3.36.31 \(\int (1-4 x+(1-4 x) \log (\frac {e}{625})+(1+\log (\frac {e}{625})) \log (\log (2))) \, dx\)

Optimal. Leaf size=18 \[ x \left (1+\log \left (\frac {e}{625}\right )\right ) (1-2 x+\log (\log (2))) \]

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Rubi [B]  time = 0.01, antiderivative size = 37, normalized size of antiderivative = 2.06, number of steps used = 1, number of rules used = 0, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} -2 x^2-\frac {1}{8} (1-4 x)^2 \log \left (\frac {e}{625}\right )+x (1+(2-\log (625)) \log (\log (2))) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1 - 4*x + (1 - 4*x)*Log[E/625] + (1 + Log[E/625])*Log[Log[2]],x]

[Out]

-2*x^2 - ((1 - 4*x)^2*Log[E/625])/8 + x*(1 + (2 - Log[625])*Log[Log[2]])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-2 x^2-\frac {1}{8} (1-4 x)^2 \log \left (\frac {e}{625}\right )+x (1+(2-\log (625)) \log (\log (2)))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 20, normalized size = 1.11 \begin {gather*} (-2+\log (625)) \left (-x+2 x^2-x \log (\log (2))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1 - 4*x + (1 - 4*x)*Log[E/625] + (1 + Log[E/625])*Log[Log[2]],x]

[Out]

(-2 + Log[625])*(-x + 2*x^2 - x*Log[Log[2]])

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fricas [A]  time = 0.51, size = 36, normalized size = 2.00 \begin {gather*} -4 \, x^{2} + 4 \, {\left (2 \, x^{2} - x\right )} \log \relax (5) - 2 \, {\left (2 \, x \log \relax (5) - x\right )} \log \left (\log \relax (2)\right ) + 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(1/625*exp(1))+1)*log(log(2))+(-4*x+1)*log(1/625*exp(1))-4*x+1,x, algorithm="fricas")

[Out]

-4*x^2 + 4*(2*x^2 - x)*log(5) - 2*(2*x*log(5) - x)*log(log(2)) + 2*x

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giac [A]  time = 0.20, size = 35, normalized size = 1.94 \begin {gather*} x {\left (\log \left (\frac {1}{625} \, e\right ) + 1\right )} \log \left (\log \relax (2)\right ) - 2 \, x^{2} - {\left (2 \, x^{2} - x\right )} \log \left (\frac {1}{625} \, e\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(1/625*exp(1))+1)*log(log(2))+(-4*x+1)*log(1/625*exp(1))-4*x+1,x, algorithm="giac")

[Out]

x*(log(1/625*e) + 1)*log(log(2)) - 2*x^2 - (2*x^2 - x)*log(1/625*e) + x

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maple [A]  time = 0.02, size = 18, normalized size = 1.00

method	result	size
gosper	\(\left (1-2 x +\ln \left (\ln \relax (2)\right )\right ) \left (\ln \left (\frac {{\mathrm e}}{625}\right )+1\right ) x\)	\(18\)
norman	\(\left (8 \ln \relax (5)-4\right ) x^{2}+\left (-4 \ln \left (\ln \relax (2)\right ) \ln \relax (5)-4 \ln \relax (5)+2 \ln \left (\ln \relax (2)\right )+2\right ) x\)	\(32\)
default	\(\left (\ln \left (\frac {{\mathrm e}}{625}\right )+1\right ) \ln \left (\ln \relax (2)\right ) x +\ln \left (\frac {{\mathrm e}}{625}\right ) \left (-2 x^{2}+x \right )-2 x^{2}+x\)	\(33\)
risch	\(-4 \ln \left (\ln \relax (2)\right ) x \ln \relax (5)+8 x^{2} \ln \relax (5)+2 x \ln \left (\ln \relax (2)\right )-4 x \ln \relax (5)-4 x^{2}+2 x\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((ln(1/625*exp(1))+1)*ln(ln(2))+(-4*x+1)*ln(1/625*exp(1))-4*x+1,x,method=_RETURNVERBOSE)

[Out]

(1-2*x+ln(ln(2)))*(ln(1/625*exp(1))+1)*x

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maxima [A]  time = 0.40, size = 35, normalized size = 1.94 \begin {gather*} x {\left (\log \left (\frac {1}{625} \, e\right ) + 1\right )} \log \left (\log \relax (2)\right ) - 2 \, x^{2} - {\left (2 \, x^{2} - x\right )} \log \left (\frac {1}{625} \, e\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(1/625*exp(1))+1)*log(log(2))+(-4*x+1)*log(1/625*exp(1))-4*x+1,x, algorithm="maxima")

[Out]

x*(log(1/625*e) + 1)*log(log(2)) - 2*x^2 - (2*x^2 - x)*log(1/625*e) + x

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mupad [B]  time = 2.04, size = 29, normalized size = 1.61 \begin {gather*} \left (8\,\ln \relax (5)-4\right )\,x^2+\left (\ln \left (\frac {{\ln \relax (2)}^2}{625}\right )-4\,\ln \relax (5)\,\ln \left (\ln \relax (2)\right )+2\right )\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(log(2))*(log(exp(1)/625) + 1) - log(exp(1)/625)*(4*x - 1) - 4*x + 1,x)

[Out]

x^2*(8*log(5) - 4) + x*(log(log(2)^2/625) - 4*log(5)*log(log(2)) + 2)

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sympy [A]  time = 0.06, size = 34, normalized size = 1.89 \begin {gather*} x^{2} \left (-4 + 8 \log {\relax (5 )}\right ) + x \left (- 4 \log {\relax (5 )} + 2 \log {\left (\log {\relax (2 )} \right )} + 2 - 4 \log {\relax (5 )} \log {\left (\log {\relax (2 )} \right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((ln(1/625*exp(1))+1)*ln(ln(2))+(-4*x+1)*ln(1/625*exp(1))-4*x+1,x)

[Out]

x**2*(-4 + 8*log(5)) + x*(-4*log(5) + 2*log(log(2)) + 2 - 4*log(5)*log(log(2)))

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