\(\int \frac {e^{\frac {e^{-\frac {5 x}{3 x+15 \log (4)}} (4+8 x)}{x}-\frac {5 x}{3 x+15 \log (4)}} (-12 x^2+(-220 x-200 x^2) \log (4)-300 \log ^2(4))}{3 x^4+30 x^3 \log (4)+75 x^2 \log ^2(4)} \, dx\) [5355]

Optimal result

Integrand size = 89, antiderivative size = 26 \[ \int \frac {e^{\frac {e^{-\frac {5 x}{3 x+15 \log (4)}} (4+8 x)}{x}-\frac {5 x}{3 x+15 \log (4)}} \left (-12 x^2+\left (-220 x-200 x^2\right ) \log (4)-300 \log ^2(4)\right )}{3 x^4+30 x^3 \log (4)+75 x^2 \log ^2(4)} \, dx=e^{2 e^{-\frac {5 x}{3 (x+5 \log (4))}} \left (4+\frac {2}{x}\right )} \]

[Out]

exp(2/exp(x/(3/5*x+6*ln(2)))*(2/x+4))

Rubi [F]

\[ \int \frac {e^{\frac {e^{-\frac {5 x}{3 x+15 \log (4)}} (4+8 x)}{x}-\frac {5 x}{3 x+15 \log (4)}} \left (-12 x^2+\left (-220 x-200 x^2\right ) \log (4)-300 \log ^2(4)\right )}{3 x^4+30 x^3 \log (4)+75 x^2 \log ^2(4)} \, dx=\int \frac {\exp \left (\frac {e^{-\frac {5 x}{3 x+15 \log (4)}} (4+8 x)}{x}-\frac {5 x}{3 x+15 \log (4)}\right ) \left (-12 x^2+\left (-220 x-200 x^2\right ) \log (4)-300 \log ^2(4)\right )}{3 x^4+30 x^3 \log (4)+75 x^2 \log ^2(4)} \, dx \]

[In]

Int[(E^((4 + 8*x)/(E^((5*x)/(3*x + 15*Log[4]))*x) - (5*x)/(3*x + 15*Log[4]))*(-12*x^2 + (-220*x - 200*x^2)*Log
[4] - 300*Log[4]^2))/(3*x^4 + 30*x^3*Log[4] + 75*x^2*Log[4]^2),x]

[Out]

-4*Defer[Int][E^((4 + 8*x)/(E^((5*x)/(3*x + 15*Log[4]))*x) - (5*x)/(3*x + 15*Log[4]))/x^2, x] - (4*Defer[Int][
E^((4 + 8*x)/(E^((5*x)/(3*x + 15*Log[4]))*x) - (5*x)/(3*x + 15*Log[4]))/x, x])/(3*Log[4]) + (20*(1 - 10*Log[4]
)*Defer[Int][E^((4 + 8*x)/(E^((5*x)/(3*x + 15*Log[4]))*x) - (5*x)/(3*x + 15*Log[4]))/(x + 5*Log[4])^2, x])/3 +
(4*Defer[Int][E^((4 + 8*x)/(E^((5*x)/(3*x + 15*Log[4]))*x) - (5*x)/(3*x + 15*Log[4]))/(x + 5*Log[4]), x])/(3*
Log[4])

Rubi steps \begin{align*} \text {integral}& = \int \frac {\exp \left (\frac {e^{-\frac {5 x}{3 x+15 \log (4)}} (4+8 x)}{x}-\frac {5 x}{3 x+15 \log (4)}\right ) \left (-12 x^2+\left (-220 x-200 x^2\right ) \log (4)-300 \log ^2(4)\right )}{x^2 \left (3 x^2+30 x \log (4)+75 \log ^2(4)\right )} \, dx \\ & = \int \frac {\exp \left (\frac {e^{-\frac {5 x}{3 x+15 \log (4)}} (4+8 x)}{x}-\frac {5 x}{3 x+15 \log (4)}\right ) \left (-12 x^2+\left (-220 x-200 x^2\right ) \log (4)-300 \log ^2(4)\right )}{3 x^2 (x+5 \log (4))^2} \, dx \\ & = \frac {1}{3} \int \frac {\exp \left (\frac {e^{-\frac {5 x}{3 x+15 \log (4)}} (4+8 x)}{x}-\frac {5 x}{3 x+15 \log (4)}\right ) \left (-12 x^2+\left (-220 x-200 x^2\right ) \log (4)-300 \log ^2(4)\right )}{x^2 (x+5 \log (4))^2} \, dx \\ & = \frac {1}{3} \int \frac {\exp \left (\frac {e^{-\frac {5 x}{3 x+15 \log (4)}} (4+8 x)}{x}-\frac {5 x}{3 x+15 \log (4)}\right ) \left (-220 x \log (4)-300 \log ^2(4)-4 x^2 (3+50 \log (4))\right )}{x^2 (x+5 \log (4))^2} \, dx \\ & = \frac {1}{3} \int \left (-\frac {12 \exp \left (\frac {e^{-\frac {5 x}{3 x+15 \log (4)}} (4+8 x)}{x}-\frac {5 x}{3 x+15 \log (4)}\right )}{x^2}-\frac {4 \exp \left (\frac {e^{-\frac {5 x}{3 x+15 \log (4)}} (4+8 x)}{x}-\frac {5 x}{3 x+15 \log (4)}\right )}{x \log (4)}+\frac {4 \exp \left (\frac {e^{-\frac {5 x}{3 x+15 \log (4)}} (4+8 x)}{x}-\frac {5 x}{3 x+15 \log (4)}\right )}{\log (4) (x+5 \log (4))}-\frac {20 \exp \left (\frac {e^{-\frac {5 x}{3 x+15 \log (4)}} (4+8 x)}{x}-\frac {5 x}{3 x+15 \log (4)}\right ) (-1+10 \log (4))}{(x+5 \log (4))^2}\right ) \, dx \\ & = -\left (4 \int \frac {\exp \left (\frac {e^{-\frac {5 x}{3 x+15 \log (4)}} (4+8 x)}{x}-\frac {5 x}{3 x+15 \log (4)}\right )}{x^2} \, dx\right )+\frac {1}{3} (20 (1-10 \log (4))) \int \frac {\exp \left (\frac {e^{-\frac {5 x}{3 x+15 \log (4)}} (4+8 x)}{x}-\frac {5 x}{3 x+15 \log (4)}\right )}{(x+5 \log (4))^2} \, dx-\frac {4 \int \frac {\exp \left (\frac {e^{-\frac {5 x}{3 x+15 \log (4)}} (4+8 x)}{x}-\frac {5 x}{3 x+15 \log (4)}\right )}{x} \, dx}{3 \log (4)}+\frac {4 \int \frac {\exp \left (\frac {e^{-\frac {5 x}{3 x+15 \log (4)}} (4+8 x)}{x}-\frac {5 x}{3 x+15 \log (4)}\right )}{x+5 \log (4)} \, dx}{3 \log (4)} \\ \end{align*}

Mathematica [A] (warning: unable to verify)

Time = 1.88 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {e^{\frac {e^{-\frac {5 x}{3 x+15 \log (4)}} (4+8 x)}{x}-\frac {5 x}{3 x+15 \log (4)}} \left (-12 x^2+\left (-220 x-200 x^2\right ) \log (4)-300 \log ^2(4)\right )}{3 x^4+30 x^3 \log (4)+75 x^2 \log ^2(4)} \, dx=e^{\frac {2^{2+\frac {50}{3 (x+5 \log (4))}} (1+2 x)}{e^{5/3} x}} \]

[In]

Integrate[(E^((4 + 8*x)/(E^((5*x)/(3*x + 15*Log[4]))*x) - (5*x)/(3*x + 15*Log[4]))*(-12*x^2 + (-220*x - 200*x^
2)*Log[4] - 300*Log[4]^2))/(3*x^4 + 30*x^3*Log[4] + 75*x^2*Log[4]^2),x]

[Out]

E^((2^(2 + 50/(3*(x + 5*Log[4])))*(1 + 2*x))/(E^(5/3)*x))

Maple [A] (verified)

Time = 2.72 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92

method	result	size
risch	\({\mathrm e}^{\frac {4 \left (1+2 x \right ) {\mathrm e}^{-\frac {5 x}{3 \left (10 \ln \left (2\right )+x \right )}}}{x}}\)	\(24\)
parallelrisch	\(\frac {30 \ln \left (2\right ) {\mathrm e}^{\frac {4 \left (1+2 x \right ) {\mathrm e}^{-\frac {5 x}{3 \left (10 \ln \left (2\right )+x \right )}}}{x}}+3 \,{\mathrm e}^{\frac {4 \left (1+2 x \right ) {\mathrm e}^{-\frac {5 x}{3 \left (10 \ln \left (2\right )+x \right )}}}{x}} x}{30 \ln \left (2\right )+3 x}\)	\(69\)
norman	\(\frac {\left (x^{2} {\mathrm e}^{\frac {5 x}{30 \ln \left (2\right )+3 x}} {\mathrm e}^{\frac {\left (8 x +4\right ) {\mathrm e}^{-\frac {5 x}{30 \ln \left (2\right )+3 x}}}{x}}+10 x \ln \left (2\right ) {\mathrm e}^{\frac {5 x}{30 \ln \left (2\right )+3 x}} {\mathrm e}^{\frac {\left (8 x +4\right ) {\mathrm e}^{-\frac {5 x}{30 \ln \left (2\right )+3 x}}}{x}}\right ) {\mathrm e}^{-\frac {5 x}{30 \ln \left (2\right )+3 x}}}{x \left (10 \ln \left (2\right )+x \right )}\)	\(119\)

[In]

int((-1200*ln(2)^2+2*(-200*x^2-220*x)*ln(2)-12*x^2)*exp((8*x+4)/x/exp(5*x/(30*ln(2)+3*x)))/(300*x^2*ln(2)^2+60
*x^3*ln(2)+3*x^4)/exp(5*x/(30*ln(2)+3*x)),x,method=_RETURNVERBOSE)

[Out]

exp(4*(1+2*x)*exp(-5/3*x/(10*ln(2)+x))/x)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 62 vs. \(2 (20) = 40\).

Time = 0.26 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.38 \[ \int \frac {e^{\frac {e^{-\frac {5 x}{3 x+15 \log (4)}} (4+8 x)}{x}-\frac {5 x}{3 x+15 \log (4)}} \left (-12 x^2+\left (-220 x-200 x^2\right ) \log (4)-300 \log ^2(4)\right )}{3 x^4+30 x^3 \log (4)+75 x^2 \log ^2(4)} \, dx=e^{\left (-\frac {5 \, x^{2} - 12 \, {\left (2 \, x^{2} + 10 \, {\left (2 \, x + 1\right )} \log \left (2\right ) + x\right )} e^{\left (-\frac {5 \, x}{3 \, {\left (x + 10 \, \log \left (2\right )\right )}}\right )}}{3 \, {\left (x^{2} + 10 \, x \log \left (2\right )\right )}} + \frac {5 \, x}{3 \, {\left (x + 10 \, \log \left (2\right )\right )}}\right )} \]

[In]

integrate((-1200*log(2)^2+2*(-200*x^2-220*x)*log(2)-12*x^2)*exp((8*x+4)/x/exp(5*x/(30*log(2)+3*x)))/(300*x^2*l
og(2)^2+60*x^3*log(2)+3*x^4)/exp(5*x/(30*log(2)+3*x)),x, algorithm="fricas")

[Out]

e^(-1/3*(5*x^2 - 12*(2*x^2 + 10*(2*x + 1)*log(2) + x)*e^(-5/3*x/(x + 10*log(2))))/(x^2 + 10*x*log(2)) + 5/3*x/
(x + 10*log(2)))

Sympy [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {e^{\frac {e^{-\frac {5 x}{3 x+15 \log (4)}} (4+8 x)}{x}-\frac {5 x}{3 x+15 \log (4)}} \left (-12 x^2+\left (-220 x-200 x^2\right ) \log (4)-300 \log ^2(4)\right )}{3 x^4+30 x^3 \log (4)+75 x^2 \log ^2(4)} \, dx=e^{\frac {\left (8 x + 4\right ) e^{- \frac {5 x}{3 x + 30 \log {\left (2 \right )}}}}{x}} \]

[In]

integrate((-1200*ln(2)**2+2*(-200*x**2-220*x)*ln(2)-12*x**2)*exp((8*x+4)/x/exp(5*x/(30*ln(2)+3*x)))/(300*x**2*
ln(2)**2+60*x**3*ln(2)+3*x**4)/exp(5*x/(30*ln(2)+3*x)),x)

[Out]

exp((8*x + 4)*exp(-5*x/(3*x + 30*log(2)))/x)

Maxima [A] (verification not implemented)

none

Time = 0.45 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.50 \[ \int \frac {e^{\frac {e^{-\frac {5 x}{3 x+15 \log (4)}} (4+8 x)}{x}-\frac {5 x}{3 x+15 \log (4)}} \left (-12 x^2+\left (-220 x-200 x^2\right ) \log (4)-300 \log ^2(4)\right )}{3 x^4+30 x^3 \log (4)+75 x^2 \log ^2(4)} \, dx=e^{\left (\frac {4 \, e^{\left (\frac {50 \, \log \left (2\right )}{3 \, {\left (x + 10 \, \log \left (2\right )\right )}} - \frac {5}{3}\right )}}{x} + 8 \, e^{\left (\frac {50 \, \log \left (2\right )}{3 \, {\left (x + 10 \, \log \left (2\right )\right )}} - \frac {5}{3}\right )}\right )} \]

[In]

integrate((-1200*log(2)^2+2*(-200*x^2-220*x)*log(2)-12*x^2)*exp((8*x+4)/x/exp(5*x/(30*log(2)+3*x)))/(300*x^2*l
og(2)^2+60*x^3*log(2)+3*x^4)/exp(5*x/(30*log(2)+3*x)),x, algorithm="maxima")

[Out]

e^(4*e^(50/3*log(2)/(x + 10*log(2)) - 5/3)/x + 8*e^(50/3*log(2)/(x + 10*log(2)) - 5/3))

Giac [F]

\[ \int \frac {e^{\frac {e^{-\frac {5 x}{3 x+15 \log (4)}} (4+8 x)}{x}-\frac {5 x}{3 x+15 \log (4)}} \left (-12 x^2+\left (-220 x-200 x^2\right ) \log (4)-300 \log ^2(4)\right )}{3 x^4+30 x^3 \log (4)+75 x^2 \log ^2(4)} \, dx=\int { -\frac {4 \, {\left (3 \, x^{2} + 10 \, {\left (10 \, x^{2} + 11 \, x\right )} \log \left (2\right ) + 300 \, \log \left (2\right )^{2}\right )} e^{\left (\frac {4 \, {\left (2 \, x + 1\right )} e^{\left (-\frac {5 \, x}{3 \, {\left (x + 10 \, \log \left (2\right )\right )}}\right )}}{x} - \frac {5 \, x}{3 \, {\left (x + 10 \, \log \left (2\right )\right )}}\right )}}{3 \, {\left (x^{4} + 20 \, x^{3} \log \left (2\right ) + 100 \, x^{2} \log \left (2\right )^{2}\right )}} \,d x } \]

[In]

integrate((-1200*log(2)^2+2*(-200*x^2-220*x)*log(2)-12*x^2)*exp((8*x+4)/x/exp(5*x/(30*log(2)+3*x)))/(300*x^2*l
og(2)^2+60*x^3*log(2)+3*x^4)/exp(5*x/(30*log(2)+3*x)),x, algorithm="giac")

[Out]

integrate(-4/3*(3*x^2 + 10*(10*x^2 + 11*x)*log(2) + 300*log(2)^2)*e^(4*(2*x + 1)*e^(-5/3*x/(x + 10*log(2)))/x
- 5/3*x/(x + 10*log(2)))/(x^4 + 20*x^3*log(2) + 100*x^2*log(2)^2), x)

Mupad [B] (verification not implemented)

Time = 12.73 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.46 \[ \int \frac {e^{\frac {e^{-\frac {5 x}{3 x+15 \log (4)}} (4+8 x)}{x}-\frac {5 x}{3 x+15 \log (4)}} \left (-12 x^2+\left (-220 x-200 x^2\right ) \log (4)-300 \log ^2(4)\right )}{3 x^4+30 x^3 \log (4)+75 x^2 \log ^2(4)} \, dx={\mathrm {e}}^{\frac {4\,{\mathrm {e}}^{-\frac {5\,x}{3\,x+30\,\ln \left (2\right )}}}{x}}\,{\mathrm {e}}^{8\,{\mathrm {e}}^{-\frac {5\,x}{3\,x+30\,\ln \left (2\right )}}} \]

[In]

int(-(exp((exp(-(5*x)/(3*x + 30*log(2)))*(8*x + 4))/x)*exp(-(5*x)/(3*x + 30*log(2)))*(2*log(2)*(220*x + 200*x^
2) + 1200*log(2)^2 + 12*x^2))/(300*x^2*log(2)^2 + 60*x^3*log(2) + 3*x^4),x)

[Out]

exp((4*exp(-(5*x)/(3*x + 30*log(2))))/x)*exp(8*exp(-(5*x)/(3*x + 30*log(2))))