3.103.69 \(\int \frac {e^4 (-3+2 x+x^2+4 \log (x)-\log ^2(x)) (-12-6 x-4 x^2-6 x^3-4 x^4+(6-6 x^2) \log (x)+2 x^2 \log ^2(x))}{3 x-2 x^2-x^3-4 x \log (x)+x \log ^2(x)} \, dx\)

Optimal. Leaf size=25 \[ e^4 \left (3+x^2\right ) (1-x-\log (x)) (-3-x+\log (x)) \]

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Rubi [B]  time = 0.32, antiderivative size = 60, normalized size of antiderivative = 2.40, number of steps used = 9, number of rules used = 7, integrand size = 88, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {12, 6688, 14, 2351, 2301, 2304, 2305} \begin {gather*} e^4 x^4+2 e^4 x^3-e^4 x^2 \log ^2(x)+4 e^4 x^2 \log (x)+6 e^4 x-3 e^4 \log ^2(x)+12 e^4 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

6*E^4*x + 2*E^4*x^3 + E^4*x^4 + 12*E^4*Log[x] + 4*E^4*x^2*Log[x] - 3*E^4*Log[x]^2 - E^4*x^2*Log[x]^2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2305

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

Rule 2351

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Wit
h[{u = ExpandIntegrand[a + b*Log[c*x^n], (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c,
d, e, f, m, n, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[m] && IntegerQ[r]))

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} &=e^4 \int \frac {\left (-3+2 x+x^2+4 \log (x)-\log ^2(x)\right ) \left (-12-6 x-4 x^2-6 x^3-4 x^4+\left (6-6 x^2\right ) \log (x)+2 x^2 \log ^2(x)\right )}{3 x-2 x^2-x^3-4 x \log (x)+x \log ^2(x)} \, dx\\ &=e^4 \int \left (6+\frac {12}{x}+4 x+6 x^2+4 x^3+\frac {6 \left (-1+x^2\right ) \log (x)}{x}-2 x \log ^2(x)\right ) \, dx\\ &=6 e^4 x+2 e^4 x^2+2 e^4 x^3+e^4 x^4+12 e^4 \log (x)-\left (2 e^4\right ) \int x \log ^2(x) \, dx+\left (6 e^4\right ) \int \frac {\left (-1+x^2\right ) \log (x)}{x} \, dx\\ &=6 e^4 x+2 e^4 x^2+2 e^4 x^3+e^4 x^4+12 e^4 \log (x)-e^4 x^2 \log ^2(x)+\left (2 e^4\right ) \int x \log (x) \, dx+\left (6 e^4\right ) \int \left (-\frac {\log (x)}{x}+x \log (x)\right ) \, dx\\ &=6 e^4 x+\frac {3 e^4 x^2}{2}+2 e^4 x^3+e^4 x^4+12 e^4 \log (x)+e^4 x^2 \log (x)-e^4 x^2 \log ^2(x)-\left (6 e^4\right ) \int \frac {\log (x)}{x} \, dx+\left (6 e^4\right ) \int x \log (x) \, dx\\ &=6 e^4 x+2 e^4 x^3+e^4 x^4+12 e^4 \log (x)+4 e^4 x^2 \log (x)-3 e^4 \log ^2(x)-e^4 x^2 \log ^2(x)\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.01, size = 60, normalized size = 2.40 \begin {gather*} 6 e^4 x+2 e^4 x^3+e^4 x^4+12 e^4 \log (x)+4 e^4 x^2 \log (x)-3 e^4 \log ^2(x)-e^4 x^2 \log ^2(x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

6*E^4*x + 2*E^4*x^3 + E^4*x^4 + 12*E^4*Log[x] + 4*E^4*x^2*Log[x] - 3*E^4*Log[x]^2 - E^4*x^2*Log[x]^2

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fricas [A]  time = 0.74, size = 40, normalized size = 1.60 \begin {gather*} -{\left (x^{2} + 3\right )} e^{4} \log \relax (x)^{2} + 4 \, {\left (x^{2} + 3\right )} e^{4} \log \relax (x) + {\left (x^{4} + 2 \, x^{3} + 6 \, x\right )} e^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2*log(x)^2+(-6*x^2+6)*log(x)-4*x^4-6*x^3-4*x^2-6*x-12)*exp(log(-log(x)^2+4*log(x)+x^2+2*x-3)+4)
/(x*log(x)^2-4*x*log(x)-x^3-2*x^2+3*x),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.20, size = 53, normalized size = 2.12 \begin {gather*} x^{4} e^{4} - x^{2} e^{4} \log \relax (x)^{2} + 2 \, x^{3} e^{4} + 4 \, x^{2} e^{4} \log \relax (x) - 3 \, e^{4} \log \relax (x)^{2} + 6 \, x e^{4} + 12 \, e^{4} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2*log(x)^2+(-6*x^2+6)*log(x)-4*x^4-6*x^3-4*x^2-6*x-12)*exp(log(-log(x)^2+4*log(x)+x^2+2*x-3)+4)
/(x*log(x)^2-4*x*log(x)-x^3-2*x^2+3*x),x, algorithm="giac")

[Out]

x^4*e^4 - x^2*e^4*log(x)^2 + 2*x^3*e^4 + 4*x^2*e^4*log(x) - 3*e^4*log(x)^2 + 6*x*e^4 + 12*e^4*log(x)

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maple [A]  time = 0.05, size = 42, normalized size = 1.68

method	result	size
default	\({\mathrm e}^{4} \left (-x^{2} \ln \relax (x )^{2}+4 x^{2} \ln \relax (x )+x^{4}+2 x^{3}-3 \ln \relax (x )^{2}+6 x +12 \ln \relax (x )\right )\)	\(42\)
risch	\({\mathrm e}^{4} \left (-x^{2}-3\right ) \ln \relax (x )^{2}+4 x^{2} {\mathrm e}^{4} \ln \relax (x )+x^{4} {\mathrm e}^{4}+2 x^{3} {\mathrm e}^{4}+6 x \,{\mathrm e}^{4}+12 \,{\mathrm e}^{4} \ln \relax (x )\)	\(49\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^2*ln(x)^2+(-6*x^2+6)*ln(x)-4*x^4-6*x^3-4*x^2-6*x-12)*exp(ln(-ln(x)^2+4*ln(x)+x^2+2*x-3)+4)/(x*ln(x)^2
-4*x*ln(x)-x^3-2*x^2+3*x),x,method=_RETURNVERBOSE)

[Out]

exp(4)*(-x^2*ln(x)^2+4*x^2*ln(x)+x^4+2*x^3-3*ln(x)^2+6*x+12*ln(x))

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maxima [A]  time = 0.41, size = 35, normalized size = 1.40 \begin {gather*} {\left (x^{4} + 2 \, x^{3} - {\left (x^{2} + 3\right )} \log \relax (x)^{2} + 4 \, {\left (x^{2} + 3\right )} \log \relax (x) + 6 \, x\right )} e^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2*log(x)^2+(-6*x^2+6)*log(x)-4*x^4-6*x^3-4*x^2-6*x-12)*exp(log(-log(x)^2+4*log(x)+x^2+2*x-3)+4)
/(x*log(x)^2-4*x*log(x)-x^3-2*x^2+3*x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 7.10, size = 41, normalized size = 1.64 \begin {gather*} {\mathrm {e}}^4\,\left (x^4+2\,x^3-x^2\,{\ln \relax (x)}^2+4\,x^2\,\ln \relax (x)+6\,x-3\,{\ln \relax (x)}^2+12\,\ln \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(log(2*x + 4*log(x) - log(x)^2 + x^2 - 3) + 4)*(6*x - 2*x^2*log(x)^2 + 4*x^2 + 6*x^3 + 4*x^4 + log(x)*
(6*x^2 - 6) + 12))/(4*x*log(x) - x*log(x)^2 - 3*x + 2*x^2 + x^3),x)

[Out]

exp(4)*(6*x + 12*log(x) + 4*x^2*log(x) - 3*log(x)^2 - x^2*log(x)^2 + 2*x^3 + x^4)

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sympy [B]  time = 0.31, size = 60, normalized size = 2.40 \begin {gather*} x^{4} e^{4} + 2 x^{3} e^{4} + 4 x^{2} e^{4} \log {\relax (x )} + 6 x e^{4} + \left (- x^{2} e^{4} - 3 e^{4}\right ) \log {\relax (x )}^{2} + 12 e^{4} \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**2*ln(x)**2+(-6*x**2+6)*ln(x)-4*x**4-6*x**3-4*x**2-6*x-12)*exp(ln(-ln(x)**2+4*ln(x)+x**2+2*x-3)
+4)/(x*ln(x)**2-4*x*ln(x)-x**3-2*x**2+3*x),x)

[Out]

x**4*exp(4) + 2*x**3*exp(4) + 4*x**2*exp(4)*log(x) + 6*x*exp(4) + (-x**2*exp(4) - 3*exp(4))*log(x)**2 + 12*exp
(4)*log(x)

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