\(\int \frac {-4 x+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} (20-4 x-4 x^2+8 x^3)+(-4 x-4 e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} x) \log (1+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}})}{x^3+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} x^3} \, dx\) [2287]

Optimal result

Integrand size = 138, antiderivative size = 31 \[ \int \frac {-4 x+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} \left (20-4 x-4 x^2+8 x^3\right )+\left (-4 x-4 e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} x\right ) \log \left (1+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}}\right )}{x^3+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} x^3} \, dx=\frac {4-x+4 \log \left (1+e^{-\frac {1}{4}-\frac {5}{x}-x+x^2}\right )}{x} \]

[Out]

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

Rubi [F]

\[ \int \frac {-4 x+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} \left (20-4 x-4 x^2+8 x^3\right )+\left (-4 x-4 e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} x\right ) \log \left (1+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}}\right )}{x^3+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} x^3} \, dx=\int \frac {-4 x+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} \left (20-4 x-4 x^2+8 x^3\right )+\left (-4 x-4 e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} x\right ) \log \left (1+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}}\right )}{x^3+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} x^3} \, dx \]

[In]

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

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {-4 e^{\frac {1}{4}+\frac {5}{x}+x} x+4 e^{x^2} \left (5-x-x^2+2 x^3\right )-4 \left (e^{x^2}+e^{\frac {1}{4}+\frac {5}{x}+x}\right ) x \log \left (1+e^{-\frac {1}{4}-\frac {5}{x}-x+x^2}\right )}{\left (e^{x^2}+e^{\frac {1}{4}+\frac {5}{x}+x}\right ) x^3} \, dx \\ & = \int \left (-\frac {4 e^{\frac {1}{4}+\frac {5}{x}+x} \left (5-x^2+2 x^3\right )}{\left (e^{x^2}+e^{\frac {1}{4}+\frac {5}{x}+x}\right ) x^3}+\frac {4 \left (5-x-x^2+2 x^3-x \log \left (1+e^{-\frac {1}{4}-\frac {5}{x}-x+x^2}\right )\right )}{x^3}\right ) \, dx \\ & = -\left (4 \int \frac {e^{\frac {1}{4}+\frac {5}{x}+x} \left (5-x^2+2 x^3\right )}{\left (e^{x^2}+e^{\frac {1}{4}+\frac {5}{x}+x}\right ) x^3} \, dx\right )+4 \int \frac {5-x-x^2+2 x^3-x \log \left (1+e^{-\frac {1}{4}-\frac {5}{x}-x+x^2}\right )}{x^3} \, dx \\ & = -\left (4 \int \left (\frac {2 e^{\frac {1}{4}+\frac {5}{x}+x}}{e^{x^2}+e^{\frac {1}{4}+\frac {5}{x}+x}}+\frac {5 e^{\frac {1}{4}+\frac {5}{x}+x}}{\left (e^{x^2}+e^{\frac {1}{4}+\frac {5}{x}+x}\right ) x^3}-\frac {e^{\frac {1}{4}+\frac {5}{x}+x}}{\left (e^{x^2}+e^{\frac {1}{4}+\frac {5}{x}+x}\right ) x}\right ) \, dx\right )+4 \int \left (\frac {5-x-x^2+2 x^3}{x^3}-\frac {\log \left (1+e^{-\frac {1}{4}-\frac {5}{x}-x+x^2}\right )}{x^2}\right ) \, dx \\ & = 4 \int \frac {e^{\frac {1}{4}+\frac {5}{x}+x}}{\left (e^{x^2}+e^{\frac {1}{4}+\frac {5}{x}+x}\right ) x} \, dx+4 \int \frac {5-x-x^2+2 x^3}{x^3} \, dx-4 \int \frac {\log \left (1+e^{-\frac {1}{4}-\frac {5}{x}-x+x^2}\right )}{x^2} \, dx-8 \int \frac {e^{\frac {1}{4}+\frac {5}{x}+x}}{e^{x^2}+e^{\frac {1}{4}+\frac {5}{x}+x}} \, dx-20 \int \frac {e^{\frac {1}{4}+\frac {5}{x}+x}}{\left (e^{x^2}+e^{\frac {1}{4}+\frac {5}{x}+x}\right ) x^3} \, dx \\ & = \frac {4 \log \left (1+e^{-\frac {1}{4}-\frac {5}{x}-x+x^2}\right )}{x}+4 \int \left (2+\frac {5}{x^3}-\frac {1}{x^2}-\frac {1}{x}\right ) \, dx+4 \int \frac {e^{\frac {1}{4}+\frac {5}{x}+x}}{\left (e^{x^2}+e^{\frac {1}{4}+\frac {5}{x}+x}\right ) x} \, dx-4 \int \frac {e^{x^2} \left (5-x^2+2 x^3\right )}{\left (e^{x^2}+e^{\frac {1}{4}+\frac {5}{x}+x}\right ) x^3} \, dx-8 \int \frac {e^{\frac {1}{4}+\frac {5}{x}+x}}{e^{x^2}+e^{\frac {1}{4}+\frac {5}{x}+x}} \, dx-20 \int \frac {e^{\frac {1}{4}+\frac {5}{x}+x}}{\left (e^{x^2}+e^{\frac {1}{4}+\frac {5}{x}+x}\right ) x^3} \, dx \\ & = -\frac {10}{x^2}+\frac {4}{x}+8 x+\frac {4 \log \left (1+e^{-\frac {1}{4}-\frac {5}{x}-x+x^2}\right )}{x}-4 \log (x)-4 \int \left (\frac {2 e^{x^2}}{e^{x^2}+e^{\frac {1}{4}+\frac {5}{x}+x}}+\frac {5 e^{x^2}}{\left (e^{x^2}+e^{\frac {1}{4}+\frac {5}{x}+x}\right ) x^3}-\frac {e^{x^2}}{\left (e^{x^2}+e^{\frac {1}{4}+\frac {5}{x}+x}\right ) x}\right ) \, dx+4 \int \frac {e^{\frac {1}{4}+\frac {5}{x}+x}}{\left (e^{x^2}+e^{\frac {1}{4}+\frac {5}{x}+x}\right ) x} \, dx-8 \int \frac {e^{\frac {1}{4}+\frac {5}{x}+x}}{e^{x^2}+e^{\frac {1}{4}+\frac {5}{x}+x}} \, dx-20 \int \frac {e^{\frac {1}{4}+\frac {5}{x}+x}}{\left (e^{x^2}+e^{\frac {1}{4}+\frac {5}{x}+x}\right ) x^3} \, dx \\ & = -\frac {10}{x^2}+\frac {4}{x}+8 x+\frac {4 \log \left (1+e^{-\frac {1}{4}-\frac {5}{x}-x+x^2}\right )}{x}-4 \log (x)+4 \int \frac {e^{x^2}}{\left (e^{x^2}+e^{\frac {1}{4}+\frac {5}{x}+x}\right ) x} \, dx+4 \int \frac {e^{\frac {1}{4}+\frac {5}{x}+x}}{\left (e^{x^2}+e^{\frac {1}{4}+\frac {5}{x}+x}\right ) x} \, dx-8 \int \frac {e^{x^2}}{e^{x^2}+e^{\frac {1}{4}+\frac {5}{x}+x}} \, dx-8 \int \frac {e^{\frac {1}{4}+\frac {5}{x}+x}}{e^{x^2}+e^{\frac {1}{4}+\frac {5}{x}+x}} \, dx-20 \int \frac {e^{x^2}}{\left (e^{x^2}+e^{\frac {1}{4}+\frac {5}{x}+x}\right ) x^3} \, dx-20 \int \frac {e^{\frac {1}{4}+\frac {5}{x}+x}}{\left (e^{x^2}+e^{\frac {1}{4}+\frac {5}{x}+x}\right ) x^3} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {-4 x+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} \left (20-4 x-4 x^2+8 x^3\right )+\left (-4 x-4 e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} x\right ) \log \left (1+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}}\right )}{x^3+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} x^3} \, dx=\frac {4 \left (1+x+\log \left (1+e^{-\frac {1}{4}-\frac {5}{x}-x+x^2}\right )\right )}{x} \]

[In]

Integrate[(-4*x + E^((-20 - x - 4*x^2 + 4*x^3)/(4*x))*(20 - 4*x - 4*x^2 + 8*x^3) + (-4*x - 4*E^((-20 - x - 4*x
^2 + 4*x^3)/(4*x))*x)*Log[1 + E^((-20 - x - 4*x^2 + 4*x^3)/(4*x))])/(x^3 + E^((-20 - x - 4*x^2 + 4*x^3)/(4*x))
*x^3),x]

[Out]

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

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.06

method	result	size
parallelrisch	\(\frac {4+4 \ln \left ({\mathrm e}^{\frac {4 x^{3}-4 x^{2}-x -20}{4 x}}+1\right )}{x}\)	\(33\)
norman	\(\frac {4 x +4 x \ln \left ({\mathrm e}^{\frac {4 x^{3}-4 x^{2}-x -20}{4 x}}+1\right )}{x^{2}}\)	\(36\)
risch	\(\frac {4 \ln \left ({\mathrm e}^{\frac {4 x^{3}-4 x^{2}-x -20}{4 x}}+1\right )}{x}+\frac {4}{x}\)	\(36\)

[In]

int(((-4*x*exp(1/4*(4*x^3-4*x^2-x-20)/x)-4*x)*ln(exp(1/4*(4*x^3-4*x^2-x-20)/x)+1)+(8*x^3-4*x^2-4*x+20)*exp(1/4
*(4*x^3-4*x^2-x-20)/x)-4*x)/(x^3*exp(1/4*(4*x^3-4*x^2-x-20)/x)+x^3),x,method=_RETURNVERBOSE)

[Out]

(4+4*ln(exp(1/4*(4*x^3-4*x^2-x-20)/x)+1))/x

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {-4 x+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} \left (20-4 x-4 x^2+8 x^3\right )+\left (-4 x-4 e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} x\right ) \log \left (1+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}}\right )}{x^3+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} x^3} \, dx=\frac {4 \, {\left (\log \left (e^{\left (\frac {4 \, x^{3} - 4 \, x^{2} - x - 20}{4 \, x}\right )} + 1\right ) + 1\right )}}{x} \]

[In]

integrate(((-4*x*exp(1/4*(4*x^3-4*x^2-x-20)/x)-4*x)*log(exp(1/4*(4*x^3-4*x^2-x-20)/x)+1)+(8*x^3-4*x^2-4*x+20)*
exp(1/4*(4*x^3-4*x^2-x-20)/x)-4*x)/(x^3*exp(1/4*(4*x^3-4*x^2-x-20)/x)+x^3),x, algorithm="fricas")

[Out]

4*(log(e^(1/4*(4*x^3 - 4*x^2 - x - 20)/x) + 1) + 1)/x

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {-4 x+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} \left (20-4 x-4 x^2+8 x^3\right )+\left (-4 x-4 e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} x\right ) \log \left (1+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}}\right )}{x^3+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} x^3} \, dx=\frac {4 \log {\left (e^{\frac {x^{3} - x^{2} - \frac {x}{4} - 5}{x}} + 1 \right )}}{x} + \frac {4}{x} \]

[In]

integrate(((-4*x*exp(1/4*(4*x**3-4*x**2-x-20)/x)-4*x)*ln(exp(1/4*(4*x**3-4*x**2-x-20)/x)+1)+(8*x**3-4*x**2-4*x
+20)*exp(1/4*(4*x**3-4*x**2-x-20)/x)-4*x)/(x**3*exp(1/4*(4*x**3-4*x**2-x-20)/x)+x**3),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {-4 x+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} \left (20-4 x-4 x^2+8 x^3\right )+\left (-4 x-4 e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} x\right ) \log \left (1+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}}\right )}{x^3+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} x^3} \, dx=\frac {4 \, x \log \left (e^{\left (x^{2}\right )} + e^{\left (x + \frac {5}{x} + \frac {1}{4}\right )}\right ) + 3 \, x - 20}{x^{2}} \]

[In]

integrate(((-4*x*exp(1/4*(4*x^3-4*x^2-x-20)/x)-4*x)*log(exp(1/4*(4*x^3-4*x^2-x-20)/x)+1)+(8*x^3-4*x^2-4*x+20)*
exp(1/4*(4*x^3-4*x^2-x-20)/x)-4*x)/(x^3*exp(1/4*(4*x^3-4*x^2-x-20)/x)+x^3),x, algorithm="maxima")

[Out]

(4*x*log(e^(x^2) + e^(x + 5/x + 1/4)) + 3*x - 20)/x^2

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {-4 x+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} \left (20-4 x-4 x^2+8 x^3\right )+\left (-4 x-4 e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} x\right ) \log \left (1+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}}\right )}{x^3+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} x^3} \, dx=\frac {4 \, {\left (\log \left (e^{\left (\frac {4 \, x^{3} - 4 \, x^{2} - x - 20}{4 \, x}\right )} + 1\right ) + 1\right )}}{x} \]

[In]

integrate(((-4*x*exp(1/4*(4*x^3-4*x^2-x-20)/x)-4*x)*log(exp(1/4*(4*x^3-4*x^2-x-20)/x)+1)+(8*x^3-4*x^2-4*x+20)*
exp(1/4*(4*x^3-4*x^2-x-20)/x)-4*x)/(x^3*exp(1/4*(4*x^3-4*x^2-x-20)/x)+x^3),x, algorithm="giac")

[Out]

4*(log(e^(1/4*(4*x^3 - 4*x^2 - x - 20)/x) + 1) + 1)/x

Mupad [B] (verification not implemented)

Time = 8.50 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {-4 x+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} \left (20-4 x-4 x^2+8 x^3\right )+\left (-4 x-4 e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} x\right ) \log \left (1+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}}\right )}{x^3+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} x^3} \, dx=\frac {4\,\left (\ln \left ({\mathrm {e}}^{-x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{-\frac {1}{4}}\,{\mathrm {e}}^{-\frac {5}{x}}+1\right )+1\right )}{x} \]

[In]

int(-(4*x + exp(-(x/4 + x^2 - x^3 + 5)/x)*(4*x + 4*x^2 - 8*x^3 - 20) + log(exp(-(x/4 + x^2 - x^3 + 5)/x) + 1)*
(4*x + 4*x*exp(-(x/4 + x^2 - x^3 + 5)/x)))/(x^3*exp(-(x/4 + x^2 - x^3 + 5)/x) + x^3),x)

[Out]

(4*(log(exp(-x)*exp(x^2)*exp(-1/4)*exp(-5/x) + 1) + 1))/x