3.11.49 \(\int \frac {(\frac {x+4 \log (\frac {(-4+4 x) \log (4)}{5 x})}{\log (\frac {(-4+4 x) \log (4)}{5 x})})^{\frac {1}{\log (x^2)}} (-x \log (x^2)+(-x+x^2) \log (x^2) \log (\frac {(-4+4 x) \log (4)}{5 x})+((2 x-2 x^2) \log (\frac {(-4+4 x) \log (4)}{5 x})+(8-8 x) \log ^2(\frac {(-4+4 x) \log (4)}{5 x})) \log (\frac {x+4 \log (\frac {(-4+4 x) \log (4)}{5 x})}{\log (\frac {(-4+4 x) \log (4)}{5 x})}))}{(-x^2+x^3) \log ^2(x^2) \log (\frac {(-4+4 x) \log (4)}{5 x})+(-4 x+4 x^2) \log ^2(x^2) \log ^2(\frac {(-4+4 x) \log (4)}{5 x})} \, dx\)

Optimal. Leaf size=26 \[ \left (4+\frac {x}{\log \left (\frac {4 (-1+x) \log (4)}{5 x}\right )}\right )^{\frac {1}{\log \left (x^2\right )}} \]

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Rubi [F]  time = 6.44, 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 {\left (\frac {x+4 \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}{\log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}\right )^{\frac {1}{\log \left (x^2\right )}} \left (-x \log \left (x^2\right )+\left (-x+x^2\right ) \log \left (x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+\left (\left (2 x-2 x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+(8-8 x) \log ^2\left (\frac {(-4+4 x) \log (4)}{5 x}\right )\right ) \log \left (\frac {x+4 \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}{\log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}\right )\right )}{\left (-x^2+x^3\right ) \log ^2\left (x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+\left (-4 x+4 x^2\right ) \log ^2\left (x^2\right ) \log ^2\left (\frac {(-4+4 x) \log (4)}{5 x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(((x + 4*Log[((-4 + 4*x)*Log[4])/(5*x)])/Log[((-4 + 4*x)*Log[4])/(5*x)])^Log[x^2]^(-1)*(-(x*Log[x^2]) + (-
x + x^2)*Log[x^2]*Log[((-4 + 4*x)*Log[4])/(5*x)] + ((2*x - 2*x^2)*Log[((-4 + 4*x)*Log[4])/(5*x)] + (8 - 8*x)*L
og[((-4 + 4*x)*Log[4])/(5*x)]^2)*Log[(x + 4*Log[((-4 + 4*x)*Log[4])/(5*x)])/Log[((-4 + 4*x)*Log[4])/(5*x)]]))/
((-x^2 + x^3)*Log[x^2]^2*Log[((-4 + 4*x)*Log[4])/(5*x)] + (-4*x + 4*x^2)*Log[x^2]^2*Log[((-4 + 4*x)*Log[4])/(5
*x)]^2),x]

[Out]

Defer[Int][(4 + x/Log[(4*(-1 + x)*Log[4])/(5*x)])^Log[x^2]^(-1)/(Log[x^2]*(x + 4*Log[(4*(-1 + x)*Log[4])/(5*x)
])), x] + Defer[Int][(4 + x/Log[(4*(-1 + x)*Log[4])/(5*x)])^Log[x^2]^(-1)/((1 - x)*Log[x^2]*(x + 4*Log[(4*(-1
+ x)*Log[4])/(5*x)])*Log[(4*Log[4])/5 - (4*Log[4])/(5*x)]), x] - 2*Defer[Int][((4 + x/Log[(4*(-1 + x)*Log[4])/
(5*x)])^Log[x^2]^(-1)*Log[4 + x/Log[(4*(-1 + x)*Log[4])/(5*x)]])/(x*Log[x^2]^2), x]

Rubi steps

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

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Mathematica [A]  time = 0.19, size = 26, normalized size = 1.00 \begin {gather*} \left (4+\frac {x}{\log \left (\frac {4 (-1+x) \log (4)}{5 x}\right )}\right )^{\frac {1}{\log \left (x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(((x + 4*Log[((-4 + 4*x)*Log[4])/(5*x)])/Log[((-4 + 4*x)*Log[4])/(5*x)])^Log[x^2]^(-1)*(-(x*Log[x^2]
) + (-x + x^2)*Log[x^2]*Log[((-4 + 4*x)*Log[4])/(5*x)] + ((2*x - 2*x^2)*Log[((-4 + 4*x)*Log[4])/(5*x)] + (8 -
8*x)*Log[((-4 + 4*x)*Log[4])/(5*x)]^2)*Log[(x + 4*Log[((-4 + 4*x)*Log[4])/(5*x)])/Log[((-4 + 4*x)*Log[4])/(5*x
)]]))/((-x^2 + x^3)*Log[x^2]^2*Log[((-4 + 4*x)*Log[4])/(5*x)] + (-4*x + 4*x^2)*Log[x^2]^2*Log[((-4 + 4*x)*Log[
4])/(5*x)]^2),x]

[Out]

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

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fricas [A]  time = 0.57, size = 36, normalized size = 1.38 \begin {gather*} \left (\frac {x + 4 \, \log \left (\frac {8 \, {\left (x - 1\right )} \log \relax (2)}{5 \, x}\right )}{\log \left (\frac {8 \, {\left (x - 1\right )} \log \relax (2)}{5 \, x}\right )}\right )^{\left (\frac {1}{\log \left (x^{2}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-8*x+8)*log(2/5*(4*x-4)*log(2)/x)^2+(-2*x^2+2*x)*log(2/5*(4*x-4)*log(2)/x))*log((4*log(2/5*(4*x-4
)*log(2)/x)+x)/log(2/5*(4*x-4)*log(2)/x))+(x^2-x)*log(x^2)*log(2/5*(4*x-4)*log(2)/x)-x*log(x^2))*exp(log((4*lo
g(2/5*(4*x-4)*log(2)/x)+x)/log(2/5*(4*x-4)*log(2)/x))/log(x^2))/((4*x^2-4*x)*log(x^2)^2*log(2/5*(4*x-4)*log(2)
/x)^2+(x^3-x^2)*log(x^2)^2*log(2/5*(4*x-4)*log(2)/x)),x, algorithm="fricas")

[Out]

((x + 4*log(8/5*(x - 1)*log(2)/x))/log(8/5*(x - 1)*log(2)/x))^(1/log(x^2))

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-8*x+8)*log(2/5*(4*x-4)*log(2)/x)^2+(-2*x^2+2*x)*log(2/5*(4*x-4)*log(2)/x))*log((4*log(2/5*(4*x-4
)*log(2)/x)+x)/log(2/5*(4*x-4)*log(2)/x))+(x^2-x)*log(x^2)*log(2/5*(4*x-4)*log(2)/x)-x*log(x^2))*exp(log((4*lo
g(2/5*(4*x-4)*log(2)/x)+x)/log(2/5*(4*x-4)*log(2)/x))/log(x^2))/((4*x^2-4*x)*log(x^2)^2*log(2/5*(4*x-4)*log(2)
/x)^2+(x^3-x^2)*log(x^2)^2*log(2/5*(4*x-4)*log(2)/x)),x, algorithm="giac")

[Out]

Timed out

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maple [C]  time = 3.80, size = 1648, normalized size = 63.38

method	result	size
risch	\(\text {Expression too large to display}\)	\(1648\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-8*x+8)*ln(2/5*(4*x-4)*ln(2)/x)^2+(-2*x^2+2*x)*ln(2/5*(4*x-4)*ln(2)/x))*ln((4*ln(2/5*(4*x-4)*ln(2)/x)+x
)/ln(2/5*(4*x-4)*ln(2)/x))+(x^2-x)*ln(x^2)*ln(2/5*(4*x-4)*ln(2)/x)-x*ln(x^2))*exp(ln((4*ln(2/5*(4*x-4)*ln(2)/x
)+x)/ln(2/5*(4*x-4)*ln(2)/x))/ln(x^2))/((4*x^2-4*x)*ln(x^2)^2*ln(2/5*(4*x-4)*ln(2)/x)^2+(x^3-x^2)*ln(x^2)^2*ln
(2/5*(4*x-4)*ln(2)/x)),x,method=_RETURNVERBOSE)

[Out]

exp((-I*Pi*csgn(I/(-Pi*csgn(I*(x-1)/x)^3+Pi*csgn(I/x)*csgn(I*(x-1)/x)^2+Pi*csgn(I*(x-1))*csgn(I*(x-1)/x)^2-Pi*
csgn(I/x)*csgn(I*(x-1))*csgn(I*(x-1)/x)+2*I*ln(x)-2*I*ln(x-1)-2*I*ln(8/5*ln(2)))*(-1/2*I*x-Pi*csgn(I*(x-1)/x)^
3-Pi*csgn(I/x)*csgn(I*(x-1))*csgn(I*(x-1)/x)+2*I*ln(x)-2*I*ln(x-1)+Pi*csgn(I/x)*csgn(I*(x-1)/x)^2+Pi*csgn(I*(x
-1))*csgn(I*(x-1)/x)^2-2*I*ln(8/5*ln(2))))^3-I*Pi*csgn(I/(-Pi*csgn(I*(x-1)/x)^3+Pi*csgn(I/x)*csgn(I*(x-1)/x)^2
+Pi*csgn(I*(x-1))*csgn(I*(x-1)/x)^2-Pi*csgn(I/x)*csgn(I*(x-1))*csgn(I*(x-1)/x)+2*I*ln(x)-2*I*ln(x-1)-2*I*ln(8/
5*ln(2)))*(-1/2*I*x-Pi*csgn(I*(x-1)/x)^3-Pi*csgn(I/x)*csgn(I*(x-1))*csgn(I*(x-1)/x)+2*I*ln(x)-2*I*ln(x-1)+Pi*c
sgn(I/x)*csgn(I*(x-1)/x)^2+Pi*csgn(I*(x-1))*csgn(I*(x-1)/x)^2-2*I*ln(8/5*ln(2))))^2*csgn(I/(-Pi*csgn(I*(x-1)/x
)^3+Pi*csgn(I/x)*csgn(I*(x-1)/x)^2+Pi*csgn(I*(x-1))*csgn(I*(x-1)/x)^2-Pi*csgn(I/x)*csgn(I*(x-1))*csgn(I*(x-1)/
x)+2*I*ln(x)-2*I*ln(x-1)-2*I*ln(8/5*ln(2))))-I*Pi*csgn(I/(-Pi*csgn(I*(x-1)/x)^3+Pi*csgn(I/x)*csgn(I*(x-1)/x)^2
+Pi*csgn(I*(x-1))*csgn(I*(x-1)/x)^2-Pi*csgn(I/x)*csgn(I*(x-1))*csgn(I*(x-1)/x)+2*I*ln(x)-2*I*ln(x-1)-2*I*ln(8/
5*ln(2)))*(-1/2*I*x-Pi*csgn(I*(x-1)/x)^3-Pi*csgn(I/x)*csgn(I*(x-1))*csgn(I*(x-1)/x)+2*I*ln(x)-2*I*ln(x-1)+Pi*c
sgn(I/x)*csgn(I*(x-1)/x)^2+Pi*csgn(I*(x-1))*csgn(I*(x-1)/x)^2-2*I*ln(8/5*ln(2))))^2*csgn(I*(-1/2*I*x-Pi*csgn(I
*(x-1)/x)^3-Pi*csgn(I/x)*csgn(I*(x-1))*csgn(I*(x-1)/x)+2*I*ln(x)-2*I*ln(x-1)+Pi*csgn(I/x)*csgn(I*(x-1)/x)^2+Pi
*csgn(I*(x-1))*csgn(I*(x-1)/x)^2-2*I*ln(8/5*ln(2))))-I*Pi*csgn(I/(-Pi*csgn(I*(x-1)/x)^3+Pi*csgn(I/x)*csgn(I*(x
-1)/x)^2+Pi*csgn(I*(x-1))*csgn(I*(x-1)/x)^2-Pi*csgn(I/x)*csgn(I*(x-1))*csgn(I*(x-1)/x)+2*I*ln(x)-2*I*ln(x-1)-2
*I*ln(8/5*ln(2)))*(-1/2*I*x-Pi*csgn(I*(x-1)/x)^3-Pi*csgn(I/x)*csgn(I*(x-1))*csgn(I*(x-1)/x)+2*I*ln(x)-2*I*ln(x
-1)+Pi*csgn(I/x)*csgn(I*(x-1)/x)^2+Pi*csgn(I*(x-1))*csgn(I*(x-1)/x)^2-2*I*ln(8/5*ln(2))))*csgn(I/(-Pi*csgn(I*(
x-1)/x)^3+Pi*csgn(I/x)*csgn(I*(x-1)/x)^2+Pi*csgn(I*(x-1))*csgn(I*(x-1)/x)^2-Pi*csgn(I/x)*csgn(I*(x-1))*csgn(I*
(x-1)/x)+2*I*ln(x)-2*I*ln(x-1)-2*I*ln(8/5*ln(2))))*csgn(I*(-1/2*I*x-Pi*csgn(I*(x-1)/x)^3-Pi*csgn(I/x)*csgn(I*(
x-1))*csgn(I*(x-1)/x)+2*I*ln(x)-2*I*ln(x-1)+Pi*csgn(I/x)*csgn(I*(x-1)/x)^2+Pi*csgn(I*(x-1))*csgn(I*(x-1)/x)^2-
2*I*ln(8/5*ln(2))))+4*ln(2)-2*ln(Pi*csgn(I*(x-1)/x)^3+Pi*csgn(I/x)*csgn(I*(x-1))*csgn(I*(x-1)/x)-2*I*ln(x)+2*I
*ln(x-1)-Pi*csgn(I/x)*csgn(I*(x-1)/x)^2-Pi*csgn(I*(x-1))*csgn(I*(x-1)/x)^2+2*I*(3*ln(2)-ln(5)+ln(ln(2))))+2*ln
(1/2*I*x+Pi*csgn(I*(x-1)/x)^3+Pi*csgn(I/x)*csgn(I*(x-1))*csgn(I*(x-1)/x)-2*I*ln(x)+2*I*ln(x-1)-Pi*csgn(I/x)*cs
gn(I*(x-1)/x)^2-Pi*csgn(I*(x-1))*csgn(I*(x-1)/x)^2+2*I*(3*ln(2)-ln(5)+ln(ln(2)))))/(-I*Pi*csgn(I*x^2)^3+2*I*Pi
*csgn(I*x^2)^2*csgn(I*x)-I*Pi*csgn(I*x)^2*csgn(I*x^2)+4*ln(x)))

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maxima [B]  time = 1.09, size = 61, normalized size = 2.35 \begin {gather*} e^{\left (\frac {\log \left (x - 4 \, \log \relax (5) + 12 \, \log \relax (2) + 4 \, \log \left (x - 1\right ) - 4 \, \log \relax (x) + 4 \, \log \left (\log \relax (2)\right )\right )}{2 \, \log \relax (x)} - \frac {\log \left (-\log \relax (5) + 3 \, \log \relax (2) + \log \left (x - 1\right ) - \log \relax (x) + \log \left (\log \relax (2)\right )\right )}{2 \, \log \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-8*x+8)*log(2/5*(4*x-4)*log(2)/x)^2+(-2*x^2+2*x)*log(2/5*(4*x-4)*log(2)/x))*log((4*log(2/5*(4*x-4
)*log(2)/x)+x)/log(2/5*(4*x-4)*log(2)/x))+(x^2-x)*log(x^2)*log(2/5*(4*x-4)*log(2)/x)-x*log(x^2))*exp(log((4*lo
g(2/5*(4*x-4)*log(2)/x)+x)/log(2/5*(4*x-4)*log(2)/x))/log(x^2))/((4*x^2-4*x)*log(x^2)^2*log(2/5*(4*x-4)*log(2)
/x)^2+(x^3-x^2)*log(x^2)^2*log(2/5*(4*x-4)*log(2)/x)),x, algorithm="maxima")

[Out]

e^(1/2*log(x - 4*log(5) + 12*log(2) + 4*log(x - 1) - 4*log(x) + 4*log(log(2)))/log(x) - 1/2*log(-log(5) + 3*lo
g(2) + log(x - 1) - log(x) + log(log(2)))/log(x))

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mupad [B]  time = 2.02, size = 29, normalized size = 1.12 \begin {gather*} {\left (\frac {x}{\ln \left (-\frac {8\,\ln \relax (2)-8\,x\,\ln \relax (2)}{5\,x}\right )}+4\right )}^{\frac {1}{\ln \left (x^2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(log((x + 4*log((2*log(2)*(4*x - 4))/(5*x)))/log((2*log(2)*(4*x - 4))/(5*x)))/log(x^2))*(x*log(x^2) -
log((x + 4*log((2*log(2)*(4*x - 4))/(5*x)))/log((2*log(2)*(4*x - 4))/(5*x)))*(log((2*log(2)*(4*x - 4))/(5*x))*
(2*x - 2*x^2) - log((2*log(2)*(4*x - 4))/(5*x))^2*(8*x - 8)) + log(x^2)*log((2*log(2)*(4*x - 4))/(5*x))*(x - x
^2)))/(log(x^2)^2*log((2*log(2)*(4*x - 4))/(5*x))^2*(4*x - 4*x^2) + log(x^2)^2*log((2*log(2)*(4*x - 4))/(5*x))
*(x^2 - x^3)),x)

[Out]

(x/log(-(8*log(2) - 8*x*log(2))/(5*x)) + 4)^(1/log(x^2))

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sympy [A]  time = 47.67, size = 41, normalized size = 1.58 \begin {gather*} e^{\frac {\log {\left (\frac {x + 4 \log {\left (\frac {\left (\frac {8 x}{5} - \frac {8}{5}\right ) \log {\relax (2 )}}{x} \right )}}{\log {\left (\frac {\left (\frac {8 x}{5} - \frac {8}{5}\right ) \log {\relax (2 )}}{x} \right )}} \right )}}{\log {\left (x^{2} \right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-8*x+8)*ln(2/5*(4*x-4)*ln(2)/x)**2+(-2*x**2+2*x)*ln(2/5*(4*x-4)*ln(2)/x))*ln((4*ln(2/5*(4*x-4)*ln
(2)/x)+x)/ln(2/5*(4*x-4)*ln(2)/x))+(x**2-x)*ln(x**2)*ln(2/5*(4*x-4)*ln(2)/x)-x*ln(x**2))*exp(ln((4*ln(2/5*(4*x
-4)*ln(2)/x)+x)/ln(2/5*(4*x-4)*ln(2)/x))/ln(x**2))/((4*x**2-4*x)*ln(x**2)**2*ln(2/5*(4*x-4)*ln(2)/x)**2+(x**3-
x**2)*ln(x**2)**2*ln(2/5*(4*x-4)*ln(2)/x)),x)

[Out]

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

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