3.5.15 $\int \frac{-16e^5-16x+e^5(-16+8x)\log (-\frac{x}{-2+x})}{e^{10}(-2+x)-2x^2+x^3+e^5(-4x+2x^2)}dx$

Optimal. Leaf size=24 $$\frac{1}{3}+\frac{8x\log \left(\frac{x}{2-x}\right)}{e^5+x}$$

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Rubi [B]  time = 0.28, antiderivative size = 61, normalized size of antiderivative = 2.54, number of steps used = 11, number of rules used = 7, integrand size = 59, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.119, Rules used = {6688, 12, 6742, 36, 31, 2490, 29} $$\begin{array}{c}-\frac{16\log (2-x)}{2+e^5}+\frac{16\log (x)}{2+e^5}-\frac{8e^5(2-x)\log \left(\frac{x}{2-x}\right)}{(2+e^5)(x+e^5)}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(-16*E^5 - 16*x + E^5*(-16 + 8*x)*Log[-(x/(-2 + x))])/(E^10*(-2 + x) - 2*x^2 + x^3 + E^5*(-4*x + 2*x^2)),x
]

[Out]

(-16*Log[2 - x])/(2 + E^5) + (16*Log[x])/(2 + E^5) - (8*E^5*(2 - x)*Log[x/(2 - x)])/((2 + E^5)*(E^5 + x))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 2490

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]^(s_.)/((g_.) + (h_.)*(x_))^
2, x_Symbol] :> Simp[((a + b*x)*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^s)/((b*g - a*h)*(g + h*x)), x] - Dist[(p*
r*s*(b*c - a*d))/(b*g - a*h), Int[Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^(s - 1)/((c + d*x)*(g + h*x)), x], x] /
; FreeQ[{a, b, c, d, e, f, g, h, p, q, r, s}, x] && NeQ[b*c - a*d, 0] && EqQ[p + q, 0] && NeQ[b*g - a*h, 0] &&
IGtQ[s, 0]

Rule 6688

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

Rule 6742

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

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{8(-\frac{2(e^5+x)}{-2+x}+e^5\log (-\frac{x}{-2+x}))}{{(e^5+x)}^2}dx\\ & =8\int \frac{-\frac{2(e^5+x)}{-2+x}+e^5\log (-\frac{x}{-2+x})}{{(e^5+x)}^2}dx\\ & =8\int (-\frac{2}{(-2+x)(e^5+x)}+\frac{e^5\log (-\frac{x}{-2+x})}{{(e^5+x)}^2})dx\\ & =-(16\int \frac{1}{(-2+x)(e^5+x)}dx)+\left(8e^5\right)\int \frac{\log (-\frac{x}{-2+x})}{{(e^5+x)}^2}dx\\ & =-\frac{8e^5(2-x)\log \left(\frac{x}{2-x}\right)}{(2+e^5)(e^5+x)}-\frac{16\int \frac{1}{-2+x}dx}{2+e^5}+\frac{16\int \frac{1}{e^5+x}dx}{2+e^5}+\frac{\left(16e^5\right)\int \frac{1}{x(e^5+x)}dx}{2+e^5}\\ & =-\frac{16\log (2-x)}{2+e^5}-\frac{8e^5(2-x)\log \left(\frac{x}{2-x}\right)}{(2+e^5)(e^5+x)}+\frac{16\log (e^5+x)}{2+e^5}+\frac{16\int \frac{1}{x}dx}{2+e^5}-\frac{16\int \frac{1}{e^5+x}dx}{2+e^5}\\ & =-\frac{16\log (2-x)}{2+e^5}+\frac{16\log (x)}{2+e^5}-\frac{8e^5(2-x)\log \left(\frac{x}{2-x}\right)}{(2+e^5)(e^5+x)}\end{array}\end{array}$$

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Mathematica [A]  time = 0.08, size = 35, normalized size = 1.46 $$\begin{array}{c}8(-\log (2-x)+\log (x)-\frac{e^5\log \left(\frac{x}{2-x}\right)}{e^5+x})\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-16*E^5 - 16*x + E^5*(-16 + 8*x)*Log[-(x/(-2 + x))])/(E^10*(-2 + x) - 2*x^2 + x^3 + E^5*(-4*x + 2*x
^2)),x]

[Out]

8*(-Log[2 - x] + Log[x] - (E^5*Log[x/(2 - x)])/(E^5 + x))

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fricas [A]  time = 0.59, size = 18, normalized size = 0.75 $$\begin{array}{c}\frac{8x\log (-\frac{x}{x-2})}{x+e^5}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x-16)*exp(5)*log(-x/(x-2))-16*exp(5)-16*x)/((x-2)*exp(5)^2+(2*x^2-4*x)*exp(5)+x^3-2*x^2),x, algo
rithm="fricas")

[Out]

8*x*log(-x/(x - 2))/(x + e^5)

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giac [B]  time = 0.79, size = 41, normalized size = 1.71 $$\begin{array}{c}\frac{16x\log (-\frac{x}{x-2})}{(x-2)(\frac{x{e}^5}{x-2}+\frac{2x}{x-2}-e^5)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x-16)*exp(5)*log(-x/(x-2))-16*exp(5)-16*x)/((x-2)*exp(5)^2+(2*x^2-4*x)*exp(5)+x^3-2*x^2),x, algo
rithm="giac")

[Out]

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

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maple [A]  time = 0.20, size = 19, normalized size = 0.79

method	result	size
norman	$\frac{8x\ln (-\frac{x}{x-2})}{{\mathrm{e}}^5+x}$	$19$
risch	$-\frac{8{\mathrm{e}}^5\ln (-\frac{x}{x-2})}{{\mathrm{e}}^5+x}+8\ln (x)-8\ln (x-2)$	$31$
derivativedivides	$\frac{16\ln (({\mathrm{e}}^5+2)(-1-\frac{2}{x-2})+{\mathrm{e}}^5)}{{\mathrm{e}}^5+2}+\frac{4{\mathrm{e}}^5\ln (-1-\frac{2}{x-2})\ln \left(\frac{4{\mathrm{e}}^5(-1-\frac{2}{x-2})+{\mathrm{e}}^{10}(-1-\frac{2}{x-2})+2{\mathrm{e}}^5-2\sqrt{{\left({\mathrm{e}}^5\right)}^2-{\mathrm{e}}^{10}}+{\mathrm{e}}^{10}-\frac{8}{x-2}-4}{{\mathrm{e}}^{10}+2{\mathrm{e}}^5-2\sqrt{{\left({\mathrm{e}}^5\right)}^2-{\mathrm{e}}^{10}}}\right)}{\sqrt{{\left({\mathrm{e}}^5\right)}^2-{\mathrm{e}}^{10}}}-\frac{4{\mathrm{e}}^5\ln (-1-\frac{2}{x-2})\ln \left(\frac{4{\mathrm{e}}^5(-1-\frac{2}{x-2})+{\mathrm{e}}^{10}(-1-\frac{2}{x-2})+2{\mathrm{e}}^5+2\sqrt{{\left({\mathrm{e}}^5\right)}^2-{\mathrm{e}}^{10}}+{\mathrm{e}}^{10}-\frac{8}{x-2}-4}{{\mathrm{e}}^{10}+2{\mathrm{e}}^5+2\sqrt{{\left({\mathrm{e}}^5\right)}^2-{\mathrm{e}}^{10}}}\right)}{\sqrt{{\left({\mathrm{e}}^5\right)}^2-{\mathrm{e}}^{10}}}+\frac{4{\mathrm{e}}^5\mathrm{dilog}\left(\frac{4{\mathrm{e}}^5(-1-\frac{2}{x-2})+{\mathrm{e}}^{10}(-1-\frac{2}{x-2})+2{\mathrm{e}}^5-2\sqrt{{\left({\mathrm{e}}^5\right)}^2-{\mathrm{e}}^{10}}+{\mathrm{e}}^{10}-\frac{8}{x-2}-4}{{\mathrm{e}}^{10}+2{\mathrm{e}}^5-2\sqrt{{\left({\mathrm{e}}^5\right)}^2-{\mathrm{e}}^{10}}}\right)}{\sqrt{{\left({\mathrm{e}}^5\right)}^2-{\mathrm{e}}^{10}}}-\frac{4{\mathrm{e}}^5\mathrm{dilog}\left(\frac{4{\mathrm{e}}^5(-1-\frac{2}{x-2})+{\mathrm{e}}^{10}(-1-\frac{2}{x-2})+2{\mathrm{e}}^5+2\sqrt{{\left({\mathrm{e}}^5\right)}^2-{\mathrm{e}}^{10}}+{\mathrm{e}}^{10}-\frac{8}{x-2}-4}{{\mathrm{e}}^{10}+2{\mathrm{e}}^5+2\sqrt{{\left({\mathrm{e}}^5\right)}^2-{\mathrm{e}}^{10}}}\right)}{\sqrt{{\left({\mathrm{e}}^5\right)}^2-{\mathrm{e}}^{10}}}$	$416$
default	$\frac{16\ln (({\mathrm{e}}^5+2)(-1-\frac{2}{x-2})+{\mathrm{e}}^5)}{{\mathrm{e}}^5+2}+\frac{4{\mathrm{e}}^5\ln (-1-\frac{2}{x-2})\ln \left(\frac{4{\mathrm{e}}^5(-1-\frac{2}{x-2})+{\mathrm{e}}^{10}(-1-\frac{2}{x-2})+2{\mathrm{e}}^5-2\sqrt{{\left({\mathrm{e}}^5\right)}^2-{\mathrm{e}}^{10}}+{\mathrm{e}}^{10}-\frac{8}{x-2}-4}{{\mathrm{e}}^{10}+2{\mathrm{e}}^5-2\sqrt{{\left({\mathrm{e}}^5\right)}^2-{\mathrm{e}}^{10}}}\right)}{\sqrt{{\left({\mathrm{e}}^5\right)}^2-{\mathrm{e}}^{10}}}-\frac{4{\mathrm{e}}^5\ln (-1-\frac{2}{x-2})\ln \left(\frac{4{\mathrm{e}}^5(-1-\frac{2}{x-2})+{\mathrm{e}}^{10}(-1-\frac{2}{x-2})+2{\mathrm{e}}^5+2\sqrt{{\left({\mathrm{e}}^5\right)}^2-{\mathrm{e}}^{10}}+{\mathrm{e}}^{10}-\frac{8}{x-2}-4}{{\mathrm{e}}^{10}+2{\mathrm{e}}^5+2\sqrt{{\left({\mathrm{e}}^5\right)}^2-{\mathrm{e}}^{10}}}\right)}{\sqrt{{\left({\mathrm{e}}^5\right)}^2-{\mathrm{e}}^{10}}}+\frac{4{\mathrm{e}}^5\mathrm{dilog}\left(\frac{4{\mathrm{e}}^5(-1-\frac{2}{x-2})+{\mathrm{e}}^{10}(-1-\frac{2}{x-2})+2{\mathrm{e}}^5-2\sqrt{{\left({\mathrm{e}}^5\right)}^2-{\mathrm{e}}^{10}}+{\mathrm{e}}^{10}-\frac{8}{x-2}-4}{{\mathrm{e}}^{10}+2{\mathrm{e}}^5-2\sqrt{{\left({\mathrm{e}}^5\right)}^2-{\mathrm{e}}^{10}}}\right)}{\sqrt{{\left({\mathrm{e}}^5\right)}^2-{\mathrm{e}}^{10}}}-\frac{4{\mathrm{e}}^5\mathrm{dilog}\left(\frac{4{\mathrm{e}}^5(-1-\frac{2}{x-2})+{\mathrm{e}}^{10}(-1-\frac{2}{x-2})+2{\mathrm{e}}^5+2\sqrt{{\left({\mathrm{e}}^5\right)}^2-{\mathrm{e}}^{10}}+{\mathrm{e}}^{10}-\frac{8}{x-2}-4}{{\mathrm{e}}^{10}+2{\mathrm{e}}^5+2\sqrt{{\left({\mathrm{e}}^5\right)}^2-{\mathrm{e}}^{10}}}\right)}{\sqrt{{\left({\mathrm{e}}^5\right)}^2-{\mathrm{e}}^{10}}}$	$416$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((8*x-16)*exp(5)*ln(-x/(x-2))-16*exp(5)-16*x)/((x-2)*exp(5)^2+(2*x^2-4*x)*exp(5)+x^3-2*x^2),x,method=_RETU
RNVERBOSE)

[Out]

8*x*ln(-x/(x-2))/(exp(5)+x)

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maxima [B]  time = 0.98, size = 168, normalized size = 7.00 $$\begin{array}{c}16(\frac{\log (x+e^5)}{e^{10}+4e^5+4}-\frac{\log (x-2)}{e^{10}+4e^5+4}-\frac{1}{x(e^5+2)+e^{10}+2e^5})e^5-\frac{8((e^{10}+2e^5)\log (x)+(x{e}^5-2e^5)\log (-x+2))}{x(e^5+2)+e^{10}+2e^5}+\frac{16e^5}{x(e^5+2)+e^{10}+2e^5}+\frac{32\log (x+e^5)}{e^{10}+4e^5+4}-\frac{16\log (x+e^5)}{e^5+2}-\frac{32\log (x-2)}{e^{10}+4e^5+4}+8\log (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x-16)*exp(5)*log(-x/(x-2))-16*exp(5)-16*x)/((x-2)*exp(5)^2+(2*x^2-4*x)*exp(5)+x^3-2*x^2),x, algo
rithm="maxima")

[Out]

16*(log(x + e^5)/(e^10 + 4*e^5 + 4) - log(x - 2)/(e^10 + 4*e^5 + 4) - 1/(x*(e^5 + 2) + e^10 + 2*e^5))*e^5 - 8*
((e^10 + 2*e^5)*log(x) + (x*e^5 - 2*e^5)*log(-x + 2))/(x*(e^5 + 2) + e^10 + 2*e^5) + 16*e^5/(x*(e^5 + 2) + e^1
0 + 2*e^5) + 32*log(x + e^5)/(e^10 + 4*e^5 + 4) - 16*log(x + e^5)/(e^5 + 2) - 32*log(x - 2)/(e^10 + 4*e^5 + 4)
+ 8*log(x)

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mupad [B]  time = 1.08, size = 26, normalized size = 1.08 $$\begin{array}{c}16\mathrm{atanh}(x-1)-\frac{8{\mathrm{e}}^5\ln (-\frac{x}{x-2})}{x+{\mathrm{e}}^5}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*x + 16*exp(5) - exp(5)*log(-x/(x - 2))*(8*x - 16))/(exp(5)*(4*x - 2*x^2) - exp(10)*(x - 2) + 2*x^2 - x
^3),x)

[Out]

16*atanh(x - 1) - (8*exp(5)*log(-x/(x - 2)))/(x + exp(5))

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sympy [A]  time = 0.21, size = 29, normalized size = 1.21 $$\begin{array}{c}8\log (x)-8\log (x-2)-\frac{8e^5\log (-\frac{x}{x-2})}{x+e^5}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x-16)*exp(5)*ln(-x/(x-2))-16*exp(5)-16*x)/((x-2)*exp(5)**2+(2*x**2-4*x)*exp(5)+x**3-2*x**2),x)

[Out]

8*log(x) - 8*log(x - 2) - 8*exp(5)*log(-x/(x - 2))/(x + exp(5))

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