3.11.97 \(\int \frac {e^{-x^2} (e^{e^{e^{-x^2} x}} (e^{x^2} (-15+12 x+3 x^2)+e^{e^{-x^2} x} (-15 x+12 x^2+33 x^3-24 x^4-6 x^5))+(\frac {-5+4 x+x^2}{x})^{5 x/3} (e^{x^2} (25+5 x^2)+e^{x^2} (-25+20 x+5 x^2) \log (\frac {-5+4 x+x^2}{x})))}{-15+12 x+3 x^2} \, dx\)

Optimal. Leaf size=30 \[ e^{e^{e^{-x^2} x}} x+\left (4-\frac {5}{x}+x\right )^{5 x/3} \]

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Rubi [F]  time = 2.63, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {e^{-x^2} \left (e^{e^{e^{-x^2} x}} \left (e^{x^2} \left (-15+12 x+3 x^2\right )+e^{e^{-x^2} x} \left (-15 x+12 x^2+33 x^3-24 x^4-6 x^5\right )\right )+\left (\frac {-5+4 x+x^2}{x}\right )^{5 x/3} \left (e^{x^2} \left (25+5 x^2\right )+e^{x^2} \left (-25+20 x+5 x^2\right ) \log \left (\frac {-5+4 x+x^2}{x}\right )\right )\right )}{-15+12 x+3 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^E^(x/E^x^2)*(E^x^2*(-15 + 12*x + 3*x^2) + E^(x/E^x^2)*(-15*x + 12*x^2 + 33*x^3 - 24*x^4 - 6*x^5)) + ((-
5 + 4*x + x^2)/x)^((5*x)/3)*(E^x^2*(25 + 5*x^2) + E^x^2*(-25 + 20*x + 5*x^2)*Log[(-5 + 4*x + x^2)/x]))/(E^x^2*
(-15 + 12*x + 3*x^2)),x]

[Out]

(4 - 5/x + x)^((5*x)/3) + Defer[Int][E^E^(x/E^x^2), x] + Defer[Int][E^(E^(x/E^x^2) + x/E^x^2 - x^2)*x, x] - 2*
Defer[Int][E^(E^(x/E^x^2) + x/E^x^2 - x^2)*x^3, x] + (5*Defer[Int][(4 - 5/x + x)^((5*x)/3), x])/3 + (5*Defer[I
nt][(4 - 5/x + x)^((5*x)/3)/(-1 + x), x])/3 - (25*Defer[Int][(4 - 5/x + x)^((5*x)/3)/(5 + x), x])/3 - (25*Defe
r[Int][(4 - 5/x + x)^(-1 + (5*x)/3)/x, x])/3 - (5*Defer[Int][x*(4 - 5/x + x)^(-1 + (5*x)/3), x])/3

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (e^{e^{e^{-x^2} x}-x^2} \left (e^{x^2}+e^{e^{-x^2} x} \left (x-2 x^3\right )\right )+\frac {5 \left (4-\frac {5}{x}+x\right )^{5 x/3} \left (5+x^2+\left (-5+4 x+x^2\right ) \log \left (4-\frac {5}{x}+x\right )\right )}{3 \left (-5+4 x+x^2\right )}\right ) \, dx\\ &=\frac {5}{3} \int \frac {\left (4-\frac {5}{x}+x\right )^{5 x/3} \left (5+x^2+\left (-5+4 x+x^2\right ) \log \left (4-\frac {5}{x}+x\right )\right )}{-5+4 x+x^2} \, dx+\int e^{e^{e^{-x^2} x}-x^2} \left (e^{x^2}+e^{e^{-x^2} x} \left (x-2 x^3\right )\right ) \, dx\\ &=\frac {5}{3} \int \left (\frac {\left (4-\frac {5}{x}+x\right )^{5 x/3} \left (5+x^2\right )}{(-1+x) (5+x)}+\left (4-\frac {5}{x}+x\right )^{5 x/3} \log \left (4-\frac {5}{x}+x\right )\right ) \, dx+\int \left (e^{e^{e^{-x^2} x}}-e^{e^{e^{-x^2} x}+e^{-x^2} x-x^2} x \left (-1+2 x^2\right )\right ) \, dx\\ &=\frac {5}{3} \int \frac {\left (4-\frac {5}{x}+x\right )^{5 x/3} \left (5+x^2\right )}{(-1+x) (5+x)} \, dx+\frac {5}{3} \int \left (4-\frac {5}{x}+x\right )^{5 x/3} \log \left (4-\frac {5}{x}+x\right ) \, dx+\int e^{e^{e^{-x^2} x}} \, dx-\int e^{e^{e^{-x^2} x}+e^{-x^2} x-x^2} x \left (-1+2 x^2\right ) \, dx\\ &=\left (4-\frac {5}{x}+x\right )^{5 x/3}-\frac {5}{3} \int \left (1+\frac {5}{x^2}\right ) x \left (4-\frac {5}{x}+x\right )^{-1+\frac {5 x}{3}} \, dx+\frac {5}{3} \int \left (\left (4-\frac {5}{x}+x\right )^{5 x/3}+\frac {\left (4-\frac {5}{x}+x\right )^{5 x/3}}{-1+x}-\frac {5 \left (4-\frac {5}{x}+x\right )^{5 x/3}}{5+x}\right ) \, dx+\int e^{e^{e^{-x^2} x}} \, dx-\int \left (-e^{e^{e^{-x^2} x}+e^{-x^2} x-x^2} x+2 e^{e^{e^{-x^2} x}+e^{-x^2} x-x^2} x^3\right ) \, dx\\ &=\left (4-\frac {5}{x}+x\right )^{5 x/3}+\frac {5}{3} \int \left (4-\frac {5}{x}+x\right )^{5 x/3} \, dx+\frac {5}{3} \int \frac {\left (4-\frac {5}{x}+x\right )^{5 x/3}}{-1+x} \, dx-\frac {5}{3} \int \left (\frac {5 \left (4-\frac {5}{x}+x\right )^{-1+\frac {5 x}{3}}}{x}+x \left (4-\frac {5}{x}+x\right )^{-1+\frac {5 x}{3}}\right ) \, dx-2 \int e^{e^{e^{-x^2} x}+e^{-x^2} x-x^2} x^3 \, dx-\frac {25}{3} \int \frac {\left (4-\frac {5}{x}+x\right )^{5 x/3}}{5+x} \, dx+\int e^{e^{e^{-x^2} x}} \, dx+\int e^{e^{e^{-x^2} x}+e^{-x^2} x-x^2} x \, dx\\ &=\left (4-\frac {5}{x}+x\right )^{5 x/3}+\frac {5}{3} \int \left (4-\frac {5}{x}+x\right )^{5 x/3} \, dx+\frac {5}{3} \int \frac {\left (4-\frac {5}{x}+x\right )^{5 x/3}}{-1+x} \, dx-\frac {5}{3} \int x \left (4-\frac {5}{x}+x\right )^{-1+\frac {5 x}{3}} \, dx-2 \int e^{e^{e^{-x^2} x}+e^{-x^2} x-x^2} x^3 \, dx-\frac {25}{3} \int \frac {\left (4-\frac {5}{x}+x\right )^{5 x/3}}{5+x} \, dx-\frac {25}{3} \int \frac {\left (4-\frac {5}{x}+x\right )^{-1+\frac {5 x}{3}}}{x} \, dx+\int e^{e^{e^{-x^2} x}} \, dx+\int e^{e^{e^{-x^2} x}+e^{-x^2} x-x^2} x \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.47, size = 30, normalized size = 1.00 \begin {gather*} e^{e^{e^{-x^2} x}} x+\left (4-\frac {5}{x}+x\right )^{5 x/3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^E^(x/E^x^2)*(E^x^2*(-15 + 12*x + 3*x^2) + E^(x/E^x^2)*(-15*x + 12*x^2 + 33*x^3 - 24*x^4 - 6*x^5))
+ ((-5 + 4*x + x^2)/x)^((5*x)/3)*(E^x^2*(25 + 5*x^2) + E^x^2*(-25 + 20*x + 5*x^2)*Log[(-5 + 4*x + x^2)/x]))/(
E^x^2*(-15 + 12*x + 3*x^2)),x]

[Out]

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

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fricas [A]  time = 0.73, size = 29, normalized size = 0.97 \begin {gather*} x e^{\left (e^{\left (x e^{\left (-x^{2}\right )}\right )}\right )} + \left (\frac {x^{2} + 4 \, x - 5}{x}\right )^{\frac {5}{3} \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-6*x^5-24*x^4+33*x^3+12*x^2-15*x)*exp(x/exp(x^2))+(3*x^2+12*x-15)*exp(x^2))*exp(exp(x/exp(x^2)))+
((5*x^2+20*x-25)*exp(x^2)*log((x^2+4*x-5)/x)+(5*x^2+25)*exp(x^2))*exp(5/3*x*log((x^2+4*x-5)/x)))/(3*x^2+12*x-1
5)/exp(x^2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (5 \, {\left ({\left (x^{2} + 4 \, x - 5\right )} e^{\left (x^{2}\right )} \log \left (\frac {x^{2} + 4 \, x - 5}{x}\right ) + {\left (x^{2} + 5\right )} e^{\left (x^{2}\right )}\right )} \left (\frac {x^{2} + 4 \, x - 5}{x}\right )^{\frac {5}{3} \, x} + 3 \, {\left ({\left (x^{2} + 4 \, x - 5\right )} e^{\left (x^{2}\right )} - {\left (2 \, x^{5} + 8 \, x^{4} - 11 \, x^{3} - 4 \, x^{2} + 5 \, x\right )} e^{\left (x e^{\left (-x^{2}\right )}\right )}\right )} e^{\left (e^{\left (x e^{\left (-x^{2}\right )}\right )}\right )}\right )} e^{\left (-x^{2}\right )}}{3 \, {\left (x^{2} + 4 \, x - 5\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-6*x^5-24*x^4+33*x^3+12*x^2-15*x)*exp(x/exp(x^2))+(3*x^2+12*x-15)*exp(x^2))*exp(exp(x/exp(x^2)))+
((5*x^2+20*x-25)*exp(x^2)*log((x^2+4*x-5)/x)+(5*x^2+25)*exp(x^2))*exp(5/3*x*log((x^2+4*x-5)/x)))/(3*x^2+12*x-1
5)/exp(x^2),x, algorithm="giac")

[Out]

integrate(1/3*(5*((x^2 + 4*x - 5)*e^(x^2)*log((x^2 + 4*x - 5)/x) + (x^2 + 5)*e^(x^2))*((x^2 + 4*x - 5)/x)^(5/3
*x) + 3*((x^2 + 4*x - 5)*e^(x^2) - (2*x^5 + 8*x^4 - 11*x^3 - 4*x^2 + 5*x)*e^(x*e^(-x^2)))*e^(e^(x*e^(-x^2))))*
e^(-x^2)/(x^2 + 4*x - 5), x)

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maple [C]  time = 0.28, size = 154, normalized size = 5.13

method	result	size
risch	\(x \,{\mathrm e}^{{\mathrm e}^{x \,{\mathrm e}^{-x^{2}}}}+{\mathrm e}^{-\frac {5 x \left (i \pi \mathrm {csgn}\left (\frac {i \left (x^{2}+4 x -5\right )}{x}\right )^{3}-i \pi \mathrm {csgn}\left (\frac {i \left (x^{2}+4 x -5\right )}{x}\right )^{2} \mathrm {csgn}\left (\frac {i}{x}\right )-i \pi \mathrm {csgn}\left (\frac {i \left (x^{2}+4 x -5\right )}{x}\right )^{2} \mathrm {csgn}\left (i \left (x^{2}+4 x -5\right )\right )+i \pi \,\mathrm {csgn}\left (\frac {i \left (x^{2}+4 x -5\right )}{x}\right ) \mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left (x^{2}+4 x -5\right )\right )+2 \ln \relax (x )-2 \ln \left (x^{2}+4 x -5\right )\right )}{6}}\)	\(154\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-6*x^5-24*x^4+33*x^3+12*x^2-15*x)*exp(x/exp(x^2))+(3*x^2+12*x-15)*exp(x^2))*exp(exp(x/exp(x^2)))+((5*x^
2+20*x-25)*exp(x^2)*ln((x^2+4*x-5)/x)+(5*x^2+25)*exp(x^2))*exp(5/3*x*ln((x^2+4*x-5)/x)))/(3*x^2+12*x-15)/exp(x
^2),x,method=_RETURNVERBOSE)

[Out]

x*exp(exp(x*exp(-x^2)))+exp(-5/6*x*(I*Pi*csgn(I/x*(x^2+4*x-5))^3-I*Pi*csgn(I/x*(x^2+4*x-5))^2*csgn(I/x)-I*Pi*c
sgn(I/x*(x^2+4*x-5))^2*csgn(I*(x^2+4*x-5))+I*Pi*csgn(I/x*(x^2+4*x-5))*csgn(I/x)*csgn(I*(x^2+4*x-5))+2*ln(x)-2*
ln(x^2+4*x-5)))

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maxima [A]  time = 0.65, size = 43, normalized size = 1.43 \begin {gather*} \frac {x e^{\left (\frac {5}{3} \, x \log \relax (x) + e^{\left (x e^{\left (-x^{2}\right )}\right )}\right )} + e^{\left (\frac {5}{3} \, x \log \left (x + 5\right ) + \frac {5}{3} \, x \log \left (x - 1\right )\right )}}{x^{\frac {5}{3} \, x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-6*x^5-24*x^4+33*x^3+12*x^2-15*x)*exp(x/exp(x^2))+(3*x^2+12*x-15)*exp(x^2))*exp(exp(x/exp(x^2)))+
((5*x^2+20*x-25)*exp(x^2)*log((x^2+4*x-5)/x)+(5*x^2+25)*exp(x^2))*exp(5/3*x*log((x^2+4*x-5)/x)))/(3*x^2+12*x-1
5)/exp(x^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.68, size = 25, normalized size = 0.83 \begin {gather*} x\,{\mathrm {e}}^{{\mathrm {e}}^{x\,{\mathrm {e}}^{-x^2}}}+{\left (x-\frac {5}{x}+4\right )}^{\frac {5\,x}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-x^2)*(exp(exp(x*exp(-x^2)))*(exp(x^2)*(12*x + 3*x^2 - 15) - exp(x*exp(-x^2))*(15*x - 12*x^2 - 33*x^3
+ 24*x^4 + 6*x^5)) + exp((5*x*log((4*x + x^2 - 5)/x))/3)*(exp(x^2)*(5*x^2 + 25) + exp(x^2)*log((4*x + x^2 - 5
)/x)*(20*x + 5*x^2 - 25))))/(12*x + 3*x^2 - 15),x)

[Out]

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

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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((((-6*x**5-24*x**4+33*x**3+12*x**2-15*x)*exp(x/exp(x**2))+(3*x**2+12*x-15)*exp(x**2))*exp(exp(x/exp(
x**2)))+((5*x**2+20*x-25)*exp(x**2)*ln((x**2+4*x-5)/x)+(5*x**2+25)*exp(x**2))*exp(5/3*x*ln((x**2+4*x-5)/x)))/(
3*x**2+12*x-15)/exp(x**2),x)

[Out]

Timed out

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