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

Optimal. Leaf size=23 \[ \frac {\log \left (\frac {1}{4 x \log \left (x^2 (x+\log (x))\right )}\right )}{x} \]

________________________________________________________________________________________

Rubi [F]  time = 2.30, 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 {-1-3 x-2 \log (x)+(-x-\log (x)) \log \left (x^3+x^2 \log (x)\right )+(-x-\log (x)) \log \left (x^3+x^2 \log (x)\right ) \log \left (\frac {1}{4 x \log \left (x^3+x^2 \log (x)\right )}\right )}{\left (x^3+x^2 \log (x)\right ) \log \left (x^3+x^2 \log (x)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.34, size = 23, normalized size = 1.00 \begin {gather*} \frac {\log \left (\frac {1}{4 x \log \left (x^2 (x+\log (x))\right )}\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[1/(4*x*Log[x^2*(x + Log[x])])]/x

________________________________________________________________________________________

fricas [A]  time = 0.57, size = 23, normalized size = 1.00 \begin {gather*} \frac {\log \left (\frac {1}{4 \, x \log \left (x^{3} + x^{2} \log \relax (x)\right )}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(1/4/(x*log(x^3 + x^2*log(x))))/x

________________________________________________________________________________________

giac [A]  time = 0.53, size = 26, normalized size = 1.13 \begin {gather*} -\frac {\log \relax (x)}{x} - \frac {\log \left (4 \, \log \left (x + \log \relax (x)\right ) + 8 \, \log \relax (x)\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(x)/x - log(4*log(x + log(x)) + 8*log(x))/x

________________________________________________________________________________________

maple [C]  time = 0.61, size = 2102, normalized size = 91.39




method result size



risch \(\text {Expression too large to display}\) \(2102\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/x*ln(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3+Pi*csgn(I*x^2)*csgn(I*(x+ln(x
)))*csgn(I*x^2*(x+ln(x)))-Pi*csgn(I*x^2)*csgn(I*x^2*(x+ln(x)))^2-Pi*csgn(I*(x+ln(x)))*csgn(I*x^2*(x+ln(x)))^2+
Pi*csgn(I*x^2*(x+ln(x)))^3+4*I*ln(x)+2*I*ln(x+ln(x)))-1/2*(I*Pi*csgn(I/x/(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn
(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3+Pi*csgn(I*x^2)*csgn(I*(x+ln(x)))*csgn(I*x^2*(x+ln(x)))-Pi*csgn(I*x^2)*csg
n(I*x^2*(x+ln(x)))^2-Pi*csgn(I*(x+ln(x)))*csgn(I*x^2*(x+ln(x)))^2+Pi*csgn(I*x^2*(x+ln(x)))^3+4*I*ln(x)+2*I*ln(
x+ln(x))))*csgn(1/x/(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3+Pi*csgn(I*x^2)*c
sgn(I*(x+ln(x)))*csgn(I*x^2*(x+ln(x)))-Pi*csgn(I*x^2)*csgn(I*x^2*(x+ln(x)))^2-Pi*csgn(I*(x+ln(x)))*csgn(I*x^2*
(x+ln(x)))^2+Pi*csgn(I*x^2*(x+ln(x)))^3+4*I*ln(x)+2*I*ln(x+ln(x))))+I*Pi*csgn(1/x/(Pi*csgn(I*x)^2*csgn(I*x^2)-
2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3+Pi*csgn(I*x^2)*csgn(I*(x+ln(x)))*csgn(I*x^2*(x+ln(x)))-Pi*csgn(I
*x^2)*csgn(I*x^2*(x+ln(x)))^2-Pi*csgn(I*(x+ln(x)))*csgn(I*x^2*(x+ln(x)))^2+Pi*csgn(I*x^2*(x+ln(x)))^3+4*I*ln(x
)+2*I*ln(x+ln(x))))^2-I*Pi*csgn(I/(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3+Pi
*csgn(I*x^2)*csgn(I*(x+ln(x)))*csgn(I*x^2*(x+ln(x)))-Pi*csgn(I*x^2)*csgn(I*x^2*(x+ln(x)))^2-Pi*csgn(I*(x+ln(x)
))*csgn(I*x^2*(x+ln(x)))^2+Pi*csgn(I*x^2*(x+ln(x)))^3+4*I*ln(x)+2*I*ln(x+ln(x))))*csgn(I/x/(Pi*csgn(I*x)^2*csg
n(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3+Pi*csgn(I*x^2)*csgn(I*(x+ln(x)))*csgn(I*x^2*(x+ln(x)))-
Pi*csgn(I*x^2)*csgn(I*x^2*(x+ln(x)))^2-Pi*csgn(I*(x+ln(x)))*csgn(I*x^2*(x+ln(x)))^2+Pi*csgn(I*x^2*(x+ln(x)))^3
+4*I*ln(x)+2*I*ln(x+ln(x))))^2+I*Pi*csgn(I/(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*
x^2)^3+Pi*csgn(I*x^2)*csgn(I*(x+ln(x)))*csgn(I*x^2*(x+ln(x)))-Pi*csgn(I*x^2)*csgn(I*x^2*(x+ln(x)))^2-Pi*csgn(I
*(x+ln(x)))*csgn(I*x^2*(x+ln(x)))^2+Pi*csgn(I*x^2*(x+ln(x)))^3+4*I*ln(x)+2*I*ln(x+ln(x))))*csgn(I/x/(Pi*csgn(I
*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3+Pi*csgn(I*x^2)*csgn(I*(x+ln(x)))*csgn(I*x^2*(x
+ln(x)))-Pi*csgn(I*x^2)*csgn(I*x^2*(x+ln(x)))^2-Pi*csgn(I*(x+ln(x)))*csgn(I*x^2*(x+ln(x)))^2+Pi*csgn(I*x^2*(x+
ln(x)))^3+4*I*ln(x)+2*I*ln(x+ln(x))))*csgn(I/x)+I*Pi*csgn(I/x/(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(
I*x^2)^2+Pi*csgn(I*x^2)^3+Pi*csgn(I*x^2)*csgn(I*(x+ln(x)))*csgn(I*x^2*(x+ln(x)))-Pi*csgn(I*x^2)*csgn(I*x^2*(x+
ln(x)))^2-Pi*csgn(I*(x+ln(x)))*csgn(I*x^2*(x+ln(x)))^2+Pi*csgn(I*x^2*(x+ln(x)))^3+4*I*ln(x)+2*I*ln(x+ln(x))))^
3-I*Pi*csgn(I/x/(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3+Pi*csgn(I*x^2)*csgn(
I*(x+ln(x)))*csgn(I*x^2*(x+ln(x)))-Pi*csgn(I*x^2)*csgn(I*x^2*(x+ln(x)))^2-Pi*csgn(I*(x+ln(x)))*csgn(I*x^2*(x+l
n(x)))^2+Pi*csgn(I*x^2*(x+ln(x)))^3+4*I*ln(x)+2*I*ln(x+ln(x))))^2*csgn(I/x)-I*Pi*csgn(I/x/(Pi*csgn(I*x)^2*csgn
(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3+Pi*csgn(I*x^2)*csgn(I*(x+ln(x)))*csgn(I*x^2*(x+ln(x)))-P
i*csgn(I*x^2)*csgn(I*x^2*(x+ln(x)))^2-Pi*csgn(I*(x+ln(x)))*csgn(I*x^2*(x+ln(x)))^2+Pi*csgn(I*x^2*(x+ln(x)))^3+
4*I*ln(x)+2*I*ln(x+ln(x))))*csgn(1/x/(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3
+Pi*csgn(I*x^2)*csgn(I*(x+ln(x)))*csgn(I*x^2*(x+ln(x)))-Pi*csgn(I*x^2)*csgn(I*x^2*(x+ln(x)))^2-Pi*csgn(I*(x+ln
(x)))*csgn(I*x^2*(x+ln(x)))^2+Pi*csgn(I*x^2*(x+ln(x)))^3+4*I*ln(x)+2*I*ln(x+ln(x))))^2-I*Pi*csgn(1/x/(Pi*csgn(
I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3+Pi*csgn(I*x^2)*csgn(I*(x+ln(x)))*csgn(I*x^2*(
x+ln(x)))-Pi*csgn(I*x^2)*csgn(I*x^2*(x+ln(x)))^2-Pi*csgn(I*(x+ln(x)))*csgn(I*x^2*(x+ln(x)))^2+Pi*csgn(I*x^2*(x
+ln(x)))^3+4*I*ln(x)+2*I*ln(x+ln(x))))^3-I*Pi+2*ln(2)+2*ln(x))/x

________________________________________________________________________________________

maxima [A]  time = 0.49, size = 23, normalized size = 1.00 \begin {gather*} -\frac {2 \, \log \relax (2) + \log \relax (x) + \log \left (\log \left (x + \log \relax (x)\right ) + 2 \, \log \relax (x)\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*log(2) + log(x) + log(log(x + log(x)) + 2*log(x)))/x

________________________________________________________________________________________

mupad [B]  time = 5.65, size = 23, normalized size = 1.00 \begin {gather*} \frac {\ln \left (\frac {1}{4\,x\,\ln \left (x^2\,\ln \relax (x)+x^3\right )}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(1/(4*x*log(x^2*log(x) + x^3)))/x

________________________________________________________________________________________

sympy [A]  time = 1.23, size = 19, normalized size = 0.83 \begin {gather*} \frac {\log {\left (\frac {1}{4 x \log {\left (x^{3} + x^{2} \log {\relax (x )} \right )}} \right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________