3.50.13 \(\int \frac {8 x^2+(2 e^3-4 x^2) \log (e^3-2 x^2)+(4 e^3-8 x^2) \log ^2(e^3-2 x^2)}{(e^3-2 x^2) \log ^2(e^3-2 x^2)} \, dx\)

Optimal. Leaf size=25 \[ -2+2 \left (4+2 x+\log (4)+\frac {x}{\log \left (e^3-2 x^2\right )}\right ) \]

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Rubi [F]  time = 0.21, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {8 x^2+\left (2 e^3-4 x^2\right ) \log \left (e^3-2 x^2\right )+\left (4 e^3-8 x^2\right ) \log ^2\left (e^3-2 x^2\right )}{\left (e^3-2 x^2\right ) \log ^2\left (e^3-2 x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(8*x^2 + (2*E^3 - 4*x^2)*Log[E^3 - 2*x^2] + (4*E^3 - 8*x^2)*Log[E^3 - 2*x^2]^2)/((E^3 - 2*x^2)*Log[E^3 - 2
*x^2]^2),x]

[Out]

4*x + 8*Defer[Int][x^2/((E^3 - 2*x^2)*Log[E^3 - 2*x^2]^2), x] + 2*Defer[Int][Log[E^3 - 2*x^2]^(-1), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (4+\frac {8 x^2}{\left (e^3-2 x^2\right ) \log ^2\left (e^3-2 x^2\right )}+\frac {2}{\log \left (e^3-2 x^2\right )}\right ) \, dx\\ &=4 x+2 \int \frac {1}{\log \left (e^3-2 x^2\right )} \, dx+8 \int \frac {x^2}{\left (e^3-2 x^2\right ) \log ^2\left (e^3-2 x^2\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.93, size = 19, normalized size = 0.76 \begin {gather*} 4 x+\frac {2 x}{\log \left (e^3-2 x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(8*x^2 + (2*E^3 - 4*x^2)*Log[E^3 - 2*x^2] + (4*E^3 - 8*x^2)*Log[E^3 - 2*x^2]^2)/((E^3 - 2*x^2)*Log[E
^3 - 2*x^2]^2),x]

[Out]

4*x + (2*x)/Log[E^3 - 2*x^2]

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fricas [A]  time = 1.19, size = 27, normalized size = 1.08 \begin {gather*} \frac {2 \, {\left (2 \, x \log \left (-2 \, x^{2} + e^{3}\right ) + x\right )}}{\log \left (-2 \, x^{2} + e^{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(3)-8*x^2)*log(exp(3)-2*x^2)^2+(2*exp(3)-4*x^2)*log(exp(3)-2*x^2)+8*x^2)/(exp(3)-2*x^2)/log(e
xp(3)-2*x^2)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.20, size = 27, normalized size = 1.08 \begin {gather*} \frac {2 \, {\left (2 \, x \log \left (-2 \, x^{2} + e^{3}\right ) + x\right )}}{\log \left (-2 \, x^{2} + e^{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(3)-8*x^2)*log(exp(3)-2*x^2)^2+(2*exp(3)-4*x^2)*log(exp(3)-2*x^2)+8*x^2)/(exp(3)-2*x^2)/log(e
xp(3)-2*x^2)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.14, size = 19, normalized size = 0.76

method	result	size
risch	\(4 x +\frac {2 x}{\ln \left ({\mathrm e}^{3}-2 x^{2}\right )}\)	\(19\)
norman	\(\frac {2 x +4 x \ln \left ({\mathrm e}^{3}-2 x^{2}\right )}{\ln \left ({\mathrm e}^{3}-2 x^{2}\right )}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*exp(3)-8*x^2)*ln(exp(3)-2*x^2)^2+(2*exp(3)-4*x^2)*ln(exp(3)-2*x^2)+8*x^2)/(exp(3)-2*x^2)/ln(exp(3)-2*x
^2)^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.40, size = 27, normalized size = 1.08 \begin {gather*} \frac {2 \, {\left (2 \, x \log \left (-2 \, x^{2} + e^{3}\right ) + x\right )}}{\log \left (-2 \, x^{2} + e^{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(3)-8*x^2)*log(exp(3)-2*x^2)^2+(2*exp(3)-4*x^2)*log(exp(3)-2*x^2)+8*x^2)/(exp(3)-2*x^2)/log(e
xp(3)-2*x^2)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.46, size = 18, normalized size = 0.72 \begin {gather*} 4\,x+\frac {2\,x}{\ln \left ({\mathrm {e}}^3-2\,x^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(exp(3) - 2*x^2)*(2*exp(3) - 4*x^2) + log(exp(3) - 2*x^2)^2*(4*exp(3) - 8*x^2) + 8*x^2)/(log(exp(3) -
2*x^2)^2*(exp(3) - 2*x^2)),x)

[Out]

4*x + (2*x)/log(exp(3) - 2*x^2)

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sympy [A]  time = 0.13, size = 15, normalized size = 0.60 \begin {gather*} 4 x + \frac {2 x}{\log {\left (- 2 x^{2} + e^{3} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(3)-8*x**2)*ln(exp(3)-2*x**2)**2+(2*exp(3)-4*x**2)*ln(exp(3)-2*x**2)+8*x**2)/(exp(3)-2*x**2)/
ln(exp(3)-2*x**2)**2,x)

[Out]

4*x + 2*x/log(-2*x**2 + exp(3))

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