Integrand size = 81, antiderivative size = 22 \[ \int \frac {1+4 x-16 x^2-16 x^3+\left (-8 x-12 x^2\right ) \log (x)+\left (-x+4 x^2+4 x^3+\left (2 x+3 x^2\right ) \log (x)\right ) \log (2 x)+(-4+\log (2 x)) \log (-4+\log (2 x))}{-4+\log (2 x)} \, dx=x (-x+x (1+x) (x+\log (x))+\log (-4+\log (2 x))) \] Output:
((1+x)*(x+ln(x))*x+ln(ln(2*x)-4)-x)*x
Time = 0.12 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {1+4 x-16 x^2-16 x^3+\left (-8 x-12 x^2\right ) \log (x)+\left (-x+4 x^2+4 x^3+\left (2 x+3 x^2\right ) \log (x)\right ) \log (2 x)+(-4+\log (2 x)) \log (-4+\log (2 x))}{-4+\log (2 x)} \, dx=x \left (x \left (-1+x+x^2\right )+x (1+x) \log (x)+\log (-4+\log (2 x))\right ) \] Input:
Integrate[(1 + 4*x - 16*x^2 - 16*x^3 + (-8*x - 12*x^2)*Log[x] + (-x + 4*x^ 2 + 4*x^3 + (2*x + 3*x^2)*Log[x])*Log[2*x] + (-4 + Log[2*x])*Log[-4 + Log[ 2*x]])/(-4 + Log[2*x]),x]
Output:
x*(x*(-1 + x + x^2) + x*(1 + x)*Log[x] + Log[-4 + Log[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.
\(\displaystyle \int \frac {-16 x^3-16 x^2+\left (-12 x^2-8 x\right ) \log (x)+\left (4 x^3+4 x^2+\left (3 x^2+2 x\right ) \log (x)-x\right ) \log (2 x)+4 x+(\log (2 x)-4) \log (\log (2 x)-4)+1}{\log (2 x)-4} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {-16 x^3+4 x^3 \log (2 x)-16 x^2-12 x^2 \log (x)+3 x^2 \log (x) \log (2 x)+4 x^2 \log (2 x)+4 x-8 x \log (x)+2 x \log (x) \log (2 x)-x \log (2 x)+1}{\log (2 x)-4}+\log (\log (2 x)-4)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \text {Subst}(\int \log (\log (x)-4)dx,x,2 x)+\frac {1}{2} e^4 \operatorname {ExpIntegralEi}(\log (2 x)-4)+x^4+x^3-x^2+\left (x^3+x^2\right ) \log (x)\) |
Input:
Int[(1 + 4*x - 16*x^2 - 16*x^3 + (-8*x - 12*x^2)*Log[x] + (-x + 4*x^2 + 4* x^3 + (2*x + 3*x^2)*Log[x])*Log[2*x] + (-4 + Log[2*x])*Log[-4 + Log[2*x]]) /(-4 + Log[2*x]),x]
Output:
$Aborted
Time = 0.81 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.55
method | result | size |
risch | \(x^{4}+x^{3} \ln \left (x \right )+x^{3}+x^{2} \ln \left (x \right )-x^{2}+x \ln \left (\ln \left (2\right )+\ln \left (x \right )-4\right )\) | \(34\) |
parallelrisch | \(x^{3} \ln \left (x \right )+x^{2} \ln \left (x \right )+x^{4}+x^{3}-x^{2}+\ln \left (\ln \left (2 x \right )-4\right ) x\) | \(34\) |
default | \(x^{2} \ln \left (x \right )-x^{2}+x^{4}-{\mathrm e}^{4-\ln \left (2\right )} \operatorname {expIntegral}_{1}\left (-\ln \left (2\right )-\ln \left (x \right )+4\right )+x^{3} \ln \left (x \right )+x^{3}+\ln \left (\ln \left (2 x \right )-4\right ) x +\frac {{\mathrm e}^{4} \operatorname {expIntegral}_{1}\left (-\ln \left (2 x \right )+4\right )}{2}\) | \(69\) |
parts | \(x^{2} \ln \left (x \right )-x^{2}+x^{4}-{\mathrm e}^{4-\ln \left (2\right )} \operatorname {expIntegral}_{1}\left (-\ln \left (2\right )-\ln \left (x \right )+4\right )+x^{3} \ln \left (x \right )+x^{3}+\ln \left (\ln \left (2 x \right )-4\right ) x +\frac {{\mathrm e}^{4} \operatorname {expIntegral}_{1}\left (-\ln \left (2 x \right )+4\right )}{2}\) | \(69\) |
Input:
int(((ln(2*x)-4)*ln(ln(2*x)-4)+((3*x^2+2*x)*ln(x)+4*x^3+4*x^2-x)*ln(2*x)+( -12*x^2-8*x)*ln(x)-16*x^3-16*x^2+4*x+1)/(ln(2*x)-4),x,method=_RETURNVERBOS E)
Output:
x^4+x^3*ln(x)+x^3+x^2*ln(x)-x^2+x*ln(ln(2)+ln(x)-4)
Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41 \[ \int \frac {1+4 x-16 x^2-16 x^3+\left (-8 x-12 x^2\right ) \log (x)+\left (-x+4 x^2+4 x^3+\left (2 x+3 x^2\right ) \log (x)\right ) \log (2 x)+(-4+\log (2 x)) \log (-4+\log (2 x))}{-4+\log (2 x)} \, dx=x^{4} + x^{3} - x^{2} + {\left (x^{3} + x^{2}\right )} \log \left (x\right ) + x \log \left (\log \left (2\right ) + \log \left (x\right ) - 4\right ) \] Input:
integrate(((log(2*x)-4)*log(log(2*x)-4)+((3*x^2+2*x)*log(x)+4*x^3+4*x^2-x) *log(2*x)+(-12*x^2-8*x)*log(x)-16*x^3-16*x^2+4*x+1)/(log(2*x)-4),x, algori thm="fricas")
Output:
x^4 + x^3 - x^2 + (x^3 + x^2)*log(x) + x*log(log(2) + log(x) - 4)
Time = 0.18 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41 \[ \int \frac {1+4 x-16 x^2-16 x^3+\left (-8 x-12 x^2\right ) \log (x)+\left (-x+4 x^2+4 x^3+\left (2 x+3 x^2\right ) \log (x)\right ) \log (2 x)+(-4+\log (2 x)) \log (-4+\log (2 x))}{-4+\log (2 x)} \, dx=x^{4} + x^{3} - x^{2} + x \log {\left (\log {\left (x \right )} - 4 + \log {\left (2 \right )} \right )} + \left (x^{3} + x^{2}\right ) \log {\left (x \right )} \] Input:
integrate(((ln(2*x)-4)*ln(ln(2*x)-4)+((3*x**2+2*x)*ln(x)+4*x**3+4*x**2-x)* ln(2*x)+(-12*x**2-8*x)*ln(x)-16*x**3-16*x**2+4*x+1)/(ln(2*x)-4),x)
Output:
x**4 + x**3 - x**2 + x*log(log(x) - 4 + log(2)) + (x**3 + x**2)*log(x)
\[ \int \frac {1+4 x-16 x^2-16 x^3+\left (-8 x-12 x^2\right ) \log (x)+\left (-x+4 x^2+4 x^3+\left (2 x+3 x^2\right ) \log (x)\right ) \log (2 x)+(-4+\log (2 x)) \log (-4+\log (2 x))}{-4+\log (2 x)} \, dx=\int { -\frac {16 \, x^{3} + 16 \, x^{2} - {\left (4 \, x^{3} + 4 \, x^{2} + {\left (3 \, x^{2} + 2 \, x\right )} \log \left (x\right ) - x\right )} \log \left (2 \, x\right ) + 4 \, {\left (3 \, x^{2} + 2 \, x\right )} \log \left (x\right ) - {\left (\log \left (2 \, x\right ) - 4\right )} \log \left (\log \left (2 \, x\right ) - 4\right ) - 4 \, x - 1}{\log \left (2 \, x\right ) - 4} \,d x } \] Input:
integrate(((log(2*x)-4)*log(log(2*x)-4)+((3*x^2+2*x)*log(x)+4*x^3+4*x^2-x) *log(2*x)+(-12*x^2-8*x)*log(x)-16*x^3-16*x^2+4*x+1)/(log(2*x)-4),x, algori thm="maxima")
Output:
11/3*x^3 + 1/4*e^8*exp_integral_e(1, -2*log(2*x) + 8)*log(2*x) - 1/2*e^12* exp_integral_e(1, -3*log(2*x) + 12)*log(2*x) - 1/4*e^16*exp_integral_e(1, -4*log(2*x) + 16)*log(2*x) + 2*e^8*exp_integral_e(1, -2*log(2*x) + 8)*log( x) + 3/2*e^12*exp_integral_e(1, -3*log(2*x) + 12)*log(x) + 7/2*x^2 - 9/8*e ^8*exp_integral_e(2, -2*log(2*x) + 8) - 1/3*e^12*exp_integral_e(2, -3*log( 2*x) + 12) + 1/16*e^16*exp_integral_e(2, -4*log(2*x) + 16) - 1/2*e^4*exp_i ntegral_e(1, -log(2*x) + 4) - e^8*exp_integral_e(1, -2*log(2*x) + 8) + 2*e ^12*exp_integral_e(1, -3*log(2*x) + 12) + e^16*exp_integral_e(1, -4*log(2* x) + 16) + (x^3 + x^2)*log(x) + x*log(log(2) + log(x) - 4) - integrate((12 *x^2*(log(2) - 4) + 8*x*(log(2) - 4) + 1)/(log(2) + log(x) - 4), x)
Time = 0.13 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.50 \[ \int \frac {1+4 x-16 x^2-16 x^3+\left (-8 x-12 x^2\right ) \log (x)+\left (-x+4 x^2+4 x^3+\left (2 x+3 x^2\right ) \log (x)\right ) \log (2 x)+(-4+\log (2 x)) \log (-4+\log (2 x))}{-4+\log (2 x)} \, dx=x^{4} + x^{3} \log \left (x\right ) + x^{3} + x^{2} \log \left (x\right ) - x^{2} + x \log \left (\log \left (2\right ) + \log \left (x\right ) - 4\right ) \] Input:
integrate(((log(2*x)-4)*log(log(2*x)-4)+((3*x^2+2*x)*log(x)+4*x^3+4*x^2-x) *log(2*x)+(-12*x^2-8*x)*log(x)-16*x^3-16*x^2+4*x+1)/(log(2*x)-4),x, algori thm="giac")
Output:
x^4 + x^3*log(x) + x^3 + x^2*log(x) - x^2 + x*log(log(2) + log(x) - 4)
Time = 1.96 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41 \[ \int \frac {1+4 x-16 x^2-16 x^3+\left (-8 x-12 x^2\right ) \log (x)+\left (-x+4 x^2+4 x^3+\left (2 x+3 x^2\right ) \log (x)\right ) \log (2 x)+(-4+\log (2 x)) \log (-4+\log (2 x))}{-4+\log (2 x)} \, dx=\ln \left (x\right )\,\left (x^3+x^2\right )-x^2+x^3+x^4+x\,\ln \left (\ln \left (2\,x\right )-4\right ) \] Input:
int((4*x + log(log(2*x) - 4)*(log(2*x) - 4) - log(x)*(8*x + 12*x^2) - 16*x ^2 - 16*x^3 + log(2*x)*(log(x)*(2*x + 3*x^2) - x + 4*x^2 + 4*x^3) + 1)/(lo g(2*x) - 4),x)
Output:
log(x)*(x^2 + x^3) - x^2 + x^3 + x^4 + x*log(log(2*x) - 4)
Time = 0.18 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {1+4 x-16 x^2-16 x^3+\left (-8 x-12 x^2\right ) \log (x)+\left (-x+4 x^2+4 x^3+\left (2 x+3 x^2\right ) \log (x)\right ) \log (2 x)+(-4+\log (2 x)) \log (-4+\log (2 x))}{-4+\log (2 x)} \, dx=x \left (\mathrm {log}\left (\mathrm {log}\left (2 x \right )-4\right )+\mathrm {log}\left (x \right ) x^{2}+\mathrm {log}\left (x \right ) x +x^{3}+x^{2}-x \right ) \] Input:
int(((log(2*x)-4)*log(log(2*x)-4)+((3*x^2+2*x)*log(x)+4*x^3+4*x^2-x)*log(2 *x)+(-12*x^2-8*x)*log(x)-16*x^3-16*x^2+4*x+1)/(log(2*x)-4),x)
Output:
x*(log(log(2*x) - 4) + log(x)*x**2 + log(x)*x + x**3 + x**2 - x)