3.36.53 \(\int \frac {-3 e^2 x^2-6 e x^3-3 x^4+(16 e+2 e^2 (-8 e-12 x)+24 x) \log ^2(x)}{(e^2 x^3+2 e x^4+x^5) \log ^2(x)} \, dx\)

Optimal. Leaf size=24 \[ \frac {4 \left (-2+2 e^2\right )}{x^2 (e+x)}+\frac {3}{\log (x)} \]

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Rubi [A]  time = 0.37, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 68, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {1594, 27, 6688, 74, 2302, 30} \begin {gather*} \frac {3}{\log (x)}-\frac {8 \left (1-e^2\right )}{x^2 (x+e)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-3*E^2*x^2 - 6*E*x^3 - 3*x^4 + (16*E + 2*E^2*(-8*E - 12*x) + 24*x)*Log[x]^2)/((E^2*x^3 + 2*E*x^4 + x^5)*L
og[x]^2),x]

[Out]

(-8*(1 - E^2))/(x^2*(E + x)) + 3/Log[x]

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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 74

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] &
& EqQ[a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 0]

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 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

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 e^2 x^2-6 e x^3-3 x^4+\left (16 e+2 e^2 (-8 e-12 x)+24 x\right ) \log ^2(x)}{x^3 \left (e^2+2 e x+x^2\right ) \log ^2(x)} \, dx\\ &=\int \frac {-3 e^2 x^2-6 e x^3-3 x^4+\left (16 e+2 e^2 (-8 e-12 x)+24 x\right ) \log ^2(x)}{x^3 (e+x)^2 \log ^2(x)} \, dx\\ &=\int \left (-\frac {8 \left (-1+e^2\right ) (2 e+3 x)}{x^3 (e+x)^2}-\frac {3}{x \log ^2(x)}\right ) \, dx\\ &=-\left (3 \int \frac {1}{x \log ^2(x)} \, dx\right )+\left (8 \left (1-e^2\right )\right ) \int \frac {2 e+3 x}{x^3 (e+x)^2} \, dx\\ &=-\frac {8 \left (1-e^2\right )}{x^2 (e+x)}-3 \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (x)\right )\\ &=-\frac {8 \left (1-e^2\right )}{x^2 (e+x)}+\frac {3}{\log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 22, normalized size = 0.92 \begin {gather*} \frac {8 \left (-1+e^2\right )}{x^2 (e+x)}+\frac {3}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3*E^2*x^2 - 6*E*x^3 - 3*x^4 + (16*E + 2*E^2*(-8*E - 12*x) + 24*x)*Log[x]^2)/((E^2*x^3 + 2*E*x^4 +
x^5)*Log[x]^2),x]

[Out]

(8*(-1 + E^2))/(x^2*(E + x)) + 3/Log[x]

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fricas [A]  time = 0.95, size = 38, normalized size = 1.58 \begin {gather*} \frac {3 \, x^{3} + 3 \, x^{2} e + 8 \, {\left (e^{2} - 1\right )} \log \relax (x)}{{\left (x^{3} + x^{2} e\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-8*exp(1)-12*x)*exp(log(2)+2)+16*exp(1)+24*x)*log(x)^2-3*x^2*exp(1)^2-6*x^3*exp(1)-3*x^4)/(x^3*ex
p(1)^2+2*x^4*exp(1)+x^5)/log(x)^2,x, algorithm="fricas")

[Out]

(3*x^3 + 3*x^2*e + 8*(e^2 - 1)*log(x))/((x^3 + x^2*e)*log(x))

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giac [A]  time = 0.15, size = 41, normalized size = 1.71 \begin {gather*} \frac {3 \, x^{3} + 3 \, x^{2} e + 8 \, e^{2} \log \relax (x) - 8 \, \log \relax (x)}{x^{3} \log \relax (x) + x^{2} e \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-8*exp(1)-12*x)*exp(log(2)+2)+16*exp(1)+24*x)*log(x)^2-3*x^2*exp(1)^2-6*x^3*exp(1)-3*x^4)/(x^3*ex
p(1)^2+2*x^4*exp(1)+x^5)/log(x)^2,x, algorithm="giac")

[Out]

(3*x^3 + 3*x^2*e + 8*e^2*log(x) - 8*log(x))/(x^3*log(x) + x^2*e*log(x))

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maple [A]  time = 0.22, size = 23, normalized size = 0.96

method	result	size
risch	\(\frac {3}{\ln \relax (x )}+\frac {8 \,{\mathrm e}^{2}-8}{x^{2} \left (x +{\mathrm e}\right )}\)	\(23\)
default	\(\frac {3}{\ln \relax (x )}+\frac {8 \,{\mathrm e}^{2}-8}{x^{2} \left (x +{\mathrm e}\right )}\)	\(24\)
norman	\(\frac {\left (8 \,{\mathrm e}^{2}-8\right ) \ln \relax (x )+3 x^{3}+3 x^{2} {\mathrm e}}{x^{2} \left (x +{\mathrm e}\right ) \ln \relax (x )}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-8*exp(1)-12*x)*exp(ln(2)+2)+16*exp(1)+24*x)*ln(x)^2-3*x^2*exp(1)^2-6*x^3*exp(1)-3*x^4)/(x^3*exp(1)^2+2
*x^4*exp(1)+x^5)/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.70, size = 38, normalized size = 1.58 \begin {gather*} \frac {3 \, x^{3} + 3 \, x^{2} e + 8 \, {\left (e^{2} - 1\right )} \log \relax (x)}{{\left (x^{3} + x^{2} e\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-8*exp(1)-12*x)*exp(log(2)+2)+16*exp(1)+24*x)*log(x)^2-3*x^2*exp(1)^2-6*x^3*exp(1)-3*x^4)/(x^3*ex
p(1)^2+2*x^4*exp(1)+x^5)/log(x)^2,x, algorithm="maxima")

[Out]

(3*x^3 + 3*x^2*e + 8*(e^2 - 1)*log(x))/((x^3 + x^2*e)*log(x))

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mupad [B]  time = 2.44, size = 44, normalized size = 1.83 \begin {gather*} \frac {8\,{\mathrm {e}}^2-8}{x^2\,\left (x+\mathrm {e}\right )}+\frac {3\,x^3+3\,\mathrm {e}\,x^2}{x^2\,\ln \relax (x)\,\left (x+\mathrm {e}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*x^2*exp(2) - log(x)^2*(24*x + 16*exp(1) - exp(log(2) + 2)*(12*x + 8*exp(1))) + 6*x^3*exp(1) + 3*x^4)/(
log(x)^2*(x^3*exp(2) + 2*x^4*exp(1) + x^5)),x)

[Out]

(8*exp(2) - 8)/(x^2*(x + exp(1))) + (3*x^2*exp(1) + 3*x^3)/(x^2*log(x)*(x + exp(1)))

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sympy [A]  time = 0.37, size = 20, normalized size = 0.83 \begin {gather*} \frac {3}{\log {\relax (x )}} + \frac {-8 + 8 e^{2}}{x^{3} + e x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-8*exp(1)-12*x)*exp(ln(2)+2)+16*exp(1)+24*x)*ln(x)**2-3*x**2*exp(1)**2-6*x**3*exp(1)-3*x**4)/(x**
3*exp(1)**2+2*x**4*exp(1)+x**5)/ln(x)**2,x)

[Out]

3/log(x) + (-8 + 8*exp(2))/(x**3 + E*x**2)

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