(x^3 - 2*E^3*x^3 + E^6*x^3) + E^x^2*(x^2 - 3*E^3*x^2 + 3*E^6*x^2 - E^9*x^2)*Log[x]))
/(x*(125*x^4 + 75*(x^3 - E^3*x^3)*Log[x] + 15*(x^2 - 2*E^3*x^2 + E^6*x^2)*Log[x]^2 + (1 - E^3)^3*x*Log[x]^3))

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 {e^{e^{x^2}-x^2} \left (5 e^{x^2} \left (x^3-2 e^3 x^3+e^6 x^3\right )+e^{x^2} \left (x^2-3 e^3 x^2+3 e^6 x^2-e^9 x^2\right ) \log (x)\right )}{x \left (125 x^4+75 \left (x^3-e^3 x^3\right ) \log (x)+15 \left (x^2-2 e^3 x^2+e^6 x^2\right ) \log ^2(x)+\left (1-e^3\right )^3 x \log ^3(x)\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.34, size = 30, normalized size = 1.30 \begin {gather*} \frac {e^{e^{x^2}} \left (-1+e^3\right )^2}{\left (5 x+\log (x)-e^3 \log (x)\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^E^x^2*(2 - 2*E^9 + E^3*(-6 - 20*x) + 10*x + E^6*(6 + 10*x) + E^x^2*(-10*x^3 + 20*E^3*x^3 - 10*E^6
*x^3) + E^x^2*(-2*x^2 + 6*E^3*x^2 - 6*E^6*x^2 + 2*E^9*x^2)*Log[x]))/(-125*x^4 + (-75*x^3 + 75*E^3*x^3)*Log[x]
+ (-15*x^2 + 30*E^3*x^2 - 15*E^6*x^2)*Log[x]^2 + (-x + 3*E^3*x - 3*E^6*x + E^9*x)*Log[x]^3),x]

[Out]

(E^E^x^2*(-1 + E^3)^2)/(5*x + Log[x] - E^3*Log[x])^2

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fricas [B]  time = 0.58, size = 47, normalized size = 2.04 \begin {gather*} \frac {{\left (e^{6} - 2 \, e^{3} + 1\right )} e^{\left (e^{\left (x^{2}\right )}\right )}}{{\left (e^{6} - 2 \, e^{3} + 1\right )} \log \relax (x)^{2} + 25 \, x^{2} - 10 \, {\left (x e^{3} - x\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2*exp(3)^3-6*x^2*exp(3)^2+6*x^2*exp(3)-2*x^2)*exp(x^2)*log(x)+(-10*x^3*exp(3)^2+20*x^3*exp(3)-
10*x^3)*exp(x^2)-2*exp(3)^3+(10*x+6)*exp(3)^2+(-20*x-6)*exp(3)+10*x+2)*exp(exp(x^2))/((x*exp(3)^3-3*x*exp(3)^2
+3*x*exp(3)-x)*log(x)^3+(-15*x^2*exp(3)^2+30*x^2*exp(3)-15*x^2)*log(x)^2+(75*x^3*exp(3)-75*x^3)*log(x)-125*x^4
),x, algorithm="fricas")

[Out]

(e^6 - 2*e^3 + 1)*e^(e^(x^2))/((e^6 - 2*e^3 + 1)*log(x)^2 + 25*x^2 - 10*(x*e^3 - x)*log(x))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {2 \, {\left ({\left (x^{2} e^{9} - 3 \, x^{2} e^{6} + 3 \, x^{2} e^{3} - x^{2}\right )} e^{\left (x^{2}\right )} \log \relax (x) + {\left (5 \, x + 3\right )} e^{6} - {\left (10 \, x + 3\right )} e^{3} - 5 \, {\left (x^{3} e^{6} - 2 \, x^{3} e^{3} + x^{3}\right )} e^{\left (x^{2}\right )} + 5 \, x - e^{9} + 1\right )} e^{\left (e^{\left (x^{2}\right )}\right )}}{125 \, x^{4} - {\left (x e^{9} - 3 \, x e^{6} + 3 \, x e^{3} - x\right )} \log \relax (x)^{3} + 15 \, {\left (x^{2} e^{6} - 2 \, x^{2} e^{3} + x^{2}\right )} \log \relax (x)^{2} - 75 \, {\left (x^{3} e^{3} - x^{3}\right )} \log \relax (x)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2*exp(3)^3-6*x^2*exp(3)^2+6*x^2*exp(3)-2*x^2)*exp(x^2)*log(x)+(-10*x^3*exp(3)^2+20*x^3*exp(3)-
10*x^3)*exp(x^2)-2*exp(3)^3+(10*x+6)*exp(3)^2+(-20*x-6)*exp(3)+10*x+2)*exp(exp(x^2))/((x*exp(3)^3-3*x*exp(3)^2
+3*x*exp(3)-x)*log(x)^3+(-15*x^2*exp(3)^2+30*x^2*exp(3)-15*x^2)*log(x)^2+(75*x^3*exp(3)-75*x^3)*log(x)-125*x^4
),x, algorithm="giac")

[Out]

integrate(-2*((x^2*e^9 - 3*x^2*e^6 + 3*x^2*e^3 - x^2)*e^(x^2)*log(x) + (5*x + 3)*e^6 - (10*x + 3)*e^3 - 5*(x^3
*e^6 - 2*x^3*e^3 + x^3)*e^(x^2) + 5*x - e^9 + 1)*e^(e^(x^2))/(125*x^4 - (x*e^9 - 3*x*e^6 + 3*x*e^3 - x)*log(x)
^3 + 15*(x^2*e^6 - 2*x^2*e^3 + x^2)*log(x)^2 - 75*(x^3*e^3 - x^3)*log(x)), x)

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maple [A]  time = 0.10, size = 30, normalized size = 1.30

method	result	size
risch	\(\frac {\left ({\mathrm e}^{6}-2 \,{\mathrm e}^{3}+1\right ) {\mathrm e}^{{\mathrm e}^{x^{2}}}}{\left (\ln \relax (x ) {\mathrm e}^{3}-\ln \relax (x )-5 x \right )^{2}}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^2*exp(3)^3-6*x^2*exp(3)^2+6*x^2*exp(3)-2*x^2)*exp(x^2)*ln(x)+(-10*x^3*exp(3)^2+20*x^3*exp(3)-10*x^3)
*exp(x^2)-2*exp(3)^3+(10*x+6)*exp(3)^2+(-20*x-6)*exp(3)+10*x+2)*exp(exp(x^2))/((x*exp(3)^3-3*x*exp(3)^2+3*x*ex
p(3)-x)*ln(x)^3+(-15*x^2*exp(3)^2+30*x^2*exp(3)-15*x^2)*ln(x)^2+(75*x^3*exp(3)-75*x^3)*ln(x)-125*x^4),x,method
=_RETURNVERBOSE)

[Out]

(exp(6)-2*exp(3)+1)/(ln(x)*exp(3)-ln(x)-5*x)^2*exp(exp(x^2))

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maxima [B]  time = 0.41, size = 46, normalized size = 2.00 \begin {gather*} -\frac {{\left (e^{6} - 2 \, e^{3} + 1\right )} e^{\left (e^{\left (x^{2}\right )}\right )}}{10 \, x {\left (e^{3} - 1\right )} \log \relax (x) - {\left (e^{6} - 2 \, e^{3} + 1\right )} \log \relax (x)^{2} - 25 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2*exp(3)^3-6*x^2*exp(3)^2+6*x^2*exp(3)-2*x^2)*exp(x^2)*log(x)+(-10*x^3*exp(3)^2+20*x^3*exp(3)-
10*x^3)*exp(x^2)-2*exp(3)^3+(10*x+6)*exp(3)^2+(-20*x-6)*exp(3)+10*x+2)*exp(exp(x^2))/((x*exp(3)^3-3*x*exp(3)^2
+3*x*exp(3)-x)*log(x)^3+(-15*x^2*exp(3)^2+30*x^2*exp(3)-15*x^2)*log(x)^2+(75*x^3*exp(3)-75*x^3)*log(x)-125*x^4
),x, algorithm="maxima")

[Out]

-(e^6 - 2*e^3 + 1)*e^(e^(x^2))/(10*x*(e^3 - 1)*log(x) - (e^6 - 2*e^3 + 1)*log(x)^2 - 25*x^2)

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mupad [B]  time = 9.79, size = 26, normalized size = 1.13 \begin {gather*} \frac {{\mathrm {e}}^{{\mathrm {e}}^{x^2}}\,{\left ({\mathrm {e}}^3-1\right )}^2}{{\left (5\,x-\ln \relax (x)\,\left ({\mathrm {e}}^3-1\right )\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp(x^2))*(10*x - 2*exp(9) - exp(x^2)*(10*x^3*exp(6) - 20*x^3*exp(3) + 10*x^3) + exp(6)*(10*x + 6) -
exp(3)*(20*x + 6) + exp(x^2)*log(x)*(6*x^2*exp(3) - 6*x^2*exp(6) + 2*x^2*exp(9) - 2*x^2) + 2))/(log(x)^3*(x -
3*x*exp(3) + 3*x*exp(6) - x*exp(9)) + log(x)^2*(15*x^2*exp(6) - 30*x^2*exp(3) + 15*x^2) - log(x)*(75*x^3*exp(
3) - 75*x^3) + 125*x^4),x)

[Out]

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

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sympy [B]  time = 0.55, size = 61, normalized size = 2.65 \begin {gather*} \frac {\left (- 2 e^{3} + 1 + e^{6}\right ) e^{e^{x^{2}}}}{25 x^{2} - 10 x e^{3} \log {\relax (x )} + 10 x \log {\relax (x )} - 2 e^{3} \log {\relax (x )}^{2} + \log {\relax (x )}^{2} + e^{6} \log {\relax (x )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**2*exp(3)**3-6*x**2*exp(3)**2+6*x**2*exp(3)-2*x**2)*exp(x**2)*ln(x)+(-10*x**3*exp(3)**2+20*x**
3*exp(3)-10*x**3)*exp(x**2)-2*exp(3)**3+(10*x+6)*exp(3)**2+(-20*x-6)*exp(3)+10*x+2)*exp(exp(x**2))/((x*exp(3)*
*3-3*x*exp(3)**2+3*x*exp(3)-x)*ln(x)**3+(-15*x**2*exp(3)**2+30*x**2*exp(3)-15*x**2)*ln(x)**2+(75*x**3*exp(3)-7
5*x**3)*ln(x)-125*x**4),x)

[Out]

(-2*exp(3) + 1 + exp(6))*exp(exp(x**2))/(25*x**2 - 10*x*exp(3)*log(x) + 10*x*log(x) - 2*exp(3)*log(x)**2 + log
(x)**2 + exp(6)*log(x)**2)

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