3.80.21 $\int \frac{10x+7x^2+x^3+e^4(2x+x^2)+(4-6x+x^3)\log (\frac{-2+x}{e^2})+(-2x-x^2)\log (x)+(2x+x^2)\log (4+2x)}{-4x+x^3}dx$

Optimal. Leaf size=25 $$\log \left(\frac{-2+x}{e^2}\right)(5+e^4+x-\log (x)+\log (4+2x))$$

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Rubi [C]  time = 0.52, antiderivative size = 123, normalized size of antiderivative = 4.92, number of steps used = 21, number of rules used = 13, integrand size = 76, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.171, Rules used = {1593, 6725, 207, 260, 321, 2418, 2394, 2315, 2393, 2391, 2389, 2295, 2316} $$\begin{array}{c}{\text{Li}}_2\left(\frac{2-x}{4}\right)+{\text{Li}}_2\left(\frac{x+2}{4}\right)+\frac{7}{2}\log (4-x^2)-2x+e^4\log (2-x)-(2-x)\log (x-2)-\log (2)\log (x-2)+(2-\log (x-2))\log \left(\frac{x}{2}\right)-(2-\log (x-2))\log \left(\frac{x+2}{4}\right)+\log \left(\frac{2-x}{4}\right)\log (2x+4)-7{\tanh }^{-1}\left(\frac{x}{2}\right)\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(10*x + 7*x^2 + x^3 + E^4*(2*x + x^2) + (4 - 6*x + x^3)*Log[(-2 + x)/E^2] + (-2*x - x^2)*Log[x] + (2*x + x
^2)*Log[4 + 2*x])/(-4*x + x^3),x]

[Out]

-2*x - 7*ArcTanh[x/2] + E^4*Log[2 - x] - (2 - x)*Log[-2 + x] - Log[2]*Log[-2 + x] + (2 - Log[-2 + x])*Log[x/2]
- (2 - Log[-2 + x])*Log[(2 + x)/4] + Log[(2 - x)/4]*Log[4 + 2*x] + (7*Log[4 - x^2])/2 + PolyLog[2, (2 - x)/4]
+ PolyLog[2, (2 + x)/4]

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 260

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

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2316

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

Rule 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2393

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[1/g, Subst[Int[(a +
b*Log[1 + (c*e*x)/g])/x, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g
+ c*(e*f - d*g), 0]

Rule 2394

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[(Log[(e*(f +
g*x))/(e*f - d*g)]*(a + b*Log[c*(d + e*x)^n]))/g, x] - Dist[(b*e*n)/g, Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]

Rule 2418

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[
(a + b*Log[c*(d + e*x)^n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n}, x] && RationalFunct
ionQ[RFx, x] && IntegerQ[p]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{10x+7x^2+x^3+e^4(2x+x^2)+(4-6x+x^3)\log \left(\frac{-2+x}{e^2}\right)+(-2x-x^2)\log (x)+(2x+x^2)\log (4+2x)}{x(-4+x^2)}dx\\ & =\int (\frac{e^4}{-2+x}+\frac{10}{-4+x^2}+\frac{7x}{-4+x^2}+\frac{x^2}{-4+x^2}+\frac{(-2+2x+x^2)(-2+\log (-2+x))}{x(2+x)}-\frac{\log (x)}{-2+x}+\frac{\log (4+2x)}{-2+x})dx\\ & =e^4\log (2-x)+7\int \frac{x}{-4+x^2}dx+10\int \frac{1}{-4+x^2}dx+\int \frac{x^2}{-4+x^2}dx+\int \frac{(-2+2x+x^2)(-2+\log (-2+x))}{x(2+x)}dx-\int \frac{\log (x)}{-2+x}dx+\int \frac{\log (4+2x)}{-2+x}dx\\ & =x-5{\tanh }^{-1}\left(\frac{x}{2}\right)+e^4\log (2-x)-\log (2)\log (-2+x)+\log \left(\frac{2-x}{4}\right)\log (4+2x)+\frac{7}{2}\log (4-x^2)-2\int \frac{\log \left(\frac{2-x}{4}\right)}{4+2x}dx+4\int \frac{1}{-4+x^2}dx+\int (-2-\frac{-2+\log (-2+x)}{x}+\frac{-2+\log (-2+x)}{2+x}+\log (-2+x))dx-\int \frac{\log \left(\frac{x}{2}\right)}{-2+x}dx\\ & =-x-7{\tanh }^{-1}\left(\frac{x}{2}\right)+e^4\log (2-x)-\log (2)\log (-2+x)+\log \left(\frac{2-x}{4}\right)\log (4+2x)+\frac{7}{2}\log (4-x^2)+{\text{Li}}_2(1-\frac{x}{2})-\int \frac{-2+\log (-2+x)}{x}dx+\int \frac{-2+\log (-2+x)}{2+x}dx+\int \log (-2+x)dx-\mathrm{Subst}(\int \frac{\log (1-\frac{x}{8})}{x}dx,x,4+2x)\\ & =-x-7{\tanh }^{-1}\left(\frac{x}{2}\right)+e^4\log (2-x)-\log (2)\log (-2+x)+(2-\log (-2+x))\log \left(\frac{x}{2}\right)-(2-\log (-2+x))\log \left(\frac{2+x}{4}\right)+\log \left(\frac{2-x}{4}\right)\log (4+2x)+\frac{7}{2}\log (4-x^2)+{\text{Li}}_2(1-\frac{x}{2})+{\text{Li}}_2\left(\frac{2+x}{4}\right)+\int \frac{\log \left(\frac{x}{2}\right)}{-2+x}dx-\int \frac{\log \left(\frac{2+x}{4}\right)}{-2+x}dx+\mathrm{Subst}(\int \log (x)dx,x,-2+x)\\ & =-2x-7{\tanh }^{-1}\left(\frac{x}{2}\right)+e^4\log (2-x)+(-2+x)\log (-2+x)-\log (2)\log (-2+x)+(2-\log (-2+x))\log \left(\frac{x}{2}\right)-(2-\log (-2+x))\log \left(\frac{2+x}{4}\right)+\log \left(\frac{2-x}{4}\right)\log (4+2x)+\frac{7}{2}\log (4-x^2)+{\text{Li}}_2\left(\frac{2+x}{4}\right)-\mathrm{Subst}(\int \frac{\log (1+\frac{x}{4})}{x}dx,x,-2+x)\\ & =-2x-7{\tanh }^{-1}\left(\frac{x}{2}\right)+e^4\log (2-x)+(-2+x)\log (-2+x)-\log (2)\log (-2+x)+(2-\log (-2+x))\log \left(\frac{x}{2}\right)-(2-\log (-2+x))\log \left(\frac{2+x}{4}\right)+\log \left(\frac{2-x}{4}\right)\log (4+2x)+\frac{7}{2}\log (4-x^2)+{\text{Li}}_2\left(\frac{2-x}{4}\right)+{\text{Li}}_2\left(\frac{2+x}{4}\right)\end{array}\end{array}$$

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Mathematica [A]  time = 0.12, size = 46, normalized size = 1.84 $$\begin{array}{c}-2x+\log (4)+(7+e^4)\log (2-x)+2\log (x)-2\log (2+x)+\log (-2+x)(-2+x-\log (x)+\log (2(2+x)))\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(10*x + 7*x^2 + x^3 + E^4*(2*x + x^2) + (4 - 6*x + x^3)*Log[(-2 + x)/E^2] + (-2*x - x^2)*Log[x] + (2
*x + x^2)*Log[4 + 2*x])/(-4*x + x^3),x]

[Out]

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

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fricas [A]  time = 0.66, size = 39, normalized size = 1.56 $$\begin{array}{c}(x+e^4+5)\log \left((x-2)e^{(-2)}\right)+\log \left((x-2)e^{(-2)}\right)\log (2x+4)-\log \left((x-2)e^{(-2)}\right)\log (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+2*x)*log(2*x+4)+(-x^2-2*x)*log(x)+(x^3-6*x+4)*log((x-2)/exp(2))+(x^2+2*x)*exp(4)+x^3+7*x^2+10*
x)/(x^3-4*x),x, algorithm="fricas")

[Out]

(x + e^4 + 5)*log((x - 2)*e^(-2)) + log((x - 2)*e^(-2))*log(2*x + 4) - log((x - 2)*e^(-2))*log(x)

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giac [B]  time = 0.21, size = 52, normalized size = 2.08 $$\begin{array}{c}x\log (x-2)+e^4\log (x-2)+\log (2x+4)\log (x-2)-\log (x-2)\log (x)-2x-2\log (x+2)+5\log (x-2)+2\log (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+2*x)*log(2*x+4)+(-x^2-2*x)*log(x)+(x^3-6*x+4)*log((x-2)/exp(2))+(x^2+2*x)*exp(4)+x^3+7*x^2+10*
x)/(x^3-4*x),x, algorithm="giac")

[Out]

x*log(x - 2) + e^4*log(x - 2) + log(2*x + 4)*log(x - 2) - log(x - 2)*log(x) - 2*x - 2*log(x + 2) + 5*log(x - 2
) + 2*log(x)

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maple [A]  time = 0.25, size = 35, normalized size = 1.40

method	result	size
risch	$(x-\ln (x)+\ln (2x+4))\ln \left((x-2){\mathrm{e}}^{-2}\right)+\ln (x-2){\mathrm{e}}^4+5\ln (x-2)$	$35$
default	$\ln (2)\ln (x-2)+(\ln (2+x)-\ln (\frac{1}{2}+\frac{x}{4}))\ln (\frac{1}{2}-\frac{x}{4})-2x+2\ln (x)-2\ln (2+x)+\ln (x-2){\mathrm{e}}^4+7\ln (x-2)+(x-2)\ln (x-2)+2-\ln (x-2)\ln \left(\frac{x}{2}\right)+\ln (x-2)\ln (\frac{1}{2}+\frac{x}{4})-(\ln (x)-\ln \left(\frac{x}{2}\right))\ln (1-\frac{x}{2})$	$102$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^2+2*x)*ln(2*x+4)+(-x^2-2*x)*ln(x)+(x^3-6*x+4)*ln((x-2)/exp(2))+(x^2+2*x)*exp(4)+x^3+7*x^2+10*x)/(x^3-4
*x),x,method=_RETURNVERBOSE)

[Out]

(x-ln(x)+ln(2*x+4))*ln((x-2)*exp(-2))+ln(x-2)*exp(4)+5*ln(x-2)

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maxima [B]  time = 0.49, size = 67, normalized size = 2.68 $$\begin{array}{c}\frac{1}{2}(\log (x+2)+\log (x-2))e^4-\frac{1}{2}(\log (x+2)-\log (x-2))e^4+(\log (x-2)-2)\log (x+2)+(x+\log (2)-\log (x)-2)\log (x-2)-2x+7\log (x-2)+2\log (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+2*x)*log(2*x+4)+(-x^2-2*x)*log(x)+(x^3-6*x+4)*log((x-2)/exp(2))+(x^2+2*x)*exp(4)+x^3+7*x^2+10*
x)/(x^3-4*x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.17, size = 48, normalized size = 1.92 $$\begin{array}{c}5\ln (x-2)+\ln (x-2){\mathrm{e}}^4+\ln \left({\mathrm{e}}^{-2}(x-2)\right)\ln (2x+4)-\ln \left({\mathrm{e}}^{-2}(x-2)\right)\ln (x)+x\ln \left({\mathrm{e}}^{-2}(x-2)\right)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(10*x + log(2*x + 4)*(2*x + x^2) - log(x)*(2*x + x^2) + log(exp(-2)*(x - 2))*(x^3 - 6*x + 4) + exp(4)*(2*
x + x^2) + 7*x^2 + x^3)/(4*x - x^3),x)

[Out]

5*log(x - 2) + log(x - 2)*exp(4) + log(exp(-2)*(x - 2))*log(2*x + 4) - log(exp(-2)*(x - 2))*log(x) + x*log(exp
(-2)*(x - 2))

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sympy [A]  time = 0.53, size = 29, normalized size = 1.16 $$\begin{array}{c}(x-\log (x)+\log (2x+4))\log \left(\frac{x-2}{e^2}\right)+(5+e^4)\log (x-2)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**2+2*x)*ln(2*x+4)+(-x**2-2*x)*ln(x)+(x**3-6*x+4)*ln((x-2)/exp(2))+(x**2+2*x)*exp(4)+x**3+7*x**2+
10*x)/(x**3-4*x),x)

[Out]

(x - log(x) + log(2*x + 4))*log((x - 2)*exp(-2)) + (5 + exp(4))*log(x - 2)

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