3.90.54 \(\int \frac {-2+e^{\frac {1}{2} (x+x^3)} (-6+3 x-x^2+(9-3 x) x^3)+2 \log (x)+(e^{\frac {1}{2} (x+x^3)} (-6+2 x)+2 \log (x)) \log (\frac {x}{e^{\frac {1}{2} (x+x^3)} (-3+x)+\log (x)})}{e^{\frac {1}{2} (x+x^3)} (-6+2 x)+2 \log (x)} \, dx\)

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

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Rubi [A]  time = 50.98, antiderivative size = 30, normalized size of antiderivative = 1.20, number of steps used = 83, number of rules used = 5, integrand size = 108, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.046, Rules used = {6742, 6688, 1850, 2549, 12} \begin {gather*} x \log \left (-\frac {x}{e^{\frac {1}{2} \left (x^3+x\right )} (3-x)-\log (x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2 + E^((x + x^3)/2)*(-6 + 3*x - x^2 + (9 - 3*x)*x^3) + 2*Log[x] + (E^((x + x^3)/2)*(-6 + 2*x) + 2*Log[x]
)*Log[x/(E^((x + x^3)/2)*(-3 + x) + Log[x])])/(E^((x + x^3)/2)*(-6 + 2*x) + 2*Log[x]),x]

[Out]

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

Rule 12

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

Rule 1850

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rule 2549

Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[x*Simplify[D[u, x]/u], x], x] /; ProductQ[
u]

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 \left (\frac {6-2 x-x \log (x)+x^2 \log (x)-9 x^3 \log (x)+3 x^4 \log (x)}{2 (-3+x) \left (-3 e^{\frac {x}{2}+\frac {x^3}{2}}+e^{\frac {x}{2}+\frac {x^3}{2}} x+\log (x)\right )}+\frac {-6+3 x-x^2+9 x^3-3 x^4-6 \log \left (\frac {x}{e^{\frac {1}{2} \left (x+x^3\right )} (-3+x)+\log (x)}\right )+2 x \log \left (\frac {x}{e^{\frac {1}{2} \left (x+x^3\right )} (-3+x)+\log (x)}\right )}{2 (-3+x)}\right ) \, dx\\ &=\frac {1}{2} \int \frac {6-2 x-x \log (x)+x^2 \log (x)-9 x^3 \log (x)+3 x^4 \log (x)}{(-3+x) \left (-3 e^{\frac {x}{2}+\frac {x^3}{2}}+e^{\frac {x}{2}+\frac {x^3}{2}} x+\log (x)\right )} \, dx+\frac {1}{2} \int \frac {-6+3 x-x^2+9 x^3-3 x^4-6 \log \left (\frac {x}{e^{\frac {1}{2} \left (x+x^3\right )} (-3+x)+\log (x)}\right )+2 x \log \left (\frac {x}{e^{\frac {1}{2} \left (x+x^3\right )} (-3+x)+\log (x)}\right )}{-3+x} \, dx\\ &=\frac {1}{2} \int \frac {-6+2 x-x \left (-1+x-9 x^2+3 x^3\right ) \log (x)}{(3-x) \left (e^{\frac {1}{2} \left (x+x^3\right )} (-3+x)+\log (x)\right )} \, dx+\frac {1}{2} \int \frac {6-3 x+x^2-9 x^3+3 x^4-2 (-3+x) \log \left (\frac {x}{e^{\frac {1}{2} \left (x+x^3\right )} (-3+x)+\log (x)}\right )}{3-x} \, dx\\ &=\frac {1}{2} \int \left (\frac {6}{(-3+x) \left (-3 e^{\frac {x}{2}+\frac {x^3}{2}}+e^{\frac {x}{2}+\frac {x^3}{2}} x+\log (x)\right )}-\frac {2 x}{(-3+x) \left (-3 e^{\frac {x}{2}+\frac {x^3}{2}}+e^{\frac {x}{2}+\frac {x^3}{2}} x+\log (x)\right )}-\frac {x \log (x)}{(-3+x) \left (-3 e^{\frac {x}{2}+\frac {x^3}{2}}+e^{\frac {x}{2}+\frac {x^3}{2}} x+\log (x)\right )}+\frac {x^2 \log (x)}{(-3+x) \left (-3 e^{\frac {x}{2}+\frac {x^3}{2}}+e^{\frac {x}{2}+\frac {x^3}{2}} x+\log (x)\right )}-\frac {9 x^3 \log (x)}{(-3+x) \left (-3 e^{\frac {x}{2}+\frac {x^3}{2}}+e^{\frac {x}{2}+\frac {x^3}{2}} x+\log (x)\right )}+\frac {3 x^4 \log (x)}{(-3+x) \left (-3 e^{\frac {x}{2}+\frac {x^3}{2}}+e^{\frac {x}{2}+\frac {x^3}{2}} x+\log (x)\right )}\right ) \, dx+\frac {1}{2} \int \left (\frac {-6+3 x-x^2+9 x^3-3 x^4}{-3+x}+2 \log \left (\frac {x}{e^{\frac {1}{2} \left (x+x^3\right )} (-3+x)+\log (x)}\right )\right ) \, dx\\ &=\frac {1}{2} \int \frac {-6+3 x-x^2+9 x^3-3 x^4}{-3+x} \, dx-\frac {1}{2} \int \frac {x \log (x)}{(-3+x) \left (-3 e^{\frac {x}{2}+\frac {x^3}{2}}+e^{\frac {x}{2}+\frac {x^3}{2}} x+\log (x)\right )} \, dx+\frac {1}{2} \int \frac {x^2 \log (x)}{(-3+x) \left (-3 e^{\frac {x}{2}+\frac {x^3}{2}}+e^{\frac {x}{2}+\frac {x^3}{2}} x+\log (x)\right )} \, dx+\frac {3}{2} \int \frac {x^4 \log (x)}{(-3+x) \left (-3 e^{\frac {x}{2}+\frac {x^3}{2}}+e^{\frac {x}{2}+\frac {x^3}{2}} x+\log (x)\right )} \, dx+3 \int \frac {1}{(-3+x) \left (-3 e^{\frac {x}{2}+\frac {x^3}{2}}+e^{\frac {x}{2}+\frac {x^3}{2}} x+\log (x)\right )} \, dx-\frac {9}{2} \int \frac {x^3 \log (x)}{(-3+x) \left (-3 e^{\frac {x}{2}+\frac {x^3}{2}}+e^{\frac {x}{2}+\frac {x^3}{2}} x+\log (x)\right )} \, dx-\int \frac {x}{(-3+x) \left (-3 e^{\frac {x}{2}+\frac {x^3}{2}}+e^{\frac {x}{2}+\frac {x^3}{2}} x+\log (x)\right )} \, dx+\int \log \left (\frac {x}{e^{\frac {1}{2} \left (x+x^3\right )} (-3+x)+\log (x)}\right ) \, dx\\ &=x \log \left (-\frac {x}{e^{\frac {1}{2} \left (x+x^3\right )} (3-x)-\log (x)}\right )+\frac {1}{2} \int \left (-\frac {6}{-3+x}-x-3 x^3\right ) \, dx-\frac {1}{2} \int \frac {x \log (x)}{(-3+x) \left (e^{\frac {1}{2} \left (x+x^3\right )} (-3+x)+\log (x)\right )} \, dx+\frac {1}{2} \int \frac {x^2 \log (x)}{(-3+x) \left (e^{\frac {1}{2} \left (x+x^3\right )} (-3+x)+\log (x)\right )} \, dx+\frac {3}{2} \int \frac {x^4 \log (x)}{(-3+x) \left (e^{\frac {1}{2} \left (x+x^3\right )} (-3+x)+\log (x)\right )} \, dx+3 \int \frac {1}{(-3+x) \left (e^{\frac {1}{2} \left (x+x^3\right )} (-3+x)+\log (x)\right )} \, dx-\frac {9}{2} \int \frac {x^3 \log (x)}{(-3+x) \left (e^{\frac {1}{2} \left (x+x^3\right )} (-3+x)+\log (x)\right )} \, dx-\int \frac {x}{(-3+x) \left (e^{\frac {1}{2} \left (x+x^3\right )} (-3+x)+\log (x)\right )} \, dx-\int -\frac {2+e^{\frac {1}{2} \left (x+x^3\right )} \left (6-3 x+x^2-9 x^3+3 x^4\right )-2 \log (x)}{2 \left (e^{\frac {1}{2} \left (x+x^3\right )} (-3+x)+\log (x)\right )} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 29, normalized size = 1.16 \begin {gather*} x \log \left (\frac {x}{e^{\frac {x}{2}+\frac {x^3}{2}} (-3+x)+\log (x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 + E^((x + x^3)/2)*(-6 + 3*x - x^2 + (9 - 3*x)*x^3) + 2*Log[x] + (E^((x + x^3)/2)*(-6 + 2*x) + 2*
Log[x])*Log[x/(E^((x + x^3)/2)*(-3 + x) + Log[x])])/(E^((x + x^3)/2)*(-6 + 2*x) + 2*Log[x]),x]

[Out]

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

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fricas [A]  time = 0.51, size = 24, normalized size = 0.96 \begin {gather*} x \log \left (\frac {x}{{\left (x - 3\right )} e^{\left (\frac {1}{2} \, x^{3} + \frac {1}{2} \, x\right )} + \log \relax (x)}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x-6)*exp(1/2*x^3+1/2*x)+2*log(x))*log(x/((x-3)*exp(1/2*x^3+1/2*x)+log(x)))+((-3*x+9)*x^3-x^2+3*
x-6)*exp(1/2*x^3+1/2*x)+2*log(x)-2)/((2*x-6)*exp(1/2*x^3+1/2*x)+2*log(x)),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.52, size = 36, normalized size = 1.44 \begin {gather*} -x \log \left (x e^{\left (\frac {1}{2} \, x^{3} + \frac {1}{2} \, x\right )} - 3 \, e^{\left (\frac {1}{2} \, x^{3} + \frac {1}{2} \, x\right )} + \log \relax (x)\right ) + x \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x-6)*exp(1/2*x^3+1/2*x)+2*log(x))*log(x/((x-3)*exp(1/2*x^3+1/2*x)+log(x)))+((-3*x+9)*x^3-x^2+3*
x-6)*exp(1/2*x^3+1/2*x)+2*log(x)-2)/((2*x-6)*exp(1/2*x^3+1/2*x)+2*log(x)),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.20, size = 261, normalized size = 10.44

method	result	size
risch	\(-x \ln \left ({\mathrm e}^{\frac {x \left (x^{2}+1\right )}{2}} x +\ln \relax (x )-3 \,{\mathrm e}^{\frac {x \left (x^{2}+1\right )}{2}}\right )+x \ln \relax (x )-\frac {i \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{\frac {x \left (x^{2}+1\right )}{2}} x +\ln \relax (x )-3 \,{\mathrm e}^{\frac {x \left (x^{2}+1\right )}{2}}}\right ) \mathrm {csgn}\left (\frac {i x}{{\mathrm e}^{\frac {x \left (x^{2}+1\right )}{2}} x +\ln \relax (x )-3 \,{\mathrm e}^{\frac {x \left (x^{2}+1\right )}{2}}}\right )}{2}+\frac {i \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i x}{{\mathrm e}^{\frac {x \left (x^{2}+1\right )}{2}} x +\ln \relax (x )-3 \,{\mathrm e}^{\frac {x \left (x^{2}+1\right )}{2}}}\right )^{2}}{2}+\frac {i \pi x \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{\frac {x \left (x^{2}+1\right )}{2}} x +\ln \relax (x )-3 \,{\mathrm e}^{\frac {x \left (x^{2}+1\right )}{2}}}\right ) \mathrm {csgn}\left (\frac {i x}{{\mathrm e}^{\frac {x \left (x^{2}+1\right )}{2}} x +\ln \relax (x )-3 \,{\mathrm e}^{\frac {x \left (x^{2}+1\right )}{2}}}\right )^{2}}{2}-\frac {i \pi x \mathrm {csgn}\left (\frac {i x}{{\mathrm e}^{\frac {x \left (x^{2}+1\right )}{2}} x +\ln \relax (x )-3 \,{\mathrm e}^{\frac {x \left (x^{2}+1\right )}{2}}}\right )^{3}}{2}\)	\(261\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x-6)*exp(1/2*x^3+1/2*x)+2*ln(x))*ln(x/((x-3)*exp(1/2*x^3+1/2*x)+ln(x)))+((-3*x+9)*x^3-x^2+3*x-6)*exp(
1/2*x^3+1/2*x)+2*ln(x)-2)/((2*x-6)*exp(1/2*x^3+1/2*x)+2*ln(x)),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.38, size = 26, normalized size = 1.04 \begin {gather*} -x \log \left ({\left (x - 3\right )} e^{\left (\frac {1}{2} \, x^{3} + \frac {1}{2} \, x\right )} + \log \relax (x)\right ) + x \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x-6)*exp(1/2*x^3+1/2*x)+2*log(x))*log(x/((x-3)*exp(1/2*x^3+1/2*x)+log(x)))+((-3*x+9)*x^3-x^2+3*
x-6)*exp(1/2*x^3+1/2*x)+2*log(x)-2)/((2*x-6)*exp(1/2*x^3+1/2*x)+2*log(x)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 7.26, size = 24, normalized size = 0.96 \begin {gather*} x\,\ln \left (\frac {x}{\ln \relax (x)+{\mathrm {e}}^{x/2}\,{\mathrm {e}}^{\frac {x^3}{2}}\,\left (x-3\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*log(x) + log(x/(log(x) + exp(x/2 + x^3/2)*(x - 3)))*(2*log(x) + exp(x/2 + x^3/2)*(2*x - 6)) - exp(x/2 +
x^3/2)*(x^3*(3*x - 9) - 3*x + x^2 + 6) - 2)/(2*log(x) + exp(x/2 + x^3/2)*(2*x - 6)),x)

[Out]

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

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sympy [A]  time = 1.84, size = 20, normalized size = 0.80 \begin {gather*} x \log {\left (\frac {x}{\left (x - 3\right ) e^{\frac {x^{3}}{2} + \frac {x}{2}} + \log {\relax (x )}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x-6)*exp(1/2*x**3+1/2*x)+2*ln(x))*ln(x/((x-3)*exp(1/2*x**3+1/2*x)+ln(x)))+((-3*x+9)*x**3-x**2+3
*x-6)*exp(1/2*x**3+1/2*x)+2*ln(x)-2)/((2*x-6)*exp(1/2*x**3+1/2*x)+2*ln(x)),x)

[Out]

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

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