\(\int \frac {-12-4 x-3 x^2+e^{5 x} (2+13 x)}{-4+e^{5 x} x-x^2} \, dx\) [8889]

Optimal result

Integrand size = 38, antiderivative size = 23 \[ \int \frac {-12-4 x-3 x^2+e^{5 x} (2+13 x)}{-4+e^{5 x} x-x^2} \, dx=-4+x+\log (3)+2 \left (x+\log \left (4+x \left (-e^{5 x}+x\right )\right )\right ) \]

[Out]

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

Rubi [F]

\[ \int \frac {-12-4 x-3 x^2+e^{5 x} (2+13 x)}{-4+e^{5 x} x-x^2} \, dx=\int \frac {-12-4 x-3 x^2+e^{5 x} (2+13 x)}{-4+e^{5 x} x-x^2} \, dx \]

[In]

Int[(-12 - 4*x - 3*x^2 + E^(5*x)*(2 + 13*x))/(-4 + E^(5*x)*x - x^2),x]

[Out]

13*x + 2*Log[x] + 40*Defer[Int][(-4 + E^(5*x)*x - x^2)^(-1), x] - 8*Defer[Int][1/(x*(4 - E^(5*x)*x + x^2)), x]
+ 2*Defer[Int][x/(4 - E^(5*x)*x + x^2), x] - 10*Defer[Int][x^2/(4 - E^(5*x)*x + x^2), x]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {2+13 x}{x}-\frac {2 \left (4+20 x-x^2+5 x^3\right )}{x \left (4-e^{5 x} x+x^2\right )}\right ) \, dx \\ & = -\left (2 \int \frac {4+20 x-x^2+5 x^3}{x \left (4-e^{5 x} x+x^2\right )} \, dx\right )+\int \frac {2+13 x}{x} \, dx \\ & = -\left (2 \int \left (-\frac {20}{-4+e^{5 x} x-x^2}+\frac {4}{x \left (4-e^{5 x} x+x^2\right )}-\frac {x}{4-e^{5 x} x+x^2}+\frac {5 x^2}{4-e^{5 x} x+x^2}\right ) \, dx\right )+\int \left (13+\frac {2}{x}\right ) \, dx \\ & = 13 x+2 \log (x)+2 \int \frac {x}{4-e^{5 x} x+x^2} \, dx-8 \int \frac {1}{x \left (4-e^{5 x} x+x^2\right )} \, dx-10 \int \frac {x^2}{4-e^{5 x} x+x^2} \, dx+40 \int \frac {1}{-4+e^{5 x} x-x^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.68 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {-12-4 x-3 x^2+e^{5 x} (2+13 x)}{-4+e^{5 x} x-x^2} \, dx=3 x+2 \log \left (4-e^{5 x} x+x^2\right ) \]

[In]

Integrate[(-12 - 4*x - 3*x^2 + E^(5*x)*(2 + 13*x))/(-4 + E^(5*x)*x - x^2),x]

[Out]

3*x + 2*Log[4 - E^(5*x)*x + x^2]

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87

method	result	size
norman	\(3 x +2 \ln \left (x^{2}-x \,{\mathrm e}^{5 x}+4\right )\)	\(20\)
parallelrisch	\(3 x +2 \ln \left (x^{2}-x \,{\mathrm e}^{5 x}+4\right )\)	\(20\)
risch	\(3 x +2 \ln \left (x \right )+2 \ln \left ({\mathrm e}^{5 x}-\frac {x^{2}+4}{x}\right )\)	\(27\)

[In]

int(((13*x+2)*exp(5*x)-3*x^2-4*x-12)/(x*exp(5*x)-x^2-4),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {-12-4 x-3 x^2+e^{5 x} (2+13 x)}{-4+e^{5 x} x-x^2} \, dx=3 \, x + 2 \, \log \left (x\right ) + 2 \, \log \left (-\frac {x^{2} - x e^{\left (5 \, x\right )} + 4}{x}\right ) \]

[In]

integrate(((13*x+2)*exp(5*x)-3*x^2-4*x-12)/(x*exp(5*x)-x^2-4),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {-12-4 x-3 x^2+e^{5 x} (2+13 x)}{-4+e^{5 x} x-x^2} \, dx=3 x + 2 \log {\left (x \right )} + 2 \log {\left (e^{5 x} + \frac {- x^{2} - 4}{x} \right )} \]

[In]

integrate(((13*x+2)*exp(5*x)-3*x**2-4*x-12)/(x*exp(5*x)-x**2-4),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {-12-4 x-3 x^2+e^{5 x} (2+13 x)}{-4+e^{5 x} x-x^2} \, dx=3 \, x + 2 \, \log \left (x\right ) + 2 \, \log \left (-\frac {x^{2} - x e^{\left (5 \, x\right )} + 4}{x}\right ) \]

[In]

integrate(((13*x+2)*exp(5*x)-3*x^2-4*x-12)/(x*exp(5*x)-x^2-4),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {-12-4 x-3 x^2+e^{5 x} (2+13 x)}{-4+e^{5 x} x-x^2} \, dx=3 \, x + 2 \, \log \left (-x^{2} + x e^{\left (5 \, x\right )} - 4\right ) \]

[In]

integrate(((13*x+2)*exp(5*x)-3*x^2-4*x-12)/(x*exp(5*x)-x^2-4),x, algorithm="giac")

[Out]

3*x + 2*log(-x^2 + x*e^(5*x) - 4)

Mupad [B] (verification not implemented)

Time = 13.35 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {-12-4 x-3 x^2+e^{5 x} (2+13 x)}{-4+e^{5 x} x-x^2} \, dx=3\,x+2\,\ln \left (x^2-x\,{\mathrm {e}}^{5\,x}+4\right ) \]

[In]

int((4*x - exp(5*x)*(13*x + 2) + 3*x^2 + 12)/(x^2 - x*exp(5*x) + 4),x)

[Out]

3*x + 2*log(x^2 - x*exp(5*x) + 4)