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

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

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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 {gather*} \text {Li}_2\left (\frac {2-x}{4}\right )+\text {Li}_2\left (\frac {x+2}{4}\right )+\frac {7}{2} \log \left (4-x^2\right )-2 x+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 (2 x+4)-7 \tanh ^{-1}\left (\frac {x}{2}\right ) \end {gather*}

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 {gather*} \begin {aligned} \text {integral} &=\int \frac {10 x+7 x^2+x^3+e^4 \left (2 x+x^2\right )+\left (4-6 x+x^3\right ) \log \left (\frac {-2+x}{e^2}\right )+\left (-2 x-x^2\right ) \log (x)+\left (2 x+x^2\right ) \log (4+2 x)}{x \left (-4+x^2\right )} \, dx\\ &=\int \left (\frac {e^4}{-2+x}+\frac {10}{-4+x^2}+\frac {7 x}{-4+x^2}+\frac {x^2}{-4+x^2}+\frac {\left (-2+2 x+x^2\right ) (-2+\log (-2+x))}{x (2+x)}-\frac {\log (x)}{-2+x}+\frac {\log (4+2 x)}{-2+x}\right ) \, 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 {\left (-2+2 x+x^2\right ) (-2+\log (-2+x))}{x (2+x)} \, dx-\int \frac {\log (x)}{-2+x} \, dx+\int \frac {\log (4+2 x)}{-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+2 x)+\frac {7}{2} \log \left (4-x^2\right )-2 \int \frac {\log \left (\frac {2-x}{4}\right )}{4+2 x} \, dx+4 \int \frac {1}{-4+x^2} \, dx+\int \left (-2-\frac {-2+\log (-2+x)}{x}+\frac {-2+\log (-2+x)}{2+x}+\log (-2+x)\right ) \, 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+2 x)+\frac {7}{2} \log \left (4-x^2\right )+\text {Li}_2\left (1-\frac {x}{2}\right )-\int \frac {-2+\log (-2+x)}{x} \, dx+\int \frac {-2+\log (-2+x)}{2+x} \, dx+\int \log (-2+x) \, dx-\operatorname {Subst}\left (\int \frac {\log \left (1-\frac {x}{8}\right )}{x} \, dx,x,4+2 x\right )\\ &=-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+2 x)+\frac {7}{2} \log \left (4-x^2\right )+\text {Li}_2\left (1-\frac {x}{2}\right )+\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+\operatorname {Subst}(\int \log (x) \, dx,x,-2+x)\\ &=-2 x-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+2 x)+\frac {7}{2} \log \left (4-x^2\right )+\text {Li}_2\left (\frac {2+x}{4}\right )-\operatorname {Subst}\left (\int \frac {\log \left (1+\frac {x}{4}\right )}{x} \, dx,x,-2+x\right )\\ &=-2 x-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+2 x)+\frac {7}{2} \log \left (4-x^2\right )+\text {Li}_2\left (\frac {2-x}{4}\right )+\text {Li}_2\left (\frac {2+x}{4}\right )\\ \end {aligned} \end {gather*}

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

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 {gather*} {\left (x + e^{4} + 5\right )} \log \left ({\left (x - 2\right )} e^{\left (-2\right )}\right ) + \log \left ({\left (x - 2\right )} e^{\left (-2\right )}\right ) \log \left (2 \, x + 4\right ) - \log \left ({\left (x - 2\right )} e^{\left (-2\right )}\right ) \log \relax (x) \end {gather*}

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 {gather*} x \log \left (x - 2\right ) + e^{4} \log \left (x - 2\right ) + \log \left (2 \, x + 4\right ) \log \left (x - 2\right ) - \log \left (x - 2\right ) \log \relax (x) - 2 \, x - 2 \, \log \left (x + 2\right ) + 5 \, \log \left (x - 2\right ) + 2 \, \log \relax (x) \end {gather*}

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	\(\left (x -\ln \relax (x )+\ln \left (2 x +4\right )\right ) \ln \left (\left (x -2\right ) {\mathrm e}^{-2}\right )+\ln \left (x -2\right ) {\mathrm e}^{4}+5 \ln \left (x -2\right )\)	\(35\)
default	\(\ln \relax (2) \ln \left (x -2\right )+\left (\ln \left (2+x \right )-\ln \left (\frac {1}{2}+\frac {x}{4}\right )\right ) \ln \left (\frac {1}{2}-\frac {x}{4}\right )-2 x +2 \ln \relax (x )-2 \ln \left (2+x \right )+\ln \left (x -2\right ) {\mathrm e}^{4}+7 \ln \left (x -2\right )+\left (x -2\right ) \ln \left (x -2\right )+2-\ln \left (x -2\right ) \ln \left (\frac {x}{2}\right )+\ln \left (x -2\right ) \ln \left (\frac {1}{2}+\frac {x}{4}\right )-\left (\ln \relax (x )-\ln \left (\frac {x}{2}\right )\right ) \ln \left (1-\frac {x}{2}\right )\)	\(102\)

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 {gather*} \frac {1}{2} \, {\left (\log \left (x + 2\right ) + \log \left (x - 2\right )\right )} e^{4} - \frac {1}{2} \, {\left (\log \left (x + 2\right ) - \log \left (x - 2\right )\right )} e^{4} + {\left (\log \left (x - 2\right ) - 2\right )} \log \left (x + 2\right ) + {\left (x + \log \relax (2) - \log \relax (x) - 2\right )} \log \left (x - 2\right ) - 2 \, x + 7 \, \log \left (x - 2\right ) + 2 \, \log \relax (x) \end {gather*}

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 {gather*} 5\,\ln \left (x-2\right )+\ln \left (x-2\right )\,{\mathrm {e}}^4+\ln \left ({\mathrm {e}}^{-2}\,\left (x-2\right )\right )\,\ln \left (2\,x+4\right )-\ln \left ({\mathrm {e}}^{-2}\,\left (x-2\right )\right )\,\ln \relax (x)+x\,\ln \left ({\mathrm {e}}^{-2}\,\left (x-2\right )\right ) \end {gather*}

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 {gather*} \left (x - \log {\relax (x )} + \log {\left (2 x + 4 \right )}\right ) \log {\left (\frac {x - 2}{e^{2}} \right )} + \left (5 + e^{4}\right ) \log {\left (x - 2 \right )} \end {gather*}

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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