Integrand size = 101, antiderivative size = 35 \[ \int \frac {-40+70 x+(10-20 x) \log (2 x)+e^{2 x^2+2 x \log \left (\frac {4}{x}\right )} \left (-36+72 x+36 \log \left (\frac {4}{x}\right )+\left (24-48 x-24 \log \left (\frac {4}{x}\right )\right ) \log (2 x)+\left (-4+8 x+4 \log \left (\frac {4}{x}\right )\right ) \log ^2(2 x)\right )}{9-6 \log (2 x)+\log ^2(2 x)} \, dx=2 \left (e^{2 x \left (x+\log \left (\frac {4}{x}\right )\right )}+\frac {5 \left (-x+x^2\right )}{3-\log (2 x)}\right ) \] Output:
10*(x^2-x)/(3-ln(2*x))+2*exp((ln(4/x)+x)*x)^2
\[ \int \frac {-40+70 x+(10-20 x) \log (2 x)+e^{2 x^2+2 x \log \left (\frac {4}{x}\right )} \left (-36+72 x+36 \log \left (\frac {4}{x}\right )+\left (24-48 x-24 \log \left (\frac {4}{x}\right )\right ) \log (2 x)+\left (-4+8 x+4 \log \left (\frac {4}{x}\right )\right ) \log ^2(2 x)\right )}{9-6 \log (2 x)+\log ^2(2 x)} \, dx=\int \frac {-40+70 x+(10-20 x) \log (2 x)+e^{2 x^2+2 x \log \left (\frac {4}{x}\right )} \left (-36+72 x+36 \log \left (\frac {4}{x}\right )+\left (24-48 x-24 \log \left (\frac {4}{x}\right )\right ) \log (2 x)+\left (-4+8 x+4 \log \left (\frac {4}{x}\right )\right ) \log ^2(2 x)\right )}{9-6 \log (2 x)+\log ^2(2 x)} \, dx \] Input:
Integrate[(-40 + 70*x + (10 - 20*x)*Log[2*x] + E^(2*x^2 + 2*x*Log[4/x])*(- 36 + 72*x + 36*Log[4/x] + (24 - 48*x - 24*Log[4/x])*Log[2*x] + (-4 + 8*x + 4*Log[4/x])*Log[2*x]^2))/(9 - 6*Log[2*x] + Log[2*x]^2),x]
Output:
Integrate[(-40 + 70*x + (10 - 20*x)*Log[2*x] + E^(2*x^2 + 2*x*Log[4/x])*(- 36 + 72*x + 36*Log[4/x] + (24 - 48*x - 24*Log[4/x])*Log[2*x] + (-4 + 8*x + 4*Log[4/x])*Log[2*x]^2))/(9 - 6*Log[2*x] + Log[2*x]^2), x]
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.
| \(\displaystyle \int \frac {e^{2 x^2+2 x \log \left (\frac {4}{x}\right )} \left (72 x+\left (8 x+4 \log \left (\frac {4}{x}\right )-4\right ) \log ^2(2 x)+\left (-48 x-24 \log \left (\frac {4}{x}\right )+24\right ) \log (2 x)+36 \log \left (\frac {4}{x}\right )-36\right )+70 x+(10-20 x) \log (2 x)-40}{\log ^2(2 x)-6 \log (2 x)+9} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {e^{2 x^2+2 x \log \left (\frac {4}{x}\right )} \left (72 x+\left (8 x+4 \log \left (\frac {4}{x}\right )-4\right ) \log ^2(2 x)+\left (-48 x-24 \log \left (\frac {4}{x}\right )+24\right ) \log (2 x)+36 \log \left (\frac {4}{x}\right )-36\right )+70 x+(10-20 x) \log (2 x)-40}{(3-\log (2 x))^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (4^{2 x+1} e^{2 x^2} \left (\frac {1}{x}\right )^{2 x} \left (2 x+\log \left (\frac {4}{x}\right )-1\right )-\frac {10 (-7 x+2 x \log (2 x)-\log (2 x)+4)}{(\log (2 x)-3)^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 4 \int \frac {\int 16^x e^{2 x^2} \left (\frac {1}{x}\right )^{2 x}dx}{x}dx+4 \log \left (\frac {4}{x}\right ) \int e^{2 x^2+2 \log (4) x} \left (\frac {1}{x}\right )^{2 x}dx-4 \int e^{2 x^2+2 \log (4) x} \left (\frac {1}{x}\right )^{2 x}dx+8 \int e^{2 x^2+2 \log (4) x} \left (\frac {1}{x}\right )^{2 x-1}dx-\frac {10 (1-x) x}{3-\log (2 x)}\) |
Input:
Int[(-40 + 70*x + (10 - 20*x)*Log[2*x] + E^(2*x^2 + 2*x*Log[4/x])*(-36 + 7 2*x + 36*Log[4/x] + (24 - 48*x - 24*Log[4/x])*Log[2*x] + (-4 + 8*x + 4*Log [4/x])*Log[2*x]^2))/(9 - 6*Log[2*x] + Log[2*x]^2),x]
Output:
$Aborted
Time = 1.99 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.14
| method | result | size |
| risch | \(-\frac {20 x \left (-1+x \right )}{-6+2 \ln \left (2\right )+2 \ln \left (x \right )}+2 x^{-2 x} 4^{2 x} {\mathrm e}^{2 x^{2}}\) | \(40\) |
| parallelrisch | \(-\frac {-300 x -60 \,{\mathrm e}^{2 \left (\ln \left (\frac {4}{x}\right )+x \right ) x} \ln \left (2 x \right )+180 \,{\mathrm e}^{2 \left (\ln \left (\frac {4}{x}\right )+x \right ) x}+300 x^{2}}{30 \left (\ln \left (2 x \right )-3\right )}\) | \(54\) |
| default | \(\frac {10 x}{\ln \left (2\right )+\ln \left (x \right )-3}-\frac {10 x^{2}}{\ln \left (2\right )+\ln \left (x \right )-3}+\frac {\left (2 \ln \left (2\right )-6\right ) {\mathrm e}^{2 \left (\ln \left (\frac {4}{x}\right )+x \right ) x}+2 \ln \left (x \right ) {\mathrm e}^{2 \left (\ln \left (\frac {4}{x}\right )+x \right ) x}}{\ln \left (2\right )+\ln \left (x \right )-3}\) | \(71\) |
Input:
int((((4*ln(4/x)+8*x-4)*ln(2*x)^2+(-24*ln(4/x)-48*x+24)*ln(2*x)+36*ln(4/x) +72*x-36)*exp(x*ln(4/x)+x^2)^2+(-20*x+10)*ln(2*x)+70*x-40)/(ln(2*x)^2-6*ln (2*x)+9),x,method=_RETURNVERBOSE)
Output:
-20*x*(-1+x)/(-6+2*ln(2)+2*ln(x))+2*(x^(-x))^2*(4^x)^2*exp(2*x^2)
Time = 0.13 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.69 \[ \int \frac {-40+70 x+(10-20 x) \log (2 x)+e^{2 x^2+2 x \log \left (\frac {4}{x}\right )} \left (-36+72 x+36 \log \left (\frac {4}{x}\right )+\left (24-48 x-24 \log \left (\frac {4}{x}\right )\right ) \log (2 x)+\left (-4+8 x+4 \log \left (\frac {4}{x}\right )\right ) \log ^2(2 x)\right )}{9-6 \log (2 x)+\log ^2(2 x)} \, dx=-\frac {2 \, {\left (5 \, x^{2} - {\left (3 \, \log \left (2\right ) - \log \left (\frac {4}{x}\right ) - 3\right )} e^{\left (2 \, x^{2} + 2 \, x \log \left (\frac {4}{x}\right )\right )} - 5 \, x\right )}}{3 \, \log \left (2\right ) - \log \left (\frac {4}{x}\right ) - 3} \] Input:
integrate((((4*log(4/x)+8*x-4)*log(2*x)^2+(-24*log(4/x)-48*x+24)*log(2*x)+ 36*log(4/x)+72*x-36)*exp(x*log(4/x)+x^2)^2+(-20*x+10)*log(2*x)+70*x-40)/(l og(2*x)^2-6*log(2*x)+9),x, algorithm="fricas")
Output:
-2*(5*x^2 - (3*log(2) - log(4/x) - 3)*e^(2*x^2 + 2*x*log(4/x)) - 5*x)/(3*l og(2) - log(4/x) - 3)
Time = 0.22 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.97 \[ \int \frac {-40+70 x+(10-20 x) \log (2 x)+e^{2 x^2+2 x \log \left (\frac {4}{x}\right )} \left (-36+72 x+36 \log \left (\frac {4}{x}\right )+\left (24-48 x-24 \log \left (\frac {4}{x}\right )\right ) \log (2 x)+\left (-4+8 x+4 \log \left (\frac {4}{x}\right )\right ) \log ^2(2 x)\right )}{9-6 \log (2 x)+\log ^2(2 x)} \, dx=\frac {- 10 x^{2} + 10 x}{\log {\left (2 x \right )} - 3} + 2 e^{2 x^{2} + 2 x \left (- \log {\left (2 x \right )} + \log {\left (8 \right )}\right )} \] Input:
integrate((((4*ln(4/x)+8*x-4)*ln(2*x)**2+(-24*ln(4/x)-48*x+24)*ln(2*x)+36* ln(4/x)+72*x-36)*exp(x*ln(4/x)+x**2)**2+(-20*x+10)*ln(2*x)+70*x-40)/(ln(2* x)**2-6*ln(2*x)+9),x)
Output:
(-10*x**2 + 10*x)/(log(2*x) - 3) + 2*exp(2*x**2 + 2*x*(-log(2*x) + log(8)) )
\[ \int \frac {-40+70 x+(10-20 x) \log (2 x)+e^{2 x^2+2 x \log \left (\frac {4}{x}\right )} \left (-36+72 x+36 \log \left (\frac {4}{x}\right )+\left (24-48 x-24 \log \left (\frac {4}{x}\right )\right ) \log (2 x)+\left (-4+8 x+4 \log \left (\frac {4}{x}\right )\right ) \log ^2(2 x)\right )}{9-6 \log (2 x)+\log ^2(2 x)} \, dx=\int { \frac {2 \, {\left (2 \, {\left ({\left (2 \, x + \log \left (\frac {4}{x}\right ) - 1\right )} \log \left (2 \, x\right )^{2} - 6 \, {\left (2 \, x + \log \left (\frac {4}{x}\right ) - 1\right )} \log \left (2 \, x\right ) + 18 \, x + 9 \, \log \left (\frac {4}{x}\right ) - 9\right )} e^{\left (2 \, x^{2} + 2 \, x \log \left (\frac {4}{x}\right )\right )} - 5 \, {\left (2 \, x - 1\right )} \log \left (2 \, x\right ) + 35 \, x - 20\right )}}{\log \left (2 \, x\right )^{2} - 6 \, \log \left (2 \, x\right ) + 9} \,d x } \] Input:
integrate((((4*log(4/x)+8*x-4)*log(2*x)^2+(-24*log(4/x)-48*x+24)*log(2*x)+ 36*log(4/x)+72*x-36)*exp(x*log(4/x)+x^2)^2+(-20*x+10)*log(2*x)+70*x-40)/(l og(2*x)^2-6*log(2*x)+9),x, algorithm="maxima")
Output:
20*e^3*exp_integral_e(2, -log(2*x) + 3)/(log(2*x) - 3) - 35/2*e^6*exp_inte gral_e(2, -2*log(2*x) + 6)/(log(2*x) - 3) + 2*(30*x^2 + (log(2) + log(x) - 3)*e^(2*x^2 + 4*x*log(2) - 2*x*log(x)) - 15*x)/(log(2) + log(x) - 3) - 2* integrate(10*(7*x - 2)/(log(2) + log(x) - 3), x)
Time = 0.20 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.74 \[ \int \frac {-40+70 x+(10-20 x) \log (2 x)+e^{2 x^2+2 x \log \left (\frac {4}{x}\right )} \left (-36+72 x+36 \log \left (\frac {4}{x}\right )+\left (24-48 x-24 \log \left (\frac {4}{x}\right )\right ) \log (2 x)+\left (-4+8 x+4 \log \left (\frac {4}{x}\right )\right ) \log ^2(2 x)\right )}{9-6 \log (2 x)+\log ^2(2 x)} \, dx=-\frac {2 \, {\left (5 \, x^{2} - e^{\left (2 \, x^{2} + 4 \, x \log \left (2\right ) - 2 \, x \log \left (x\right )\right )} \log \left (2 \, x\right ) - 5 \, x + 3 \, e^{\left (2 \, x^{2} + 4 \, x \log \left (2\right ) - 2 \, x \log \left (x\right )\right )}\right )}}{\log \left (2 \, x\right ) - 3} \] Input:
integrate((((4*log(4/x)+8*x-4)*log(2*x)^2+(-24*log(4/x)-48*x+24)*log(2*x)+ 36*log(4/x)+72*x-36)*exp(x*log(4/x)+x^2)^2+(-20*x+10)*log(2*x)+70*x-40)/(l og(2*x)^2-6*log(2*x)+9),x, algorithm="giac")
Output:
-2*(5*x^2 - e^(2*x^2 + 4*x*log(2) - 2*x*log(x))*log(2*x) - 5*x + 3*e^(2*x^ 2 + 4*x*log(2) - 2*x*log(x)))/(log(2*x) - 3)
Time = 4.26 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.71 \[ \int \frac {-40+70 x+(10-20 x) \log (2 x)+e^{2 x^2+2 x \log \left (\frac {4}{x}\right )} \left (-36+72 x+36 \log \left (\frac {4}{x}\right )+\left (24-48 x-24 \log \left (\frac {4}{x}\right )\right ) \log (2 x)+\left (-4+8 x+4 \log \left (\frac {4}{x}\right )\right ) \log ^2(2 x)\right )}{9-6 \log (2 x)+\log ^2(2 x)} \, dx=10\,x-\frac {10\,x\,\left (7\,x-4\right )-10\,x\,\ln \left (2\,x\right )\,\left (2\,x-1\right )}{\ln \left (2\,x\right )-3}-20\,x^2+2\,2^{4\,x}\,{\mathrm {e}}^{2\,x^2}\,{\left (\frac {1}{x}\right )}^{2\,x} \] Input:
int((70*x + exp(2*x*log(4/x) + 2*x^2)*(72*x + 36*log(4/x) + log(2*x)^2*(8* x + 4*log(4/x) - 4) - log(2*x)*(48*x + 24*log(4/x) - 24) - 36) - log(2*x)* (20*x - 10) - 40)/(log(2*x)^2 - 6*log(2*x) + 9),x)
Output:
10*x - (10*x*(7*x - 4) - 10*x*log(2*x)*(2*x - 1))/(log(2*x) - 3) - 20*x^2 + 2*2^(4*x)*exp(2*x^2)*(1/x)^(2*x)
\[ \int \frac {-40+70 x+(10-20 x) \log (2 x)+e^{2 x^2+2 x \log \left (\frac {4}{x}\right )} \left (-36+72 x+36 \log \left (\frac {4}{x}\right )+\left (24-48 x-24 \log \left (\frac {4}{x}\right )\right ) \log (2 x)+\left (-4+8 x+4 \log \left (\frac {4}{x}\right )\right ) \log ^2(2 x)\right )}{9-6 \log (2 x)+\log ^2(2 x)} \, dx=\frac {4 \left (\int \frac {e^{2 x^{2}} 4^{2 x} \mathrm {log}\left (\frac {4}{x}\right )}{x^{2 x}}d x \right ) \mathrm {log}\left (2 x \right )-12 \left (\int \frac {e^{2 x^{2}} 4^{2 x} \mathrm {log}\left (\frac {4}{x}\right )}{x^{2 x}}d x \right )+8 \left (\int \frac {e^{2 x^{2}} 4^{2 x} x}{x^{2 x}}d x \right ) \mathrm {log}\left (2 x \right )-24 \left (\int \frac {e^{2 x^{2}} 4^{2 x} x}{x^{2 x}}d x \right )-4 \left (\int \frac {e^{2 x^{2}} 4^{2 x}}{x^{2 x}}d x \right ) \mathrm {log}\left (2 x \right )+12 \left (\int \frac {e^{2 x^{2}} 4^{2 x}}{x^{2 x}}d x \right )-10 x^{2}+10 x}{\mathrm {log}\left (2 x \right )-3} \] Input:
int((((4*log(4/x)+8*x-4)*log(2*x)^2+(-24*log(4/x)-48*x+24)*log(2*x)+36*log (4/x)+72*x-36)*exp(x*log(4/x)+x^2)^2+(-20*x+10)*log(2*x)+70*x-40)/(log(2*x )^2-6*log(2*x)+9),x)
Output:
(2*(2*int((e**(2*x**2)*4**(2*x)*log(4/x))/x**(2*x),x)*log(2*x) - 6*int((e* *(2*x**2)*4**(2*x)*log(4/x))/x**(2*x),x) + 4*int((e**(2*x**2)*4**(2*x)*x)/ x**(2*x),x)*log(2*x) - 12*int((e**(2*x**2)*4**(2*x)*x)/x**(2*x),x) - 2*int ((e**(2*x**2)*4**(2*x))/x**(2*x),x)*log(2*x) + 6*int((e**(2*x**2)*4**(2*x) )/x**(2*x),x) - 5*x**2 + 5*x))/(log(2*x) - 3)