3.19.5 \(\int \frac {e^{\frac {-18 x+3 x^2+3 \log (3)+(-5 x+x^2) \log (e^{e^x x^2}-x)}{-5 x+x^2}} (-25 x^2+7 x^3-x^4+(-15 x+6 x^2) \log (3)+e^{e^x x^2} (3 x^2+e^x (50 x^3+5 x^4-8 x^5+x^6)+(15-6 x) \log (3)))}{-25 x^3+10 x^4-x^5+e^{e^x x^2} (25 x^2-10 x^3+x^4)} \, dx\)

Optimal. Leaf size=34 \[ e^{3+\frac {3 (-x+\log (3))}{(-5+x) x}} \left (e^{e^x x^2}-x\right ) \]

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Rubi [F]  time = 40.68, 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 {\exp \left (\frac {-18 x+3 x^2+3 \log (3)+\left (-5 x+x^2\right ) \log \left (e^{e^x x^2}-x\right )}{-5 x+x^2}\right ) \left (-25 x^2+7 x^3-x^4+\left (-15 x+6 x^2\right ) \log (3)+e^{e^x x^2} \left (3 x^2+e^x \left (50 x^3+5 x^4-8 x^5+x^6\right )+(15-6 x) \log (3)\right )\right )}{-25 x^3+10 x^4-x^5+e^{e^x x^2} \left (25 x^2-10 x^3+x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((-18*x + 3*x^2 + 3*Log[3] + (-5*x + x^2)*Log[E^(E^x*x^2) - x])/(-5*x + x^2))*(-25*x^2 + 7*x^3 - x^4 +
(-15*x + 6*x^2)*Log[3] + E^(E^x*x^2)*(3*x^2 + E^x*(50*x^3 + 5*x^4 - 8*x^5 + x^6) + (15 - 6*x)*Log[3])))/(-25*x
^3 + 10*x^4 - x^5 + E^(E^x*x^2)*(25*x^2 - 10*x^3 + x^4)),x]

[Out]

(E^((3*(6 - x))/(5 - x) + E^x*x^2)*(5 - 2*x))/(3^(3/((5 - x)*x))*(1/((5 - x)*x^2) - 1/((5 - x)^2*x))*(5 - x)^2
*x^2) - Defer[Int][3^(3/((-5 + x)*x))*E^((3*(-6 + x))/(-5 + x)), x] + Log[3]*Defer[Int][(3^(1 + 3/((-5 + x)*x)
)*E^((3*(-6 + x))/(-5 + x)))/(-5 + x)^2, x] - 15*Defer[Int][(3^(3/((-5 + x)*x))*E^((3*(-6 + x))/(-5 + x)))/(-5
+ x)^2, x] + (Log[3]*Defer[Int][(3^(1 + 3/((-5 + x)*x))*E^((3*(-6 + x))/(-5 + x)))/(-5 + x), x])/5 - 3*Defer[
Int][(3^(3/((-5 + x)*x))*E^((3*(-6 + x))/(-5 + x)))/(-5 + x), x] - (Log[3]*Defer[Int][(3^(1 + 3/((-5 + x)*x))*
E^((3*(-6 + x))/(-5 + x)))/x, x])/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (\frac {-18 x+3 x^2+3 \log (3)+\left (-5 x+x^2\right ) \log \left (e^{e^x x^2}-x\right )}{(-5+x) x}\right ) \left (-25 x^2+7 x^3-x^4+\left (-15 x+6 x^2\right ) \log (3)+e^{e^x x^2} \left (3 x^2+e^x \left (50 x^3+5 x^4-8 x^5+x^6\right )+(15-6 x) \log (3)\right )\right )}{(5-x)^2 \left (e^{e^x x^2}-x\right ) x^2} \, dx\\ &=\int \frac {3^{\frac {3}{(-5+x) x}} e^{\frac {3 (-6+x)}{-5+x}} \left (-25 x^2+7 x^3-x^4+3 x (-5+2 x) \log (3)+e^{e^x x^2} \left (3 x^2+e^x (-5+x)^2 x^3 (2+x)+(15-6 x) \log (3)\right )\right )}{(5-x)^2 x^2} \, dx\\ &=\int \left (-\frac {25\ 3^{\frac {3}{(-5+x) x}} e^{\frac {3 (-6+x)}{-5+x}}}{(-5+x)^2}+\frac {7\ 3^{\frac {3}{(-5+x) x}} e^{\frac {3 (-6+x)}{-5+x}} x}{(-5+x)^2}-\frac {3^{\frac {3}{(-5+x) x}} e^{\frac {3 (-6+x)}{-5+x}} x^2}{(-5+x)^2}+\frac {3^{1+\frac {3}{(-5+x) x}} e^{\frac {3 (-6+x)}{-5+x}} (-5+2 x) \log (3)}{(-5+x)^2 x}+\frac {3^{\frac {3}{(-5+x) x}} e^{\frac {3 (-6+x)}{-5+x}+e^x x^2} \left (3 x^2+50 e^x x^3+5 e^x x^4-8 e^x x^5+e^x x^6+15 \log (3)-6 x \log (3)\right )}{(-5+x)^2 x^2}\right ) \, dx\\ &=7 \int \frac {3^{\frac {3}{(-5+x) x}} e^{\frac {3 (-6+x)}{-5+x}} x}{(-5+x)^2} \, dx-25 \int \frac {3^{\frac {3}{(-5+x) x}} e^{\frac {3 (-6+x)}{-5+x}}}{(-5+x)^2} \, dx+\log (3) \int \frac {3^{1+\frac {3}{(-5+x) x}} e^{\frac {3 (-6+x)}{-5+x}} (-5+2 x)}{(-5+x)^2 x} \, dx-\int \frac {3^{\frac {3}{(-5+x) x}} e^{\frac {3 (-6+x)}{-5+x}} x^2}{(-5+x)^2} \, dx+\int \frac {3^{\frac {3}{(-5+x) x}} e^{\frac {3 (-6+x)}{-5+x}+e^x x^2} \left (3 x^2+50 e^x x^3+5 e^x x^4-8 e^x x^5+e^x x^6+15 \log (3)-6 x \log (3)\right )}{(-5+x)^2 x^2} \, dx\\ &=\frac {3^{-\frac {3}{(5-x) x}} e^{\frac {3 (6-x)}{5-x}+e^x x^2} (5-2 x)}{\left (\frac {1}{(5-x) x^2}-\frac {1}{(5-x)^2 x}\right ) (5-x)^2 x^2}+7 \int \left (\frac {5\ 3^{\frac {3}{(-5+x) x}} e^{\frac {3 (-6+x)}{-5+x}}}{(-5+x)^2}+\frac {3^{\frac {3}{(-5+x) x}} e^{\frac {3 (-6+x)}{-5+x}}}{-5+x}\right ) \, dx-25 \int \frac {3^{\frac {3}{(-5+x) x}} e^{\frac {3 (-6+x)}{-5+x}}}{(-5+x)^2} \, dx+\log (3) \int \left (\frac {3^{1+\frac {3}{(-5+x) x}} e^{\frac {3 (-6+x)}{-5+x}}}{(-5+x)^2}+\frac {3^{1+\frac {3}{(-5+x) x}} e^{\frac {3 (-6+x)}{-5+x}}}{5 (-5+x)}-\frac {3^{1+\frac {3}{(-5+x) x}} e^{\frac {3 (-6+x)}{-5+x}}}{5 x}\right ) \, dx-\int \left (3^{\frac {3}{(-5+x) x}} e^{\frac {3 (-6+x)}{-5+x}}+\frac {25\ 3^{\frac {3}{(-5+x) x}} e^{\frac {3 (-6+x)}{-5+x}}}{(-5+x)^2}+\frac {10\ 3^{\frac {3}{(-5+x) x}} e^{\frac {3 (-6+x)}{-5+x}}}{-5+x}\right ) \, dx\\ &=\frac {3^{-\frac {3}{(5-x) x}} e^{\frac {3 (6-x)}{5-x}+e^x x^2} (5-2 x)}{\left (\frac {1}{(5-x) x^2}-\frac {1}{(5-x)^2 x}\right ) (5-x)^2 x^2}+7 \int \frac {3^{\frac {3}{(-5+x) x}} e^{\frac {3 (-6+x)}{-5+x}}}{-5+x} \, dx-10 \int \frac {3^{\frac {3}{(-5+x) x}} e^{\frac {3 (-6+x)}{-5+x}}}{-5+x} \, dx-2 \left (25 \int \frac {3^{\frac {3}{(-5+x) x}} e^{\frac {3 (-6+x)}{-5+x}}}{(-5+x)^2} \, dx\right )+35 \int \frac {3^{\frac {3}{(-5+x) x}} e^{\frac {3 (-6+x)}{-5+x}}}{(-5+x)^2} \, dx+\frac {1}{5} \log (3) \int \frac {3^{1+\frac {3}{(-5+x) x}} e^{\frac {3 (-6+x)}{-5+x}}}{-5+x} \, dx-\frac {1}{5} \log (3) \int \frac {3^{1+\frac {3}{(-5+x) x}} e^{\frac {3 (-6+x)}{-5+x}}}{x} \, dx+\log (3) \int \frac {3^{1+\frac {3}{(-5+x) x}} e^{\frac {3 (-6+x)}{-5+x}}}{(-5+x)^2} \, dx-\int 3^{\frac {3}{(-5+x) x}} e^{\frac {3 (-6+x)}{-5+x}} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.24, size = 37, normalized size = 1.09 \begin {gather*} 3^{\frac {3}{(-5+x) x}} e^{3-\frac {3}{-5+x}} \left (e^{e^x x^2}-x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-18*x + 3*x^2 + 3*Log[3] + (-5*x + x^2)*Log[E^(E^x*x^2) - x])/(-5*x + x^2))*(-25*x^2 + 7*x^3 -
x^4 + (-15*x + 6*x^2)*Log[3] + E^(E^x*x^2)*(3*x^2 + E^x*(50*x^3 + 5*x^4 - 8*x^5 + x^6) + (15 - 6*x)*Log[3])))/
(-25*x^3 + 10*x^4 - x^5 + E^(E^x*x^2)*(25*x^2 - 10*x^3 + x^4)),x]

[Out]

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

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fricas [A]  time = 0.76, size = 44, normalized size = 1.29 \begin {gather*} e^{\left (\frac {3 \, x^{2} + {\left (x^{2} - 5 \, x\right )} \log \left (-x + e^{\left (x^{2} e^{x}\right )}\right ) - 18 \, x + 3 \, \log \relax (3)}{x^{2} - 5 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^6-8*x^5+5*x^4+50*x^3)*exp(x)+(-6*x+15)*log(3)+3*x^2)*exp(exp(x)*x^2)+(6*x^2-15*x)*log(3)-x^4+7*
x^3-25*x^2)*exp(((x^2-5*x)*log(exp(exp(x)*x^2)-x)+3*log(3)+3*x^2-18*x)/(x^2-5*x))/((x^4-10*x^3+25*x^2)*exp(exp
(x)*x^2)-x^5+10*x^4-25*x^3),x, algorithm="fricas")

[Out]

e^((3*x^2 + (x^2 - 5*x)*log(-x + e^(x^2*e^x)) - 18*x + 3*log(3))/(x^2 - 5*x))

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giac [B]  time = 4.24, size = 90, normalized size = 2.65 \begin {gather*} e^{\left (\frac {x^{2} \log \left (-x + e^{\left (x^{2} e^{x}\right )}\right )}{x^{2} - 5 \, x} + \frac {3 \, x^{2}}{x^{2} - 5 \, x} - \frac {5 \, x \log \left (-x + e^{\left (x^{2} e^{x}\right )}\right )}{x^{2} - 5 \, x} - \frac {18 \, x}{x^{2} - 5 \, x} + \frac {3 \, \log \relax (3)}{x^{2} - 5 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^6-8*x^5+5*x^4+50*x^3)*exp(x)+(-6*x+15)*log(3)+3*x^2)*exp(exp(x)*x^2)+(6*x^2-15*x)*log(3)-x^4+7*
x^3-25*x^2)*exp(((x^2-5*x)*log(exp(exp(x)*x^2)-x)+3*log(3)+3*x^2-18*x)/(x^2-5*x))/((x^4-10*x^3+25*x^2)*exp(exp
(x)*x^2)-x^5+10*x^4-25*x^3),x, algorithm="giac")

[Out]

e^(x^2*log(-x + e^(x^2*e^x))/(x^2 - 5*x) + 3*x^2/(x^2 - 5*x) - 5*x*log(-x + e^(x^2*e^x))/(x^2 - 5*x) - 18*x/(x
^2 - 5*x) + 3*log(3)/(x^2 - 5*x))

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maple [A]  time = 0.13, size = 55, normalized size = 1.62

method	result	size
risch	\({\mathrm e}^{\frac {\ln \left ({\mathrm e}^{{\mathrm e}^{x} x^{2}}-x \right ) x^{2}-5 \ln \left ({\mathrm e}^{{\mathrm e}^{x} x^{2}}-x \right ) x +3 x^{2}+3 \ln \relax (3)-18 x}{\left (x -5\right ) x}}\)	\(55\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x^6-8*x^5+5*x^4+50*x^3)*exp(x)+(-6*x+15)*ln(3)+3*x^2)*exp(exp(x)*x^2)+(6*x^2-15*x)*ln(3)-x^4+7*x^3-25*x
^2)*exp(((x^2-5*x)*ln(exp(exp(x)*x^2)-x)+3*ln(3)+3*x^2-18*x)/(x^2-5*x))/((x^4-10*x^3+25*x^2)*exp(exp(x)*x^2)-x
^5+10*x^4-25*x^3),x,method=_RETURNVERBOSE)

[Out]

exp((ln(exp(exp(x)*x^2)-x)*x^2-5*ln(exp(exp(x)*x^2)-x)*x+3*x^2+3*ln(3)-18*x)/(x-5)/x)

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maxima [A]  time = 0.98, size = 43, normalized size = 1.26 \begin {gather*} -{\left (x e^{3} - e^{\left (x^{2} e^{x} + 3\right )}\right )} e^{\left (\frac {3 \, \log \relax (3)}{5 \, {\left (x - 5\right )}} - \frac {3 \, \log \relax (3)}{5 \, x} - \frac {3}{x - 5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^6-8*x^5+5*x^4+50*x^3)*exp(x)+(-6*x+15)*log(3)+3*x^2)*exp(exp(x)*x^2)+(6*x^2-15*x)*log(3)-x^4+7*
x^3-25*x^2)*exp(((x^2-5*x)*log(exp(exp(x)*x^2)-x)+3*log(3)+3*x^2-18*x)/(x^2-5*x))/((x^4-10*x^3+25*x^2)*exp(exp
(x)*x^2)-x^5+10*x^4-25*x^3),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} -\int \frac {{\mathrm {e}}^{\frac {18\,x-3\,\ln \relax (3)-3\,x^2+\ln \left ({\mathrm {e}}^{x^2\,{\mathrm {e}}^x}-x\right )\,\left (5\,x-x^2\right )}{5\,x-x^2}}\,\left (\ln \relax (3)\,\left (15\,x-6\,x^2\right )-{\mathrm {e}}^{x^2\,{\mathrm {e}}^x}\,\left ({\mathrm {e}}^x\,\left (x^6-8\,x^5+5\,x^4+50\,x^3\right )-\ln \relax (3)\,\left (6\,x-15\right )+3\,x^2\right )+25\,x^2-7\,x^3+x^4\right )}{{\mathrm {e}}^{x^2\,{\mathrm {e}}^x}\,\left (x^4-10\,x^3+25\,x^2\right )-25\,x^3+10\,x^4-x^5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((18*x - 3*log(3) - 3*x^2 + log(exp(x^2*exp(x)) - x)*(5*x - x^2))/(5*x - x^2))*(log(3)*(15*x - 6*x^2)
- exp(x^2*exp(x))*(exp(x)*(50*x^3 + 5*x^4 - 8*x^5 + x^6) - log(3)*(6*x - 15) + 3*x^2) + 25*x^2 - 7*x^3 + x^4)
)/(exp(x^2*exp(x))*(25*x^2 - 10*x^3 + x^4) - 25*x^3 + 10*x^4 - x^5),x)

[Out]

-int((exp((18*x - 3*log(3) - 3*x^2 + log(exp(x^2*exp(x)) - x)*(5*x - x^2))/(5*x - x^2))*(log(3)*(15*x - 6*x^2)
- exp(x^2*exp(x))*(exp(x)*(50*x^3 + 5*x^4 - 8*x^5 + x^6) - log(3)*(6*x - 15) + 3*x^2) + 25*x^2 - 7*x^3 + x^4)
)/(exp(x^2*exp(x))*(25*x^2 - 10*x^3 + x^4) - 25*x^3 + 10*x^4 - x^5), x)

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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((((x**6-8*x**5+5*x**4+50*x**3)*exp(x)+(-6*x+15)*ln(3)+3*x**2)*exp(exp(x)*x**2)+(6*x**2-15*x)*ln(3)-x
**4+7*x**3-25*x**2)*exp(((x**2-5*x)*ln(exp(exp(x)*x**2)-x)+3*ln(3)+3*x**2-18*x)/(x**2-5*x))/((x**4-10*x**3+25*
x**2)*exp(exp(x)*x**2)-x**5+10*x**4-25*x**3),x)

[Out]

Timed out

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