\(\int \frac {1260+10 x-125 x^3-x^4+e^{8+4 x+x^2} (-10-3 x^3-2 x^4)}{-126 x^3+e^{8+4 x+x^2} x^3-x^4} \, dx\) [7626]

Optimal result

Integrand size = 66, antiderivative size = 24 \[ \int \frac {1260+10 x-125 x^3-x^4+e^{8+4 x+x^2} \left (-10-3 x^3-2 x^4\right )}{-126 x^3+e^{8+4 x+x^2} x^3-x^4} \, dx=\frac {5}{x^2}+x-\log \left (126-e^{4+(2+x)^2}+x\right ) \]

[Out]

5/x^2+x-ln(x+126-exp((2+x)^2+4))

Rubi [F]

\[ \int \frac {1260+10 x-125 x^3-x^4+e^{8+4 x+x^2} \left (-10-3 x^3-2 x^4\right )}{-126 x^3+e^{8+4 x+x^2} x^3-x^4} \, dx=\int \frac {1260+10 x-125 x^3-x^4+e^{8+4 x+x^2} \left (-10-3 x^3-2 x^4\right )}{-126 x^3+e^{8+4 x+x^2} x^3-x^4} \, dx \]

[In]

Int[(1260 + 10*x - 125*x^3 - x^4 + E^(8 + 4*x + x^2)*(-10 - 3*x^3 - 2*x^4))/(-126*x^3 + E^(8 + 4*x + x^2)*x^3
- x^4),x]

[Out]

5/x^2 - 3*x - x^2 - 503*Defer[Int][(-126 + E^(8 + 4*x + x^2) - x)^(-1), x] + 256*Defer[Int][x/(126 - E^(8 + 4*
x + x^2) + x), x] + 2*Defer[Int][x^2/(126 - E^(8 + 4*x + x^2) + x), x]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {503+256 x+2 x^2}{-126+e^{8+4 x+x^2}-x}+\frac {-10-3 x^3-2 x^4}{x^3}\right ) \, dx \\ & = -\int \frac {503+256 x+2 x^2}{-126+e^{8+4 x+x^2}-x} \, dx+\int \frac {-10-3 x^3-2 x^4}{x^3} \, dx \\ & = \int \left (-3-\frac {10}{x^3}-2 x\right ) \, dx-\int \left (\frac {503}{-126+e^{8+4 x+x^2}-x}-\frac {256 x}{126-e^{8+4 x+x^2}+x}-\frac {2 x^2}{126-e^{8+4 x+x^2}+x}\right ) \, dx \\ & = \frac {5}{x^2}-3 x-x^2+2 \int \frac {x^2}{126-e^{8+4 x+x^2}+x} \, dx+256 \int \frac {x}{126-e^{8+4 x+x^2}+x} \, dx-503 \int \frac {1}{-126+e^{8+4 x+x^2}-x} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 4.43 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {1260+10 x-125 x^3-x^4+e^{8+4 x+x^2} \left (-10-3 x^3-2 x^4\right )}{-126 x^3+e^{8+4 x+x^2} x^3-x^4} \, dx=\frac {5}{x^2}+x-\log \left (126-e^{8+4 x+x^2}+x\right ) \]

[In]

Integrate[(1260 + 10*x - 125*x^3 - x^4 + E^(8 + 4*x + x^2)*(-10 - 3*x^3 - 2*x^4))/(-126*x^3 + E^(8 + 4*x + x^2
)*x^3 - x^4),x]

[Out]

5/x^2 + x - Log[126 - E^(8 + 4*x + x^2) + x]

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08

method	result	size
risch	\(x +\frac {5}{x^{2}}+8-\ln \left ({\mathrm e}^{x^{2}+4 x +8}-x -126\right )\)	\(26\)
norman	\(\frac {x^{3}+5}{x^{2}}-\ln \left (x -{\mathrm e}^{x^{2}+4 x +8}+126\right )\)	\(28\)
parallelrisch	\(-\frac {\ln \left (x -{\mathrm e}^{x^{2}+4 x +8}+126\right ) x^{2}-x^{3}-5}{x^{2}}\)	\(32\)

[In]

int(((-2*x^4-3*x^3-10)*exp(x^2+4*x+8)-x^4-125*x^3+10*x+1260)/(x^3*exp(x^2+4*x+8)-x^4-126*x^3),x,method=_RETURN
VERBOSE)

[Out]

x+5/x^2+8-ln(exp(x^2+4*x+8)-x-126)

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {1260+10 x-125 x^3-x^4+e^{8+4 x+x^2} \left (-10-3 x^3-2 x^4\right )}{-126 x^3+e^{8+4 x+x^2} x^3-x^4} \, dx=\frac {x^{3} - x^{2} \log \left (-x + e^{\left (x^{2} + 4 \, x + 8\right )} - 126\right ) + 5}{x^{2}} \]

[In]

integrate(((-2*x^4-3*x^3-10)*exp(x^2+4*x+8)-x^4-125*x^3+10*x+1260)/(x^3*exp(x^2+4*x+8)-x^4-126*x^3),x, algorit
hm="fricas")

[Out]

(x^3 - x^2*log(-x + e^(x^2 + 4*x + 8) - 126) + 5)/x^2

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {1260+10 x-125 x^3-x^4+e^{8+4 x+x^2} \left (-10-3 x^3-2 x^4\right )}{-126 x^3+e^{8+4 x+x^2} x^3-x^4} \, dx=x - \log {\left (- x + e^{x^{2} + 4 x + 8} - 126 \right )} + \frac {5}{x^{2}} \]

[In]

integrate(((-2*x**4-3*x**3-10)*exp(x**2+4*x+8)-x**4-125*x**3+10*x+1260)/(x**3*exp(x**2+4*x+8)-x**4-126*x**3),x
)

[Out]

x - log(-x + exp(x**2 + 4*x + 8) - 126) + 5/x**2

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.58 \[ \int \frac {1260+10 x-125 x^3-x^4+e^{8+4 x+x^2} \left (-10-3 x^3-2 x^4\right )}{-126 x^3+e^{8+4 x+x^2} x^3-x^4} \, dx=-\frac {3 \, x^{3} - 5}{x^{2}} - \log \left (-{\left (x - e^{\left (x^{2} + 4 \, x + 8\right )} + 126\right )} e^{\left (-4 \, x - 8\right )}\right ) \]

[In]

integrate(((-2*x^4-3*x^3-10)*exp(x^2+4*x+8)-x^4-125*x^3+10*x+1260)/(x^3*exp(x^2+4*x+8)-x^4-126*x^3),x, algorit
hm="maxima")

[Out]

-(3*x^3 - 5)/x^2 - log(-(x - e^(x^2 + 4*x + 8) + 126)*e^(-4*x - 8))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {1260+10 x-125 x^3-x^4+e^{8+4 x+x^2} \left (-10-3 x^3-2 x^4\right )}{-126 x^3+e^{8+4 x+x^2} x^3-x^4} \, dx=\frac {x^{3} - x^{2} \log \left (x - e^{\left (x^{2} + 4 \, x + 8\right )} + 126\right ) + 5}{x^{2}} \]

[In]

integrate(((-2*x^4-3*x^3-10)*exp(x^2+4*x+8)-x^4-125*x^3+10*x+1260)/(x^3*exp(x^2+4*x+8)-x^4-126*x^3),x, algorit
hm="giac")

[Out]

(x^3 - x^2*log(x - e^(x^2 + 4*x + 8) + 126) + 5)/x^2

Mupad [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {1260+10 x-125 x^3-x^4+e^{8+4 x+x^2} \left (-10-3 x^3-2 x^4\right )}{-126 x^3+e^{8+4 x+x^2} x^3-x^4} \, dx=\frac {x^3+5}{x^2}-\ln \left (x-{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^8+126\right ) \]

[In]

int((exp(4*x + x^2 + 8)*(3*x^3 + 2*x^4 + 10) - 10*x + 125*x^3 + x^4 - 1260)/(126*x^3 - x^3*exp(4*x + x^2 + 8)
+ x^4),x)

[Out]

(x^3 + 5)/x^2 - log(x - exp(4*x)*exp(x^2)*exp(8) + 126)