\(\int \frac {2 x-48 x^3+e (1-48 x^2)+e^{10+2 x} (1+2 x-48 x^2)}{e x+e^{10+2 x} x+x^2} \, dx\) [3279]

Optimal result

Integrand size = 55, antiderivative size = 20 \[ \int \frac {2 x-48 x^3+e \left (1-48 x^2\right )+e^{10+2 x} \left (1+2 x-48 x^2\right )}{e x+e^{10+2 x} x+x^2} \, dx=\log \left (e^{-24 x^2} x \left (e+e^{2 (5+x)}+x\right )\right ) \]

[Out]

ln(x*(exp(2*x+10)+x+exp(1))/exp(24*x^2))

Rubi [F]

\[ \int \frac {2 x-48 x^3+e \left (1-48 x^2\right )+e^{10+2 x} \left (1+2 x-48 x^2\right )}{e x+e^{10+2 x} x+x^2} \, dx=\int \frac {2 x-48 x^3+e \left (1-48 x^2\right )+e^{10+2 x} \left (1+2 x-48 x^2\right )}{e x+e^{10+2 x} x+x^2} \, dx \]

[In]

Int[(2*x - 48*x^3 + E*(1 - 48*x^2) + E^(10 + 2*x)*(1 + 2*x - 48*x^2))/(E*x + E^(10 + 2*x)*x + x^2),x]

[Out]

2*x - 24*x^2 + Log[x] + (1 - 2*E)*Defer[Int][(E + E^(10 + 2*x) + x)^(-1), x] - 2*Defer[Int][x/(E + E^(10 + 2*x
) + x), x]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {-1+2 e+2 x}{e+e^{10+2 x}+x}+\frac {1+2 x-48 x^2}{x}\right ) \, dx \\ & = -\int \frac {-1+2 e+2 x}{e+e^{10+2 x}+x} \, dx+\int \frac {1+2 x-48 x^2}{x} \, dx \\ & = \int \left (2+\frac {1}{x}-48 x\right ) \, dx-\int \left (-\frac {1-2 e}{e+e^{10+2 x}+x}+\frac {2 x}{e+e^{10+2 x}+x}\right ) \, dx \\ & = 2 x-24 x^2+\log (x)-2 \int \frac {x}{e+e^{10+2 x}+x} \, dx-(-1+2 e) \int \frac {1}{e+e^{10+2 x}+x} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 1.32 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {2 x-48 x^3+e \left (1-48 x^2\right )+e^{10+2 x} \left (1+2 x-48 x^2\right )}{e x+e^{10+2 x} x+x^2} \, dx=-24 x^2+\log (x)+\log \left (e+e^{10+2 x}+x\right ) \]

[In]

Integrate[(2*x - 48*x^3 + E*(1 - 48*x^2) + E^(10 + 2*x)*(1 + 2*x - 48*x^2))/(E*x + E^(10 + 2*x)*x + x^2),x]

[Out]

-24*x^2 + Log[x] + Log[E + E^(10 + 2*x) + x]

Maple [A] (verified)

Time = 2.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00

method	result	size
norman	\(-24 x^{2}+\ln \left (x \right )+\ln \left ({\mathrm e}^{2 x +10}+x +{\mathrm e}\right )\)	\(20\)
parallelrisch	\(-24 x^{2}+\ln \left (x \right )+\ln \left ({\mathrm e}^{2 x +10}+x +{\mathrm e}\right )\)	\(20\)
risch	\(-24 x^{2}+\ln \left (x \right )-10+\ln \left ({\mathrm e}^{2 x +10}+x +{\mathrm e}\right )\)	\(21\)

[In]

int(((-48*x^2+2*x+1)*exp(2*x+10)+(-48*x^2+1)*exp(1)-48*x^3+2*x)/(x*exp(2*x+10)+x*exp(1)+x^2),x,method=_RETURNV
ERBOSE)

[Out]

-24*x^2+ln(x)+ln(exp(2*x+10)+x+exp(1))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {2 x-48 x^3+e \left (1-48 x^2\right )+e^{10+2 x} \left (1+2 x-48 x^2\right )}{e x+e^{10+2 x} x+x^2} \, dx=-24 \, x^{2} + \log \left (x + e + e^{\left (2 \, x + 10\right )}\right ) + \log \left (x\right ) \]

[In]

integrate(((-48*x^2+2*x+1)*exp(2*x+10)+(-48*x^2+1)*exp(1)-48*x^3+2*x)/(x*exp(2*x+10)+x*exp(1)+x^2),x, algorith
m="fricas")

[Out]

-24*x^2 + log(x + e + e^(2*x + 10)) + log(x)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {2 x-48 x^3+e \left (1-48 x^2\right )+e^{10+2 x} \left (1+2 x-48 x^2\right )}{e x+e^{10+2 x} x+x^2} \, dx=- 24 x^{2} + \log {\left (x \right )} + \log {\left (x + e^{2 x + 10} + e \right )} \]

[In]

integrate(((-48*x**2+2*x+1)*exp(2*x+10)+(-48*x**2+1)*exp(1)-48*x**3+2*x)/(x*exp(2*x+10)+x*exp(1)+x**2),x)

[Out]

-24*x**2 + log(x) + log(x + exp(2*x + 10) + E)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {2 x-48 x^3+e \left (1-48 x^2\right )+e^{10+2 x} \left (1+2 x-48 x^2\right )}{e x+e^{10+2 x} x+x^2} \, dx=-24 \, x^{2} + \log \left ({\left (x + e + e^{\left (2 \, x + 10\right )}\right )} e^{\left (-10\right )}\right ) + \log \left (x\right ) \]

[In]

integrate(((-48*x^2+2*x+1)*exp(2*x+10)+(-48*x^2+1)*exp(1)-48*x^3+2*x)/(x*exp(2*x+10)+x*exp(1)+x^2),x, algorith
m="maxima")

[Out]

-24*x^2 + log((x + e + e^(2*x + 10))*e^(-10)) + log(x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.25 \[ \int \frac {2 x-48 x^3+e \left (1-48 x^2\right )+e^{10+2 x} \left (1+2 x-48 x^2\right )}{e x+e^{10+2 x} x+x^2} \, dx=-24 \, x^{2} + \log \left (x\right ) + \log \left (-x - e - e^{\left (2 \, x + 10\right )}\right ) \]

[In]

integrate(((-48*x^2+2*x+1)*exp(2*x+10)+(-48*x^2+1)*exp(1)-48*x^3+2*x)/(x*exp(2*x+10)+x*exp(1)+x^2),x, algorith
m="giac")

[Out]

-24*x^2 + log(x) + log(-x - e - e^(2*x + 10))

Mupad [B] (verification not implemented)

Time = 9.23 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {2 x-48 x^3+e \left (1-48 x^2\right )+e^{10+2 x} \left (1+2 x-48 x^2\right )}{e x+e^{10+2 x} x+x^2} \, dx=\ln \left (x+\mathrm {e}+{\mathrm {e}}^{2\,x+10}\right )+\ln \left (x\right )-24\,x^2 \]

[In]

int((2*x - exp(1)*(48*x^2 - 1) + exp(2*x + 10)*(2*x - 48*x^2 + 1) - 48*x^3)/(x*exp(1) + x*exp(2*x + 10) + x^2)
,x)

[Out]

log(x + exp(1) + exp(2*x + 10)) + log(x) - 24*x^2