3.45.55 \(\int \frac {-2+e^2+(-4-e^4+e^2 (4-4 x)+8 x-4 x^2) \log (4)}{(4+e^4-8 x+4 x^2+e^2 (-4+4 x)) \log (4)} \, dx\)

Optimal. Leaf size=24 \[ 2+\frac {1}{e^3}-x+\frac {x}{\left (-2+e^2+2 x\right ) \log (4)} \]

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 31, normalized size of antiderivative = 1.29, number of steps used = 5, number of rules used = 4, integrand size = 61, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.066, Rules used = {12, 1984, 27, 683} \begin {gather*} -x-\frac {2-e^2}{2 \left (-2 x-e^2+2\right ) \log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2 + E^2 + (-4 - E^4 + E^2*(4 - 4*x) + 8*x - 4*x^2)*Log[4])/((4 + E^4 - 8*x + 4*x^2 + E^2*(-4 + 4*x))*Log
[4]),x]

[Out]

-x - (2 - E^2)/(2*(2 - E^2 - 2*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 683

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e,
0] && IGtQ[p, 0] &&  !(EqQ[m, 3] && NeQ[p, 1])

Rule 1984

Int[(u_)^(p_.)*(v_)^(q_.), x_Symbol] :> Int[ExpandToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{p, q}, x] &&
QuadraticQ[{u, v}, x] &&  !QuadraticMatchQ[{u, v}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-2+e^2+\left (-4-e^4+e^2 (4-4 x)+8 x-4 x^2\right ) \log (4)}{4+e^4-8 x+4 x^2+e^2 (-4+4 x)} \, dx}{\log (4)}\\ &=\frac {\int \frac {4 \left (2-e^2\right ) x \log (4)-4 x^2 \log (4)-\left (2-e^2\right ) \left (1+2 \log (4)-e^2 \log (4)\right )}{\left (-2+e^2\right )^2-4 \left (2-e^2\right ) x+4 x^2} \, dx}{\log (4)}\\ &=\frac {\int \frac {4 \left (2-e^2\right ) x \log (4)-4 x^2 \log (4)-\left (2-e^2\right ) \left (1+2 \log (4)-e^2 \log (4)\right )}{\left (-2+e^2+2 x\right )^2} \, dx}{\log (4)}\\ &=\frac {\int \left (\frac {-2+e^2}{\left (-2+e^2+2 x\right )^2}-\log (4)\right ) \, dx}{\log (4)}\\ &=-x-\frac {2-e^2}{2 \left (2-e^2-2 x\right ) \log (4)}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [B]  time = 0.04, size = 49, normalized size = 2.04 \begin {gather*} -\frac {-2+e^2-4 e^2 \log (4)+\left (-2+e^2+2 x\right )^2 \log (4)+e^2 \log (256)}{2 \left (-2+e^2+2 x\right ) \log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 + E^2 + (-4 - E^4 + E^2*(4 - 4*x) + 8*x - 4*x^2)*Log[4])/((4 + E^4 - 8*x + 4*x^2 + E^2*(-4 + 4*x
))*Log[4]),x]

[Out]

-1/2*(-2 + E^2 - 4*E^2*Log[4] + (-2 + E^2 + 2*x)^2*Log[4] + E^2*Log[256])/((-2 + E^2 + 2*x)*Log[4])

________________________________________________________________________________________

fricas [A]  time = 0.49, size = 36, normalized size = 1.50 \begin {gather*} -\frac {4 \, {\left (2 \, x^{2} + x e^{2} - 2 \, x\right )} \log \relax (2) + e^{2} - 2}{4 \, {\left (2 \, x + e^{2} - 2\right )} \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2*(-exp(2)^2+(-4*x+4)*exp(2)-4*x^2+8*x-4)*log(2)+exp(2)-2)/(exp(2)^2+(4*x-4)*exp(2)+4*x^2-8*x+4
)/log(2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2*(-exp(2)^2+(-4*x+4)*exp(2)-4*x^2+8*x-4)*log(2)+exp(2)-2)/(exp(2)^2+(4*x-4)*exp(2)+4*x^2-8*x+4
)/log(2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: -ln(2)^-1/2*(8*sageVARx*ln(2)/4+(-exp(2)
+2)*1/4/sqrt(exp(2)^2-exp(4))*ln(sqrt((8*sageVARx+4*exp(2)-8)^2+(-4*sqrt(-exp(2)^2+exp(4)))^2)/sqrt((8*sageVAR
x+4*exp(2)-8)^2+(4*sq

________________________________________________________________________________________

maple [A]  time = 0.24, size = 37, normalized size = 1.54

method	result	size
risch	\(-x -\frac {{\mathrm e}^{2}}{4 \ln \relax (2) \left (2 x -2+{\mathrm e}^{2}\right )}+\frac {1}{2 \ln \relax (2) \left (2 x -2+{\mathrm e}^{2}\right )}\)	\(37\)
gosper	\(\frac {2 \,{\mathrm e}^{4} \ln \relax (2)-8 x^{2} \ln \relax (2)-8 \,{\mathrm e}^{2} \ln \relax (2)-{\mathrm e}^{2}+8 \ln \relax (2)+2}{4 \ln \relax (2) \left (2 x -2+{\mathrm e}^{2}\right )}\)	\(47\)
meijerg	\(-\frac {x}{2 \left (\frac {{\mathrm e}^{2}}{2}-1\right ) \ln \relax (2) \left (1+\frac {2 x}{{\mathrm e}^{2}-2}\right ) \left ({\mathrm e}^{2}-2\right )}-\frac {{\mathrm e}^{4} x}{2 \left (\frac {{\mathrm e}^{2}}{2}-1\right ) \left (1+\frac {2 x}{{\mathrm e}^{2}-2}\right ) \left ({\mathrm e}^{2}-2\right )}+\frac {2 \,{\mathrm e}^{2} x}{\left (\frac {{\mathrm e}^{2}}{2}-1\right ) \left (1+\frac {2 x}{{\mathrm e}^{2}-2}\right ) \left ({\mathrm e}^{2}-2\right )}+\frac {{\mathrm e}^{2} x}{4 \left (\frac {{\mathrm e}^{2}}{2}-1\right ) \ln \relax (2) \left (1+\frac {2 x}{{\mathrm e}^{2}-2}\right ) \left ({\mathrm e}^{2}-2\right )}-\frac {2 x}{\left (\frac {{\mathrm e}^{2}}{2}-1\right ) \left (1+\frac {2 x}{{\mathrm e}^{2}-2}\right ) \left ({\mathrm e}^{2}-2\right )}-\left ({\mathrm e}^{2}-2\right ) \left (-\frac {2 x}{\left (1+\frac {2 x}{{\mathrm e}^{2}-2}\right ) \left ({\mathrm e}^{2}-2\right )}+\ln \left (1+\frac {2 x}{{\mathrm e}^{2}-2}\right )\right )-\frac {\left ({\mathrm e}^{2}-2\right )^{2} \left (\frac {2 x \left (\frac {6 x}{{\mathrm e}^{2}-2}+6\right )}{3 \left ({\mathrm e}^{2}-2\right ) \left (1+\frac {2 x}{{\mathrm e}^{2}-2}\right )}-2 \ln \left (1+\frac {2 x}{{\mathrm e}^{2}-2}\right )\right )}{4 \left (\frac {{\mathrm e}^{2}}{2}-1\right )}\)	\(271\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(2*(-exp(2)^2+(-4*x+4)*exp(2)-4*x^2+8*x-4)*ln(2)+exp(2)-2)/(exp(2)^2+(4*x-4)*exp(2)+4*x^2-8*x+4)/ln(2)
,x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.37, size = 26, normalized size = 1.08 \begin {gather*} -\frac {4 \, x \log \relax (2) + \frac {e^{2} - 2}{2 \, x + e^{2} - 2}}{4 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2*(-exp(2)^2+(-4*x+4)*exp(2)-4*x^2+8*x-4)*log(2)+exp(2)-2)/(exp(2)^2+(4*x-4)*exp(2)+4*x^2-8*x+4
)/log(2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.18, size = 30, normalized size = 1.25 \begin {gather*} -x-\frac {\frac {{\mathrm {e}}^2}{2}-1}{2\,{\mathrm {e}}^2\,\ln \relax (2)-4\,\ln \relax (2)+4\,x\,\ln \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(2)*(exp(4) - 8*x + 4*x^2 + exp(2)*(4*x - 4) + 4) - exp(2)/2 + 1)/(log(2)*(exp(4) - 8*x + 4*x^2 + exp
(2)*(4*x - 4) + 4)),x)

[Out]

- x - (exp(2)/2 - 1)/(2*exp(2)*log(2) - 4*log(2) + 4*x*log(2))

________________________________________________________________________________________

sympy [A]  time = 0.28, size = 27, normalized size = 1.12 \begin {gather*} - x - \frac {-2 + e^{2}}{8 x \log {\relax (2 )} - 8 \log {\relax (2 )} + 4 e^{2} \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2*(-exp(2)**2+(-4*x+4)*exp(2)-4*x**2+8*x-4)*ln(2)+exp(2)-2)/(exp(2)**2+(4*x-4)*exp(2)+4*x**2-8*
x+4)/ln(2),x)

[Out]

-x - (-2 + exp(2))/(8*x*log(2) - 8*log(2) + 4*exp(2)*log(2))

________________________________________________________________________________________