3.96.13 \(\int \frac {x-x^2 \log ^2(2)+2 x \log ^3(2)-\log ^4(2)+((-x^2-x^3) \log ^2(2)+(1+x) \log ^4(2)) \log (\frac {x}{1+x}) \log (\log (\frac {x}{1+x})) \log (\log (\log (\frac {x}{1+x})))}{(10 x^2+10 x^3) \log (\frac {x}{1+x}) \log (\log (\frac {x}{1+x}))} \, dx\)

Optimal. Leaf size=33 \[ \frac {\left (x-(x-\log (2))^2 \log ^2(2)\right ) \log \left (\log \left (\log \left (\frac {x}{1+x}\right )\right )\right )}{10 x} \]

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Rubi [F]  time = 1.99, 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 {x-x^2 \log ^2(2)+2 x \log ^3(2)-\log ^4(2)+\left (\left (-x^2-x^3\right ) \log ^2(2)+(1+x) \log ^4(2)\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right ) \log \left (\log \left (\log \left (\frac {x}{1+x}\right )\right )\right )}{\left (10 x^2+10 x^3\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(x - x^2*Log[2]^2 + 2*x*Log[2]^3 - Log[2]^4 + ((-x^2 - x^3)*Log[2]^2 + (1 + x)*Log[2]^4)*Log[x/(1 + x)]*Lo
g[Log[x/(1 + x)]]*Log[Log[Log[x/(1 + x)]]])/((10*x^2 + 10*x^3)*Log[x/(1 + x)]*Log[Log[x/(1 + x)]]),x]

[Out]

-1/10*(Log[2]^4*Defer[Int][1/(x^2*Log[x/(1 + x)]*Log[Log[x/(1 + x)]]), x]) + ((1 + 2*Log[2]^3 + Log[2]^4)*Defe
r[Int][1/(x*Log[x/(1 + x)]*Log[Log[x/(1 + x)]]), x])/10 - ((1 + Log[2]^2 + 2*Log[2]^3 + Log[2]^4)*Defer[Int][1
/((1 + x)*Log[x/(1 + x)]*Log[Log[x/(1 + x)]]), x])/10 - (Log[2]^2*Defer[Int][Log[Log[Log[x/(1 + x)]]], x])/10
+ (Log[2]^4*Defer[Int][Log[Log[Log[x/(1 + x)]]]/x^2, x])/10

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-x^2 \log ^2(2)-\log ^4(2)+x \left (1+2 \log ^3(2)\right )+\left (\left (-x^2-x^3\right ) \log ^2(2)+(1+x) \log ^4(2)\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right ) \log \left (\log \left (\log \left (\frac {x}{1+x}\right )\right )\right )}{\left (10 x^2+10 x^3\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right )} \, dx\\ &=\int \frac {-x^2 \log ^2(2)-\log ^4(2)+x \left (1+2 \log ^3(2)\right )+\left (\left (-x^2-x^3\right ) \log ^2(2)+(1+x) \log ^4(2)\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right ) \log \left (\log \left (\log \left (\frac {x}{1+x}\right )\right )\right )}{x^2 (10+10 x) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right )} \, dx\\ &=\int \left (\frac {-x^2 \log ^2(2)-\log ^4(2)+x \left (1+2 \log ^3(2)\right )}{10 x^2 (1+x) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right )}-\frac {(x-\log (2)) \log ^2(2) (x+\log (2)) \log \left (\log \left (\log \left (\frac {x}{1+x}\right )\right )\right )}{10 x^2}\right ) \, dx\\ &=\frac {1}{10} \int \frac {-x^2 \log ^2(2)-\log ^4(2)+x \left (1+2 \log ^3(2)\right )}{x^2 (1+x) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right )} \, dx-\frac {1}{10} \log ^2(2) \int \frac {(x-\log (2)) (x+\log (2)) \log \left (\log \left (\log \left (\frac {x}{1+x}\right )\right )\right )}{x^2} \, dx\\ &=\frac {1}{10} \int \left (-\frac {\log ^4(2)}{x^2 \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right )}+\frac {-1-\log ^2(2)-2 \log ^3(2)-\log ^4(2)}{(1+x) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right )}+\frac {1+2 \log ^3(2)+\log ^4(2)}{x \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right )}\right ) \, dx-\frac {1}{10} \log ^2(2) \int \left (\log \left (\log \left (\log \left (\frac {x}{1+x}\right )\right )\right )-\frac {\log ^2(2) \log \left (\log \left (\log \left (\frac {x}{1+x}\right )\right )\right )}{x^2}\right ) \, dx\\ &=-\left (\frac {1}{10} \log ^2(2) \int \log \left (\log \left (\log \left (\frac {x}{1+x}\right )\right )\right ) \, dx\right )-\frac {1}{10} \log ^4(2) \int \frac {1}{x^2 \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right )} \, dx+\frac {1}{10} \log ^4(2) \int \frac {\log \left (\log \left (\log \left (\frac {x}{1+x}\right )\right )\right )}{x^2} \, dx+\frac {1}{10} \left (-1-\log ^2(2)-2 \log ^3(2)-\log ^4(2)\right ) \int \frac {1}{(1+x) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right )} \, dx+\frac {1}{10} \left (1+2 \log ^3(2)+\log ^4(2)\right ) \int \frac {1}{x \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 41, normalized size = 1.24 \begin {gather*} -\frac {\left (x^2 \log ^2(2)+\log ^4(2)-x \left (1+2 \log ^3(2)\right )\right ) \log \left (\log \left (\log \left (\frac {x}{1+x}\right )\right )\right )}{10 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x - x^2*Log[2]^2 + 2*x*Log[2]^3 - Log[2]^4 + ((-x^2 - x^3)*Log[2]^2 + (1 + x)*Log[2]^4)*Log[x/(1 +
x)]*Log[Log[x/(1 + x)]]*Log[Log[Log[x/(1 + x)]]])/((10*x^2 + 10*x^3)*Log[x/(1 + x)]*Log[Log[x/(1 + x)]]),x]

[Out]

-1/10*((x^2*Log[2]^2 + Log[2]^4 - x*(1 + 2*Log[2]^3))*Log[Log[Log[x/(1 + x)]]])/x

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fricas [A]  time = 0.55, size = 38, normalized size = 1.15 \begin {gather*} -\frac {{\left (x^{2} \log \relax (2)^{2} - 2 \, x \log \relax (2)^{3} + \log \relax (2)^{4} - x\right )} \log \left (\log \left (\log \left (\frac {x}{x + 1}\right )\right )\right )}{10 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*log(2)^4+(-x^3-x^2)*log(2)^2)*log(x/(x+1))*log(log(x/(x+1)))*log(log(log(x/(x+1))))-log(2)^4
+2*x*log(2)^3-x^2*log(2)^2+x)/(10*x^3+10*x^2)/log(x/(x+1))/log(log(x/(x+1))),x, algorithm="fricas")

[Out]

-1/10*(x^2*log(2)^2 - 2*x*log(2)^3 + log(2)^4 - x)*log(log(log(x/(x + 1))))/x

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giac [A]  time = 0.51, size = 49, normalized size = 1.48 \begin {gather*} \frac {1}{10} \, {\left (2 \, \log \relax (2)^{3} + 1\right )} \log \left (\log \left (-\log \left (x + 1\right ) + \log \relax (x)\right )\right ) - \frac {1}{10} \, {\left (x \log \relax (2)^{2} + \frac {\log \relax (2)^{4}}{x}\right )} \log \left (\log \left (\log \left (\frac {x}{x + 1}\right )\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*log(2)^4+(-x^3-x^2)*log(2)^2)*log(x/(x+1))*log(log(x/(x+1)))*log(log(log(x/(x+1))))-log(2)^4
+2*x*log(2)^3-x^2*log(2)^2+x)/(10*x^3+10*x^2)/log(x/(x+1))/log(log(x/(x+1))),x, algorithm="giac")

[Out]

1/10*(2*log(2)^3 + 1)*log(log(-log(x + 1) + log(x))) - 1/10*(x*log(2)^2 + log(2)^4/x)*log(log(log(x/(x + 1))))

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (x +1\right ) \ln \relax (2)^{4}+\left (-x^{3}-x^{2}\right ) \ln \relax (2)^{2}\right ) \ln \left (\frac {x}{x +1}\right ) \ln \left (\ln \left (\frac {x}{x +1}\right )\right ) \ln \left (\ln \left (\ln \left (\frac {x}{x +1}\right )\right )\right )-\ln \relax (2)^{4}+2 x \ln \relax (2)^{3}-x^{2} \ln \relax (2)^{2}+x}{\left (10 x^{3}+10 x^{2}\right ) \ln \left (\frac {x}{x +1}\right ) \ln \left (\ln \left (\frac {x}{x +1}\right )\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x+1)*ln(2)^4+(-x^3-x^2)*ln(2)^2)*ln(x/(x+1))*ln(ln(x/(x+1)))*ln(ln(ln(x/(x+1))))-ln(2)^4+2*x*ln(2)^3-x^
2*ln(2)^2+x)/(10*x^3+10*x^2)/ln(x/(x+1))/ln(ln(x/(x+1))),x)

[Out]

int((((x+1)*ln(2)^4+(-x^3-x^2)*ln(2)^2)*ln(x/(x+1))*ln(ln(x/(x+1)))*ln(ln(ln(x/(x+1))))-ln(2)^4+2*x*ln(2)^3-x^
2*ln(2)^2+x)/(10*x^3+10*x^2)/ln(x/(x+1))/ln(ln(x/(x+1))),x)

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maxima [A]  time = 0.49, size = 40, normalized size = 1.21 \begin {gather*} -\frac {{\left (x^{2} \log \relax (2)^{2} + \log \relax (2)^{4} - {\left (2 \, \log \relax (2)^{3} + 1\right )} x\right )} \log \left (\log \left (-\log \left (x + 1\right ) + \log \relax (x)\right )\right )}{10 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*log(2)^4+(-x^3-x^2)*log(2)^2)*log(x/(x+1))*log(log(x/(x+1)))*log(log(log(x/(x+1))))-log(2)^4
+2*x*log(2)^3-x^2*log(2)^2+x)/(10*x^3+10*x^2)/log(x/(x+1))/log(log(x/(x+1))),x, algorithm="maxima")

[Out]

-1/10*(x^2*log(2)^2 + log(2)^4 - (2*log(2)^3 + 1)*x)*log(log(-log(x + 1) + log(x)))/x

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mupad [B]  time = 8.94, size = 39, normalized size = 1.18 \begin {gather*} \frac {\ln \left (\ln \left (\ln \left (\frac {x}{x+1}\right )\right )\right )\,\left (x-x^2\,{\ln \relax (2)}^2+2\,x\,{\ln \relax (2)}^3-{\ln \relax (2)}^4\right )}{10\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x - x^2*log(2)^2 + 2*x*log(2)^3 - log(2)^4 + log(log(log(x/(x + 1))))*log(x/(x + 1))*log(log(x/(x + 1)))*
(log(2)^4*(x + 1) - log(2)^2*(x^2 + x^3)))/(log(x/(x + 1))*log(log(x/(x + 1)))*(10*x^2 + 10*x^3)),x)

[Out]

(log(log(log(x/(x + 1))))*(x - x^2*log(2)^2 + 2*x*log(2)^3 - log(2)^4))/(10*x)

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sympy [A]  time = 0.76, size = 48, normalized size = 1.45 \begin {gather*} \frac {\left (2 \log {\relax (2 )}^{3} + 1\right ) \log {\left (\log {\left (\log {\left (\frac {x}{x + 1} \right )} \right )} \right )}}{10} + \frac {\left (- x^{2} \log {\relax (2 )}^{2} - \log {\relax (2 )}^{4}\right ) \log {\left (\log {\left (\log {\left (\frac {x}{x + 1} \right )} \right )} \right )}}{10 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*ln(2)**4+(-x**3-x**2)*ln(2)**2)*ln(x/(x+1))*ln(ln(x/(x+1)))*ln(ln(ln(x/(x+1))))-ln(2)**4+2*x
*ln(2)**3-x**2*ln(2)**2+x)/(10*x**3+10*x**2)/ln(x/(x+1))/ln(ln(x/(x+1))),x)

[Out]

(2*log(2)**3 + 1)*log(log(log(x/(x + 1))))/10 + (-x**2*log(2)**2 - log(2)**4)*log(log(log(x/(x + 1))))/(10*x)

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