\(\int \frac {3+2 e^{2 x} x^2}{-3 x+5 x^2+e^{2 x} x^2} \, dx\) [6409]

Optimal result

Integrand size = 33, antiderivative size = 13 \[ \int \frac {3+2 e^{2 x} x^2}{-3 x+5 x^2+e^{2 x} x^2} \, dx=\log \left (5+e^{2 x}-\frac {3}{x}\right ) \]

[Out]

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

Rubi [F]

\[ \int \frac {3+2 e^{2 x} x^2}{-3 x+5 x^2+e^{2 x} x^2} \, dx=\int \frac {3+2 e^{2 x} x^2}{-3 x+5 x^2+e^{2 x} x^2} \, dx \]

[In]

Int[(3 + 2*E^(2*x)*x^2)/(-3*x + 5*x^2 + E^(2*x)*x^2),x]

[Out]

2*x + 6*Defer[Int][(-3 + 5*x + E^(2*x)*x)^(-1), x] + 3*Defer[Int][1/(x*(-3 + 5*x + E^(2*x)*x)), x] - 10*Defer[
Int][x/(-3 + 5*x + E^(2*x)*x), x]

Rubi steps \begin{align*} \text {integral}& = \int \left (2-\frac {-3-6 x+10 x^2}{x \left (-3+5 x+e^{2 x} x\right )}\right ) \, dx \\ & = 2 x-\int \frac {-3-6 x+10 x^2}{x \left (-3+5 x+e^{2 x} x\right )} \, dx \\ & = 2 x-\int \left (-\frac {6}{-3+5 x+e^{2 x} x}-\frac {3}{x \left (-3+5 x+e^{2 x} x\right )}+\frac {10 x}{-3+5 x+e^{2 x} x}\right ) \, dx \\ & = 2 x+3 \int \frac {1}{x \left (-3+5 x+e^{2 x} x\right )} \, dx+6 \int \frac {1}{-3+5 x+e^{2 x} x} \, dx-10 \int \frac {x}{-3+5 x+e^{2 x} x} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.46 \[ \int \frac {3+2 e^{2 x} x^2}{-3 x+5 x^2+e^{2 x} x^2} \, dx=-\log (x)+\log \left (3-5 x-e^{2 x} x\right ) \]

[In]

Integrate[(3 + 2*E^(2*x)*x^2)/(-3*x + 5*x^2 + E^(2*x)*x^2),x]

[Out]

-Log[x] + Log[3 - 5*x - E^(2*x)*x]

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.23

method	result	size
risch	\(\ln \left ({\mathrm e}^{2 x}+\frac {5 x -3}{x}\right )\)	\(16\)
norman	\(-\ln \left (x \right )+\ln \left (x \,{\mathrm e}^{2 x}+5 x -3\right )\)	\(18\)
parallelrisch	\(-\ln \left (x \right )+\ln \left (x \,{\mathrm e}^{2 x}+5 x -3\right )\)	\(18\)

[In]

int((2*exp(x)^2*x^2+3)/(exp(x)^2*x^2+5*x^2-3*x),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.23 \[ \int \frac {3+2 e^{2 x} x^2}{-3 x+5 x^2+e^{2 x} x^2} \, dx=\log \left (\frac {x e^{\left (2 \, x\right )} + 5 \, x - 3}{x}\right ) \]

[In]

integrate((2*exp(x)^2*x^2+3)/(exp(x)^2*x^2+5*x^2-3*x),x, algorithm="fricas")

[Out]

log((x*e^(2*x) + 5*x - 3)/x)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.92 \[ \int \frac {3+2 e^{2 x} x^2}{-3 x+5 x^2+e^{2 x} x^2} \, dx=\log {\left (e^{2 x} + \frac {5 x - 3}{x} \right )} \]

[In]

integrate((2*exp(x)**2*x**2+3)/(exp(x)**2*x**2+5*x**2-3*x),x)

[Out]

log(exp(2*x) + (5*x - 3)/x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.23 \[ \int \frac {3+2 e^{2 x} x^2}{-3 x+5 x^2+e^{2 x} x^2} \, dx=\log \left (\frac {x e^{\left (2 \, x\right )} + 5 \, x - 3}{x}\right ) \]

[In]

integrate((2*exp(x)^2*x^2+3)/(exp(x)^2*x^2+5*x^2-3*x),x, algorithm="maxima")

[Out]

log((x*e^(2*x) + 5*x - 3)/x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.31 \[ \int \frac {3+2 e^{2 x} x^2}{-3 x+5 x^2+e^{2 x} x^2} \, dx=\log \left (x e^{\left (2 \, x\right )} + 5 \, x - 3\right ) - \log \left (x\right ) \]

[In]

integrate((2*exp(x)^2*x^2+3)/(exp(x)^2*x^2+5*x^2-3*x),x, algorithm="giac")

[Out]

log(x*e^(2*x) + 5*x - 3) - log(x)

Mupad [B] (verification not implemented)

Time = 13.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.31 \[ \int \frac {3+2 e^{2 x} x^2}{-3 x+5 x^2+e^{2 x} x^2} \, dx=\ln \left (5\,x+x\,{\mathrm {e}}^{2\,x}-3\right )-\ln \left (x\right ) \]

[In]

int((2*x^2*exp(2*x) + 3)/(x^2*exp(2*x) - 3*x + 5*x^2),x)

[Out]

log(5*x + x*exp(2*x) - 3) - log(x)