3.76.53 \(\int \frac {(-2 x^2+2 x \log (x)) \log ^2(x-\log (x))+e^{\frac {2 (x+(1+x) \log (x-\log (x)))}{\log (x-\log (x))}} (-2+2 x+(-2 x+2 \log (x)) \log (x-\log (x))+(-2 x+2 \log (x)) \log ^2(x-\log (x)))+e^{\frac {x+(1+x) \log (x-\log (x))}{\log (x-\log (x))}} (2 x-2 x^2+(2 x^2-2 x \log (x)) \log (x-\log (x))+(2 x+2 x^2+(-2-2 x) \log (x)) \log ^2(x-\log (x)))}{(-x+\log (x)) \log ^2(x-\log (x))} \, dx\)

Optimal. Leaf size=22 \[ \left (e^{1+x+\frac {x}{\log (x-\log (x))}}-x\right )^2 \]

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Rubi [B]  time = 2.95, antiderivative size = 155, normalized size of antiderivative = 7.05, number of steps used = 4, number of rules used = 3, integrand size = 187, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.016, Rules used = {6742, 6706, 2288} \begin {gather*} x^2-\frac {2 e^{x+\frac {x}{\log (x-\log (x))}+1} \left (-x^2+x^2 \log ^2(x-\log (x))+x^2 \log (x-\log (x))+x-x \log (x) \log ^2(x-\log (x))-x \log (x) \log (x-\log (x))\right )}{(x-\log (x)) \left (-\frac {\left (1-\frac {1}{x}\right ) x}{(x-\log (x)) \log ^2(x-\log (x))}+\frac {1}{\log (x-\log (x))}+1\right ) \log ^2(x-\log (x))}+e^{2 x+\frac {2 x}{\log (x-\log (x))}+2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (2 x+\frac {2 e^{2+2 x+\frac {2 x}{\log (x-\log (x))}} \left (1-x+x \log (x-\log (x))-\log (x) \log (x-\log (x))+x \log ^2(x-\log (x))-\log (x) \log ^2(x-\log (x))\right )}{(x-\log (x)) \log ^2(x-\log (x))}-\frac {2 e^{1+x+\frac {x}{\log (x-\log (x))}} \left (x-x^2+x^2 \log (x-\log (x))-x \log (x) \log (x-\log (x))+x \log ^2(x-\log (x))+x^2 \log ^2(x-\log (x))-\log (x) \log ^2(x-\log (x))-x \log (x) \log ^2(x-\log (x))\right )}{(x-\log (x)) \log ^2(x-\log (x))}\right ) \, dx\\ &=x^2+2 \int \frac {e^{2+2 x+\frac {2 x}{\log (x-\log (x))}} \left (1-x+x \log (x-\log (x))-\log (x) \log (x-\log (x))+x \log ^2(x-\log (x))-\log (x) \log ^2(x-\log (x))\right )}{(x-\log (x)) \log ^2(x-\log (x))} \, dx-2 \int \frac {e^{1+x+\frac {x}{\log (x-\log (x))}} \left (x-x^2+x^2 \log (x-\log (x))-x \log (x) \log (x-\log (x))+x \log ^2(x-\log (x))+x^2 \log ^2(x-\log (x))-\log (x) \log ^2(x-\log (x))-x \log (x) \log ^2(x-\log (x))\right )}{(x-\log (x)) \log ^2(x-\log (x))} \, dx\\ &=e^{2+2 x+\frac {2 x}{\log (x-\log (x))}}+x^2-\frac {2 e^{1+x+\frac {x}{\log (x-\log (x))}} \left (x-x^2+x^2 \log (x-\log (x))-x \log (x) \log (x-\log (x))+x^2 \log ^2(x-\log (x))-x \log (x) \log ^2(x-\log (x))\right )}{(x-\log (x)) \left (1-\frac {\left (1-\frac {1}{x}\right ) x}{(x-\log (x)) \log ^2(x-\log (x))}+\frac {1}{\log (x-\log (x))}\right ) \log ^2(x-\log (x))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.30, size = 22, normalized size = 1.00 \begin {gather*} \left (e^{1+x+\frac {x}{\log (x-\log (x))}}-x\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-2*x^2 + 2*x*Log[x])*Log[x - Log[x]]^2 + E^((2*(x + (1 + x)*Log[x - Log[x]]))/Log[x - Log[x]])*(-2
+ 2*x + (-2*x + 2*Log[x])*Log[x - Log[x]] + (-2*x + 2*Log[x])*Log[x - Log[x]]^2) + E^((x + (1 + x)*Log[x - Lo
g[x]])/Log[x - Log[x]])*(2*x - 2*x^2 + (2*x^2 - 2*x*Log[x])*Log[x - Log[x]] + (2*x + 2*x^2 + (-2 - 2*x)*Log[x]
)*Log[x - Log[x]]^2))/((-x + Log[x])*Log[x - Log[x]]^2),x]

[Out]

(E^(1 + x + x/Log[x - Log[x]]) - x)^2

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fricas [B]  time = 0.66, size = 56, normalized size = 2.55 \begin {gather*} x^{2} - 2 \, x e^{\left (\frac {{\left (x + 1\right )} \log \left (x - \log \relax (x)\right ) + x}{\log \left (x - \log \relax (x)\right )}\right )} + e^{\left (\frac {2 \, {\left ({\left (x + 1\right )} \log \left (x - \log \relax (x)\right ) + x\right )}}{\log \left (x - \log \relax (x)\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*log(x)-2*x)*log(x-log(x))^2+(2*log(x)-2*x)*log(x-log(x))+2*x-2)*exp(((x+1)*log(x-log(x))+x)/log
(x-log(x)))^2+(((-2*x-2)*log(x)+2*x^2+2*x)*log(x-log(x))^2+(-2*x*log(x)+2*x^2)*log(x-log(x))-2*x^2+2*x)*exp(((
x+1)*log(x-log(x))+x)/log(x-log(x)))+(2*x*log(x)-2*x^2)*log(x-log(x))^2)/(log(x)-x)/log(x-log(x))^2,x, algorit
hm="fricas")

[Out]

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

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giac [B]  time = 3.13, size = 54, normalized size = 2.45 \begin {gather*} x^{2} - 2 \, x e^{\left (\frac {x \log \left (x - \log \relax (x)\right ) + x + \log \left (x - \log \relax (x)\right )}{\log \left (x - \log \relax (x)\right )}\right )} + e^{\left (2 \, x + \frac {2 \, x}{\log \left (x - \log \relax (x)\right )} + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*log(x)-2*x)*log(x-log(x))^2+(2*log(x)-2*x)*log(x-log(x))+2*x-2)*exp(((x+1)*log(x-log(x))+x)/log
(x-log(x)))^2+(((-2*x-2)*log(x)+2*x^2+2*x)*log(x-log(x))^2+(-2*x*log(x)+2*x^2)*log(x-log(x))-2*x^2+2*x)*exp(((
x+1)*log(x-log(x))+x)/log(x-log(x)))+(2*x*log(x)-2*x^2)*log(x-log(x))^2)/(log(x)-x)/log(x-log(x))^2,x, algorit
hm="giac")

[Out]

x^2 - 2*x*e^((x*log(x - log(x)) + x + log(x - log(x)))/log(x - log(x))) + e^(2*x + 2*x/log(x - log(x)) + 2)

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maple [B]  time = 0.08, size = 67, normalized size = 3.05

method	result	size
risch	\(x^{2}-2 x \,{\mathrm e}^{\frac {x \ln \left (x -\ln \relax (x )\right )+\ln \left (x -\ln \relax (x )\right )+x}{\ln \left (x -\ln \relax (x )\right )}}+{\mathrm e}^{\frac {2 x \ln \left (x -\ln \relax (x )\right )+2 \ln \left (x -\ln \relax (x )\right )+2 x}{\ln \left (x -\ln \relax (x )\right )}}\)	\(67\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*log(x)-2*x)*log(x-log(x))^2+(2*log(x)-2*x)*log(x-log(x))+2*x-2)*exp(((x+1)*log(x-log(x))+x)/log
(x-log(x)))^2+(((-2*x-2)*log(x)+2*x^2+2*x)*log(x-log(x))^2+(-2*x*log(x)+2*x^2)*log(x-log(x))-2*x^2+2*x)*exp(((
x+1)*log(x-log(x))+x)/log(x-log(x)))+(2*x*log(x)-2*x^2)*log(x-log(x))^2)/(log(x)-x)/log(x-log(x))^2,x, algorit
hm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
of the expected type LIST

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mupad [B]  time = 6.82, size = 43, normalized size = 1.95 \begin {gather*} x^2+{\mathrm {e}}^{\frac {2\,x}{\ln \left (x-\ln \relax (x)\right )}}\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^2-2\,x\,{\mathrm {e}}^{\frac {x}{\ln \left (x-\ln \relax (x)\right )}}\,\mathrm {e}\,{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((x + log(x - log(x))*(x + 1))/log(x - log(x)))*(2*x + log(x - log(x))^2*(2*x - log(x)*(2*x + 2) + 2*
x^2) - log(x - log(x))*(2*x*log(x) - 2*x^2) - 2*x^2) + log(x - log(x))^2*(2*x*log(x) - 2*x^2) - exp((2*(x + lo
g(x - log(x))*(x + 1)))/log(x - log(x)))*(log(x - log(x))*(2*x - 2*log(x)) - 2*x + log(x - log(x))^2*(2*x - 2*
log(x)) + 2))/(log(x - log(x))^2*(x - log(x))),x)

[Out]

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

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sympy [B]  time = 3.03, size = 49, normalized size = 2.23 \begin {gather*} x^{2} - 2 x e^{\frac {x + \left (x + 1\right ) \log {\left (x - \log {\relax (x )} \right )}}{\log {\left (x - \log {\relax (x )} \right )}}} + e^{\frac {2 \left (x + \left (x + 1\right ) \log {\left (x - \log {\relax (x )} \right )}\right )}{\log {\left (x - \log {\relax (x )} \right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**2 - 2*x*exp((x + (x + 1)*log(x - log(x)))/log(x - log(x))) + exp(2*(x + (x + 1)*log(x - log(x)))/log(x - lo
g(x)))

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