\(\int \frac {-2+3 e x^2+e (-2+6 x) \log (4)+3 e \log ^2(4)}{e x^2+2 e x \log (4)+e \log ^2(4)} \, dx\) [1258]

Optimal result

Integrand size = 45, antiderivative size = 20 \[ \int \frac {-2+3 e x^2+e (-2+6 x) \log (4)+3 e \log ^2(4)}{e x^2+2 e x \log (4)+e \log ^2(4)} \, dx=x+2 \left (x+\frac {\frac {1}{e}-x}{x+\log (4)}\right ) \]

[Out]

3*x+2*(exp(-1)-x)/(x+2*ln(2))

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {27, 12, 1864} \[ \int \frac {-2+3 e x^2+e (-2+6 x) \log (4)+3 e \log ^2(4)}{e x^2+2 e x \log (4)+e \log ^2(4)} \, dx=3 x+\frac {2 (1+e \log (4))}{e (x+\log (4))} \]

[In]

Int[(-2 + 3*E*x^2 + E*(-2 + 6*x)*Log[4] + 3*E*Log[4]^2)/(E*x^2 + 2*E*x*Log[4] + E*Log[4]^2),x]

[Out]

3*x + (2*(1 + E*Log[4]))/(E*(x + Log[4]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1864

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rubi steps \begin{align*} \text {integral}& = \int \frac {-2+3 e x^2+e (-2+6 x) \log (4)+3 e \log ^2(4)}{e (x+\log (4))^2} \, dx \\ & = \frac {\int \frac {-2+3 e x^2+e (-2+6 x) \log (4)+3 e \log ^2(4)}{(x+\log (4))^2} \, dx}{e} \\ & = \frac {\int \left (3 e-\frac {2 (1+e \log (4))}{(x+\log (4))^2}\right ) \, dx}{e} \\ & = 3 x+\frac {2 (1+e \log (4))}{e (x+\log (4))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.25 \[ \int \frac {-2+3 e x^2+e (-2+6 x) \log (4)+3 e \log ^2(4)}{e x^2+2 e x \log (4)+e \log ^2(4)} \, dx=\frac {3 e (x+\log (4))+\frac {2+e \log (16)}{x+\log (4)}}{e} \]

[In]

Integrate[(-2 + 3*E*x^2 + E*(-2 + 6*x)*Log[4] + 3*E*Log[4]^2)/(E*x^2 + 2*E*x*Log[4] + E*Log[4]^2),x]

[Out]

(3*E*(x + Log[4]) + (2 + E*Log[16])/(x + Log[4]))/E

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.45

method	result	size
risch	\(\frac {4 \ln \left (2\right )+6 x \ln \left (2\right )+3 x^{2}+2 \,{\mathrm e}^{-1}}{x +2 \ln \left (2\right )}\)	\(29\)
default	\({\mathrm e}^{-1} \left (3 x \,{\mathrm e}-\frac {-4 \,{\mathrm e} \ln \left (2\right )-2}{x +2 \ln \left (2\right )}\right )\)	\(30\)
gosper	\(-\frac {\left (12 \,{\mathrm e} \ln \left (2\right )^{2}-3 x^{2} {\mathrm e}-4 \,{\mathrm e} \ln \left (2\right )-2\right ) {\mathrm e}^{-1}}{x +2 \ln \left (2\right )}\)	\(38\)
norman	\(\frac {3 x^{2}-2 \left (6 \,{\mathrm e} \ln \left (2\right )^{2}-2 \,{\mathrm e} \ln \left (2\right )-1\right ) {\mathrm e}^{-1}}{x +2 \ln \left (2\right )}\)	\(38\)
parallelrisch	\(-\frac {\left (12 \,{\mathrm e} \ln \left (2\right )^{2}-3 x^{2} {\mathrm e}-4 \,{\mathrm e} \ln \left (2\right )-2\right ) {\mathrm e}^{-1}}{x +2 \ln \left (2\right )}\)	\(38\)
meijerg	\(\frac {3 x}{1+\frac {x}{2 \ln \left (2\right )}}-\frac {{\mathrm e}^{-1} x}{2 \ln \left (2\right )^{2} \left (1+\frac {x}{2 \ln \left (2\right )}\right )}-\frac {x}{\ln \left (2\right ) \left (1+\frac {x}{2 \ln \left (2\right )}\right )}+12 \ln \left (2\right ) \left (-\frac {x}{2 \ln \left (2\right ) \left (1+\frac {x}{2 \ln \left (2\right )}\right )}+\ln \left (1+\frac {x}{2 \ln \left (2\right )}\right )\right )+6 \ln \left (2\right ) \left (\frac {x \left (\frac {3 x}{2 \ln \left (2\right )}+6\right )}{6 \ln \left (2\right ) \left (1+\frac {x}{2 \ln \left (2\right )}\right )}-2 \ln \left (1+\frac {x}{2 \ln \left (2\right )}\right )\right )\)	\(131\)

[In]

int((12*exp(1)*ln(2)^2+2*(6*x-2)*exp(1)*ln(2)+3*x^2*exp(1)-2)/(4*exp(1)*ln(2)^2+4*x*exp(1)*ln(2)+x^2*exp(1)),x
,method=_RETURNVERBOSE)

[Out]

(4*ln(2)+6*x*ln(2)+3*x^2+2*exp(-1))/(x+2*ln(2))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.70 \[ \int \frac {-2+3 e x^2+e (-2+6 x) \log (4)+3 e \log ^2(4)}{e x^2+2 e x \log (4)+e \log ^2(4)} \, dx=\frac {3 \, x^{2} e + 2 \, {\left (3 \, x + 2\right )} e \log \left (2\right ) + 2}{x e + 2 \, e \log \left (2\right )} \]

[In]

integrate((12*exp(1)*log(2)^2+2*(6*x-2)*exp(1)*log(2)+3*x^2*exp(1)-2)/(4*exp(1)*log(2)^2+4*x*exp(1)*log(2)+x^2
*exp(1)),x, algorithm="fricas")

[Out]

(3*x^2*e + 2*(3*x + 2)*e*log(2) + 2)/(x*e + 2*e*log(2))

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.30 \[ \int \frac {-2+3 e x^2+e (-2+6 x) \log (4)+3 e \log ^2(4)}{e x^2+2 e x \log (4)+e \log ^2(4)} \, dx=3 x + \frac {2 + 4 e \log {\left (2 \right )}}{e x + 2 e \log {\left (2 \right )}} \]

[In]

integrate((12*exp(1)*ln(2)**2+2*(6*x-2)*exp(1)*ln(2)+3*x**2*exp(1)-2)/(4*exp(1)*ln(2)**2+4*x*exp(1)*ln(2)+x**2
*exp(1)),x)

[Out]

3*x + (2 + 4*E*log(2))/(E*x + 2*E*log(2))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.35 \[ \int \frac {-2+3 e x^2+e (-2+6 x) \log (4)+3 e \log ^2(4)}{e x^2+2 e x \log (4)+e \log ^2(4)} \, dx=3 \, x + \frac {2 \, {\left (2 \, e \log \left (2\right ) + 1\right )}}{x e + 2 \, e \log \left (2\right )} \]

[In]

integrate((12*exp(1)*log(2)^2+2*(6*x-2)*exp(1)*log(2)+3*x^2*exp(1)-2)/(4*exp(1)*log(2)^2+4*x*exp(1)*log(2)+x^2
*exp(1)),x, algorithm="maxima")

[Out]

3*x + 2*(2*e*log(2) + 1)/(x*e + 2*e*log(2))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20 \[ \int \frac {-2+3 e x^2+e (-2+6 x) \log (4)+3 e \log ^2(4)}{e x^2+2 e x \log (4)+e \log ^2(4)} \, dx=3 \, x + \frac {2 \, {\left (2 \, e \log \left (2\right ) + 1\right )} e^{\left (-1\right )}}{x + 2 \, \log \left (2\right )} \]

[In]

integrate((12*exp(1)*log(2)^2+2*(6*x-2)*exp(1)*log(2)+3*x^2*exp(1)-2)/(4*exp(1)*log(2)^2+4*x*exp(1)*log(2)+x^2
*exp(1)),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {-2+3 e x^2+e (-2+6 x) \log (4)+3 e \log ^2(4)}{e x^2+2 e x \log (4)+e \log ^2(4)} \, dx=\int \frac {12\,\mathrm {e}\,{\ln \left (2\right )}^2+3\,x^2\,\mathrm {e}+2\,\mathrm {e}\,\ln \left (2\right )\,\left (6\,x-2\right )-2}{\mathrm {e}\,x^2+4\,\mathrm {e}\,\ln \left (2\right )\,x+4\,\mathrm {e}\,{\ln \left (2\right )}^2} \,d x \]

[In]

int((12*exp(1)*log(2)^2 + 3*x^2*exp(1) + 2*exp(1)*log(2)*(6*x - 2) - 2)/(4*exp(1)*log(2)^2 + x^2*exp(1) + 4*x*
exp(1)*log(2)),x)

[Out]

int((12*exp(1)*log(2)^2 + 3*x^2*exp(1) + 2*exp(1)*log(2)*(6*x - 2) - 2)/(4*exp(1)*log(2)^2 + x^2*exp(1) + 4*x*
exp(1)*log(2)), x)