\(\int \frac {(4 x-10 x^2+6 x^3) \log ^2(x)+e^{\frac {-x+x^2+5 \log (x)}{(-1+x) \log (x)}} (-x+2 x^2-x^3+(x-2 x^2+x^3) \log (x)+(1-8 x+2 x^2) \log ^2(x))}{(-1+x) \log ^2(x)} \, dx\) [6770]

Optimal result

Integrand size = 96, antiderivative size = 28 \[ \int \frac {\left (4 x-10 x^2+6 x^3\right ) \log ^2(x)+e^{\frac {-x+x^2+5 \log (x)}{(-1+x) \log (x)}} \left (-x+2 x^2-x^3+\left (x-2 x^2+x^3\right ) \log (x)+\left (1-8 x+2 x^2\right ) \log ^2(x)\right )}{(-1+x) \log ^2(x)} \, dx=\left (e^{\frac {5}{-1+x}+\frac {x}{\log (x)}}+2 x\right ) \left (-x+x^2\right ) \]

[Out]

(exp(5/(-1+x)+x/ln(x))+2*x)*(x^2-x)

Rubi [F]

\[ \int \frac {\left (4 x-10 x^2+6 x^3\right ) \log ^2(x)+e^{\frac {-x+x^2+5 \log (x)}{(-1+x) \log (x)}} \left (-x+2 x^2-x^3+\left (x-2 x^2+x^3\right ) \log (x)+\left (1-8 x+2 x^2\right ) \log ^2(x)\right )}{(-1+x) \log ^2(x)} \, dx=\int \frac {\left (4 x-10 x^2+6 x^3\right ) \log ^2(x)+e^{\frac {-x+x^2+5 \log (x)}{(-1+x) \log (x)}} \left (-x+2 x^2-x^3+\left (x-2 x^2+x^3\right ) \log (x)+\left (1-8 x+2 x^2\right ) \log ^2(x)\right )}{(-1+x) \log ^2(x)} \, dx \]

[In]

Int[((4*x - 10*x^2 + 6*x^3)*Log[x]^2 + E^((-x + x^2 + 5*Log[x])/((-1 + x)*Log[x]))*(-x + 2*x^2 - x^3 + (x - 2*
x^2 + x^3)*Log[x] + (1 - 8*x + 2*x^2)*Log[x]^2))/((-1 + x)*Log[x]^2),x]

[Out]

-2*x^2 + 2*x^3 - 6*Defer[Int][E^(5/(-1 + x) + x/Log[x]), x] - 5*Defer[Int][E^(5/(-1 + x) + x/Log[x])/(-1 + x),
x] + 2*Defer[Int][E^(5/(-1 + x) + x/Log[x])*x, x] + Defer[Int][(E^(5/(-1 + x) + x/Log[x])*x)/Log[x]^2, x] - D
efer[Int][(E^(5/(-1 + x) + x/Log[x])*x^2)/Log[x]^2, x] - Defer[Int][(E^(5/(-1 + x) + x/Log[x])*x)/Log[x], x] +
Defer[Int][(E^(5/(-1 + x) + x/Log[x])*x^2)/Log[x], x]

Rubi steps \begin{align*} \text {integral}& = \int \left (2 x (-2+3 x)+\frac {e^{\frac {5}{-1+x}+\frac {x}{\log (x)}} \left (-x+2 x^2-x^3+x \log (x)-2 x^2 \log (x)+x^3 \log (x)+\log ^2(x)-8 x \log ^2(x)+2 x^2 \log ^2(x)\right )}{(-1+x) \log ^2(x)}\right ) \, dx \\ & = 2 \int x (-2+3 x) \, dx+\int \frac {e^{\frac {5}{-1+x}+\frac {x}{\log (x)}} \left (-x+2 x^2-x^3+x \log (x)-2 x^2 \log (x)+x^3 \log (x)+\log ^2(x)-8 x \log ^2(x)+2 x^2 \log ^2(x)\right )}{(-1+x) \log ^2(x)} \, dx \\ & = 2 \int \left (-2 x+3 x^2\right ) \, dx+\int \frac {e^{\frac {5}{-1+x}+\frac {x}{\log (x)}} \left ((-1+x)^2 x-(-1+x)^2 x \log (x)-\left (1-8 x+2 x^2\right ) \log ^2(x)\right )}{(1-x) \log ^2(x)} \, dx \\ & = -2 x^2+2 x^3+\int \left (\frac {e^{\frac {5}{-1+x}+\frac {x}{\log (x)}} \left (1-8 x+2 x^2\right )}{-1+x}-\frac {e^{\frac {5}{-1+x}+\frac {x}{\log (x)}} (-1+x) x}{\log ^2(x)}+\frac {e^{\frac {5}{-1+x}+\frac {x}{\log (x)}} (-1+x) x}{\log (x)}\right ) \, dx \\ & = -2 x^2+2 x^3+\int \frac {e^{\frac {5}{-1+x}+\frac {x}{\log (x)}} \left (1-8 x+2 x^2\right )}{-1+x} \, dx-\int \frac {e^{\frac {5}{-1+x}+\frac {x}{\log (x)}} (-1+x) x}{\log ^2(x)} \, dx+\int \frac {e^{\frac {5}{-1+x}+\frac {x}{\log (x)}} (-1+x) x}{\log (x)} \, dx \\ & = -2 x^2+2 x^3+\int \left (-6 e^{\frac {5}{-1+x}+\frac {x}{\log (x)}}-\frac {5 e^{\frac {5}{-1+x}+\frac {x}{\log (x)}}}{-1+x}+2 e^{\frac {5}{-1+x}+\frac {x}{\log (x)}} x\right ) \, dx-\int \left (-\frac {e^{\frac {5}{-1+x}+\frac {x}{\log (x)}} x}{\log ^2(x)}+\frac {e^{\frac {5}{-1+x}+\frac {x}{\log (x)}} x^2}{\log ^2(x)}\right ) \, dx+\int \left (-\frac {e^{\frac {5}{-1+x}+\frac {x}{\log (x)}} x}{\log (x)}+\frac {e^{\frac {5}{-1+x}+\frac {x}{\log (x)}} x^2}{\log (x)}\right ) \, dx \\ & = -2 x^2+2 x^3+2 \int e^{\frac {5}{-1+x}+\frac {x}{\log (x)}} x \, dx-5 \int \frac {e^{\frac {5}{-1+x}+\frac {x}{\log (x)}}}{-1+x} \, dx-6 \int e^{\frac {5}{-1+x}+\frac {x}{\log (x)}} \, dx+\int \frac {e^{\frac {5}{-1+x}+\frac {x}{\log (x)}} x}{\log ^2(x)} \, dx-\int \frac {e^{\frac {5}{-1+x}+\frac {x}{\log (x)}} x^2}{\log ^2(x)} \, dx-\int \frac {e^{\frac {5}{-1+x}+\frac {x}{\log (x)}} x}{\log (x)} \, dx+\int \frac {e^{\frac {5}{-1+x}+\frac {x}{\log (x)}} x^2}{\log (x)} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 2.28 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {\left (4 x-10 x^2+6 x^3\right ) \log ^2(x)+e^{\frac {-x+x^2+5 \log (x)}{(-1+x) \log (x)}} \left (-x+2 x^2-x^3+\left (x-2 x^2+x^3\right ) \log (x)+\left (1-8 x+2 x^2\right ) \log ^2(x)\right )}{(-1+x) \log ^2(x)} \, dx=(-1+x) x \left (e^{\frac {5}{-1+x}+\frac {x}{\log (x)}}+2 x\right ) \]

[In]

Integrate[((4*x - 10*x^2 + 6*x^3)*Log[x]^2 + E^((-x + x^2 + 5*Log[x])/((-1 + x)*Log[x]))*(-x + 2*x^2 - x^3 + (
x - 2*x^2 + x^3)*Log[x] + (1 - 8*x + 2*x^2)*Log[x]^2))/((-1 + x)*Log[x]^2),x]

[Out]

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

Maple [A] (verified)

Time = 0.69 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50

method	result	size
risch	\(2 x^{3}-2 x^{2}+\left (x^{2}-x \right ) {\mathrm e}^{\frac {5 \ln \left (x \right )+x^{2}-x}{\left (-1+x \right ) \ln \left (x \right )}}\)	\(42\)
parallelrisch	\(2 x^{3}+{\mathrm e}^{\frac {5 \ln \left (x \right )+x^{2}-x}{\left (-1+x \right ) \ln \left (x \right )}} x^{2}-2 x^{2}-x \,{\mathrm e}^{\frac {5 \ln \left (x \right )+x^{2}-x}{\left (-1+x \right ) \ln \left (x \right )}}+\frac {2}{3}\)	\(64\)

[In]

int((((2*x^2-8*x+1)*ln(x)^2+(x^3-2*x^2+x)*ln(x)-x^3+2*x^2-x)*exp((5*ln(x)+x^2-x)/(-1+x)/ln(x))+(6*x^3-10*x^2+4
*x)*ln(x)^2)/(-1+x)/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

2*x^3-2*x^2+(x^2-x)*exp((5*ln(x)+x^2-x)/(-1+x)/ln(x))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.46 \[ \int \frac {\left (4 x-10 x^2+6 x^3\right ) \log ^2(x)+e^{\frac {-x+x^2+5 \log (x)}{(-1+x) \log (x)}} \left (-x+2 x^2-x^3+\left (x-2 x^2+x^3\right ) \log (x)+\left (1-8 x+2 x^2\right ) \log ^2(x)\right )}{(-1+x) \log ^2(x)} \, dx=2 \, x^{3} - 2 \, x^{2} + {\left (x^{2} - x\right )} e^{\left (\frac {x^{2} - x + 5 \, \log \left (x\right )}{{\left (x - 1\right )} \log \left (x\right )}\right )} \]

[In]

integrate((((2*x^2-8*x+1)*log(x)^2+(x^3-2*x^2+x)*log(x)-x^3+2*x^2-x)*exp((5*log(x)+x^2-x)/(-1+x)/log(x))+(6*x^
3-10*x^2+4*x)*log(x)^2)/(-1+x)/log(x)^2,x, algorithm="fricas")

[Out]

2*x^3 - 2*x^2 + (x^2 - x)*e^((x^2 - x + 5*log(x))/((x - 1)*log(x)))

Sympy [A] (verification not implemented)

Time = 8.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {\left (4 x-10 x^2+6 x^3\right ) \log ^2(x)+e^{\frac {-x+x^2+5 \log (x)}{(-1+x) \log (x)}} \left (-x+2 x^2-x^3+\left (x-2 x^2+x^3\right ) \log (x)+\left (1-8 x+2 x^2\right ) \log ^2(x)\right )}{(-1+x) \log ^2(x)} \, dx=2 x^{3} - 2 x^{2} + \left (x^{2} - x\right ) e^{\frac {x^{2} - x + 5 \log {\left (x \right )}}{\left (x - 1\right ) \log {\left (x \right )}}} \]

[In]

integrate((((2*x**2-8*x+1)*ln(x)**2+(x**3-2*x**2+x)*ln(x)-x**3+2*x**2-x)*exp((5*ln(x)+x**2-x)/(-1+x)/ln(x))+(6
*x**3-10*x**2+4*x)*ln(x)**2)/(-1+x)/ln(x)**2,x)

[Out]

2*x**3 - 2*x**2 + (x**2 - x)*exp((x**2 - x + 5*log(x))/((x - 1)*log(x)))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {\left (4 x-10 x^2+6 x^3\right ) \log ^2(x)+e^{\frac {-x+x^2+5 \log (x)}{(-1+x) \log (x)}} \left (-x+2 x^2-x^3+\left (x-2 x^2+x^3\right ) \log (x)+\left (1-8 x+2 x^2\right ) \log ^2(x)\right )}{(-1+x) \log ^2(x)} \, dx=2 \, x^{3} - 2 \, x^{2} + {\left (x^{2} - x\right )} e^{\left (\frac {x}{\log \left (x\right )} + \frac {5}{x - 1}\right )} \]

[In]

integrate((((2*x^2-8*x+1)*log(x)^2+(x^3-2*x^2+x)*log(x)-x^3+2*x^2-x)*exp((5*log(x)+x^2-x)/(-1+x)/log(x))+(6*x^
3-10*x^2+4*x)*log(x)^2)/(-1+x)/log(x)^2,x, algorithm="maxima")

[Out]

2*x^3 - 2*x^2 + (x^2 - x)*e^(x/log(x) + 5/(x - 1))

Giac [F(-2)]

Exception generated. \[ \int \frac {\left (4 x-10 x^2+6 x^3\right ) \log ^2(x)+e^{\frac {-x+x^2+5 \log (x)}{(-1+x) \log (x)}} \left (-x+2 x^2-x^3+\left (x-2 x^2+x^3\right ) \log (x)+\left (1-8 x+2 x^2\right ) \log ^2(x)\right )}{(-1+x) \log ^2(x)} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((((2*x^2-8*x+1)*log(x)^2+(x^3-2*x^2+x)*log(x)-x^3+2*x^2-x)*exp((5*log(x)+x^2-x)/(-1+x)/log(x))+(6*x^
3-10*x^2+4*x)*log(x)^2)/(-1+x)/log(x)^2,x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Unable to divide, perhaps due to rounding error%%%{4,[1,28]%%%}+%%%{-74,[1,27]%%%}+%%%{630,[1,26]%%%}+%%%{-
3290,[1,25]

Mupad [B] (verification not implemented)

Time = 12.39 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.96 \[ \int \frac {\left (4 x-10 x^2+6 x^3\right ) \log ^2(x)+e^{\frac {-x+x^2+5 \log (x)}{(-1+x) \log (x)}} \left (-x+2 x^2-x^3+\left (x-2 x^2+x^3\right ) \log (x)+\left (1-8 x+2 x^2\right ) \log ^2(x)\right )}{(-1+x) \log ^2(x)} \, dx=x\,\left (2\,x+\frac {{\mathrm {e}}^{\frac {x}{\ln \left (x\right )-x\,\ln \left (x\right )}-\frac {x^2}{\ln \left (x\right )-x\,\ln \left (x\right )}}}{x^{\frac {5}{\ln \left (x\right )-x\,\ln \left (x\right )}}}\right )\,\left (x-1\right ) \]

[In]

int((log(x)^2*(4*x - 10*x^2 + 6*x^3) + exp((5*log(x) - x + x^2)/(log(x)*(x - 1)))*(log(x)^2*(2*x^2 - 8*x + 1)
- x + log(x)*(x - 2*x^2 + x^3) + 2*x^2 - x^3))/(log(x)^2*(x - 1)),x)

[Out]

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