3.10.74 \(\int \frac {10 e+2 x^2-5 x^3+(-2 e+x^3) \log (\frac {e^2}{3 x^2})}{-5 x^3+x^3 \log (\frac {e^2}{3 x^2})} \, dx\)

Optimal. Leaf size=26 \[ 3+\frac {e}{x^2}+x-\log \left (5-\log \left (\frac {e^2}{3 x^2}\right )\right ) \]

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Rubi [A]  time = 0.33, antiderivative size = 22, normalized size of antiderivative = 0.85, number of steps used = 7, number of rules used = 5, integrand size = 57, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {2561, 6742, 14, 2302, 29} \begin {gather*} \frac {e}{x^2}-\log \left (3-\log \left (\frac {1}{3 x^2}\right )\right )+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(10*E + 2*x^2 - 5*x^3 + (-2*E + x^3)*Log[E^2/(3*x^2)])/(-5*x^3 + x^3*Log[E^2/(3*x^2)]),x]

[Out]

E/x^2 + x - Log[3 - Log[1/(3*x^2)]]

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 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

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 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 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 {10 e+2 x^2-5 x^3+\left (-2 e+x^3\right ) \log \left (\frac {e^2}{3 x^2}\right )}{x^3 \left (-5+\log \left (\frac {e^2}{3 x^2}\right )\right )} \, dx\\ &=\int \left (\frac {-2 e+x^3}{x^3}+\frac {2}{x \left (-3+\log \left (\frac {1}{3 x^2}\right )\right )}\right ) \, dx\\ &=2 \int \frac {1}{x \left (-3+\log \left (\frac {1}{3 x^2}\right )\right )} \, dx+\int \frac {-2 e+x^3}{x^3} \, dx\\ &=\int \left (1-\frac {2 e}{x^3}\right ) \, dx-\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,-3+\log \left (\frac {1}{3 x^2}\right )\right )\\ &=\frac {e}{x^2}+x-\log \left (3-\log \left (\frac {1}{3 x^2}\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 22, normalized size = 0.85 \begin {gather*} \frac {e}{x^2}+x-\log \left (3-\log \left (\frac {1}{3 x^2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10*E + 2*x^2 - 5*x^3 + (-2*E + x^3)*Log[E^2/(3*x^2)])/(-5*x^3 + x^3*Log[E^2/(3*x^2)]),x]

[Out]

E/x^2 + x - Log[3 - Log[1/(3*x^2)]]

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fricas [A]  time = 1.20, size = 26, normalized size = 1.00 \begin {gather*} \frac {x^{3} - x^{2} \log \left (\log \left (\frac {e^{2}}{3 \, x^{2}}\right ) - 5\right ) + e}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(1)+x^3)*log(1/3*exp(2)/x^2)+10*exp(1)-5*x^3+2*x^2)/(x^3*log(1/3*exp(2)/x^2)-5*x^3),x, algor
ithm="fricas")

[Out]

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

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giac [A]  time = 0.50, size = 24, normalized size = 0.92 \begin {gather*} \frac {x^{3} - x^{2} \log \left (\log \left (3 \, x^{2}\right ) + 3\right ) + e}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(1)+x^3)*log(1/3*exp(2)/x^2)+10*exp(1)-5*x^3+2*x^2)/(x^3*log(1/3*exp(2)/x^2)-5*x^3),x, algor
ithm="giac")

[Out]

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

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maple [A]  time = 0.11, size = 25, normalized size = 0.96

method	result	size
norman	\(\frac {x^{3}+{\mathrm e}}{x^{2}}-\ln \left (\ln \left (\frac {{\mathrm e}^{2}}{3 x^{2}}\right )-5\right )\)	\(25\)
risch	\(\frac {x^{3}+{\mathrm e}}{x^{2}}-\ln \left (\ln \left (\frac {{\mathrm e}^{2}}{3 x^{2}}\right )-5\right )\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*exp(1)+x^3)*ln(1/3*exp(2)/x^2)+10*exp(1)-5*x^3+2*x^2)/(x^3*ln(1/3*exp(2)/x^2)-5*x^3),x,method=_RETURN
VERBOSE)

[Out]

(x^3+exp(1))/x^2-ln(ln(1/3*exp(2)/x^2)-5)

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maxima [A]  time = 0.56, size = 22, normalized size = 0.85 \begin {gather*} \frac {x^{3} + e}{x^{2}} - \log \left (\frac {1}{2} \, \log \relax (3) + \log \relax (x) + \frac {3}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(1)+x^3)*log(1/3*exp(2)/x^2)+10*exp(1)-5*x^3+2*x^2)/(x^3*log(1/3*exp(2)/x^2)-5*x^3),x, algor
ithm="maxima")

[Out]

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

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mupad [B]  time = 0.87, size = 19, normalized size = 0.73 \begin {gather*} x-\ln \left (\ln \left (\frac {1}{3\,x^2}\right )-3\right )+\frac {\mathrm {e}}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*exp(1) - log(exp(2)/(3*x^2))*(2*exp(1) - x^3) + 2*x^2 - 5*x^3)/(x^3*log(exp(2)/(3*x^2)) - 5*x^3),x)

[Out]

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

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sympy [A]  time = 0.15, size = 20, normalized size = 0.77 \begin {gather*} x - \log {\left (\log {\left (\frac {e^{2}}{3 x^{2}} \right )} - 5 \right )} + \frac {e}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(1)+x**3)*ln(1/3*exp(2)/x**2)+10*exp(1)-5*x**3+2*x**2)/(x**3*ln(1/3*exp(2)/x**2)-5*x**3),x)

[Out]

x - log(log(exp(2)/(3*x**2)) - 5) + E/x**2

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