3.1.60 \(\int \frac {2 x-12 e x^2-12 x^3+(-3 e-3 x) \log (e^2+2 e x+x^2)}{12 e x^3+12 x^4+(e x+x^2) \log (e^2+2 e x+x^2)} \, dx\)

Optimal. Leaf size=24 \[ -4+\log \left (\frac {\log (5) \left (3+\frac {\log \left ((e+x)^2\right )}{4 x^2}\right )}{x}\right ) \]

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Rubi [A]  time = 0.54, antiderivative size = 18, normalized size of antiderivative = 0.75, number of steps used = 4, number of rules used = 3, integrand size = 70, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {6688, 6742, 6684} \begin {gather*} \log \left (12 x^2+\log \left ((x+e)^2\right )\right )-3 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*x - 12*E*x^2 - 12*x^3 + (-3*E - 3*x)*Log[E^2 + 2*E*x + x^2])/(12*E*x^3 + 12*x^4 + (E*x + x^2)*Log[E^2 +
2*E*x + x^2]),x]

[Out]

-3*Log[x] + Log[12*x^2 + Log[(E + x)^2]]

Rule 6684

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

Rule 6688

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-2 x \left (-1+6 e x+6 x^2\right )-3 (e+x) \log \left ((e+x)^2\right )}{x (e+x) \left (12 x^2+\log \left ((e+x)^2\right )\right )} \, dx\\ &=\int \left (-\frac {3}{x}+\frac {2 \left (1+12 e x+12 x^2\right )}{(e+x) \left (12 x^2+\log \left ((e+x)^2\right )\right )}\right ) \, dx\\ &=-3 \log (x)+2 \int \frac {1+12 e x+12 x^2}{(e+x) \left (12 x^2+\log \left ((e+x)^2\right )\right )} \, dx\\ &=-3 \log (x)+\log \left (12 x^2+\log \left ((e+x)^2\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.20, size = 18, normalized size = 0.75 \begin {gather*} -3 \log (x)+\log \left (12 x^2+\log \left ((e+x)^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*x - 12*E*x^2 - 12*x^3 + (-3*E - 3*x)*Log[E^2 + 2*E*x + x^2])/(12*E*x^3 + 12*x^4 + (E*x + x^2)*Log
[E^2 + 2*E*x + x^2]),x]

[Out]

-3*Log[x] + Log[12*x^2 + Log[(E + x)^2]]

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fricas [A]  time = 0.67, size = 24, normalized size = 1.00 \begin {gather*} \log \left (12 \, x^{2} + \log \left (x^{2} + 2 \, x e + e^{2}\right )\right ) - 3 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(12*x^2 + log(x^2 + 2*x*e + e^2)) - 3*log(x)

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giac [A]  time = 0.43, size = 26, normalized size = 1.08 \begin {gather*} \log \left (-12 \, x^{2} - \log \left (x^{2} + 2 \, x e + e^{2}\right )\right ) - 3 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(-12*x^2 - log(x^2 + 2*x*e + e^2)) - 3*log(x)

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maple [A]  time = 0.10, size = 25, normalized size = 1.04

method	result	size
risch	\(-3 \ln \relax (x )+\ln \left (12 x^{2}+\ln \left ({\mathrm e}^{2}+2 x \,{\mathrm e}+x^{2}\right )\right )\)	\(25\)
norman	\(-3 \ln \relax (x )+\ln \left (12 x^{2}+\ln \left ({\mathrm e}^{2}+2 x \,{\mathrm e}+x^{2}\right )\right )\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*ln(x)+ln(12*x^2+ln(exp(2)+2*x*exp(1)+x^2))

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maxima [A]  time = 0.55, size = 17, normalized size = 0.71 \begin {gather*} \log \left (6 \, x^{2} + \log \left (x + e\right )\right ) - 3 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(6*x^2 + log(x + e)) - 3*log(x)

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mupad [B]  time = 0.55, size = 19, normalized size = 0.79 \begin {gather*} \ln \left (\frac {\ln \left ({\left (x+\mathrm {e}\right )}^2\right )}{12}+x^2\right )-3\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(log((x + exp(1))^2)/12 + x^2) - 3*log(x)

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sympy [A]  time = 0.20, size = 26, normalized size = 1.08 \begin {gather*} - 3 \log {\relax (x )} + \log {\left (12 x^{2} + \log {\left (x^{2} + 2 e x + e^{2} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*log(x) + log(12*x**2 + log(x**2 + 2*E*x + exp(2)))

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