3.35.30 \(\int \frac {-58 x+19 x^2+64 x^3+20 x^4+(4 x-3 x^2-x^3) \log (x)+(-40 x+30 x^3+10 x^4) \log (\frac {-8+8 x}{x})}{-20+15 x^2+5 x^3} \, dx\)

Optimal. Leaf size=28 \[ x^2 \left (2-\frac {\log (x)}{5 (2+x)}+\log \left (\frac {4 (-2+2 x)}{x}\right )\right ) \]

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Rubi [A]  time = 0.27, antiderivative size = 37, normalized size of antiderivative = 1.32, number of steps used = 19, number of rules used = 10, integrand size = 76, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {6742, 77, 88, 2455, 193, 43, 2357, 2295, 2314, 31} \begin {gather*} 2 x^2+x^2 \log \left (8-\frac {8}{x}\right )+\frac {2 x \log (x)}{5 (x+2)}-\frac {1}{5} x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-58*x + 19*x^2 + 64*x^3 + 20*x^4 + (4*x - 3*x^2 - x^3)*Log[x] + (-40*x + 30*x^3 + 10*x^4)*Log[(-8 + 8*x)/
x])/(-20 + 15*x^2 + 5*x^3),x]

[Out]

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

Rule 31

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

Rule 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 193

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b}, x] && LtQ[n, 0]
&& IntegerQ[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 2314

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp[(x*(d + e*x^r)^(q
+ 1)*(a + b*Log[c*x^n]))/d, x] - Dist[(b*n)/d, Int[(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, n, q,
r}, x] && EqQ[r*(q + 1) + 1, 0]

Rule 2357

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]

Rule 2455

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {58 x}{5 (-1+x) (2+x)^2}+\frac {19 x^2}{5 (-1+x) (2+x)^2}+\frac {64 x^3}{5 (-1+x) (2+x)^2}+\frac {4 x^4}{(-1+x) (2+x)^2}+2 x \log \left (8-\frac {8}{x}\right )-\frac {x (4+x) \log (x)}{5 (2+x)^2}\right ) \, dx\\ &=-\left (\frac {1}{5} \int \frac {x (4+x) \log (x)}{(2+x)^2} \, dx\right )+2 \int x \log \left (8-\frac {8}{x}\right ) \, dx+\frac {19}{5} \int \frac {x^2}{(-1+x) (2+x)^2} \, dx+4 \int \frac {x^4}{(-1+x) (2+x)^2} \, dx-\frac {58}{5} \int \frac {x}{(-1+x) (2+x)^2} \, dx+\frac {64}{5} \int \frac {x^3}{(-1+x) (2+x)^2} \, dx\\ &=x^2 \log \left (8-\frac {8}{x}\right )-\frac {1}{5} \int \left (\log (x)-\frac {4 \log (x)}{(2+x)^2}\right ) \, dx+\frac {19}{5} \int \left (\frac {1}{9 (-1+x)}-\frac {4}{3 (2+x)^2}+\frac {8}{9 (2+x)}\right ) \, dx+4 \int \left (-3+\frac {1}{9 (-1+x)}+x-\frac {16}{3 (2+x)^2}+\frac {80}{9 (2+x)}\right ) \, dx-8 \int \frac {1}{8-\frac {8}{x}} \, dx-\frac {58}{5} \int \left (\frac {1}{9 (-1+x)}+\frac {2}{3 (2+x)^2}-\frac {1}{9 (2+x)}\right ) \, dx+\frac {64}{5} \int \left (1+\frac {1}{9 (-1+x)}+\frac {8}{3 (2+x)^2}-\frac {28}{9 (2+x)}\right ) \, dx\\ &=\frac {4 x}{5}+2 x^2+x^2 \log \left (8-\frac {8}{x}\right )+\log (1-x)+\frac {2}{5} \log (2+x)-\frac {1}{5} \int \log (x) \, dx+\frac {4}{5} \int \frac {\log (x)}{(2+x)^2} \, dx-8 \int \frac {x}{-8+8 x} \, dx\\ &=x+2 x^2+x^2 \log \left (8-\frac {8}{x}\right )+\log (1-x)-\frac {1}{5} x \log (x)+\frac {2 x \log (x)}{5 (2+x)}+\frac {2}{5} \log (2+x)-\frac {2}{5} \int \frac {1}{2+x} \, dx-8 \int \left (\frac {1}{8}+\frac {1}{8 (-1+x)}\right ) \, dx\\ &=2 x^2+x^2 \log \left (8-\frac {8}{x}\right )-\frac {1}{5} x \log (x)+\frac {2 x \log (x)}{5 (2+x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 28, normalized size = 1.00 \begin {gather*} \frac {1}{5} x^2 \left (10+5 \log \left (8-\frac {8}{x}\right )-\frac {\log (x)}{2+x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-58*x + 19*x^2 + 64*x^3 + 20*x^4 + (4*x - 3*x^2 - x^3)*Log[x] + (-40*x + 30*x^3 + 10*x^4)*Log[(-8 +
8*x)/x])/(-20 + 15*x^2 + 5*x^3),x]

[Out]

(x^2*(10 + 5*Log[8 - 8/x] - Log[x]/(2 + x)))/5

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fricas [A]  time = 0.54, size = 45, normalized size = 1.61 \begin {gather*} \frac {10 \, x^{3} - x^{2} \log \relax (x) + 20 \, x^{2} + 5 \, {\left (x^{3} + 2 \, x^{2}\right )} \log \left (\frac {8 \, {\left (x - 1\right )}}{x}\right )}{5 \, {\left (x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^3-3*x^2+4*x)*log(x)+(10*x^4+30*x^3-40*x)*log((8*x-8)/x)+20*x^4+64*x^3+19*x^2-58*x)/(5*x^3+15*x^
2-20),x, algorithm="fricas")

[Out]

1/5*(10*x^3 - x^2*log(x) + 20*x^2 + 5*(x^3 + 2*x^2)*log(8*(x - 1)/x))/(x + 2)

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giac [A]  time = 0.13, size = 38, normalized size = 1.36 \begin {gather*} x^{2} \log \left (8 \, x - 8\right ) + 2 \, x^{2} - \frac {1}{5} \, {\left (5 \, x^{2} + x + \frac {4}{x + 2}\right )} \log \relax (x) + \frac {2}{5} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^3-3*x^2+4*x)*log(x)+(10*x^4+30*x^3-40*x)*log((8*x-8)/x)+20*x^4+64*x^3+19*x^2-58*x)/(5*x^3+15*x^
2-20),x, algorithm="giac")

[Out]

x^2*log(8*x - 8) + 2*x^2 - 1/5*(5*x^2 + x + 4/(x + 2))*log(x) + 2/5*log(x)

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maple [B]  time = 0.10, size = 66, normalized size = 2.36

method	result	size
default	\(2 x^{2}+\ln \left (x -1\right )+3 x^{2} \ln \relax (2)+\ln \left (-\frac {1}{x}\right )-\ln \left (1-\frac {1}{x}\right ) \left (1-\frac {1}{x}\right ) \left (-\frac {1}{x}-1\right ) x^{2}-\frac {x \ln \relax (x )}{5}+\frac {2 \ln \relax (x ) x}{5 \left (2+x \right )}\)	\(66\)
risch	\(x^{2} \ln \left (x -1\right )+\frac {-22 x^{2} \ln \relax (x )+20 x^{3}+40 x^{2}-10 x^{3} \ln \relax (x )+60 x^{2} \ln \relax (2)+30 x^{3} \ln \relax (2)+5 i \pi \,x^{3} \mathrm {csgn}\left (i \left (x -1\right )\right ) \mathrm {csgn}\left (\frac {i \left (x -1\right )}{x}\right )^{2}+5 i \pi \,x^{3} \mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left (x -1\right )}{x}\right )^{2}-10 i \pi \,x^{2} \mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left (x -1\right )\right ) \mathrm {csgn}\left (\frac {i \left (x -1\right )}{x}\right )-5 i \pi \,x^{3} \mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left (x -1\right )\right ) \mathrm {csgn}\left (\frac {i \left (x -1\right )}{x}\right )-5 i \pi \,x^{3} \mathrm {csgn}\left (\frac {i \left (x -1\right )}{x}\right )^{3}-10 i \pi \,x^{2} \mathrm {csgn}\left (\frac {i \left (x -1\right )}{x}\right )^{3}+10 i \pi \,x^{2} \mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left (x -1\right )}{x}\right )^{2}+10 i \pi \,x^{2} \mathrm {csgn}\left (i \left (x -1\right )\right ) \mathrm {csgn}\left (\frac {i \left (x -1\right )}{x}\right )^{2}}{10 x +20}\)	\(260\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^3-3*x^2+4*x)*ln(x)+(10*x^4+30*x^3-40*x)*ln((8*x-8)/x)+20*x^4+64*x^3+19*x^2-58*x)/(5*x^3+15*x^2-20),x,
method=_RETURNVERBOSE)

[Out]

2*x^2+ln(x-1)+3*x^2*ln(2)+ln(-1/x)-ln(1-1/x)*(1-1/x)*(-1/x-1)*x^2-1/5*x*ln(x)+2/5*ln(x)*x/(2+x)

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maxima [B]  time = 0.72, size = 80, normalized size = 2.86 \begin {gather*} \frac {45 \, x^{3} {\left (3 \, \log \relax (2) + 2\right )} + 90 \, x^{2} {\left (3 \, \log \relax (2) + 2\right )} + {\left (45 \, x^{3} + 90 \, x^{2} + 58 \, x + 116\right )} \log \left (x - 1\right ) - 9 \, {\left (5 \, x^{3} + 11 \, x^{2}\right )} \log \relax (x) - 348}{45 \, {\left (x + 2\right )}} + \frac {116}{15 \, {\left (x + 2\right )}} - \frac {58}{45} \, \log \left (x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^3-3*x^2+4*x)*log(x)+(10*x^4+30*x^3-40*x)*log((8*x-8)/x)+20*x^4+64*x^3+19*x^2-58*x)/(5*x^3+15*x^
2-20),x, algorithm="maxima")

[Out]

1/45*(45*x^3*(3*log(2) + 2) + 90*x^2*(3*log(2) + 2) + (45*x^3 + 90*x^2 + 58*x + 116)*log(x - 1) - 9*(5*x^3 + 1
1*x^2)*log(x) - 348)/(x + 2) + 116/15/(x + 2) - 58/45*log(x - 1)

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mupad [B]  time = 2.36, size = 34, normalized size = 1.21 \begin {gather*} x^2\,\ln \left (\frac {8\,x-8}{x}\right )+2\,x^2-\frac {x^2\,\ln \relax (x)}{5\,\left (x+2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((19*x^2 - log(x)*(3*x^2 - 4*x + x^3) - 58*x + 64*x^3 + 20*x^4 + log((8*x - 8)/x)*(30*x^3 - 40*x + 10*x^4))
/(15*x^2 + 5*x^3 - 20),x)

[Out]

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

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sympy [B]  time = 0.65, size = 51, normalized size = 1.82 \begin {gather*} 2 x^{2} + \left (x^{2} - \frac {1}{6}\right ) \log {\left (\frac {8 x - 8}{x} \right )} + \frac {7 \log {\relax (x )}}{30} + \frac {\log {\left (x - 1 \right )}}{6} + \frac {\left (- x^{2} - 2 x - 4\right ) \log {\relax (x )}}{5 x + 10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**3-3*x**2+4*x)*ln(x)+(10*x**4+30*x**3-40*x)*ln((8*x-8)/x)+20*x**4+64*x**3+19*x**2-58*x)/(5*x**3
+15*x**2-20),x)

[Out]

2*x**2 + (x**2 - 1/6)*log((8*x - 8)/x) + 7*log(x)/30 + log(x - 1)/6 + (-x**2 - 2*x - 4)*log(x)/(5*x + 10)

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