Integrand size = 79, antiderivative size = 19 \[ \int \frac {-64 x+32 e^{2 x} x+\left (-64 x+e^{2 x} \left (32 x+64 x^2\right )\right ) \log (x)+\left (-128 x+64 e^{2 x} x\right ) \log (x) \log \left (\left (2 x-e^{2 x} x\right ) \log (x)\right )}{\left (-2+e^{2 x}\right ) \log (x)} \, dx=32 x^2 \log \left (\left (2-e^{2 x}\right ) x \log (x)\right ) \] Output:
32*ln(ln(x)*(-exp(x)^2+2)*x)*x^2
Time = 0.20 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {-64 x+32 e^{2 x} x+\left (-64 x+e^{2 x} \left (32 x+64 x^2\right )\right ) \log (x)+\left (-128 x+64 e^{2 x} x\right ) \log (x) \log \left (\left (2 x-e^{2 x} x\right ) \log (x)\right )}{\left (-2+e^{2 x}\right ) \log (x)} \, dx=32 x^2 \log \left (-\left (\left (-2+e^{2 x}\right ) x \log (x)\right )\right ) \] Input:
Integrate[(-64*x + 32*E^(2*x)*x + (-64*x + E^(2*x)*(32*x + 64*x^2))*Log[x] + (-128*x + 64*E^(2*x)*x)*Log[x]*Log[(2*x - E^(2*x)*x)*Log[x]])/((-2 + E^ (2*x))*Log[x]),x]
Output:
32*x^2*Log[-((-2 + E^(2*x))*x*Log[x])]
Time = 1.21 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {7293, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (e^{2 x} \left (64 x^2+32 x\right )-64 x\right ) \log (x)+32 e^{2 x} x-64 x+\left (64 e^{2 x} x-128 x\right ) \log (x) \log \left (\left (2 x-e^{2 x} x\right ) \log (x)\right )}{\left (e^{2 x}-2\right ) \log (x)} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {128 x^2}{e^{2 x}-2}+\frac {32 x \left (2 x \log (x)+2 \log \left (-\left (\left (e^{2 x}-2\right ) x \log (x)\right )\right ) \log (x)+\log (x)+1\right )}{\log (x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 32 x^2 \log \left (\left (2-e^{2 x}\right ) x \log (x)\right )\) |
Input:
Int[(-64*x + 32*E^(2*x)*x + (-64*x + E^(2*x)*(32*x + 64*x^2))*Log[x] + (-1 28*x + 64*E^(2*x)*x)*Log[x]*Log[(2*x - E^(2*x)*x)*Log[x]])/((-2 + E^(2*x)) *Log[x]),x]
Output:
32*x^2*Log[(2 - E^(2*x))*x*Log[x]]
Time = 1.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11
method | result | size |
parallelrisch | \(32 x^{2} \ln \left (\left (-x \,{\mathrm e}^{2 x}+2 x \right ) \ln \left (x \right )\right )\) | \(21\) |
risch | \(32 x^{2} \ln \left ({\mathrm e}^{2 x}-2\right )+32 x^{2} \ln \left (\ln \left (x \right )\right )+16 i \pi \,x^{2} {\operatorname {csgn}\left (i x \ln \left (x \right ) \left ({\mathrm e}^{2 x}-2\right )\right )}^{3}-16 i \pi \,x^{2} \operatorname {csgn}\left (i \left ({\mathrm e}^{2 x}-2\right ) \ln \left (x \right )\right ) \operatorname {csgn}\left (i x \ln \left (x \right ) \left ({\mathrm e}^{2 x}-2\right )\right ) \operatorname {csgn}\left (i x \right )+16 i \pi \,x^{2} \operatorname {csgn}\left (i \ln \left (x \right )\right ) {\operatorname {csgn}\left (i \left ({\mathrm e}^{2 x}-2\right ) \ln \left (x \right )\right )}^{2}+16 i \pi \,x^{2} \operatorname {csgn}\left (i \left ({\mathrm e}^{2 x}-2\right ) \ln \left (x \right )\right ) {\operatorname {csgn}\left (i x \ln \left (x \right ) \left ({\mathrm e}^{2 x}-2\right )\right )}^{2}-16 i \pi \,x^{2} {\operatorname {csgn}\left (i \left ({\mathrm e}^{2 x}-2\right ) \ln \left (x \right )\right )}^{3}+16 i \pi \,x^{2} {\operatorname {csgn}\left (i x \ln \left (x \right ) \left ({\mathrm e}^{2 x}-2\right )\right )}^{2} \operatorname {csgn}\left (i x \right )-16 i \pi \,x^{2} \operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i \left ({\mathrm e}^{2 x}-2\right )\right ) \operatorname {csgn}\left (i \left ({\mathrm e}^{2 x}-2\right ) \ln \left (x \right )\right )+16 i \pi \,x^{2} \operatorname {csgn}\left (i \left ({\mathrm e}^{2 x}-2\right )\right ) {\operatorname {csgn}\left (i \left ({\mathrm e}^{2 x}-2\right ) \ln \left (x \right )\right )}^{2}+32 i \pi \,x^{2}-32 i \pi \,x^{2} {\operatorname {csgn}\left (i x \ln \left (x \right ) \left ({\mathrm e}^{2 x}-2\right )\right )}^{2}+32 x^{2} \ln \left (x \right )\) | \(292\) |
Input:
int(((64*x*exp(x)^2-128*x)*ln(x)*ln((-x*exp(x)^2+2*x)*ln(x))+((64*x^2+32*x )*exp(x)^2-64*x)*ln(x)+32*x*exp(x)^2-64*x)/(exp(x)^2-2)/ln(x),x,method=_RE TURNVERBOSE)
Output:
32*x^2*ln((-x*exp(x)^2+2*x)*ln(x))
Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {-64 x+32 e^{2 x} x+\left (-64 x+e^{2 x} \left (32 x+64 x^2\right )\right ) \log (x)+\left (-128 x+64 e^{2 x} x\right ) \log (x) \log \left (\left (2 x-e^{2 x} x\right ) \log (x)\right )}{\left (-2+e^{2 x}\right ) \log (x)} \, dx=32 \, x^{2} \log \left (-{\left (x e^{\left (2 \, x\right )} - 2 \, x\right )} \log \left (x\right )\right ) \] Input:
integrate(((64*x*exp(x)^2-128*x)*log(x)*log((-x*exp(x)^2+2*x)*log(x))+((64 *x^2+32*x)*exp(x)^2-64*x)*log(x)+32*x*exp(x)^2-64*x)/(exp(x)^2-2)/log(x),x , algorithm="fricas")
Output:
32*x^2*log(-(x*e^(2*x) - 2*x)*log(x))
Time = 0.35 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {-64 x+32 e^{2 x} x+\left (-64 x+e^{2 x} \left (32 x+64 x^2\right )\right ) \log (x)+\left (-128 x+64 e^{2 x} x\right ) \log (x) \log \left (\left (2 x-e^{2 x} x\right ) \log (x)\right )}{\left (-2+e^{2 x}\right ) \log (x)} \, dx=32 x^{2} \log {\left (\left (- x e^{2 x} + 2 x\right ) \log {\left (x \right )} \right )} \] Input:
integrate(((64*x*exp(x)**2-128*x)*ln(x)*ln((-x*exp(x)**2+2*x)*ln(x))+((64* x**2+32*x)*exp(x)**2-64*x)*ln(x)+32*x*exp(x)**2-64*x)/(exp(x)**2-2)/ln(x), x)
Output:
32*x**2*log((-x*exp(2*x) + 2*x)*log(x))
Time = 0.08 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.58 \[ \int \frac {-64 x+32 e^{2 x} x+\left (-64 x+e^{2 x} \left (32 x+64 x^2\right )\right ) \log (x)+\left (-128 x+64 e^{2 x} x\right ) \log (x) \log \left (\left (2 x-e^{2 x} x\right ) \log (x)\right )}{\left (-2+e^{2 x}\right ) \log (x)} \, dx=32 \, x^{2} \log \left (x\right ) + 32 \, x^{2} \log \left (-e^{\left (2 \, x\right )} + 2\right ) + 32 \, x^{2} \log \left (\log \left (x\right )\right ) \] Input:
integrate(((64*x*exp(x)^2-128*x)*log(x)*log((-x*exp(x)^2+2*x)*log(x))+((64 *x^2+32*x)*exp(x)^2-64*x)*log(x)+32*x*exp(x)^2-64*x)/(exp(x)^2-2)/log(x),x , algorithm="maxima")
Output:
32*x^2*log(x) + 32*x^2*log(-e^(2*x) + 2) + 32*x^2*log(log(x))
Time = 0.18 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.42 \[ \int \frac {-64 x+32 e^{2 x} x+\left (-64 x+e^{2 x} \left (32 x+64 x^2\right )\right ) \log (x)+\left (-128 x+64 e^{2 x} x\right ) \log (x) \log \left (\left (2 x-e^{2 x} x\right ) \log (x)\right )}{\left (-2+e^{2 x}\right ) \log (x)} \, dx=32 \, x^{2} \log \left (-e^{\left (2 \, x\right )} \log \left (x\right ) + 2 \, \log \left (x\right )\right ) + 32 \, x^{2} \log \left (x\right ) \] Input:
integrate(((64*x*exp(x)^2-128*x)*log(x)*log((-x*exp(x)^2+2*x)*log(x))+((64 *x^2+32*x)*exp(x)^2-64*x)*log(x)+32*x*exp(x)^2-64*x)/(exp(x)^2-2)/log(x),x , algorithm="giac")
Output:
32*x^2*log(-e^(2*x)*log(x) + 2*log(x)) + 32*x^2*log(x)
Time = 7.47 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {-64 x+32 e^{2 x} x+\left (-64 x+e^{2 x} \left (32 x+64 x^2\right )\right ) \log (x)+\left (-128 x+64 e^{2 x} x\right ) \log (x) \log \left (\left (2 x-e^{2 x} x\right ) \log (x)\right )}{\left (-2+e^{2 x}\right ) \log (x)} \, dx=32\,x^2\,\ln \left (\ln \left (x\right )\,\left (2\,x-x\,{\mathrm {e}}^{2\,x}\right )\right ) \] Input:
int(-(64*x - 32*x*exp(2*x) + log(x)*(64*x - exp(2*x)*(32*x + 64*x^2)) + lo g(x)*log(log(x)*(2*x - x*exp(2*x)))*(128*x - 64*x*exp(2*x)))/(log(x)*(exp( 2*x) - 2)),x)
Output:
32*x^2*log(log(x)*(2*x - x*exp(2*x)))
Time = 0.17 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16 \[ \int \frac {-64 x+32 e^{2 x} x+\left (-64 x+e^{2 x} \left (32 x+64 x^2\right )\right ) \log (x)+\left (-128 x+64 e^{2 x} x\right ) \log (x) \log \left (\left (2 x-e^{2 x} x\right ) \log (x)\right )}{\left (-2+e^{2 x}\right ) \log (x)} \, dx=32 \,\mathrm {log}\left (-e^{2 x} \mathrm {log}\left (x \right ) x +2 \,\mathrm {log}\left (x \right ) x \right ) x^{2} \] Input:
int(((64*x*exp(x)^2-128*x)*log(x)*log((-x*exp(x)^2+2*x)*log(x))+((64*x^2+3 2*x)*exp(x)^2-64*x)*log(x)+32*x*exp(x)^2-64*x)/(exp(x)^2-2)/log(x),x)
Output:
32*log( - e**(2*x)*log(x)*x + 2*log(x)*x)*x**2