3.73.20 $\int \frac{-14+4e^3+4e^4+e^x(-20+8e^3+8e^4-8x)+e^{2x}(-6+4e^3+4e^4-4x)-4x+(2x+2e^{2x}x)\log (x)}{(x+2e^{x}x+e^{2x}x){\log }^3(x)}dx$

Optimal. Leaf size=33 $$\frac{2(-e^3-e^4+\frac{1}{2}(3+\frac{4}{1+e^x})+x)}{{\log }^2(x)}$$

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Rubi [F]  time = 2.28, 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 = {} $$\begin{array}{c}\int \frac{-14+4e^3+4e^4+e^x(-20+8e^3+8e^4-8x)+e^{2x}(-6+4e^3+4e^4-4x)-4x+(2x+2e^{2x}x)\log (x)}{(x+2e^{x}x+e^{2x}x){\log }^3(x)}dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[(-14 + 4*E^3 + 4*E^4 + E^x*(-20 + 8*E^3 + 8*E^4 - 8*x) + E^(2*x)*(-6 + 4*E^3 + 4*E^4 - 4*x) - 4*x + (2*x +
2*E^(2*x)*x)*Log[x])/((x + 2*E^x*x + E^(2*x)*x)*Log[x]^3),x]

[Out]

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

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{-14(1-\frac{2}{7}{e}^3(1+e))+e^x(-20+8e^3+8e^4-8x)+e^{2x}(-6+4e^3+4e^4-4x)-4x+(2x+2e^{2x}x)\log (x)}{{(1+e^x)}^{2}x{\log }^3(x)}dx\\ & =\int (\frac{4}{{(1+e^x)}^2{\log }^2(x)}-\frac{4(2+x\log (x))}{(1+e^x)x{\log }^3(x)}+\frac{2(-3(1-\frac{2}{3}{e}^3(1+e))-2x+x\log (x))}{x{\log }^3(x)})dx\\ & =2\int \frac{-3(1-\frac{2}{3}{e}^3(1+e))-2x+x\log (x)}{x{\log }^3(x)}dx+4\int \frac{1}{{(1+e^x)}^2{\log }^2(x)}dx-4\int \frac{2+x\log (x)}{(1+e^x)x{\log }^3(x)}dx\\ & =2\int (\frac{-3+2e^3+2e^4-2x}{x{\log }^3(x)}+\frac{1}{{\log }^2(x)})dx-4\int (\frac{2}{(1+e^x)x{\log }^3(x)}+\frac{1}{(1+e^x){\log }^2(x)})dx+4\int \frac{1}{{(1+e^x)}^2{\log }^2(x)}dx\\ & =2\int \frac{-3+2e^3+2e^4-2x}{x{\log }^3(x)}dx+2\int \frac{1}{{\log }^2(x)}dx+4\int \frac{1}{{(1+e^x)}^2{\log }^2(x)}dx-4\int \frac{1}{(1+e^x){\log }^2(x)}dx-8\int \frac{1}{(1+e^x)x{\log }^3(x)}dx\\ & =-\frac{2x}{\log (x)}+2\int (-\frac{2}{{\log }^3(x)}+\frac{-3+2e^3+2e^4}{x{\log }^3(x)})dx+2\int \frac{1}{\log (x)}dx+4\int \frac{1}{{(1+e^x)}^2{\log }^2(x)}dx-4\int \frac{1}{(1+e^x){\log }^2(x)}dx-8\int \frac{1}{(1+e^x)x{\log }^3(x)}dx\\ & =-\frac{2x}{\log (x)}+2\text{li}(x)-4\int \frac{1}{{\log }^3(x)}dx+4\int \frac{1}{{(1+e^x)}^2{\log }^2(x)}dx-4\int \frac{1}{(1+e^x){\log }^2(x)}dx-8\int \frac{1}{(1+e^x)x{\log }^3(x)}dx-\left(2(3-2e^3-2e^4)\right)\int \frac{1}{x{\log }^3(x)}dx\\ & =\frac{2x}{{\log }^2(x)}-\frac{2x}{\log (x)}+2\text{li}(x)-2\int \frac{1}{{\log }^2(x)}dx+4\int \frac{1}{{(1+e^x)}^2{\log }^2(x)}dx-4\int \frac{1}{(1+e^x){\log }^2(x)}dx-8\int \frac{1}{(1+e^x)x{\log }^3(x)}dx-\left(2(3-2e^3-2e^4)\right)\mathrm{Subst}(\int \frac{1}{x^3}dx,x,\log (x))\\ & =\frac{3-2e^3-2e^4}{{\log }^2(x)}+\frac{2x}{{\log }^2(x)}+2\text{li}(x)-2\int \frac{1}{\log (x)}dx+4\int \frac{1}{{(1+e^x)}^2{\log }^2(x)}dx-4\int \frac{1}{(1+e^x){\log }^2(x)}dx-8\int \frac{1}{(1+e^x)x{\log }^3(x)}dx\\ & =\frac{3-2e^3-2e^4}{{\log }^2(x)}+\frac{2x}{{\log }^2(x)}+4\int \frac{1}{{(1+e^x)}^2{\log }^2(x)}dx-4\int \frac{1}{(1+e^x){\log }^2(x)}dx-8\int \frac{1}{(1+e^x)x{\log }^3(x)}dx\end{array}\end{array}$$

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Mathematica [A]  time = 0.36, size = 53, normalized size = 1.61 $$\begin{array}{c}-\frac{-7+2e^3+2e^4-3e^x+2e^{3+x}+2e^{4+x}-2x-2e^{x}x}{(1+e^x){\log }^2(x)}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-14 + 4*E^3 + 4*E^4 + E^x*(-20 + 8*E^3 + 8*E^4 - 8*x) + E^(2*x)*(-6 + 4*E^3 + 4*E^4 - 4*x) - 4*x +
(2*x + 2*E^(2*x)*x)*Log[x])/((x + 2*E^x*x + E^(2*x)*x)*Log[x]^3),x]

[Out]

-((-7 + 2*E^3 + 2*E^4 - 3*E^x + 2*E^(3 + x) + 2*E^(4 + x) - 2*x - 2*E^x*x)/((1 + E^x)*Log[x]^2))

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fricas [A]  time = 0.63, size = 40, normalized size = 1.21 $$\begin{array}{c}\frac{(2x-2e^4-2e^3+3)e^x+2x-2e^4-2e^3+7}{(e^x+1)\log (x{)}^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(x)^2+2*x)*log(x)+(4*exp(4)+4*exp(3)-4*x-6)*exp(x)^2+(8*exp(4)+8*exp(3)-8*x-20)*exp(x)+4*ex
p(4)+4*exp(3)-4*x-14)/(x*exp(x)^2+2*exp(x)*x+x)/log(x)^3,x, algorithm="fricas")

[Out]

((2*x - 2*e^4 - 2*e^3 + 3)*e^x + 2*x - 2*e^4 - 2*e^3 + 7)/((e^x + 1)*log(x)^2)

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giac [A]  time = 0.18, size = 49, normalized size = 1.48 $$\begin{array}{c}\frac{2x{e}^x+2x-2e^4-2e^3-2e^{(x+4)}-2e^{(x+3)}+3e^x+7}{e^x\log (x{)}^2+\log (x{)}^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(x)^2+2*x)*log(x)+(4*exp(4)+4*exp(3)-4*x-6)*exp(x)^2+(8*exp(4)+8*exp(3)-8*x-20)*exp(x)+4*ex
p(4)+4*exp(3)-4*x-14)/(x*exp(x)^2+2*exp(x)*x+x)/log(x)^3,x, algorithm="giac")

[Out]

(2*x*e^x + 2*x - 2*e^4 - 2*e^3 - 2*e^(x + 4) - 2*e^(x + 3) + 3*e^x + 7)/(e^x*log(x)^2 + log(x)^2)

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maple [A]  time = 0.04, size = 47, normalized size = 1.42

method	result	size
risch	$-\frac{2{\mathrm{e}}^{4+x}+2{\mathrm{e}}^4+2{\mathrm{e}}^{3+x}+2{\mathrm{e}}^3-2{\mathrm{e}}^{x}x-2x-3{\mathrm{e}}^x-7}{({\mathrm{e}}^x+1)\ln (x{)}^2}$	$47$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x*exp(x)^2+2*x)*ln(x)+(4*exp(4)+4*exp(3)-4*x-6)*exp(x)^2+(8*exp(4)+8*exp(3)-8*x-20)*exp(x)+4*exp(4)+4*
exp(3)-4*x-14)/(x*exp(x)^2+2*exp(x)*x+x)/ln(x)^3,x,method=_RETURNVERBOSE)

[Out]

-(2*exp(4+x)+2*exp(4)+2*exp(3+x)+2*exp(3)-2*exp(x)*x-2*x-3*exp(x)-7)/(exp(x)+1)/ln(x)^2

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maxima [A]  time = 0.39, size = 44, normalized size = 1.33 $$\begin{array}{c}\frac{(2x-2e^4-2e^3+3)e^x+2x-2e^4-2e^3+7}{e^x\log (x{)}^2+\log (x{)}^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(x)^2+2*x)*log(x)+(4*exp(4)+4*exp(3)-4*x-6)*exp(x)^2+(8*exp(4)+8*exp(3)-8*x-20)*exp(x)+4*ex
p(4)+4*exp(3)-4*x-14)/(x*exp(x)^2+2*exp(x)*x+x)/log(x)^3,x, algorithm="maxima")

[Out]

((2*x - 2*e^4 - 2*e^3 + 3)*e^x + 2*x - 2*e^4 - 2*e^3 + 7)/(e^x*log(x)^2 + log(x)^2)

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mupad [B]  time = 4.62, size = 213, normalized size = 6.45 $$\begin{array}{c}x-\frac{2x^2+2x}{{\mathrm{e}}^{2x}+2{\mathrm{e}}^x+1}+\frac{2x+{\mathrm{e}}^{2x}(\frac{2x^2}{3}-2x+\frac{2}{3})-{\mathrm{e}}^x(\frac{8x^2}{3}-\frac{4}{3})+\frac{2x^2}{3}+\frac{2}{3}}{3{\mathrm{e}}^{2x}+{\mathrm{e}}^{3x}+3{\mathrm{e}}^x+1}+\frac{\frac{4x^2}{3}-\frac{2}{3}}{{\mathrm{e}}^x+1}+\frac{\frac{2x-2{\mathrm{e}}^{x+3}-2{\mathrm{e}}^{x+4}-2{\mathrm{e}}^3-2{\mathrm{e}}^4+3{\mathrm{e}}^x+2x{\mathrm{e}}^x+7}{{\mathrm{e}}^x+1}-\frac{x\ln (x)({\mathrm{e}}^{2x}+1)}{{({\mathrm{e}}^x+1)}^2}}{{\ln (x)}^2}+\frac{\frac{x({\mathrm{e}}^{2x}+1)}{{({\mathrm{e}}^x+1)}^2}-\frac{x\ln (x)({\mathrm{e}}^{2x}+{\mathrm{e}}^{3x}+{\mathrm{e}}^x+2x{\mathrm{e}}^{2x}-2x{\mathrm{e}}^x+1)}{{({\mathrm{e}}^x+1)}^3}}{\ln (x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*x - 4*exp(3) - 4*exp(4) + exp(2*x)*(4*x - 4*exp(3) - 4*exp(4) + 6) - log(x)*(2*x + 2*x*exp(2*x)) + exp
(x)*(8*x - 8*exp(3) - 8*exp(4) + 20) + 14)/(log(x)^3*(x + x*exp(2*x) + 2*x*exp(x))),x)

[Out]

x - (2*x + 2*x^2)/(exp(2*x) + 2*exp(x) + 1) + (2*x + exp(2*x)*((2*x^2)/3 - 2*x + 2/3) - exp(x)*((8*x^2)/3 - 4/
3) + (2*x^2)/3 + 2/3)/(3*exp(2*x) + exp(3*x) + 3*exp(x) + 1) + ((4*x^2)/3 - 2/3)/(exp(x) + 1) + ((2*x - 2*exp(
x + 3) - 2*exp(x + 4) - 2*exp(3) - 2*exp(4) + 3*exp(x) + 2*x*exp(x) + 7)/(exp(x) + 1) - (x*log(x)*(exp(2*x) +
1))/(exp(x) + 1)^2)/log(x)^2 + ((x*(exp(2*x) + 1))/(exp(x) + 1)^2 - (x*log(x)*(exp(2*x) + exp(3*x) + exp(x) +
2*x*exp(2*x) - 2*x*exp(x) + 1))/(exp(x) + 1)^3)/log(x)

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sympy [A]  time = 0.31, size = 34, normalized size = 1.03 $$\begin{array}{c}\frac{2x-2e^4-2e^3+3}{\log {(x)}^2}+\frac{4}{e^x\log {(x)}^2+\log {(x)}^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(x)**2+2*x)*ln(x)+(4*exp(4)+4*exp(3)-4*x-6)*exp(x)**2+(8*exp(4)+8*exp(3)-8*x-20)*exp(x)+4*e
xp(4)+4*exp(3)-4*x-14)/(x*exp(x)**2+2*exp(x)*x+x)/ln(x)**3,x)

[Out]

(2*x - 2*exp(4) - 2*exp(3) + 3)/log(x)**2 + 4/(exp(x)*log(x)**2 + log(x)**2)

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