3.102.52 \(\int \frac {-3 x+(48-144 x+108 x^2-24 x^3+4 e^5 (-48+48 x-12 x^2)) \log (x)+(-48 x+48 x^2-12 x^3) \log ^2(x)}{4 x-4 x^2+x^3} \, dx\)

Optimal. Leaf size=28 \[ 4+\frac {3}{-2+x}-2 \left (-3+3 \left (4 e^5+2 x\right )\right ) \log ^2(x) \]

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Rubi [A]  time = 0.22, antiderivative size = 30, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 7, integrand size = 71, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.099, Rules used = {1594, 27, 6688, 2346, 2301, 2295, 2296} \begin {gather*} -\frac {3}{2-x}-12 x \log ^2(x)+6 \left (1-4 e^5\right ) \log ^2(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-3*x + (48 - 144*x + 108*x^2 - 24*x^3 + 4*E^5*(-48 + 48*x - 12*x^2))*Log[x] + (-48*x + 48*x^2 - 12*x^3)*L
og[x]^2)/(4*x - 4*x^2 + x^3),x]

[Out]

-3/(2 - x) + 6*(1 - 4*E^5)*Log[x]^2 - 12*x*Log[x]^2

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - 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 2296

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2346

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.))/(x_), x_Symbol] :> Dist[d, Int[((d
+ e*x)^(q - 1)*(a + b*Log[c*x^n])^p)/x, x], x] + Dist[e, Int[(d + e*x)^(q - 1)*(a + b*Log[c*x^n])^p, x], x] /
; FreeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && GtQ[q, 0] && IntegerQ[2*q]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-3 x+\left (48-144 x+108 x^2-24 x^3+4 e^5 \left (-48+48 x-12 x^2\right )\right ) \log (x)+\left (-48 x+48 x^2-12 x^3\right ) \log ^2(x)}{x \left (4-4 x+x^2\right )} \, dx\\ &=\int \frac {-3 x+\left (48-144 x+108 x^2-24 x^3+4 e^5 \left (-48+48 x-12 x^2\right )\right ) \log (x)+\left (-48 x+48 x^2-12 x^3\right ) \log ^2(x)}{(-2+x)^2 x} \, dx\\ &=\int \left (-\frac {3}{(-2+x)^2}-\frac {12 \left (-1+4 e^5+2 x\right ) \log (x)}{x}-12 \log ^2(x)\right ) \, dx\\ &=-\frac {3}{2-x}-12 \int \frac {\left (-1+4 e^5+2 x\right ) \log (x)}{x} \, dx-12 \int \log ^2(x) \, dx\\ &=-\frac {3}{2-x}-12 x \log ^2(x)+\left (12 \left (1-4 e^5\right )\right ) \int \frac {\log (x)}{x} \, dx\\ &=-\frac {3}{2-x}+6 \left (1-4 e^5\right ) \log ^2(x)-12 x \log ^2(x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 28, normalized size = 1.00 \begin {gather*} \frac {3}{-2+x}+6 \left (1-4 e^5\right ) \log ^2(x)-12 x \log ^2(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3*x + (48 - 144*x + 108*x^2 - 24*x^3 + 4*E^5*(-48 + 48*x - 12*x^2))*Log[x] + (-48*x + 48*x^2 - 12*
x^3)*Log[x]^2)/(4*x - 4*x^2 + x^3),x]

[Out]

3/(-2 + x) + 6*(1 - 4*E^5)*Log[x]^2 - 12*x*Log[x]^2

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fricas [A]  time = 0.52, size = 36, normalized size = 1.29 \begin {gather*} -\frac {3 \, {\left (2 \, {\left (2 \, x^{2} + {\left (x - 2\right )} e^{\left (2 \, \log \relax (2) + 5\right )} - 5 \, x + 2\right )} \log \relax (x)^{2} - 1\right )}}{x - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-12*x^3+48*x^2-48*x)*log(x)^2+((-12*x^2+48*x-48)*exp(2*log(2)+5)-24*x^3+108*x^2-144*x+48)*log(x)-3
*x)/(x^3-4*x^2+4*x),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.14, size = 48, normalized size = 1.71 \begin {gather*} -\frac {3 \, {\left (4 \, x^{2} \log \relax (x)^{2} + 8 \, x e^{5} \log \relax (x)^{2} - 10 \, x \log \relax (x)^{2} - 16 \, e^{5} \log \relax (x)^{2} + 4 \, \log \relax (x)^{2} - 1\right )}}{x - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-12*x^3+48*x^2-48*x)*log(x)^2+((-12*x^2+48*x-48)*exp(2*log(2)+5)-24*x^3+108*x^2-144*x+48)*log(x)-3
*x)/(x^3-4*x^2+4*x),x, algorithm="giac")

[Out]

-3*(4*x^2*log(x)^2 + 8*x*e^5*log(x)^2 - 10*x*log(x)^2 - 16*e^5*log(x)^2 + 4*log(x)^2 - 1)/(x - 2)

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maple [A]  time = 0.06, size = 23, normalized size = 0.82

method	result	size
risch	\(\left (-12 x -24 \,{\mathrm e}^{5}+6\right ) \ln \relax (x )^{2}+\frac {3}{x -2}\)	\(23\)
default	\(\frac {3}{x -2}-12 x \ln \relax (x )^{2}-24 \,{\mathrm e}^{5} \ln \relax (x )^{2}+6 \ln \relax (x )^{2}\)	\(30\)
norman	\(\frac {\left (-12+48 \,{\mathrm e}^{5}\right ) \ln \relax (x )^{2}+\left (30-24 \,{\mathrm e}^{5}\right ) x \ln \relax (x )^{2}-12 x^{2} \ln \relax (x )^{2}+3}{x -2}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-12*x^3+48*x^2-48*x)*ln(x)^2+((-12*x^2+48*x-48)*exp(2*ln(2)+5)-24*x^3+108*x^2-144*x+48)*ln(x)-3*x)/(x^3-
4*x^2+4*x),x,method=_RETURNVERBOSE)

[Out]

(-12*x-24*exp(5)+6)*ln(x)^2+3/(x-2)

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maxima [A]  time = 0.39, size = 23, normalized size = 0.82 \begin {gather*} -6 \, {\left (2 \, x + 4 \, e^{5} - 1\right )} \log \relax (x)^{2} + \frac {3}{x - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-12*x^3+48*x^2-48*x)*log(x)^2+((-12*x^2+48*x-48)*exp(2*log(2)+5)-24*x^3+108*x^2-144*x+48)*log(x)-3
*x)/(x^3-4*x^2+4*x),x, algorithm="maxima")

[Out]

-6*(2*x + 4*e^5 - 1)*log(x)^2 + 3/(x - 2)

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mupad [B]  time = 6.64, size = 29, normalized size = 1.04 \begin {gather*} \frac {3}{x-2}-12\,x\,{\ln \relax (x)}^2-24\,{\mathrm {e}}^5\,{\ln \relax (x)}^2+6\,{\ln \relax (x)}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*x + log(x)*(144*x + exp(2*log(2) + 5)*(12*x^2 - 48*x + 48) - 108*x^2 + 24*x^3 - 48) + log(x)^2*(48*x -
48*x^2 + 12*x^3))/(4*x - 4*x^2 + x^3),x)

[Out]

3/(x - 2) - 12*x*log(x)^2 - 24*exp(5)*log(x)^2 + 6*log(x)^2

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sympy [A]  time = 0.24, size = 19, normalized size = 0.68 \begin {gather*} \left (- 12 x - 24 e^{5} + 6\right ) \log {\relax (x )}^{2} + \frac {3}{x - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-12*x**3+48*x**2-48*x)*ln(x)**2+((-12*x**2+48*x-48)*exp(2*ln(2)+5)-24*x**3+108*x**2-144*x+48)*ln(x
)-3*x)/(x**3-4*x**2+4*x),x)

[Out]

(-12*x - 24*exp(5) + 6)*log(x)**2 + 3/(x - 2)

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