\(\int \frac {5^{\frac {5 x^2}{(-80+25 x) \log (5)+e^x (5 x^3+(3 x^2+5 x^3) \log (5))}} ((-800 x+125 x^2) \log ^2(5)+e^x ((-25 x^4-25 x^5) \log (5)+(-40 x^4-25 x^5) \log ^2(5)))}{(6400-4000 x+625 x^2) \log ^2(5)+e^x ((-800 x^3+250 x^4) \log (5)+(-480 x^2-650 x^3+250 x^4) \log ^2(5))+e^{2 x} (25 x^6+(30 x^5+50 x^6) \log (5)+(9 x^4+30 x^5+25 x^6) \log ^2(5))} \, dx\) [9823]

Optimal result

Integrand size = 198, antiderivative size = 30 \[ \int \frac {5^{\frac {5 x^2}{(-80+25 x) \log (5)+e^x \left (5 x^3+\left (3 x^2+5 x^3\right ) \log (5)\right )}} \left (\left (-800 x+125 x^2\right ) \log ^2(5)+e^x \left (\left (-25 x^4-25 x^5\right ) \log (5)+\left (-40 x^4-25 x^5\right ) \log ^2(5)\right )\right )}{\left (6400-4000 x+625 x^2\right ) \log ^2(5)+e^x \left (\left (-800 x^3+250 x^4\right ) \log (5)+\left (-480 x^2-650 x^3+250 x^4\right ) \log ^2(5)\right )+e^{2 x} \left (25 x^6+\left (30 x^5+50 x^6\right ) \log (5)+\left (9 x^4+30 x^5+25 x^6\right ) \log ^2(5)\right )} \, dx=e^{\frac {x^2}{-16+x \left (5+e^x x \left (\frac {3}{5}+x+\frac {x}{\log (5)}\right )\right )}} \]

[Out]

exp(x^2/((x*(x+x/ln(5)+3/5)*exp(x)+5)*x-16))

Rubi [F(-1)]

Timed out. \[ \int \frac {5^{\frac {5 x^2}{(-80+25 x) \log (5)+e^x \left (5 x^3+\left (3 x^2+5 x^3\right ) \log (5)\right )}} \left (\left (-800 x+125 x^2\right ) \log ^2(5)+e^x \left (\left (-25 x^4-25 x^5\right ) \log (5)+\left (-40 x^4-25 x^5\right ) \log ^2(5)\right )\right )}{\left (6400-4000 x+625 x^2\right ) \log ^2(5)+e^x \left (\left (-800 x^3+250 x^4\right ) \log (5)+\left (-480 x^2-650 x^3+250 x^4\right ) \log ^2(5)\right )+e^{2 x} \left (25 x^6+\left (30 x^5+50 x^6\right ) \log (5)+\left (9 x^4+30 x^5+25 x^6\right ) \log ^2(5)\right )} \, dx=\text {\$Aborted} \]

[In]

Int[(5^((5*x^2)/((-80 + 25*x)*Log[5] + E^x*(5*x^3 + (3*x^2 + 5*x^3)*Log[5])))*((-800*x + 125*x^2)*Log[5]^2 + E
^x*((-25*x^4 - 25*x^5)*Log[5] + (-40*x^4 - 25*x^5)*Log[5]^2)))/((6400 - 4000*x + 625*x^2)*Log[5]^2 + E^x*((-80
0*x^3 + 250*x^4)*Log[5] + (-480*x^2 - 650*x^3 + 250*x^4)*Log[5]^2) + E^(2*x)*(25*x^6 + (30*x^5 + 50*x^6)*Log[5
] + (9*x^4 + 30*x^5 + 25*x^6)*Log[5]^2)),x]

[Out]

$Aborted

Rubi steps Aborted

Mathematica [F]

\[ \int \frac {5^{\frac {5 x^2}{(-80+25 x) \log (5)+e^x \left (5 x^3+\left (3 x^2+5 x^3\right ) \log (5)\right )}} \left (\left (-800 x+125 x^2\right ) \log ^2(5)+e^x \left (\left (-25 x^4-25 x^5\right ) \log (5)+\left (-40 x^4-25 x^5\right ) \log ^2(5)\right )\right )}{\left (6400-4000 x+625 x^2\right ) \log ^2(5)+e^x \left (\left (-800 x^3+250 x^4\right ) \log (5)+\left (-480 x^2-650 x^3+250 x^4\right ) \log ^2(5)\right )+e^{2 x} \left (25 x^6+\left (30 x^5+50 x^6\right ) \log (5)+\left (9 x^4+30 x^5+25 x^6\right ) \log ^2(5)\right )} \, dx=\int \frac {5^{\frac {5 x^2}{(-80+25 x) \log (5)+e^x \left (5 x^3+\left (3 x^2+5 x^3\right ) \log (5)\right )}} \left (\left (-800 x+125 x^2\right ) \log ^2(5)+e^x \left (\left (-25 x^4-25 x^5\right ) \log (5)+\left (-40 x^4-25 x^5\right ) \log ^2(5)\right )\right )}{\left (6400-4000 x+625 x^2\right ) \log ^2(5)+e^x \left (\left (-800 x^3+250 x^4\right ) \log (5)+\left (-480 x^2-650 x^3+250 x^4\right ) \log ^2(5)\right )+e^{2 x} \left (25 x^6+\left (30 x^5+50 x^6\right ) \log (5)+\left (9 x^4+30 x^5+25 x^6\right ) \log ^2(5)\right )} \, dx \]

[In]

Integrate[(5^((5*x^2)/((-80 + 25*x)*Log[5] + E^x*(5*x^3 + (3*x^2 + 5*x^3)*Log[5])))*((-800*x + 125*x^2)*Log[5]
^2 + E^x*((-25*x^4 - 25*x^5)*Log[5] + (-40*x^4 - 25*x^5)*Log[5]^2)))/((6400 - 4000*x + 625*x^2)*Log[5]^2 + E^x
*((-800*x^3 + 250*x^4)*Log[5] + (-480*x^2 - 650*x^3 + 250*x^4)*Log[5]^2) + E^(2*x)*(25*x^6 + (30*x^5 + 50*x^6)
*Log[5] + (9*x^4 + 30*x^5 + 25*x^6)*Log[5]^2)),x]

[Out]

Integrate[(5^((5*x^2)/((-80 + 25*x)*Log[5] + E^x*(5*x^3 + (3*x^2 + 5*x^3)*Log[5])))*((-800*x + 125*x^2)*Log[5]
^2 + E^x*((-25*x^4 - 25*x^5)*Log[5] + (-40*x^4 - 25*x^5)*Log[5]^2)))/((6400 - 4000*x + 625*x^2)*Log[5]^2 + E^x
*((-800*x^3 + 250*x^4)*Log[5] + (-480*x^2 - 650*x^3 + 250*x^4)*Log[5]^2) + E^(2*x)*(25*x^6 + (30*x^5 + 50*x^6)
*Log[5] + (9*x^4 + 30*x^5 + 25*x^6)*Log[5]^2)), x]

Maple [A] (verified)

Time = 60.96 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.47

method	result	size
risch	\(3125^{\frac {x^{2}}{5 \ln \left (5\right ) {\mathrm e}^{x} x^{3}+3 x^{2} \ln \left (5\right ) {\mathrm e}^{x}+5 \,{\mathrm e}^{x} x^{3}+25 x \ln \left (5\right )-80 \ln \left (5\right )}}\)	\(44\)
parallelrisch	\({\mathrm e}^{\frac {5 x^{2} \ln \left (5\right )}{5 \ln \left (5\right ) {\mathrm e}^{x} x^{3}+3 x^{2} \ln \left (5\right ) {\mathrm e}^{x}+5 \,{\mathrm e}^{x} x^{3}+25 x \ln \left (5\right )-80 \ln \left (5\right )}}\)	\(46\)

[In]

int((((-25*x^5-40*x^4)*ln(5)^2+(-25*x^5-25*x^4)*ln(5))*exp(x)+(125*x^2-800*x)*ln(5)^2)*exp(5*x^2*ln(5)/(((5*x^
3+3*x^2)*ln(5)+5*x^3)*exp(x)+(25*x-80)*ln(5)))/(((25*x^6+30*x^5+9*x^4)*ln(5)^2+(50*x^6+30*x^5)*ln(5)+25*x^6)*e
xp(x)^2+((250*x^4-650*x^3-480*x^2)*ln(5)^2+(250*x^4-800*x^3)*ln(5))*exp(x)+(625*x^2-4000*x+6400)*ln(5)^2),x,me
thod=_RETURNVERBOSE)

[Out]

3125^(x^2/(5*ln(5)*exp(x)*x^3+3*x^2*ln(5)*exp(x)+5*exp(x)*x^3+25*x*ln(5)-80*ln(5)))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.40 \[ \int \frac {5^{\frac {5 x^2}{(-80+25 x) \log (5)+e^x \left (5 x^3+\left (3 x^2+5 x^3\right ) \log (5)\right )}} \left (\left (-800 x+125 x^2\right ) \log ^2(5)+e^x \left (\left (-25 x^4-25 x^5\right ) \log (5)+\left (-40 x^4-25 x^5\right ) \log ^2(5)\right )\right )}{\left (6400-4000 x+625 x^2\right ) \log ^2(5)+e^x \left (\left (-800 x^3+250 x^4\right ) \log (5)+\left (-480 x^2-650 x^3+250 x^4\right ) \log ^2(5)\right )+e^{2 x} \left (25 x^6+\left (30 x^5+50 x^6\right ) \log (5)+\left (9 x^4+30 x^5+25 x^6\right ) \log ^2(5)\right )} \, dx=5^{\frac {5 \, x^{2}}{{\left (5 \, x^{3} + {\left (5 \, x^{3} + 3 \, x^{2}\right )} \log \left (5\right )\right )} e^{x} + 5 \, {\left (5 \, x - 16\right )} \log \left (5\right )}} \]

[In]

integrate((((-25*x^5-40*x^4)*log(5)^2+(-25*x^5-25*x^4)*log(5))*exp(x)+(125*x^2-800*x)*log(5)^2)*exp(5*x^2*log(
5)/(((5*x^3+3*x^2)*log(5)+5*x^3)*exp(x)+(25*x-80)*log(5)))/(((25*x^6+30*x^5+9*x^4)*log(5)^2+(50*x^6+30*x^5)*lo
g(5)+25*x^6)*exp(x)^2+((250*x^4-650*x^3-480*x^2)*log(5)^2+(250*x^4-800*x^3)*log(5))*exp(x)+(625*x^2-4000*x+640
0)*log(5)^2),x, algorithm="fricas")

[Out]

5^(5*x^2/((5*x^3 + (5*x^3 + 3*x^2)*log(5))*e^x + 5*(5*x - 16)*log(5)))

Sympy [A] (verification not implemented)

Time = 0.71 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.30 \[ \int \frac {5^{\frac {5 x^2}{(-80+25 x) \log (5)+e^x \left (5 x^3+\left (3 x^2+5 x^3\right ) \log (5)\right )}} \left (\left (-800 x+125 x^2\right ) \log ^2(5)+e^x \left (\left (-25 x^4-25 x^5\right ) \log (5)+\left (-40 x^4-25 x^5\right ) \log ^2(5)\right )\right )}{\left (6400-4000 x+625 x^2\right ) \log ^2(5)+e^x \left (\left (-800 x^3+250 x^4\right ) \log (5)+\left (-480 x^2-650 x^3+250 x^4\right ) \log ^2(5)\right )+e^{2 x} \left (25 x^6+\left (30 x^5+50 x^6\right ) \log (5)+\left (9 x^4+30 x^5+25 x^6\right ) \log ^2(5)\right )} \, dx=e^{\frac {5 x^{2} \log {\left (5 \right )}}{\left (25 x - 80\right ) \log {\left (5 \right )} + \left (5 x^{3} + \left (5 x^{3} + 3 x^{2}\right ) \log {\left (5 \right )}\right ) e^{x}}} \]

[In]

integrate((((-25*x**5-40*x**4)*ln(5)**2+(-25*x**5-25*x**4)*ln(5))*exp(x)+(125*x**2-800*x)*ln(5)**2)*exp(5*x**2
*ln(5)/(((5*x**3+3*x**2)*ln(5)+5*x**3)*exp(x)+(25*x-80)*ln(5)))/(((25*x**6+30*x**5+9*x**4)*ln(5)**2+(50*x**6+3
0*x**5)*ln(5)+25*x**6)*exp(x)**2+((250*x**4-650*x**3-480*x**2)*ln(5)**2+(250*x**4-800*x**3)*ln(5))*exp(x)+(625
*x**2-4000*x+6400)*ln(5)**2),x)

[Out]

exp(5*x**2*log(5)/((25*x - 80)*log(5) + (5*x**3 + (5*x**3 + 3*x**2)*log(5))*exp(x)))

Maxima [A] (verification not implemented)

none

Time = 0.51 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.30 \[ \int \frac {5^{\frac {5 x^2}{(-80+25 x) \log (5)+e^x \left (5 x^3+\left (3 x^2+5 x^3\right ) \log (5)\right )}} \left (\left (-800 x+125 x^2\right ) \log ^2(5)+e^x \left (\left (-25 x^4-25 x^5\right ) \log (5)+\left (-40 x^4-25 x^5\right ) \log ^2(5)\right )\right )}{\left (6400-4000 x+625 x^2\right ) \log ^2(5)+e^x \left (\left (-800 x^3+250 x^4\right ) \log (5)+\left (-480 x^2-650 x^3+250 x^4\right ) \log ^2(5)\right )+e^{2 x} \left (25 x^6+\left (30 x^5+50 x^6\right ) \log (5)+\left (9 x^4+30 x^5+25 x^6\right ) \log ^2(5)\right )} \, dx=5^{\frac {5 \, x^{2}}{{\left (5 \, x^{3} {\left (\log \left (5\right ) + 1\right )} + 3 \, x^{2} \log \left (5\right )\right )} e^{x} + 25 \, x \log \left (5\right ) - 80 \, \log \left (5\right )}} \]

[In]

integrate((((-25*x^5-40*x^4)*log(5)^2+(-25*x^5-25*x^4)*log(5))*exp(x)+(125*x^2-800*x)*log(5)^2)*exp(5*x^2*log(
5)/(((5*x^3+3*x^2)*log(5)+5*x^3)*exp(x)+(25*x-80)*log(5)))/(((25*x^6+30*x^5+9*x^4)*log(5)^2+(50*x^6+30*x^5)*lo
g(5)+25*x^6)*exp(x)^2+((250*x^4-650*x^3-480*x^2)*log(5)^2+(250*x^4-800*x^3)*log(5))*exp(x)+(625*x^2-4000*x+640
0)*log(5)^2),x, algorithm="maxima")

[Out]

5^(5*x^2/((5*x^3*(log(5) + 1) + 3*x^2*log(5))*e^x + 25*x*log(5) - 80*log(5)))

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.47 \[ \int \frac {5^{\frac {5 x^2}{(-80+25 x) \log (5)+e^x \left (5 x^3+\left (3 x^2+5 x^3\right ) \log (5)\right )}} \left (\left (-800 x+125 x^2\right ) \log ^2(5)+e^x \left (\left (-25 x^4-25 x^5\right ) \log (5)+\left (-40 x^4-25 x^5\right ) \log ^2(5)\right )\right )}{\left (6400-4000 x+625 x^2\right ) \log ^2(5)+e^x \left (\left (-800 x^3+250 x^4\right ) \log (5)+\left (-480 x^2-650 x^3+250 x^4\right ) \log ^2(5)\right )+e^{2 x} \left (25 x^6+\left (30 x^5+50 x^6\right ) \log (5)+\left (9 x^4+30 x^5+25 x^6\right ) \log ^2(5)\right )} \, dx=5^{\frac {5 \, x^{2}}{5 \, x^{3} e^{x} \log \left (5\right ) + 5 \, x^{3} e^{x} + 3 \, x^{2} e^{x} \log \left (5\right ) + 25 \, x \log \left (5\right ) - 80 \, \log \left (5\right )}} \]

[In]

integrate((((-25*x^5-40*x^4)*log(5)^2+(-25*x^5-25*x^4)*log(5))*exp(x)+(125*x^2-800*x)*log(5)^2)*exp(5*x^2*log(
5)/(((5*x^3+3*x^2)*log(5)+5*x^3)*exp(x)+(25*x-80)*log(5)))/(((25*x^6+30*x^5+9*x^4)*log(5)^2+(50*x^6+30*x^5)*lo
g(5)+25*x^6)*exp(x)^2+((250*x^4-650*x^3-480*x^2)*log(5)^2+(250*x^4-800*x^3)*log(5))*exp(x)+(625*x^2-4000*x+640
0)*log(5)^2),x, algorithm="giac")

[Out]

5^(5*x^2/(5*x^3*e^x*log(5) + 5*x^3*e^x + 3*x^2*e^x*log(5) + 25*x*log(5) - 80*log(5)))

Mupad [B] (verification not implemented)

Time = 15.61 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.50 \[ \int \frac {5^{\frac {5 x^2}{(-80+25 x) \log (5)+e^x \left (5 x^3+\left (3 x^2+5 x^3\right ) \log (5)\right )}} \left (\left (-800 x+125 x^2\right ) \log ^2(5)+e^x \left (\left (-25 x^4-25 x^5\right ) \log (5)+\left (-40 x^4-25 x^5\right ) \log ^2(5)\right )\right )}{\left (6400-4000 x+625 x^2\right ) \log ^2(5)+e^x \left (\left (-800 x^3+250 x^4\right ) \log (5)+\left (-480 x^2-650 x^3+250 x^4\right ) \log ^2(5)\right )+e^{2 x} \left (25 x^6+\left (30 x^5+50 x^6\right ) \log (5)+\left (9 x^4+30 x^5+25 x^6\right ) \log ^2(5)\right )} \, dx={\mathrm {e}}^{\frac {5\,x^2\,\ln \left (5\right )}{5\,x^3\,{\mathrm {e}}^x-80\,\ln \left (5\right )+25\,x\,\ln \left (5\right )+3\,x^2\,{\mathrm {e}}^x\,\ln \left (5\right )+5\,x^3\,{\mathrm {e}}^x\,\ln \left (5\right )}} \]

[In]

int(-(exp((5*x^2*log(5))/(log(5)*(25*x - 80) + exp(x)*(log(5)*(3*x^2 + 5*x^3) + 5*x^3)))*(exp(x)*(log(5)*(25*x
^4 + 25*x^5) + log(5)^2*(40*x^4 + 25*x^5)) + log(5)^2*(800*x - 125*x^2)))/(log(5)^2*(625*x^2 - 4000*x + 6400)
- exp(x)*(log(5)^2*(480*x^2 + 650*x^3 - 250*x^4) + log(5)*(800*x^3 - 250*x^4)) + exp(2*x)*(log(5)^2*(9*x^4 + 3
0*x^5 + 25*x^6) + log(5)*(30*x^5 + 50*x^6) + 25*x^6)),x)

[Out]

exp((5*x^2*log(5))/(5*x^3*exp(x) - 80*log(5) + 25*x*log(5) + 3*x^2*exp(x)*log(5) + 5*x^3*exp(x)*log(5)))