3.43.24 \(\int (1-6 x^2+5 x^4+(4-12 x^2) \log (\frac {1}{5} (-2+5 e^3))+(6-6 x^2) \log ^2(\frac {1}{5} (-2+5 e^3))+4 \log ^3(\frac {1}{5} (-2+5 e^3))+\log ^4(\frac {1}{5} (-2+5 e^3))) \, dx\)

Optimal. Leaf size=22 \[ x \left (x^2-\left (1+\log \left (-\frac {2}{5}+e^3\right )\right )^2\right )^2 \]

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Rubi [B]  time = 0.03, antiderivative size = 87, normalized size of antiderivative = 3.95, number of steps used = 3, number of rules used = 0, integrand size = 84, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} x^5-2 x^3-2 x^3 \log ^2\left (e^3-\frac {2}{5}\right )-4 x^3 \log \left (e^3-\frac {2}{5}\right )+6 x \log ^2\left (e^3-\frac {2}{5}\right )+x \left (1+\log ^4\left (e^3-\frac {2}{5}\right )+4 \log ^3\left (e^3-\frac {2}{5}\right )\right )+4 x \log \left (e^3-\frac {2}{5}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1 - 6*x^2 + 5*x^4 + (4 - 12*x^2)*Log[(-2 + 5*E^3)/5] + (6 - 6*x^2)*Log[(-2 + 5*E^3)/5]^2 + 4*Log[(-2 + 5*E
^3)/5]^3 + Log[(-2 + 5*E^3)/5]^4,x]

[Out]

-2*x^3 + x^5 + 4*x*Log[-2/5 + E^3] - 4*x^3*Log[-2/5 + E^3] + 6*x*Log[-2/5 + E^3]^2 - 2*x^3*Log[-2/5 + E^3]^2 +
x*(1 + 4*Log[-2/5 + E^3]^3 + Log[-2/5 + E^3]^4)

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-2 x^3+x^5+x \left (1+4 \log ^3\left (-\frac {2}{5}+e^3\right )+\log ^4\left (-\frac {2}{5}+e^3\right )\right )+\log \left (-\frac {2}{5}+e^3\right ) \int \left (4-12 x^2\right ) \, dx+\log ^2\left (-\frac {2}{5}+e^3\right ) \int \left (6-6 x^2\right ) \, dx\\ &=-2 x^3+x^5+4 x \log \left (-\frac {2}{5}+e^3\right )-4 x^3 \log \left (-\frac {2}{5}+e^3\right )+6 x \log ^2\left (-\frac {2}{5}+e^3\right )-2 x^3 \log ^2\left (-\frac {2}{5}+e^3\right )+x \left (1+4 \log ^3\left (-\frac {2}{5}+e^3\right )+\log ^4\left (-\frac {2}{5}+e^3\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 22, normalized size = 1.00 \begin {gather*} x \left (x^2-\left (1+\log \left (-\frac {2}{5}+e^3\right )\right )^2\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1 - 6*x^2 + 5*x^4 + (4 - 12*x^2)*Log[(-2 + 5*E^3)/5] + (6 - 6*x^2)*Log[(-2 + 5*E^3)/5]^2 + 4*Log[(-2
+ 5*E^3)/5]^3 + Log[(-2 + 5*E^3)/5]^4,x]

[Out]

x*(x^2 - (1 + Log[-2/5 + E^3])^2)^2

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fricas [B]  time = 0.74, size = 59, normalized size = 2.68 \begin {gather*} x^{5} + x \log \left (e^{3} - \frac {2}{5}\right )^{4} + 4 \, x \log \left (e^{3} - \frac {2}{5}\right )^{3} - 2 \, x^{3} - 2 \, {\left (x^{3} - 3 \, x\right )} \log \left (e^{3} - \frac {2}{5}\right )^{2} - 4 \, {\left (x^{3} - x\right )} \log \left (e^{3} - \frac {2}{5}\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(exp(3)-2/5)^4+4*log(exp(3)-2/5)^3+(-6*x^2+6)*log(exp(3)-2/5)^2+(-12*x^2+4)*log(exp(3)-2/5)+5*x^4
-6*x^2+1,x, algorithm="fricas")

[Out]

x^5 + x*log(e^3 - 2/5)^4 + 4*x*log(e^3 - 2/5)^3 - 2*x^3 - 2*(x^3 - 3*x)*log(e^3 - 2/5)^2 - 4*(x^3 - x)*log(e^3
- 2/5) + x

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giac [B]  time = 0.19, size = 59, normalized size = 2.68 \begin {gather*} x^{5} + x \log \left (e^{3} - \frac {2}{5}\right )^{4} + 4 \, x \log \left (e^{3} - \frac {2}{5}\right )^{3} - 2 \, x^{3} - 2 \, {\left (x^{3} - 3 \, x\right )} \log \left (e^{3} - \frac {2}{5}\right )^{2} - 4 \, {\left (x^{3} - x\right )} \log \left (e^{3} - \frac {2}{5}\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(exp(3)-2/5)^4+4*log(exp(3)-2/5)^3+(-6*x^2+6)*log(exp(3)-2/5)^2+(-12*x^2+4)*log(exp(3)-2/5)+5*x^4
-6*x^2+1,x, algorithm="giac")

[Out]

x^5 + x*log(e^3 - 2/5)^4 + 4*x*log(e^3 - 2/5)^3 - 2*x^3 - 2*(x^3 - 3*x)*log(e^3 - 2/5)^2 - 4*(x^3 - x)*log(e^3
- 2/5) + x

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maple [B]  time = 0.04, size = 62, normalized size = 2.82

method	result	size
default	\(\ln \left ({\mathrm e}^{3}-\frac {2}{5}\right )^{4} x +4 \ln \left ({\mathrm e}^{3}-\frac {2}{5}\right )^{3} x +\ln \left ({\mathrm e}^{3}-\frac {2}{5}\right )^{2} \left (-2 x^{3}+6 x \right )+\ln \left ({\mathrm e}^{3}-\frac {2}{5}\right ) \left (-4 x^{3}+4 x \right )+x^{5}-2 x^{3}+x\)	\(62\)
gosper	\(\left (\ln \left ({\mathrm e}^{3}-\frac {2}{5}\right )+1-x \right ) x \left (\ln \left ({\mathrm e}^{3}-\frac {2}{5}\right )^{3}+x \ln \left ({\mathrm e}^{3}-\frac {2}{5}\right )^{2}-\ln \left ({\mathrm e}^{3}-\frac {2}{5}\right ) x^{2}-x^{3}+3 \ln \left ({\mathrm e}^{3}-\frac {2}{5}\right )^{2}+2 x \ln \left ({\mathrm e}^{3}-\frac {2}{5}\right )-x^{2}+3 \ln \left ({\mathrm e}^{3}-\frac {2}{5}\right )+x +1\right )\)	\(76\)
norman	\(x^{5}+\left (-2 \ln \relax (5)^{2}+4 \ln \relax (5) \ln \left (5 \,{\mathrm e}^{3}-2\right )-2 \ln \left (5 \,{\mathrm e}^{3}-2\right )^{2}+4 \ln \relax (5)-4 \ln \left (5 \,{\mathrm e}^{3}-2\right )-2\right ) x^{3}+\left (1+\ln \relax (5)^{4}-4 \ln \relax (5)^{3}-4 \ln \relax (5)+6 \ln \relax (5)^{2}-12 \ln \relax (5) \ln \left (5 \,{\mathrm e}^{3}-2\right )+6 \ln \left (5 \,{\mathrm e}^{3}-2\right )^{2}+12 \ln \relax (5)^{2} \ln \left (5 \,{\mathrm e}^{3}-2\right )-12 \ln \relax (5) \ln \left (5 \,{\mathrm e}^{3}-2\right )^{2}-4 \ln \relax (5)^{3} \ln \left (5 \,{\mathrm e}^{3}-2\right )+6 \ln \relax (5)^{2} \ln \left (5 \,{\mathrm e}^{3}-2\right )^{2}-4 \ln \relax (5) \ln \left (5 \,{\mathrm e}^{3}-2\right )^{3}+4 \ln \left (5 \,{\mathrm e}^{3}-2\right )+4 \ln \left (5 \,{\mathrm e}^{3}-2\right )^{3}+\ln \left (5 \,{\mathrm e}^{3}-2\right )^{4}\right ) x\)	\(194\)
risch	\(x \ln \relax (5)^{4}-4 \ln \relax (5)^{3} \ln \left (5 \,{\mathrm e}^{3}-2\right ) x +6 \ln \relax (5)^{2} \ln \left (5 \,{\mathrm e}^{3}-2\right )^{2} x -2 x^{3} \ln \relax (5)^{2}-4 \ln \relax (5) \ln \left (5 \,{\mathrm e}^{3}-2\right )^{3} x +4 \ln \relax (5) \ln \left (5 \,{\mathrm e}^{3}-2\right ) x^{3}+\ln \left (5 \,{\mathrm e}^{3}-2\right )^{4} x -2 \ln \left (5 \,{\mathrm e}^{3}-2\right )^{2} x^{3}+x^{5}-4 \ln \relax (5)^{3} x +12 \ln \relax (5)^{2} \ln \left (5 \,{\mathrm e}^{3}-2\right ) x -12 \ln \relax (5) \ln \left (5 \,{\mathrm e}^{3}-2\right )^{2} x +4 x^{3} \ln \relax (5)+4 \ln \left (5 \,{\mathrm e}^{3}-2\right )^{3} x -4 x^{3} \ln \left (5 \,{\mathrm e}^{3}-2\right )+6 x \ln \relax (5)^{2}-12 \ln \relax (5) \ln \left (5 \,{\mathrm e}^{3}-2\right ) x +6 \ln \left (5 \,{\mathrm e}^{3}-2\right )^{2} x -2 x^{3}-4 x \ln \relax (5)+4 x \ln \left (5 \,{\mathrm e}^{3}-2\right )+x\)	\(221\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(exp(3)-2/5)^4+4*ln(exp(3)-2/5)^3+(-6*x^2+6)*ln(exp(3)-2/5)^2+(-12*x^2+4)*ln(exp(3)-2/5)+5*x^4-6*x^2+1,x
,method=_RETURNVERBOSE)

[Out]

ln(exp(3)-2/5)^4*x+4*ln(exp(3)-2/5)^3*x+ln(exp(3)-2/5)^2*(-2*x^3+6*x)+ln(exp(3)-2/5)*(-4*x^3+4*x)+x^5-2*x^3+x

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maxima [B]  time = 0.34, size = 59, normalized size = 2.68 \begin {gather*} x^{5} + x \log \left (e^{3} - \frac {2}{5}\right )^{4} + 4 \, x \log \left (e^{3} - \frac {2}{5}\right )^{3} - 2 \, x^{3} - 2 \, {\left (x^{3} - 3 \, x\right )} \log \left (e^{3} - \frac {2}{5}\right )^{2} - 4 \, {\left (x^{3} - x\right )} \log \left (e^{3} - \frac {2}{5}\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(exp(3)-2/5)^4+4*log(exp(3)-2/5)^3+(-6*x^2+6)*log(exp(3)-2/5)^2+(-12*x^2+4)*log(exp(3)-2/5)+5*x^4
-6*x^2+1,x, algorithm="maxima")

[Out]

x^5 + x*log(e^3 - 2/5)^4 + 4*x*log(e^3 - 2/5)^3 - 2*x^3 - 2*(x^3 - 3*x)*log(e^3 - 2/5)^2 - 4*(x^3 - x)*log(e^3
- 2/5) + x

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mupad [B]  time = 2.90, size = 63, normalized size = 2.86 \begin {gather*} x^5+\left (-4\,\ln \left ({\mathrm {e}}^3-\frac {2}{5}\right )-2\,{\ln \left ({\mathrm {e}}^3-\frac {2}{5}\right )}^2-2\right )\,x^3+\left (4\,\ln \left ({\mathrm {e}}^3-\frac {2}{5}\right )+6\,{\ln \left ({\mathrm {e}}^3-\frac {2}{5}\right )}^2+4\,{\ln \left ({\mathrm {e}}^3-\frac {2}{5}\right )}^3+{\ln \left ({\mathrm {e}}^3-\frac {2}{5}\right )}^4+1\right )\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(4*log(exp(3) - 2/5)^3 - log(exp(3) - 2/5)^2*(6*x^2 - 6) - log(exp(3) - 2/5)*(12*x^2 - 4) + log(exp(3) - 2/
5)^4 - 6*x^2 + 5*x^4 + 1,x)

[Out]

x*(4*log(exp(3) - 2/5) + 6*log(exp(3) - 2/5)^2 + 4*log(exp(3) - 2/5)^3 + log(exp(3) - 2/5)^4 + 1) - x^3*(4*log
(exp(3) - 2/5) + 2*log(exp(3) - 2/5)^2 + 2) + x^5

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sympy [B]  time = 0.07, size = 78, normalized size = 3.55 \begin {gather*} x^{5} + x^{3} \left (- 2 \log {\left (- \frac {2}{5} + e^{3} \right )}^{2} - 4 \log {\left (- \frac {2}{5} + e^{3} \right )} - 2\right ) + x \left (1 + 4 \log {\left (- \frac {2}{5} + e^{3} \right )} + 6 \log {\left (- \frac {2}{5} + e^{3} \right )}^{2} + \log {\left (- \frac {2}{5} + e^{3} \right )}^{4} + 4 \log {\left (- \frac {2}{5} + e^{3} \right )}^{3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(exp(3)-2/5)**4+4*ln(exp(3)-2/5)**3+(-6*x**2+6)*ln(exp(3)-2/5)**2+(-12*x**2+4)*ln(exp(3)-2/5)+5*x*
*4-6*x**2+1,x)

[Out]

x**5 + x**3*(-2*log(-2/5 + exp(3))**2 - 4*log(-2/5 + exp(3)) - 2) + x*(1 + 4*log(-2/5 + exp(3)) + 6*log(-2/5 +
exp(3))**2 + log(-2/5 + exp(3))**4 + 4*log(-2/5 + exp(3))**3)

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