3.17.36 $\int \frac{e^2(-8-18x-10x^2-x^3)+e^2(8+4x)\log (\frac{4+2x}{x})}{32+32x+10x^2+x^3}dx$

Optimal. Leaf size=27 $$\frac{e^{2}x(x-\log (3-\frac{-4+x}{x}))}{-4-x}$$

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Rubi [A]  time = 0.27, antiderivative size = 50, normalized size of antiderivative = 1.85, number of steps used = 16, number of rules used = 9, integrand size = 55, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.164, Rules used = {6741, 12, 6742, 44, 77, 88, 2463, 514, 72} $$\begin{array}{c}-e^{2}x-\frac{16e^2}{x+4}-\frac{4e^2\log (\frac{4}{x}+2)}{x+4}-e^2\log (x)+e^2\log (x+2)\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(E^2*(-8 - 18*x - 10*x^2 - x^3) + E^2*(8 + 4*x)*Log[(4 + 2*x)/x])/(32 + 32*x + 10*x^2 + x^3),x]

[Out]

-(E^2*x) - (16*E^2)/(4 + x) - (4*E^2*Log[2 + 4/x])/(4 + x) - E^2*Log[x] + E^2*Log[2 + 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 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 514

Int[(x_)^(m_.)*((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[x^(m - n*q)*
(a + b*x^n)^p*(d + c*x^n)^q, x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[mn, -n] && IntegerQ[q] && (PosQ[n] |
|  !IntegerQ[p])

Rule 2463

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))*((f_.) + (g_.)*(x_))^(r_.), x_Symbol] :> Simp[((
f + g*x)^(r + 1)*(a + b*Log[c*(d + e*x^n)^p]))/(g*(r + 1)), x] - Dist[(b*e*n*p)/(g*(r + 1)), Int[(x^(n - 1)*(f
+ g*x)^(r + 1))/(d + e*x^n), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, r}, x] && (IGtQ[r, 0] || RationalQ[n
]) && NeQ[r, -1]

Rule 6741

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

Rule 6742

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

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{e^2(-8-18x-10x^2-x^3+8\log (2+\frac{4}{x})+4x\log (2+\frac{4}{x}))}{32+32x+10x^2+x^3}dx\\ & =e^2\int \frac{-8-18x-10x^2-x^3+8\log (2+\frac{4}{x})+4x\log (2+\frac{4}{x})}{32+32x+10x^2+x^3}dx\\ & =e^2\int (-\frac{8}{(2+x)(4+x{)}^2}-\frac{18x}{(2+x)(4+x{)}^2}-\frac{10x^2}{(2+x)(4+x{)}^2}-\frac{x^3}{(2+x)(4+x{)}^2}+\frac{4\log (2+\frac{4}{x})}{(4+x{)}^2})dx\\ & =-(e^2\int \frac{x^3}{(2+x)(4+x{)}^2}dx)+\left(4e^2\right)\int \frac{\log (2+\frac{4}{x})}{(4+x{)}^2}dx-\left(8e^2\right)\int \frac{1}{(2+x)(4+x{)}^2}dx-\left(10e^2\right)\int \frac{x^2}{(2+x)(4+x{)}^2}dx-\left(18e^2\right)\int \frac{x}{(2+x)(4+x{)}^2}dx\\ & =-\frac{4e^2\log (2+\frac{4}{x})}{4+x}-e^2\int (1-\frac{2}{2+x}+\frac{32}{(4+x{)}^2}-\frac{8}{4+x})dx-\left(8e^2\right)\int (\frac{1}{4(2+x)}-\frac{1}{2(4+x{)}^2}-\frac{1}{4(4+x)})dx-\left(10e^2\right)\int (\frac{1}{2+x}-\frac{8}{(4+x{)}^2})dx-\left(16e^2\right)\int \frac{1}{(2+\frac{4}{x})x^2(4+x)}dx-\left(18e^2\right)\int (-\frac{1}{2(2+x)}+\frac{2}{(4+x{)}^2}+\frac{1}{2(4+x)})dx\\ & =-e^{2}x-\frac{16e^2}{4+x}-\frac{4e^2\log (2+\frac{4}{x})}{4+x}-e^2\log (2+x)+e^2\log (4+x)-\left(16e^2\right)\int \frac{1}{x(4+x)(4+2x)}dx\\ & =-e^{2}x-\frac{16e^2}{4+x}-\frac{4e^2\log (2+\frac{4}{x})}{4+x}-e^2\log (2+x)+e^2\log (4+x)-\left(16e^2\right)\int (\frac{1}{16x}-\frac{1}{8(2+x)}+\frac{1}{16(4+x)})dx\\ & =-e^{2}x-\frac{16e^2}{4+x}-\frac{4e^2\log (2+\frac{4}{x})}{4+x}-e^2\log (x)+e^2\log (2+x)\end{array}\end{array}$$

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Mathematica [A]  time = 0.06, size = 32, normalized size = 1.19 $$\begin{array}{c}-e^2(x+\frac{4(4+\log (2+\frac{4}{x}))}{4+x}+\log (x)-\log (2+x))\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(E^2*(-8 - 18*x - 10*x^2 - x^3) + E^2*(8 + 4*x)*Log[(4 + 2*x)/x])/(32 + 32*x + 10*x^2 + x^3),x]

[Out]

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

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fricas [A]  time = 0.76, size = 32, normalized size = 1.19 $$\begin{array}{c}\frac{x{e}^2\log \left(\frac{2(x+2)}{x}\right)-(x^2+4x+16)e^2}{x+4}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x+8)*exp(2)*log((2*x+4)/x)+(-x^3-10*x^2-18*x-8)*exp(2))/(x^3+10*x^2+32*x+32),x, algorithm="frica
s")

[Out]

(x*e^2*log(2*(x + 2)/x) - (x^2 + 4*x + 16)*e^2)/(x + 4)

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giac [B]  time = 0.41, size = 59, normalized size = 2.19 $$\begin{array}{c}\frac{\frac{(x+2)e^2\log \left(\frac{2(x+2)}{x}\right)}{x}-e^2\log \left(\frac{2(x+2)}{x}\right)-2e^2}{\frac{2{(x+2)}^2}{x^2}-\frac{3(x+2)}{x}+1}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x+8)*exp(2)*log((2*x+4)/x)+(-x^3-10*x^2-18*x-8)*exp(2))/(x^3+10*x^2+32*x+32),x, algorithm="giac"
)

[Out]

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

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maple [A]  time = 0.11, size = 29, normalized size = 1.07

method	result	size
norman	$\frac{{\mathrm{e}}^{2}x\ln \left(\frac{2x+4}{x}\right)-x^2{\mathrm{e}}^2}{4+x}$	$29$
derivativedivides	$\frac{{\mathrm{e}}^2\ln (2+\frac{4}{x})(2+\frac{4}{x})}{\frac{4}{x}+1}-{\mathrm{e}}^{2}x+\frac{4{\mathrm{e}}^2}{\frac{4}{x}+1}-{\mathrm{e}}^2\ln (2+\frac{4}{x})$	$59$
default	$\frac{{\mathrm{e}}^2\ln (2+\frac{4}{x})(2+\frac{4}{x})}{\frac{4}{x}+1}-{\mathrm{e}}^{2}x+\frac{4{\mathrm{e}}^2}{\frac{4}{x}+1}-{\mathrm{e}}^2\ln (2+\frac{4}{x})$	$59$
risch	$-\frac{4{\mathrm{e}}^2\ln \left(\frac{2x+4}{x}\right)}{4+x}-\frac{{\mathrm{e}}^2(\ln (-x)x-\ln (-x-2)x+x^2+4\ln (-x)-4\ln (-x-2)+4x+16)}{4+x}$	$67$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x+8)*exp(2)*ln((2*x+4)/x)+(-x^3-10*x^2-18*x-8)*exp(2))/(x^3+10*x^2+32*x+32),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.68, size = 80, normalized size = 2.96 $$\begin{array}{c}-2(\frac{2}{x+4}-\log (x+4)+\log (x+2))e^2-2e^2\log (x+4)-\frac{x^2{e}^2+x{e}^2\log (x)+4x{e}^2+4(\log (2)+3)e^2-(3x{e}^2+8e^2)\log (x+2)}{x+4}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x+8)*exp(2)*log((2*x+4)/x)+(-x^3-10*x^2-18*x-8)*exp(2))/(x^3+10*x^2+32*x+32),x, algorithm="maxim
a")

[Out]

-2*(2/(x + 4) - log(x + 4) + log(x + 2))*e^2 - 2*e^2*log(x + 4) - (x^2*e^2 + x*e^2*log(x) + 4*x*e^2 + 4*(log(2
) + 3)*e^2 - (3*x*e^2 + 8*e^2)*log(x + 2))/(x + 4)

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mupad [B]  time = 1.31, size = 23, normalized size = 0.85 $$\begin{array}{c}-\frac{x{\mathrm{e}}^2(x-\ln \left(\frac{2(x+2)}{x}\right))}{x+4}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2)*(18*x + 10*x^2 + x^3 + 8) - exp(2)*log((2*x + 4)/x)*(4*x + 8))/(32*x + 10*x^2 + x^3 + 32),x)

[Out]

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

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sympy [B]  time = 0.33, size = 44, normalized size = 1.63 $$\begin{array}{c}-x{e}^2-e^2\log (x)+e^2\log (x+2)-\frac{4e^2\log \left(\frac{2x+4}{x}\right)}{x+4}-\frac{16e^2}{x+4}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x+8)*exp(2)*ln((2*x+4)/x)+(-x**3-10*x**2-18*x-8)*exp(2))/(x**3+10*x**2+32*x+32),x)

[Out]

-x*exp(2) - exp(2)*log(x) + exp(2)*log(x + 2) - 4*exp(2)*log((2*x + 4)/x)/(x + 4) - 16*exp(2)/(x + 4)

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