\(\int \frac {2+x+e^4 x-x^2+x \log (\frac {16}{x^2})}{e^4 x-x^2+x \log (\frac {16}{x^2})} \, dx\) [2963]

Optimal result

Integrand size = 43, antiderivative size = 21 \[ \int \frac {2+x+e^4 x-x^2+x \log \left (\frac {16}{x^2}\right )}{e^4 x-x^2+x \log \left (\frac {16}{x^2}\right )} \, dx=-5+x-\log \left (-e^4+x-\log \left (\frac {16}{x^2}\right )\right ) \]

[Out]

x-ln(x-ln(16/x^2)-exp(4))-5

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {6, 6874, 6816} \[ \int \frac {2+x+e^4 x-x^2+x \log \left (\frac {16}{x^2}\right )}{e^4 x-x^2+x \log \left (\frac {16}{x^2}\right )} \, dx=x-\log \left (\log \left (\frac {16}{x^2}\right )-x+e^4\right ) \]

[In]

Int[(2 + x + E^4*x - x^2 + x*Log[16/x^2])/(E^4*x - x^2 + x*Log[16/x^2]),x]

[Out]

x - Log[E^4 - x + Log[16/x^2]]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 6816

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps \begin{align*} \text {integral}& = \int \frac {2+\left (1+e^4\right ) x-x^2+x \log \left (\frac {16}{x^2}\right )}{e^4 x-x^2+x \log \left (\frac {16}{x^2}\right )} \, dx \\ & = \int \left (1+\frac {-2-x}{x \left (-e^4+x-\log \left (\frac {16}{x^2}\right )\right )}\right ) \, dx \\ & = x+\int \frac {-2-x}{x \left (-e^4+x-\log \left (\frac {16}{x^2}\right )\right )} \, dx \\ & = x-\log \left (e^4-x+\log \left (\frac {16}{x^2}\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {2+x+e^4 x-x^2+x \log \left (\frac {16}{x^2}\right )}{e^4 x-x^2+x \log \left (\frac {16}{x^2}\right )} \, dx=x-\log \left (e^4-x+\log \left (\frac {16}{x^2}\right )\right ) \]

[In]

Integrate[(2 + x + E^4*x - x^2 + x*Log[16/x^2])/(E^4*x - x^2 + x*Log[16/x^2]),x]

[Out]

x - Log[E^4 - x + Log[16/x^2]]

Maple [A] (verified)

Time = 2.64 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86

method	result	size
norman	\(x -\ln \left ({\mathrm e}^{4}+\ln \left (\frac {16}{x^{2}}\right )-x \right )\)	\(18\)
risch	\(x -\ln \left ({\mathrm e}^{4}+\ln \left (\frac {16}{x^{2}}\right )-x \right )\)	\(18\)
parallelrisch	\(x -\ln \left (x -\ln \left (\frac {16}{x^{2}}\right )-{\mathrm e}^{4}\right )\)	\(20\)

[In]

int((x*ln(16/x^2)+x*exp(4)-x^2+x+2)/(x*ln(16/x^2)+x*exp(4)-x^2),x,method=_RETURNVERBOSE)

[Out]

x-ln(exp(4)+ln(16/x^2)-x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {2+x+e^4 x-x^2+x \log \left (\frac {16}{x^2}\right )}{e^4 x-x^2+x \log \left (\frac {16}{x^2}\right )} \, dx=x - \log \left (-x + e^{4} + \log \left (\frac {16}{x^{2}}\right )\right ) \]

[In]

integrate((x*log(16/x^2)+x*exp(4)-x^2+x+2)/(x*log(16/x^2)+x*exp(4)-x^2),x, algorithm="fricas")

[Out]

x - log(-x + e^4 + log(16/x^2))

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67 \[ \int \frac {2+x+e^4 x-x^2+x \log \left (\frac {16}{x^2}\right )}{e^4 x-x^2+x \log \left (\frac {16}{x^2}\right )} \, dx=x - \log {\left (- x + \log {\left (\frac {16}{x^{2}} \right )} + e^{4} \right )} \]

[In]

integrate((x*ln(16/x**2)+x*exp(4)-x**2+x+2)/(x*ln(16/x**2)+x*exp(4)-x**2),x)

[Out]

x - log(-x + log(16/x**2) + exp(4))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {2+x+e^4 x-x^2+x \log \left (\frac {16}{x^2}\right )}{e^4 x-x^2+x \log \left (\frac {16}{x^2}\right )} \, dx=x - \log \left (\frac {1}{2} \, x - \frac {1}{2} \, e^{4} - 2 \, \log \left (2\right ) + \log \left (x\right )\right ) \]

[In]

integrate((x*log(16/x^2)+x*exp(4)-x^2+x+2)/(x*log(16/x^2)+x*exp(4)-x^2),x, algorithm="maxima")

[Out]

x - log(1/2*x - 1/2*e^4 - 2*log(2) + log(x))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {2+x+e^4 x-x^2+x \log \left (\frac {16}{x^2}\right )}{e^4 x-x^2+x \log \left (\frac {16}{x^2}\right )} \, dx=x - \log \left (x - e^{4} - \log \left (\frac {16}{x^{2}}\right )\right ) \]

[In]

integrate((x*log(16/x^2)+x*exp(4)-x^2+x+2)/(x*log(16/x^2)+x*exp(4)-x^2),x, algorithm="giac")

[Out]

x - log(x - e^4 - log(16/x^2))

Mupad [B] (verification not implemented)

Time = 8.99 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {2+x+e^4 x-x^2+x \log \left (\frac {16}{x^2}\right )}{e^4 x-x^2+x \log \left (\frac {16}{x^2}\right )} \, dx=x-\ln \left (x-{\mathrm {e}}^4+\ln \left (\frac {x^2}{16}\right )\right ) \]

[In]

int((x + x*exp(4) + x*log(16/x^2) - x^2 + 2)/(x*exp(4) + x*log(16/x^2) - x^2),x)

[Out]

x - log(x - exp(4) + log(x^2/16))