3.25.78 \(\int \frac {-3 e+(-12 e-8 x^3) \log (x)-12 e \log ^2(x)}{x^2+4 x^2 \log (x)+4 x^2 \log ^2(x)} \, dx\)

Optimal. Leaf size=21 \[ -3+\frac {3 e}{x}-\frac {x^2}{\frac {1}{2}+\log (x)} \]

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Rubi [A]  time = 0.35, antiderivative size = 20, normalized size of antiderivative = 0.95, number of steps used = 9, number of rules used = 6, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {6688, 2561, 6742, 2306, 2309, 2178} \begin {gather*} \frac {3 e}{x}-\frac {2 x^2}{2 \log (x)+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-3*E + (-12*E - 8*x^3)*Log[x] - 12*E*Log[x]^2)/(x^2 + 4*x^2*Log[x] + 4*x^2*Log[x]^2),x]

[Out]

(3*E)/x - (2*x^2)/(1 + 2*Log[x])

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2306

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log
[c*x^n])^(p + 1))/(b*d*n*(p + 1)), x] - Dist[(m + 1)/(b*n*(p + 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] && LtQ[p, -1]

Rule 2309

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[E^((m + 1)*x)*(a
+ b*x)^p, x], x, Log[c*x]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[m]

Rule 2561

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

Rule 6688

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

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

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Mathematica [A]  time = 0.06, size = 20, normalized size = 0.95 \begin {gather*} \frac {3 e}{x}-\frac {2 x^2}{1+2 \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3*E + (-12*E - 8*x^3)*Log[x] - 12*E*Log[x]^2)/(x^2 + 4*x^2*Log[x] + 4*x^2*Log[x]^2),x]

[Out]

(3*E)/x - (2*x^2)/(1 + 2*Log[x])

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fricas [A]  time = 0.99, size = 27, normalized size = 1.29 \begin {gather*} -\frac {2 \, x^{3} - 6 \, e \log \relax (x) - 3 \, e}{2 \, x \log \relax (x) + x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-12*exp(1)*log(x)^2+(-12*exp(1)-8*x^3)*log(x)-3*exp(1))/(4*x^2*log(x)^2+4*x^2*log(x)+x^2),x, algori
thm="fricas")

[Out]

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

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giac [A]  time = 0.43, size = 27, normalized size = 1.29 \begin {gather*} -\frac {2 \, x^{3} - 6 \, e \log \relax (x) - 3 \, e}{2 \, x \log \relax (x) + x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-12*exp(1)*log(x)^2+(-12*exp(1)-8*x^3)*log(x)-3*exp(1))/(4*x^2*log(x)^2+4*x^2*log(x)+x^2),x, algori
thm="giac")

[Out]

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

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maple [A]  time = 0.04, size = 22, normalized size = 1.05

method	result	size
risch	\(\frac {3 \,{\mathrm e}}{x}-\frac {2 x^{2}}{1+2 \ln \relax (x )}\)	\(22\)
norman	\(\frac {-2 x^{3}+6 \,{\mathrm e} \ln \relax (x )+3 \,{\mathrm e}}{x \left (1+2 \ln \relax (x )\right )}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-12*exp(1)*ln(x)^2+(-12*exp(1)-8*x^3)*ln(x)-3*exp(1))/(4*x^2*ln(x)^2+4*x^2*ln(x)+x^2),x,method=_RETURNVER
BOSE)

[Out]

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

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maxima [A]  time = 0.62, size = 27, normalized size = 1.29 \begin {gather*} -\frac {2 \, x^{3} - 6 \, e \log \relax (x) - 3 \, e}{2 \, x \log \relax (x) + x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-12*exp(1)*log(x)^2+(-12*exp(1)-8*x^3)*log(x)-3*exp(1))/(4*x^2*log(x)^2+4*x^2*log(x)+x^2),x, algori
thm="maxima")

[Out]

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

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mupad [B]  time = 1.45, size = 21, normalized size = 1.00 \begin {gather*} \frac {3\,\mathrm {e}}{x}-\frac {2\,x^2}{2\,\ln \relax (x)+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*exp(1) + 12*exp(1)*log(x)^2 + log(x)*(12*exp(1) + 8*x^3))/(4*x^2*log(x) + 4*x^2*log(x)^2 + x^2),x)

[Out]

(3*exp(1))/x - (2*x^2)/(2*log(x) + 1)

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sympy [A]  time = 0.10, size = 17, normalized size = 0.81 \begin {gather*} - \frac {2 x^{2}}{2 \log {\relax (x )} + 1} + \frac {3 e}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-12*exp(1)*ln(x)**2+(-12*exp(1)-8*x**3)*ln(x)-3*exp(1))/(4*x**2*ln(x)**2+4*x**2*ln(x)+x**2),x)

[Out]

-2*x**2/(2*log(x) + 1) + 3*E/x

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