\(\int \frac {4+4 x}{e^9-2 x-x^2+\log (5)} \, dx\) [8027]

Optimal result

Integrand size = 22, antiderivative size = 21 \[ \int \frac {4+4 x}{e^9-2 x-x^2+\log (5)} \, dx=\log \left (\frac {1}{2 \left (e^9-2 x-x^2+\log (5)\right )^2}\right ) \]

[Out]

ln(1/2/(ln(5)+exp(9)-x^2-2*x)^2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {642} \[ \int \frac {4+4 x}{e^9-2 x-x^2+\log (5)} \, dx=-2 \log \left (-x^2-2 x+e^9+\log (5)\right ) \]

[In]

Int[(4 + 4*x)/(E^9 - 2*x - x^2 + Log[5]),x]

[Out]

-2*Log[E^9 - 2*x - x^2 + Log[5]]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps \begin{align*} \text {integral}& = -2 \log \left (e^9-2 x-x^2+\log (5)\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {4+4 x}{e^9-2 x-x^2+\log (5)} \, dx=-2 \log \left (e^9-2 x-x^2+\log (5)\right ) \]

[In]

Integrate[(4 + 4*x)/(E^9 - 2*x - x^2 + Log[5]),x]

[Out]

-2*Log[E^9 - 2*x - x^2 + Log[5]]

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81

method	result	size
default	\(-2 \ln \left (\ln \left (5\right )+{\mathrm e}^{9}-x^{2}-2 x \right )\)	\(17\)
norman	\(-2 \ln \left (\ln \left (5\right )+{\mathrm e}^{9}-x^{2}-2 x \right )\)	\(17\)
risch	\(-2 \ln \left (-\ln \left (5\right )-{\mathrm e}^{9}+x^{2}+2 x \right )\)	\(19\)
parallelrisch	\(-2 \ln \left (-\ln \left (5\right )-{\mathrm e}^{9}+x^{2}+2 x \right )\)	\(19\)

[In]

int((4+4*x)/(ln(5)+exp(9)-x^2-2*x),x,method=_RETURNVERBOSE)

[Out]

-2*ln(ln(5)+exp(9)-x^2-2*x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {4+4 x}{e^9-2 x-x^2+\log (5)} \, dx=-2 \, \log \left (x^{2} + 2 \, x - e^{9} - \log \left (5\right )\right ) \]

[In]

integrate((4+4*x)/(log(5)+exp(9)-x^2-2*x),x, algorithm="fricas")

[Out]

-2*log(x^2 + 2*x - e^9 - log(5))

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {4+4 x}{e^9-2 x-x^2+\log (5)} \, dx=- 2 \log {\left (x^{2} + 2 x - e^{9} - \log {\left (5 \right )} \right )} \]

[In]

integrate((4+4*x)/(ln(5)+exp(9)-x**2-2*x),x)

[Out]

-2*log(x**2 + 2*x - exp(9) - log(5))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {4+4 x}{e^9-2 x-x^2+\log (5)} \, dx=-2 \, \log \left (x^{2} + 2 \, x - e^{9} - \log \left (5\right )\right ) \]

[In]

integrate((4+4*x)/(log(5)+exp(9)-x^2-2*x),x, algorithm="maxima")

[Out]

-2*log(x^2 + 2*x - e^9 - log(5))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {4+4 x}{e^9-2 x-x^2+\log (5)} \, dx=-2 \, \log \left ({\left | x^{2} + 2 \, x - e^{9} - \log \left (5\right ) \right |}\right ) \]

[In]

integrate((4+4*x)/(log(5)+exp(9)-x^2-2*x),x, algorithm="giac")

[Out]

-2*log(abs(x^2 + 2*x - e^9 - log(5)))

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {4+4 x}{e^9-2 x-x^2+\log (5)} \, dx=-2\,\ln \left (x^2+2\,x-{\mathrm {e}}^9-\ln \left (5\right )\right ) \]

[In]

int(-(4*x + 4)/(2*x - exp(9) - log(5) + x^2),x)

[Out]

-2*log(2*x - exp(9) - log(5) + x^2)