3.41.76 $\int \frac{(-2x-x^2+2x^3+e^x(-1-x+x^2))\log (x)+(-e^{x}x-x^2)\log (x)\log (e^{x}x+x^2)+(-2e^{x}x-2x^2+(2e^x+2x)\log (e^{x}x+x^2))\log (\log (x))+(e^x+2x-x^2)\log (x){\log }^2(\log (x))}{(2e^{x}x+2x^2)\log (x)}dx$

Optimal. Leaf size=26 $$\frac{1}{2}(-x+\log \left(x(e^x+x)\right))(-1-x+{\log }^2(\log (x)))$$

________________________________________________________________________________________

Rubi [F]  time = 3.32, 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 = {} $$\begin{array}{c}\int \frac{(-2x-x^2+2x^3+e^x(-1-x+x^2))\log (x)+(-e^{x}x-x^2)\log (x)\log (e^{x}x+x^2)+(-2e^{x}x-2x^2+(2e^x+2x)\log (e^{x}x+x^2))\log (\log (x))+(e^x+2x-x^2)\log (x){\log }^2(\log (x))}{(2e^{x}x+2x^2)\log (x)}dx\end{array}$$

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{(-2x-x^2+2x^3+e^x(-1-x+x^2))\log (x)+(-e^{x}x-x^2)\log (x)\log (e^{x}x+x^2)+(-2e^{x}x-2x^2+(2e^x+2x)\log (e^{x}x+x^2))\log (\log (x))+(e^x+2x-x^2)\log (x){\log }^2(\log (x))}{2x(e^x+x)\log (x)}dx\\ & =\frac{1}{2}\int \frac{(-2x-x^2+2x^3+e^x(-1-x+x^2))\log (x)+(-e^{x}x-x^2)\log (x)\log (e^{x}x+x^2)+(-2e^{x}x-2x^2+(2e^x+2x)\log (e^{x}x+x^2))\log (\log (x))+(e^x+2x-x^2)\log (x){\log }^2(\log (x))}{x(e^x+x)\log (x)}dx\\ & =\frac{1}{2}\int \frac{-\frac{2(x-\log \left(x(e^x+x)\right))\log (\log (x))}{\log (x)}-\frac{e^x+2x+e^{x}x+x^2-e^x{x}^2-2x^3+x(e^x+x)\log \left(x(e^x+x)\right)+(-e^x+(-2+x)x){\log }^2(\log (x))}{e^x+x}}{x}dx\\ & =\frac{1}{2}\int (\frac{(-1+x)(1+x-{\log }^2(\log (x)))}{e^x+x}+\frac{-\log (x)-x\log (x)+x^2\log (x)-x\log (x)\log \left(x(e^x+x)\right)-2x\log (\log (x))+2\log \left(x(e^x+x)\right)\log (\log (x))+\log (x){\log }^2(\log (x))}{x\log (x)})dx\\ & =\frac{1}{2}\int \frac{(-1+x)(1+x-{\log }^2(\log (x)))}{e^x+x}dx+\frac{1}{2}\int \frac{-\log (x)-x\log (x)+x^2\log (x)-x\log (x)\log \left(x(e^x+x)\right)-2x\log (\log (x))+2\log \left(x(e^x+x)\right)\log (\log (x))+\log (x){\log }^2(\log (x))}{x\log (x)}dx\\ & =\frac{1}{2}\int (\frac{-1-x+x^2-x\log \left(x(e^x+x)\right)}{x}-\frac{2(x-\log \left(x(e^x+x)\right))\log (\log (x))}{x\log (x)}+\frac{{\log }^2(\log (x))}{x})dx+\frac{1}{2}\int (-\frac{1+x-{\log }^2(\log (x))}{e^x+x}+\frac{x(1+x-{\log }^2(\log (x)))}{e^x+x})dx\\ & =\frac{1}{2}\int \frac{-1-x+x^2-x\log \left(x(e^x+x)\right)}{x}dx+\frac{1}{2}\int \frac{{\log }^2(\log (x))}{x}dx-\frac{1}{2}\int \frac{1+x-{\log }^2(\log (x))}{e^x+x}dx+\frac{1}{2}\int \frac{x(1+x-{\log }^2(\log (x)))}{e^x+x}dx-\int \frac{(x-\log \left(x(e^x+x)\right))\log (\log (x))}{x\log (x)}dx\\ & =\frac{1}{2}\int (\frac{-1-x+x^2}{x}-\log (e^{x}x+x^2))dx-\frac{1}{2}\int (\frac{1}{e^x+x}+\frac{x}{e^x+x}-\frac{{\log }^2(\log (x))}{e^x+x})dx+\frac{1}{2}\int (\frac{x}{e^x+x}+\frac{x^2}{e^x+x}-\frac{x{\log }^2(\log (x))}{e^x+x})dx+\frac{1}{2}\mathrm{Subst}(\int {\log }^2(x)dx,x,\log (x))-\int (\frac{\log (\log (x))}{\log (x)}-\frac{\log (e^{x}x+x^2)\log (\log (x))}{x\log (x)})dx\\ & =\frac{1}{2}\log (x){\log }^2(\log (x))-\frac{1}{2}\int \frac{1}{e^x+x}dx+\frac{1}{2}\int \frac{x^2}{e^x+x}dx+\frac{1}{2}\int \frac{-1-x+x^2}{x}dx-\frac{1}{2}\int \log (e^{x}x+x^2)dx+\frac{1}{2}\int \frac{{\log }^2(\log (x))}{e^x+x}dx-\frac{1}{2}\int \frac{x{\log }^2(\log (x))}{e^x+x}dx-\int \frac{\log (\log (x))}{\log (x)}dx+\int \frac{\log (e^{x}x+x^2)\log (\log (x))}{x\log (x)}dx-\mathrm{Subst}(\int \log (x)dx,x,\log (x))\\ & =\log (x)-\frac{1}{2}x\log (e^{x}x+x^2)-\log (x)\log (\log (x))+\frac{1}{2}\log (x){\log }^2(\log (x))-\frac{1}{2}\int \frac{1}{e^x+x}dx+\frac{1}{2}\int \frac{x^2}{e^x+x}dx+\frac{1}{2}\int (-1-\frac{1}{x}+x)dx+\frac{1}{2}\int \frac{2x+e^x(1+x)}{e^x+x}dx+\frac{1}{2}\int \frac{{\log }^2(\log (x))}{e^x+x}dx-\frac{1}{2}\int \frac{x{\log }^2(\log (x))}{e^x+x}dx-\int \frac{\log (\log (x))}{\log (x)}dx+\int \frac{\log (e^{x}x+x^2)\log (\log (x))}{x\log (x)}dx\\ & =-\frac{x}{2}+\frac{x^2}{4}+\frac{\log (x)}{2}-\frac{1}{2}x\log (e^{x}x+x^2)-\log (x)\log (\log (x))+\frac{1}{2}\log (x){\log }^2(\log (x))-\frac{1}{2}\int \frac{1}{e^x+x}dx+\frac{1}{2}\int \frac{x^2}{e^x+x}dx+\frac{1}{2}\int (1+x-\frac{(-1+x)x}{e^x+x})dx+\frac{1}{2}\int \frac{{\log }^2(\log (x))}{e^x+x}dx-\frac{1}{2}\int \frac{x{\log }^2(\log (x))}{e^x+x}dx-\int \frac{\log (\log (x))}{\log (x)}dx+\int \frac{\log (e^{x}x+x^2)\log (\log (x))}{x\log (x)}dx\\ & =\frac{x^2}{2}+\frac{\log (x)}{2}-\frac{1}{2}x\log (e^{x}x+x^2)-\log (x)\log (\log (x))+\frac{1}{2}\log (x){\log }^2(\log (x))-\frac{1}{2}\int \frac{1}{e^x+x}dx-\frac{1}{2}\int \frac{(-1+x)x}{e^x+x}dx+\frac{1}{2}\int \frac{x^2}{e^x+x}dx+\frac{1}{2}\int \frac{{\log }^2(\log (x))}{e^x+x}dx-\frac{1}{2}\int \frac{x{\log }^2(\log (x))}{e^x+x}dx-\int \frac{\log (\log (x))}{\log (x)}dx+\int \frac{\log (e^{x}x+x^2)\log (\log (x))}{x\log (x)}dx\\ & =\frac{x^2}{2}+\frac{\log (x)}{2}-\frac{1}{2}x\log (e^{x}x+x^2)-\log (x)\log (\log (x))+\frac{1}{2}\log (x){\log }^2(\log (x))-\frac{1}{2}\int \frac{1}{e^x+x}dx+\frac{1}{2}\int \frac{x^2}{e^x+x}dx-\frac{1}{2}\int (-\frac{x}{e^x+x}+\frac{x^2}{e^x+x})dx+\frac{1}{2}\int \frac{{\log }^2(\log (x))}{e^x+x}dx-\frac{1}{2}\int \frac{x{\log }^2(\log (x))}{e^x+x}dx-\int \frac{\log (\log (x))}{\log (x)}dx+\int \frac{\log (e^{x}x+x^2)\log (\log (x))}{x\log (x)}dx\\ & =\frac{x^2}{2}+\frac{\log (x)}{2}-\frac{1}{2}x\log (e^{x}x+x^2)-\log (x)\log (\log (x))+\frac{1}{2}\log (x){\log }^2(\log (x))-\frac{1}{2}\int \frac{1}{e^x+x}dx+\frac{1}{2}\int \frac{x}{e^x+x}dx+\frac{1}{2}\int \frac{{\log }^2(\log (x))}{e^x+x}dx-\frac{1}{2}\int \frac{x{\log }^2(\log (x))}{e^x+x}dx-\int \frac{\log (\log (x))}{\log (x)}dx+\int \frac{\log (e^{x}x+x^2)\log (\log (x))}{x\log (x)}dx\end{array}\end{array}$$

________________________________________________________________________________________

Mathematica [A]  time = 0.25, size = 51, normalized size = 1.96 $$\begin{array}{c}\frac{1}{2}(x+x^2-\log (x)-\log (e^x+x)-x\log \left(x(e^x+x)\right)-(x-\log \left(x(e^x+x)\right)){\log }^2(\log (x)))\end{array}$$

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 1.01, size = 43, normalized size = 1.65 $$\begin{array}{c}-\frac{1}{2}(x-\log (x^2+x{e}^x))\log {(\log (x))}^2+\frac{1}{2}{x}^2-\frac{1}{2}(x+1)\log (x^2+x{e}^x)+\frac{1}{2}x\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [B]  time = 0.24, size = 62, normalized size = 2.38 $$\begin{array}{c}-\frac{1}{2}x\log {(\log (x))}^2+\frac{1}{2}\log (x+e^x)\log {(\log (x))}^2+\frac{1}{2}\log (x)\log {(\log (x))}^2+\frac{1}{2}{x}^2-\frac{1}{2}x\log (x+e^x)-\frac{1}{2}x\log (x)+\frac{1}{2}x-\frac{1}{2}\log (x+e^x)-\frac{1}{2}\log (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [C]  time = 0.13, size = 252, normalized size = 9.69

method	result	size
risch	$(\frac{\ln {(\ln (x))}^2}{2}-\frac{x}{2})\ln ({\mathrm{e}}^x+x)-\frac{x\ln {(\ln (x))}^2}{2}+\frac{\ln (x)\ln {(\ln (x))}^2}{2}+\frac{ix\pi \mathrm{csgn}\left(ix\right)\mathrm{csgn}\left(i({\mathrm{e}}^x+x)\right)\mathrm{csgn}\left(ix({\mathrm{e}}^x+x)\right)}{4}+\frac{i\ln {(\ln (x))}^2\pi \mathrm{csgn}\left(ix\right)\mathrm{csgn}{\left(ix({\mathrm{e}}^x+x)\right)}^2}{4}+\frac{i\ln {(\ln (x))}^2\pi \mathrm{csgn}\left(i({\mathrm{e}}^x+x)\right)\mathrm{csgn}{\left(ix({\mathrm{e}}^x+x)\right)}^2}{4}-\frac{i\ln {(\ln (x))}^2\pi \mathrm{csgn}\left(ix\right)\mathrm{csgn}\left(i({\mathrm{e}}^x+x)\right)\mathrm{csgn}\left(ix({\mathrm{e}}^x+x)\right)}{4}-\frac{x\ln (x)}{2}+\frac{ix\pi \mathrm{csgn}{\left(ix({\mathrm{e}}^x+x)\right)}^3}{4}-\frac{ix\pi \mathrm{csgn}\left(i({\mathrm{e}}^x+x)\right)\mathrm{csgn}{\left(ix({\mathrm{e}}^x+x)\right)}^2}{4}-\frac{ix\pi \mathrm{csgn}\left(ix\right)\mathrm{csgn}{\left(ix({\mathrm{e}}^x+x)\right)}^2}{4}-\frac{i\ln {(\ln (x))}^2\pi \mathrm{csgn}{\left(ix({\mathrm{e}}^x+x)\right)}^3}{4}+\frac{x^2}{2}-\frac{\ln (x)}{2}+\frac{x}{2}-\frac{\ln ({\mathrm{e}}^x+x)}{2}$	$252$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(x)-x^2+2*x)*ln(x)*ln(ln(x))^2+((2*exp(x)+2*x)*ln(exp(x)*x+x^2)-2*exp(x)*x-2*x^2)*ln(ln(x))+(-exp(x)*
x-x^2)*ln(x)*ln(exp(x)*x+x^2)+((x^2-x-1)*exp(x)+2*x^3-x^2-2*x)*ln(x))/(2*exp(x)*x+2*x^2)/ln(x),x,method=_RETUR
NVERBOSE)

[Out]

(1/2*ln(ln(x))^2-1/2*x)*ln(exp(x)+x)-1/2*x*ln(ln(x))^2+1/2*ln(x)*ln(ln(x))^2+1/4*I*x*Pi*csgn(I*x)*csgn(I*(exp(
x)+x))*csgn(I*x*(exp(x)+x))+1/4*I*ln(ln(x))^2*Pi*csgn(I*x)*csgn(I*x*(exp(x)+x))^2+1/4*I*ln(ln(x))^2*Pi*csgn(I*
(exp(x)+x))*csgn(I*x*(exp(x)+x))^2-1/4*I*ln(ln(x))^2*Pi*csgn(I*x)*csgn(I*(exp(x)+x))*csgn(I*x*(exp(x)+x))-1/2*
x*ln(x)+1/4*I*x*Pi*csgn(I*x*(exp(x)+x))^3-1/4*I*x*Pi*csgn(I*(exp(x)+x))*csgn(I*x*(exp(x)+x))^2-1/4*I*x*Pi*csgn
(I*x)*csgn(I*x*(exp(x)+x))^2-1/4*I*ln(ln(x))^2*Pi*csgn(I*x*(exp(x)+x))^3+1/2*x^2-1/2*ln(x)+1/2*x-1/2*ln(exp(x)
+x)

________________________________________________________________________________________

maxima [A]  time = 0.42, size = 46, normalized size = 1.77 $$\begin{array}{c}-\frac{1}{2}(x-\log (x))\log {(\log (x))}^2+\frac{1}{2}{x}^2+\frac{1}{2}(\log {(\log (x))}^2-x-1)\log (x+e^x)-\frac{1}{2}(x+1)\log (x)+\frac{1}{2}x\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.04 $$\begin{array}{c}\int -\frac{-\ln (x)(2x+{\mathrm{e}}^x-x^2){\ln (\ln (x))}^2+(2x{\mathrm{e}}^x-\ln (x{\mathrm{e}}^x+x^2)(2x+2{\mathrm{e}}^x)+2x^2)\ln (\ln (x))+\ln (x)(2x+{\mathrm{e}}^x(-x^2+x+1)+x^2-2x^3)+\ln (x{\mathrm{e}}^x+x^2)\ln (x)(x{\mathrm{e}}^x+x^2)}{\ln (x)(2x{\mathrm{e}}^x+2x^2)}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [B]  time = 1.74, size = 53, normalized size = 2.04 $$\begin{array}{c}\frac{x^2}{2}-\frac{x\log {(\log (x))}^2}{2}+\frac{x}{2}+(-\frac{x}{2}+\frac{\log {(\log (x))}^2}{2})\log (x^2+x{e}^x)-\frac{\log (x)}{2}-\frac{\log (x+e^x)}{2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________