3.81.47 \(\int \frac {e (2+2 x-5 x^2+9 x^3)+e (-4 x^2+6 x^3) \log (x)+e (-x^2+x^3) \log ^2(x)+(e (-4 x+10 x^3)+8 e x^3 \log (x)+2 e x^3 \log ^2(x)) \log (\frac {2-5 x^2-4 x^2 \log (x)-x^2 \log ^2(x)}{x})}{-2+5 x^2+4 x^2 \log (x)+x^2 \log ^2(x)} \, dx\)

Optimal. Leaf size=29 \[ e \left (-x+x^2 \log \left (\frac {2}{x}-x-x (2+\log (x))^2\right )\right ) \]

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Rubi [F]  time = 0.94, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {e \left (2+2 x-5 x^2+9 x^3\right )+e \left (-4 x^2+6 x^3\right ) \log (x)+e \left (-x^2+x^3\right ) \log ^2(x)+\left (e \left (-4 x+10 x^3\right )+8 e x^3 \log (x)+2 e x^3 \log ^2(x)\right ) \log \left (\frac {2-5 x^2-4 x^2 \log (x)-x^2 \log ^2(x)}{x}\right )}{-2+5 x^2+4 x^2 \log (x)+x^2 \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E*(2 + 2*x - 5*x^2 + 9*x^3) + E*(-4*x^2 + 6*x^3)*Log[x] + E*(-x^2 + x^3)*Log[x]^2 + (E*(-4*x + 10*x^3) +
8*E*x^3*Log[x] + 2*E*x^3*Log[x]^2)*Log[(2 - 5*x^2 - 4*x^2*Log[x] - x^2*Log[x]^2)/x])/(-2 + 5*x^2 + 4*x^2*Log[x
] + x^2*Log[x]^2),x]

[Out]

-(E*x) + (E*x^2)/2 + 4*E*Defer[Int][x/(-2 + 5*x^2 + 4*x^2*Log[x] + x^2*Log[x]^2), x] + 4*E*Defer[Int][x^3/(-2
+ 5*x^2 + 4*x^2*Log[x] + x^2*Log[x]^2), x] + 2*E*Defer[Int][(x^3*Log[x])/(-2 + 5*x^2 + 4*x^2*Log[x] + x^2*Log[
x]^2), x] + 2*E*Defer[Int][x*Log[2/x - 5*x - 4*x*Log[x] - x*Log[x]^2], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {e \left (2+2 x-5 x^2+9 x^3-4 x^2 \log (x)+6 x^3 \log (x)-x^2 \log ^2(x)+x^3 \log ^2(x)\right )}{-2+5 x^2+4 x^2 \log (x)+x^2 \log ^2(x)}+2 e x \log \left (\frac {2}{x}-5 x-4 x \log (x)-x \log ^2(x)\right )\right ) \, dx\\ &=e \int \frac {2+2 x-5 x^2+9 x^3-4 x^2 \log (x)+6 x^3 \log (x)-x^2 \log ^2(x)+x^3 \log ^2(x)}{-2+5 x^2+4 x^2 \log (x)+x^2 \log ^2(x)} \, dx+(2 e) \int x \log \left (\frac {2}{x}-5 x-4 x \log (x)-x \log ^2(x)\right ) \, dx\\ &=e \int \left (-1+x+\frac {2 x \left (2+2 x^2+x^2 \log (x)\right )}{-2+5 x^2+4 x^2 \log (x)+x^2 \log ^2(x)}\right ) \, dx+(2 e) \int x \log \left (\frac {2}{x}-5 x-4 x \log (x)-x \log ^2(x)\right ) \, dx\\ &=-e x+\frac {e x^2}{2}+(2 e) \int \frac {x \left (2+2 x^2+x^2 \log (x)\right )}{-2+5 x^2+4 x^2 \log (x)+x^2 \log ^2(x)} \, dx+(2 e) \int x \log \left (\frac {2}{x}-5 x-4 x \log (x)-x \log ^2(x)\right ) \, dx\\ &=-e x+\frac {e x^2}{2}+(2 e) \int \left (\frac {2 x}{-2+5 x^2+4 x^2 \log (x)+x^2 \log ^2(x)}+\frac {2 x^3}{-2+5 x^2+4 x^2 \log (x)+x^2 \log ^2(x)}+\frac {x^3 \log (x)}{-2+5 x^2+4 x^2 \log (x)+x^2 \log ^2(x)}\right ) \, dx+(2 e) \int x \log \left (\frac {2}{x}-5 x-4 x \log (x)-x \log ^2(x)\right ) \, dx\\ &=-e x+\frac {e x^2}{2}+(2 e) \int \frac {x^3 \log (x)}{-2+5 x^2+4 x^2 \log (x)+x^2 \log ^2(x)} \, dx+(2 e) \int x \log \left (\frac {2}{x}-5 x-4 x \log (x)-x \log ^2(x)\right ) \, dx+(4 e) \int \frac {x}{-2+5 x^2+4 x^2 \log (x)+x^2 \log ^2(x)} \, dx+(4 e) \int \frac {x^3}{-2+5 x^2+4 x^2 \log (x)+x^2 \log ^2(x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 32, normalized size = 1.10 \begin {gather*} e \left (-x+x^2 \log \left (\frac {2}{x}-5 x-4 x \log (x)-x \log ^2(x)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E*(2 + 2*x - 5*x^2 + 9*x^3) + E*(-4*x^2 + 6*x^3)*Log[x] + E*(-x^2 + x^3)*Log[x]^2 + (E*(-4*x + 10*x
^3) + 8*E*x^3*Log[x] + 2*E*x^3*Log[x]^2)*Log[(2 - 5*x^2 - 4*x^2*Log[x] - x^2*Log[x]^2)/x])/(-2 + 5*x^2 + 4*x^2
*Log[x] + x^2*Log[x]^2),x]

[Out]

E*(-x + x^2*Log[2/x - 5*x - 4*x*Log[x] - x*Log[x]^2])

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fricas [A]  time = 0.71, size = 40, normalized size = 1.38 \begin {gather*} x^{2} e \log \left (-\frac {x^{2} \log \relax (x)^{2} + 4 \, x^{2} \log \relax (x) + 5 \, x^{2} - 2}{x}\right ) - x e \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3*exp(1)*log(x)^2+8*x^3*exp(1)*log(x)+(10*x^3-4*x)*exp(1))*log((-x^2*log(x)^2-4*x^2*log(x)-5*x
^2+2)/x)+(x^3-x^2)*exp(1)*log(x)^2+(6*x^3-4*x^2)*exp(1)*log(x)+(9*x^3-5*x^2+2*x+2)*exp(1))/(x^2*log(x)^2+4*x^2
*log(x)+5*x^2-2),x, algorithm="fricas")

[Out]

x^2*e*log(-(x^2*log(x)^2 + 4*x^2*log(x) + 5*x^2 - 2)/x) - x*e

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giac [A]  time = 0.26, size = 45, normalized size = 1.55 \begin {gather*} x^{2} e \log \left (-x^{2} \log \relax (x)^{2} - 4 \, x^{2} \log \relax (x) - 5 \, x^{2} + 2\right ) - x^{2} e \log \relax (x) - x e \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3*exp(1)*log(x)^2+8*x^3*exp(1)*log(x)+(10*x^3-4*x)*exp(1))*log((-x^2*log(x)^2-4*x^2*log(x)-5*x
^2+2)/x)+(x^3-x^2)*exp(1)*log(x)^2+(6*x^3-4*x^2)*exp(1)*log(x)+(9*x^3-5*x^2+2*x+2)*exp(1))/(x^2*log(x)^2+4*x^2
*log(x)+5*x^2-2),x, algorithm="giac")

[Out]

x^2*e*log(-x^2*log(x)^2 - 4*x^2*log(x) - 5*x^2 + 2) - x^2*e*log(x) - x*e

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maple [C]  time = 0.34, size = 270, normalized size = 9.31

method	result	size
risch	\(x^{2} {\mathrm e} \ln \left (-2+\left (\ln \relax (x )^{2}+4 \ln \relax (x )+5\right ) x^{2}\right )-x^{2} {\mathrm e} \ln \relax (x )+\frac {i \pi \,{\mathrm e} x^{2} \mathrm {csgn}\left (i \left (-2+\left (\ln \relax (x )^{2}+4 \ln \relax (x )+5\right ) x^{2}\right )\right ) \mathrm {csgn}\left (\frac {i \left (-2+\left (\ln \relax (x )^{2}+4 \ln \relax (x )+5\right ) x^{2}\right )}{x}\right )^{2}}{2}-\frac {i \pi \,{\mathrm e} x^{2} \mathrm {csgn}\left (i \left (-2+\left (\ln \relax (x )^{2}+4 \ln \relax (x )+5\right ) x^{2}\right )\right ) \mathrm {csgn}\left (\frac {i \left (-2+\left (\ln \relax (x )^{2}+4 \ln \relax (x )+5\right ) x^{2}\right )}{x}\right ) \mathrm {csgn}\left (\frac {i}{x}\right )}{2}+\frac {i \pi \,{\mathrm e} x^{2} \mathrm {csgn}\left (\frac {i \left (-2+\left (\ln \relax (x )^{2}+4 \ln \relax (x )+5\right ) x^{2}\right )}{x}\right )^{3}}{2}-i \pi \,{\mathrm e} x^{2} \mathrm {csgn}\left (\frac {i \left (-2+\left (\ln \relax (x )^{2}+4 \ln \relax (x )+5\right ) x^{2}\right )}{x}\right )^{2}+\frac {i \pi \,{\mathrm e} x^{2} \mathrm {csgn}\left (\frac {i \left (-2+\left (\ln \relax (x )^{2}+4 \ln \relax (x )+5\right ) x^{2}\right )}{x}\right )^{2} \mathrm {csgn}\left (\frac {i}{x}\right )}{2}+i {\mathrm e} \pi \,x^{2}-x \,{\mathrm e}\)	\(270\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^3*exp(1)*ln(x)^2+8*x^3*exp(1)*ln(x)+(10*x^3-4*x)*exp(1))*ln((-x^2*ln(x)^2-4*x^2*ln(x)-5*x^2+2)/x)+(x
^3-x^2)*exp(1)*ln(x)^2+(6*x^3-4*x^2)*exp(1)*ln(x)+(9*x^3-5*x^2+2*x+2)*exp(1))/(x^2*ln(x)^2+4*x^2*ln(x)+5*x^2-2
),x,method=_RETURNVERBOSE)

[Out]

x^2*exp(1)*ln(-2+(ln(x)^2+4*ln(x)+5)*x^2)-x^2*exp(1)*ln(x)+1/2*I*Pi*exp(1)*x^2*csgn(I*(-2+(ln(x)^2+4*ln(x)+5)*
x^2))*csgn(I/x*(-2+(ln(x)^2+4*ln(x)+5)*x^2))^2-1/2*I*Pi*exp(1)*x^2*csgn(I*(-2+(ln(x)^2+4*ln(x)+5)*x^2))*csgn(I
/x*(-2+(ln(x)^2+4*ln(x)+5)*x^2))*csgn(I/x)+1/2*I*Pi*exp(1)*x^2*csgn(I/x*(-2+(ln(x)^2+4*ln(x)+5)*x^2))^3-I*Pi*e
xp(1)*x^2*csgn(I/x*(-2+(ln(x)^2+4*ln(x)+5)*x^2))^2+1/2*I*Pi*exp(1)*x^2*csgn(I/x*(-2+(ln(x)^2+4*ln(x)+5)*x^2))^
2*csgn(I/x)+I*exp(1)*Pi*x^2-x*exp(1)

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maxima [A]  time = 0.40, size = 45, normalized size = 1.55 \begin {gather*} x^{2} e \log \left (-x^{2} \log \relax (x)^{2} - 4 \, x^{2} \log \relax (x) - 5 \, x^{2} + 2\right ) - x^{2} e \log \relax (x) - x e \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3*exp(1)*log(x)^2+8*x^3*exp(1)*log(x)+(10*x^3-4*x)*exp(1))*log((-x^2*log(x)^2-4*x^2*log(x)-5*x
^2+2)/x)+(x^3-x^2)*exp(1)*log(x)^2+(6*x^3-4*x^2)*exp(1)*log(x)+(9*x^3-5*x^2+2*x+2)*exp(1))/(x^2*log(x)^2+4*x^2
*log(x)+5*x^2-2),x, algorithm="maxima")

[Out]

x^2*e*log(-x^2*log(x)^2 - 4*x^2*log(x) - 5*x^2 + 2) - x^2*e*log(x) - x*e

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mupad [B]  time = 6.04, size = 40, normalized size = 1.38 \begin {gather*} x^2\,\mathrm {e}\,\ln \left (-\frac {x^2\,{\ln \relax (x)}^2+4\,x^2\,\ln \relax (x)+5\,x^2-2}{x}\right )-x\,\mathrm {e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(-(4*x^2*log(x) + x^2*log(x)^2 + 5*x^2 - 2)/x)*(8*x^3*exp(1)*log(x) - exp(1)*(4*x - 10*x^3) + 2*x^3*ex
p(1)*log(x)^2) + exp(1)*(2*x - 5*x^2 + 9*x^3 + 2) - exp(1)*log(x)*(4*x^2 - 6*x^3) - exp(1)*log(x)^2*(x^2 - x^3
))/(4*x^2*log(x) + x^2*log(x)^2 + 5*x^2 - 2),x)

[Out]

x^2*exp(1)*log(-(4*x^2*log(x) + x^2*log(x)^2 + 5*x^2 - 2)/x) - x*exp(1)

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sympy [A]  time = 0.64, size = 37, normalized size = 1.28 \begin {gather*} e x^{2} \log {\left (\frac {- x^{2} \log {\relax (x )}^{2} - 4 x^{2} \log {\relax (x )} - 5 x^{2} + 2}{x} \right )} - e x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**3*exp(1)*ln(x)**2+8*x**3*exp(1)*ln(x)+(10*x**3-4*x)*exp(1))*ln((-x**2*ln(x)**2-4*x**2*ln(x)-5
*x**2+2)/x)+(x**3-x**2)*exp(1)*ln(x)**2+(6*x**3-4*x**2)*exp(1)*ln(x)+(9*x**3-5*x**2+2*x+2)*exp(1))/(x**2*ln(x)
**2+4*x**2*ln(x)+5*x**2-2),x)

[Out]

E*x**2*log((-x**2*log(x)**2 - 4*x**2*log(x) - 5*x**2 + 2)/x) - E*x

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