3.20.97 \(\int \frac {(6 e^x-4 e^{2 x} x) \log (x)+(-3 e^x+2 e^{2 x} x+e^x (3+3 x) \log (x)) \log (x^2)}{(9-12 e^x x+4 e^{2 x} x^2) \log ^2(x)} \, dx\)

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

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Rubi [F]  time = 3.87, 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 {\left (6 e^x-4 e^{2 x} x\right ) \log (x)+\left (-3 e^x+2 e^{2 x} x+e^x (3+3 x) \log (x)\right ) \log \left (x^2\right )}{\left (9-12 e^x x+4 e^{2 x} x^2\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((6*E^x - 4*E^(2*x)*x)*Log[x] + (-3*E^x + 2*E^(2*x)*x + E^x*(3 + 3*x)*Log[x])*Log[x^2])/((9 - 12*E^x*x + 4
*E^(2*x)*x^2)*Log[x]^2),x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.55, size = 25, normalized size = 1.00 \begin {gather*} 1+\frac {e^x x \log \left (x^2\right )}{\left (3-2 e^x x\right ) \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((6*E^x - 4*E^(2*x)*x)*Log[x] + (-3*E^x + 2*E^(2*x)*x + E^x*(3 + 3*x)*Log[x])*Log[x^2])/((9 - 12*E^x
*x + 4*E^(2*x)*x^2)*Log[x]^2),x]

[Out]

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

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fricas [A]  time = 0.77, size = 11, normalized size = 0.44 \begin {gather*} -\frac {3}{2 \, x e^{x} - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x+3)*exp(x)*log(x)+2*x*exp(x)^2-3*exp(x))*log(x^2)+(-4*x*exp(x)^2+6*exp(x))*log(x))/(4*exp(x)^2
*x^2-12*exp(x)*x+9)/log(x)^2,x, algorithm="fricas")

[Out]

-3/(2*x*e^x - 3)

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giac [A]  time = 0.27, size = 11, normalized size = 0.44 \begin {gather*} -\frac {3}{2 \, x e^{x} - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x+3)*exp(x)*log(x)+2*x*exp(x)^2-3*exp(x))*log(x^2)+(-4*x*exp(x)^2+6*exp(x))*log(x))/(4*exp(x)^2
*x^2-12*exp(x)*x+9)/log(x)^2,x, algorithm="giac")

[Out]

-3/(2*x*e^x - 3)

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maple [C]  time = 0.12, size = 71, normalized size = 2.84

method	result	size
risch	\(-\frac {3}{2 \,{\mathrm e}^{x} x -3}+\frac {i x \pi \,\mathrm {csgn}\left (i x^{2}\right ) {\mathrm e}^{x} \left (\mathrm {csgn}\left (i x \right )^{2}-2 \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )+\mathrm {csgn}\left (i x^{2}\right )^{2}\right )}{2 \left (2 \,{\mathrm e}^{x} x -3\right ) \ln \relax (x )}\)	\(71\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((3*x+3)*exp(x)*ln(x)+2*x*exp(x)^2-3*exp(x))*ln(x^2)+(-4*x*exp(x)^2+6*exp(x))*ln(x))/(4*exp(x)^2*x^2-12*e
xp(x)*x+9)/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.54, size = 11, normalized size = 0.44 \begin {gather*} -\frac {3}{2 \, x e^{x} - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x+3)*exp(x)*log(x)+2*x*exp(x)^2-3*exp(x))*log(x^2)+(-4*x*exp(x)^2+6*exp(x))*log(x))/(4*exp(x)^2
*x^2-12*exp(x)*x+9)/log(x)^2,x, algorithm="maxima")

[Out]

-3/(2*x*e^x - 3)

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mupad [B]  time = 1.46, size = 35, normalized size = 1.40 \begin {gather*} -\frac {3\,\ln \relax (x)-2\,x\,{\mathrm {e}}^x\,\ln \relax (x)+x\,\ln \left (x^2\right )\,{\mathrm {e}}^x}{\ln \relax (x)\,\left (2\,x\,{\mathrm {e}}^x-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x^2)*(2*x*exp(2*x) - 3*exp(x) + exp(x)*log(x)*(3*x + 3)) + log(x)*(6*exp(x) - 4*x*exp(2*x)))/(log(x)^
2*(4*x^2*exp(2*x) - 12*x*exp(x) + 9)),x)

[Out]

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

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sympy [A]  time = 0.27, size = 10, normalized size = 0.40 \begin {gather*} - \frac {3}{2 x e^{x} - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x+3)*exp(x)*ln(x)+2*x*exp(x)**2-3*exp(x))*ln(x**2)+(-4*x*exp(x)**2+6*exp(x))*ln(x))/(4*exp(x)**
2*x**2-12*exp(x)*x+9)/ln(x)**2,x)

[Out]

-3/(2*x*exp(x) - 3)

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