3.97.10 \(\int \frac {e^5 (-12-2 x+2 x^2)+e^5 (1-3 x+2 x^2) \log (1-2 x+x^2) \log (\log (1-2 x+x^2))}{(-1+x) \log (1-2 x+x^2)} \, dx\)

Optimal. Leaf size=17 \[ e^5 (-3+x) (2+x) \log \left (\log \left ((-1+x)^2\right )\right ) \]

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Rubi [F]  time = 0.36, 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 {e^5 \left (-12-2 x+2 x^2\right )+e^5 \left (1-3 x+2 x^2\right ) \log \left (1-2 x+x^2\right ) \log \left (\log \left (1-2 x+x^2\right )\right )}{(-1+x) \log \left (1-2 x+x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^5*(-12 - 2*x + 2*x^2) + E^5*(1 - 3*x + 2*x^2)*Log[1 - 2*x + x^2]*Log[Log[1 - 2*x + x^2]])/((-1 + x)*Log
[1 - 2*x + x^2]),x]

[Out]

(-2*E^5*(1 - x)*ExpIntegralEi[Log[(-1 + x)^2]/2])/Sqrt[(-1 + x)^2] - 6*E^5*Log[Log[(-1 + x)^2]] + E^5*(1 - x)*
Log[Log[(-1 + x)^2]] + E^5*LogIntegral[(-1 + x)^2] + 2*E^5*Defer[Int][x*Log[Log[(-1 + x)^2]], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {2 e^5 (-3+x) (2+x)}{(-1+x) \log \left ((-1+x)^2\right )}+e^5 (-1+2 x) \log \left (\log \left ((-1+x)^2\right )\right )\right ) \, dx\\ &=e^5 \int (-1+2 x) \log \left (\log \left ((-1+x)^2\right )\right ) \, dx+\left (2 e^5\right ) \int \frac {(-3+x) (2+x)}{(-1+x) \log \left ((-1+x)^2\right )} \, dx\\ &=e^5 \int \left (-\log \left (\log \left ((-1+x)^2\right )\right )+2 x \log \left (\log \left ((-1+x)^2\right )\right )\right ) \, dx+\left (2 e^5\right ) \int \left (-\frac {6}{(-1+x) \log \left ((-1+x)^2\right )}+\frac {x}{\log \left ((-1+x)^2\right )}\right ) \, dx\\ &=-\left (e^5 \int \log \left (\log \left ((-1+x)^2\right )\right ) \, dx\right )+\left (2 e^5\right ) \int \frac {x}{\log \left ((-1+x)^2\right )} \, dx+\left (2 e^5\right ) \int x \log \left (\log \left ((-1+x)^2\right )\right ) \, dx-\left (12 e^5\right ) \int \frac {1}{(-1+x) \log \left ((-1+x)^2\right )} \, dx\\ &=-\left (e^5 \operatorname {Subst}\left (\int \log \left (\log \left (x^2\right )\right ) \, dx,x,-1+x\right )\right )+\left (2 e^5\right ) \int \left (\frac {1}{\log \left ((-1+x)^2\right )}+\frac {-1+x}{\log \left ((-1+x)^2\right )}\right ) \, dx+\left (2 e^5\right ) \int x \log \left (\log \left ((-1+x)^2\right )\right ) \, dx-\left (12 e^5\right ) \operatorname {Subst}\left (\int \frac {1}{x \log \left (x^2\right )} \, dx,x,-1+x\right )\\ &=e^5 (1-x) \log \left (\log \left ((-1+x)^2\right )\right )+\left (2 e^5\right ) \int \frac {1}{\log \left ((-1+x)^2\right )} \, dx+\left (2 e^5\right ) \int \frac {-1+x}{\log \left ((-1+x)^2\right )} \, dx+\left (2 e^5\right ) \int x \log \left (\log \left ((-1+x)^2\right )\right ) \, dx+\left (2 e^5\right ) \operatorname {Subst}\left (\int \frac {1}{\log \left (x^2\right )} \, dx,x,-1+x\right )-\left (6 e^5\right ) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left ((-1+x)^2\right )\right )\\ &=-6 e^5 \log \left (\log \left ((-1+x)^2\right )\right )+e^5 (1-x) \log \left (\log \left ((-1+x)^2\right )\right )+\left (2 e^5\right ) \int x \log \left (\log \left ((-1+x)^2\right )\right ) \, dx+\left (2 e^5\right ) \operatorname {Subst}\left (\int \frac {1}{\log \left (x^2\right )} \, dx,x,-1+x\right )+\left (2 e^5\right ) \operatorname {Subst}\left (\int \frac {x}{\log \left (x^2\right )} \, dx,x,-1+x\right )+\frac {\left (e^5 (-1+x)\right ) \operatorname {Subst}\left (\int \frac {e^{x/2}}{x} \, dx,x,\log \left ((-1+x)^2\right )\right )}{\sqrt {(-1+x)^2}}\\ &=-\frac {e^5 (1-x) \text {Ei}\left (\frac {1}{2} \log \left ((-1+x)^2\right )\right )}{\sqrt {(-1+x)^2}}-6 e^5 \log \left (\log \left ((-1+x)^2\right )\right )+e^5 (1-x) \log \left (\log \left ((-1+x)^2\right )\right )+e^5 \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,(-1+x)^2\right )+\left (2 e^5\right ) \int x \log \left (\log \left ((-1+x)^2\right )\right ) \, dx+\frac {\left (e^5 (-1+x)\right ) \operatorname {Subst}\left (\int \frac {e^{x/2}}{x} \, dx,x,\log \left ((-1+x)^2\right )\right )}{\sqrt {(-1+x)^2}}\\ &=-\frac {2 e^5 (1-x) \text {Ei}\left (\frac {1}{2} \log \left ((-1+x)^2\right )\right )}{\sqrt {(-1+x)^2}}-6 e^5 \log \left (\log \left ((-1+x)^2\right )\right )+e^5 (1-x) \log \left (\log \left ((-1+x)^2\right )\right )+e^5 \text {li}\left ((-1+x)^2\right )+\left (2 e^5\right ) \int x \log \left (\log \left ((-1+x)^2\right )\right ) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 18, normalized size = 1.06 \begin {gather*} e^5 (-6+(-1+x) x) \log \left (\log \left ((-1+x)^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^5*(-12 - 2*x + 2*x^2) + E^5*(1 - 3*x + 2*x^2)*Log[1 - 2*x + x^2]*Log[Log[1 - 2*x + x^2]])/((-1 +
x)*Log[1 - 2*x + x^2]),x]

[Out]

E^5*(-6 + (-1 + x)*x)*Log[Log[(-1 + x)^2]]

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fricas [A]  time = 0.50, size = 21, normalized size = 1.24 \begin {gather*} {\left (x^{2} - x - 6\right )} e^{5} \log \left (\log \left (x^{2} - 2 \, x + 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2-3*x+1)*exp(5)*log(x^2-2*x+1)*log(log(x^2-2*x+1))+(2*x^2-2*x-12)*exp(5))/(x-1)/log(x^2-2*x+1)
,x, algorithm="fricas")

[Out]

(x^2 - x - 6)*e^5*log(log(x^2 - 2*x + 1))

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giac [B]  time = 0.26, size = 46, normalized size = 2.71 \begin {gather*} x^{2} e^{5} \log \left (\log \left (x^{2} - 2 \, x + 1\right )\right ) - x e^{5} \log \left (\log \left (x^{2} - 2 \, x + 1\right )\right ) - 6 \, e^{5} \log \left (\log \left (x^{2} - 2 \, x + 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2-3*x+1)*exp(5)*log(x^2-2*x+1)*log(log(x^2-2*x+1))+(2*x^2-2*x-12)*exp(5))/(x-1)/log(x^2-2*x+1)
,x, algorithm="giac")

[Out]

x^2*e^5*log(log(x^2 - 2*x + 1)) - x*e^5*log(log(x^2 - 2*x + 1)) - 6*e^5*log(log(x^2 - 2*x + 1))

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maple [B]  time = 0.12, size = 47, normalized size = 2.76

method	result	size
norman	\(-6 \,{\mathrm e}^{5} \ln \left (\ln \left (x^{2}-2 x +1\right )\right )+x^{2} {\mathrm e}^{5} \ln \left (\ln \left (x^{2}-2 x +1\right )\right )-x \,{\mathrm e}^{5} \ln \left (\ln \left (x^{2}-2 x +1\right )\right )\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^2-3*x+1)*exp(5)*ln(x^2-2*x+1)*ln(ln(x^2-2*x+1))+(2*x^2-2*x-12)*exp(5))/(x-1)/ln(x^2-2*x+1),x,method=
_RETURNVERBOSE)

[Out]

-6*exp(5)*ln(ln(x^2-2*x+1))+x^2*exp(5)*ln(ln(x^2-2*x+1))-x*exp(5)*ln(ln(x^2-2*x+1))

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maxima [B]  time = 0.48, size = 43, normalized size = 2.53 \begin {gather*} x^{2} e^{5} \log \relax (2) - x e^{5} \log \relax (2) + {\left (x^{2} e^{5} - x e^{5}\right )} \log \left (\log \left (x - 1\right )\right ) - 6 \, e^{5} \log \left (\log \left (x - 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2-3*x+1)*exp(5)*log(x^2-2*x+1)*log(log(x^2-2*x+1))+(2*x^2-2*x-12)*exp(5))/(x-1)/log(x^2-2*x+1)
,x, algorithm="maxima")

[Out]

x^2*e^5*log(2) - x*e^5*log(2) + (x^2*e^5 - x*e^5)*log(log(x - 1)) - 6*e^5*log(log(x - 1))

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mupad [B]  time = 6.04, size = 22, normalized size = 1.29 \begin {gather*} -{\mathrm {e}}^5\,\ln \left (\ln \left (x^2-2\,x+1\right )\right )\,\left (-x^2+x+6\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(5)*(2*x - 2*x^2 + 12) - exp(5)*log(x^2 - 2*x + 1)*log(log(x^2 - 2*x + 1))*(2*x^2 - 3*x + 1))/(log(x^
2 - 2*x + 1)*(x - 1)),x)

[Out]

-exp(5)*log(log(x^2 - 2*x + 1))*(x - x^2 + 6)

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sympy [B]  time = 0.61, size = 46, normalized size = 2.71 \begin {gather*} \left (x^{2} e^{5} - x e^{5} + \frac {e^{5}}{6}\right ) \log {\left (\log {\left (x^{2} - 2 x + 1 \right )} \right )} - \frac {37 e^{5} \log {\left (\log {\left (x^{2} - 2 x + 1 \right )} \right )}}{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**2-3*x+1)*exp(5)*ln(x**2-2*x+1)*ln(ln(x**2-2*x+1))+(2*x**2-2*x-12)*exp(5))/(x-1)/ln(x**2-2*x+1
),x)

[Out]

(x**2*exp(5) - x*exp(5) + exp(5)/6)*log(log(x**2 - 2*x + 1)) - 37*exp(5)*log(log(x**2 - 2*x + 1))/6

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