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

Optimal. Leaf size=28 \[ \left (e^{\frac {5}{-1+x}+\frac {x}{\log (x)}}+2 x\right ) \left (-x+x^2\right ) \]

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Rubi [F]  time = 5.48, 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 (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 \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 2.07, size = 25, normalized size = 0.89 \begin {gather*} (-1+x) x \left (e^{\frac {5}{-1+x}+\frac {x}{\log (x)}}+2 x\right ) \end {gather*}

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.53, size = 41, normalized size = 1.46 \begin {gather*} 2 \, x^{3} - 2 \, x^{2} + {\left (x^{2} - x\right )} e^{\left (\frac {x^{2} - x + 5 \, \log \relax (x)}{{\left (x - 1\right )} \log \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)/(x-1)/log(x))+(6*x^3
-10*x^2+4*x)*log(x)^2)/(x-1)/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)))

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Evaluation time: 0.45Unable to divide, perhaps due to rounding error%%%{4,[1,28]%%%}+%%%{-74,[1,27]%%%}+%%%
{630,[1,26]

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maple [A]  time = 0.07, size = 42, normalized size = 1.50

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

Verification of antiderivative is not currently implemented for this CAS.

[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)/(x-1)/ln(x))+(6*x^3-10*x^2+4*
x)*ln(x)^2)/(x-1)/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 1.17, size = 34, normalized size = 1.21 \begin {gather*} 2 \, x^{3} - 2 \, x^{2} + {\left (x^{2} - x\right )} e^{\left (\frac {x}{\log \relax (x)} + \frac {5}{x - 1}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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mupad [B]  time = 4.38, size = 55, normalized size = 1.96 \begin {gather*} x\,\left (2\,x+\frac {{\mathrm {e}}^{\frac {x}{\ln \relax (x)-x\,\ln \relax (x)}-\frac {x^2}{\ln \relax (x)-x\,\ln \relax (x)}}}{x^{\frac {5}{\ln \relax (x)-x\,\ln \relax (x)}}}\right )\,\left (x-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 13.71, size = 32, normalized size = 1.14 \begin {gather*} 2 x^{3} - 2 x^{2} + \left (x^{2} - x\right ) e^{\frac {x^{2} - x + 5 \log {\relax (x )}}{\left (x - 1\right ) \log {\relax (x )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)/(x-1)/ln(x))+(6*
x**3-10*x**2+4*x)*ln(x)**2)/(x-1)/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)))

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