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

Optimal. Leaf size=20 \[ x+2 \left (x+\frac {\frac {1}{e}-x}{x+\log (4)}\right ) \]

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Rubi [A]  time = 0.04, antiderivative size = 21, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 3, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {27, 12, 1850} \begin {gather*} 3 x+\frac {2 (1+e \log (4))}{e (x+\log (4))} \end {gather*}

Antiderivative was successfully verified.

[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 1850

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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 25, normalized size = 1.25 \begin {gather*} \frac {3 e (x+\log (4))+\frac {2+e \log (16)}{x+\log (4)}}{e} \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.59, size = 34, normalized size = 1.70 \begin {gather*} \frac {3 \, x^{2} e + 2 \, {\left (3 \, x + 2\right )} e \log \relax (2) + 2}{x e + 2 \, e \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [A]  time = 0.30, size = 24, normalized size = 1.20 \begin {gather*} 3 \, x + \frac {2 \, {\left (2 \, e \log \relax (2) + 1\right )} e^{\left (-1\right )}}{x + 2 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maple [A]  time = 0.24, size = 30, normalized size = 1.50




method result size



default \({\mathrm e}^{-1} \left (3 x \,{\mathrm e}-\frac {-4 \,{\mathrm e} \ln \relax (2)-2}{x +2 \ln \relax (2)}\right )\) \(30\)
risch \(3 x +\frac {4 \,{\mathrm e}^{-1} {\mathrm e} \ln \relax (2)}{x +2 \ln \relax (2)}+\frac {2 \,{\mathrm e}^{-1}}{x +2 \ln \relax (2)}\) \(33\)
gosper \(-\frac {\left (12 \,{\mathrm e} \ln \relax (2)^{2}-3 x^{2} {\mathrm e}-4 \,{\mathrm e} \ln \relax (2)-2\right ) {\mathrm e}^{-1}}{x +2 \ln \relax (2)}\) \(38\)
norman \(\frac {3 x^{2}-2 \left (6 \,{\mathrm e} \ln \relax (2)^{2}-2 \,{\mathrm e} \ln \relax (2)-1\right ) {\mathrm e}^{-1}}{x +2 \ln \relax (2)}\) \(38\)
meijerg \(\frac {3 x}{1+\frac {x}{2 \ln \relax (2)}}-\frac {{\mathrm e}^{-1} x}{2 \ln \relax (2)^{2} \left (1+\frac {x}{2 \ln \relax (2)}\right )}-\frac {x}{\ln \relax (2) \left (1+\frac {x}{2 \ln \relax (2)}\right )}+12 \ln \relax (2) \left (-\frac {x}{2 \ln \relax (2) \left (1+\frac {x}{2 \ln \relax (2)}\right )}+\ln \left (1+\frac {x}{2 \ln \relax (2)}\right )\right )+6 \ln \relax (2) \left (\frac {x \left (\frac {3 x}{2 \ln \relax (2)}+6\right )}{6 \ln \relax (2) \left (1+\frac {x}{2 \ln \relax (2)}\right )}-2 \ln \left (1+\frac {x}{2 \ln \relax (2)}\right )\right )\) \(131\)



Verification of antiderivative is not currently implemented for this CAS.

[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]

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

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maxima [A]  time = 0.35, size = 27, normalized size = 1.35 \begin {gather*} 3 \, x + \frac {2 \, {\left (2 \, e \log \relax (2) + 1\right )}}{x e + 2 \, e \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int \frac {12\,\mathrm {e}\,{\ln \relax (2)}^2+3\,x^2\,\mathrm {e}+2\,\mathrm {e}\,\ln \relax (2)\,\left (6\,x-2\right )-2}{\mathrm {e}\,x^2+4\,\mathrm {e}\,\ln \relax (2)\,x+4\,\mathrm {e}\,{\ln \relax (2)}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.18, size = 26, normalized size = 1.30 \begin {gather*} 3 x + \frac {2 + 4 e \log {\relax (2 )}}{e x + 2 e \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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