3.91.7 \(\int \frac {-168 x^2+48 x^3+e^{3/x} (-45 x-20 x^2+10 x^3)+(72 x-24 x^2+e^{3/x} (45-5 x^2)) \log (3-x)}{72 x^3-24 x^4+e^{3/x} (15 x^3-5 x^4)+(-72 x^2+24 x^3+e^{3/x} (-15 x^2+5 x^3)) \log (3-x)} \, dx\)

Optimal. Leaf size=30 \[ \log \left (\frac {-\frac {24}{5}-e^{3/x}}{x (-x+\log (3-x))}\right ) \]

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Rubi [A]  time = 3.67, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 134, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {6688, 6742, 6715, 2282, 36, 29, 31, 43, 6684} \begin {gather*} \log \left (5 e^{3/x}+24\right )-\log (x)-\log (x-\log (3-x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-168*x^2 + 48*x^3 + E^(3/x)*(-45*x - 20*x^2 + 10*x^3) + (72*x - 24*x^2 + E^(3/x)*(45 - 5*x^2))*Log[3 - x]
)/(72*x^3 - 24*x^4 + E^(3/x)*(15*x^3 - 5*x^4) + (-72*x^2 + 24*x^3 + E^(3/x)*(-15*x^2 + 5*x^3))*Log[3 - x]),x]

[Out]

Log[24 + 5*E^(3/x)] - Log[x] - Log[x - Log[3 - x]]

Rule 29

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6688

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

Rule 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), 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 {-x \left (24 (7-2 x) x-5 e^{3/x} \left (-9-4 x+2 x^2\right )\right )-(-3+x) \left (24 x+5 e^{3/x} (3+x)\right ) \log (3-x)}{\left (24+5 e^{3/x}\right ) (3-x) x^2 (x-\log (3-x))} \, dx\\ &=\int \left (\frac {72}{\left (24+5 e^{3/x}\right ) x^2}+\frac {9 x+4 x^2-2 x^3-9 \log (3-x)+x^2 \log (3-x)}{(-3+x) x^2 (x-\log (3-x))}\right ) \, dx\\ &=72 \int \frac {1}{\left (24+5 e^{3/x}\right ) x^2} \, dx+\int \frac {9 x+4 x^2-2 x^3-9 \log (3-x)+x^2 \log (3-x)}{(-3+x) x^2 (x-\log (3-x))} \, dx\\ &=-\left (72 \operatorname {Subst}\left (\int \frac {1}{24+5 e^{3 x}} \, dx,x,\frac {1}{x}\right )\right )+\int \left (\frac {-3-x}{x^2}+\frac {4-x}{(-3+x) (x-\log (3-x))}\right ) \, dx\\ &=-\left (24 \operatorname {Subst}\left (\int \frac {1}{x (24+5 x)} \, dx,x,e^{3/x}\right )\right )+\int \frac {-3-x}{x^2} \, dx+\int \frac {4-x}{(-3+x) (x-\log (3-x))} \, dx\\ &=-\log (x-\log (3-x))+5 \operatorname {Subst}\left (\int \frac {1}{24+5 x} \, dx,x,e^{3/x}\right )+\int \left (-\frac {3}{x^2}-\frac {1}{x}\right ) \, dx-\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^{3/x}\right )\\ &=\log \left (24+5 e^{3/x}\right )-\log (x)-\log (x-\log (3-x))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 30, normalized size = 1.00 \begin {gather*} \log \left (24+5 e^{3/x}\right )-\log (x)-\log (x-\log (3-x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-168*x^2 + 48*x^3 + E^(3/x)*(-45*x - 20*x^2 + 10*x^3) + (72*x - 24*x^2 + E^(3/x)*(45 - 5*x^2))*Log[
3 - x])/(72*x^3 - 24*x^4 + E^(3/x)*(15*x^3 - 5*x^4) + (-72*x^2 + 24*x^3 + E^(3/x)*(-15*x^2 + 5*x^3))*Log[3 - x
]),x]

[Out]

Log[24 + 5*E^(3/x)] - Log[x] - Log[x - Log[3 - x]]

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fricas [A]  time = 0.45, size = 29, normalized size = 0.97 \begin {gather*} -\log \relax (x) - \log \left (-x + \log \left (-x + 3\right )\right ) + \log \left (5 \, e^{\frac {3}{x}} + 24\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x^2+45)*exp(3/x)-24*x^2+72*x)*log(3-x)+(10*x^3-20*x^2-45*x)*exp(3/x)+48*x^3-168*x^2)/(((5*x^3-
15*x^2)*exp(3/x)+24*x^3-72*x^2)*log(3-x)+(-5*x^4+15*x^3)*exp(3/x)-24*x^4+72*x^3),x, algorithm="fricas")

[Out]

-log(x) - log(-x + log(-x + 3)) + log(5*e^(3/x) + 24)

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giac [A]  time = 0.21, size = 29, normalized size = 0.97 \begin {gather*} -\log \relax (x) - \log \left (-x + \log \left (-x + 3\right )\right ) + \log \left (5 \, e^{\frac {3}{x}} + 24\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x^2+45)*exp(3/x)-24*x^2+72*x)*log(3-x)+(10*x^3-20*x^2-45*x)*exp(3/x)+48*x^3-168*x^2)/(((5*x^3-
15*x^2)*exp(3/x)+24*x^3-72*x^2)*log(3-x)+(-5*x^4+15*x^3)*exp(3/x)-24*x^4+72*x^3),x, algorithm="giac")

[Out]

-log(x) - log(-x + log(-x + 3)) + log(5*e^(3/x) + 24)

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maple [A]  time = 0.10, size = 28, normalized size = 0.93




method result size



risch \(-\ln \relax (x )+\ln \left ({\mathrm e}^{\frac {3}{x}}+\frac {24}{5}\right )-\ln \left (\ln \left (3-x \right )-x \right )\) \(28\)
norman \(-\ln \relax (x )-\ln \left (x -\ln \left (3-x \right )\right )+\ln \left (5 \,{\mathrm e}^{\frac {3}{x}}+24\right )\) \(30\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-5*x^2+45)*exp(3/x)-24*x^2+72*x)*ln(3-x)+(10*x^3-20*x^2-45*x)*exp(3/x)+48*x^3-168*x^2)/(((5*x^3-15*x^2)
*exp(3/x)+24*x^3-72*x^2)*ln(3-x)+(-5*x^4+15*x^3)*exp(3/x)-24*x^4+72*x^3),x,method=_RETURNVERBOSE)

[Out]

-ln(x)+ln(exp(3/x)+24/5)-ln(ln(3-x)-x)

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maxima [A]  time = 0.50, size = 27, normalized size = 0.90 \begin {gather*} -\log \relax (x) - \log \left (-x + \log \left (-x + 3\right )\right ) + \log \left (e^{\frac {3}{x}} + \frac {24}{5}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x^2+45)*exp(3/x)-24*x^2+72*x)*log(3-x)+(10*x^3-20*x^2-45*x)*exp(3/x)+48*x^3-168*x^2)/(((5*x^3-
15*x^2)*exp(3/x)+24*x^3-72*x^2)*log(3-x)+(-5*x^4+15*x^3)*exp(3/x)-24*x^4+72*x^3),x, algorithm="maxima")

[Out]

-log(x) - log(-x + log(-x + 3)) + log(e^(3/x) + 24/5)

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mupad [B]  time = 8.36, size = 27, normalized size = 0.90 \begin {gather*} \ln \left ({\mathrm {e}}^{3/x}+\frac {24}{5}\right )-\ln \left (\ln \left (3-x\right )-x\right )-\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(3 - x)*(exp(3/x)*(5*x^2 - 45) - 72*x + 24*x^2) + exp(3/x)*(45*x + 20*x^2 - 10*x^3) + 168*x^2 - 48*x^3
)/(log(3 - x)*(exp(3/x)*(15*x^2 - 5*x^3) + 72*x^2 - 24*x^3) - exp(3/x)*(15*x^3 - 5*x^4) - 72*x^3 + 24*x^4),x)

[Out]

log(exp(3/x) + 24/5) - log(log(3 - x) - x) - log(x)

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sympy [A]  time = 0.44, size = 20, normalized size = 0.67 \begin {gather*} - \log {\relax (x )} - \log {\left (- x + \log {\left (3 - x \right )} \right )} + \log {\left (e^{\frac {3}{x}} + \frac {24}{5} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x**2+45)*exp(3/x)-24*x**2+72*x)*ln(3-x)+(10*x**3-20*x**2-45*x)*exp(3/x)+48*x**3-168*x**2)/(((5
*x**3-15*x**2)*exp(3/x)+24*x**3-72*x**2)*ln(3-x)+(-5*x**4+15*x**3)*exp(3/x)-24*x**4+72*x**3),x)

[Out]

-log(x) - log(-x + log(3 - x)) + log(exp(3/x) + 24/5)

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