3.29.19 \(\int \frac {\log (e x) (e^{x+x^2} (-4+2 x)-2 e^{x+x^2} \log (x^2))+\log ^2(e x) (e^{x+x^2} (4-4 x-3 x^2+2 x^3)+e^{x+x^2} (1-x-2 x^2) \log (x^2))}{4 x^2-4 x^3+x^4+(4 x^2-2 x^3) \log (x^2)+x^2 \log ^2(x^2)} \, dx\)

Optimal. Leaf size=28 \[ \frac {e^{x (1+x)} \log ^2(e x)}{x \left (-2+x-\log \left (x^2\right )\right )} \]

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Rubi [F]  time = 6.91, 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 {\log (e x) \left (e^{x+x^2} (-4+2 x)-2 e^{x+x^2} \log \left (x^2\right )\right )+\log ^2(e x) \left (e^{x+x^2} \left (4-4 x-3 x^2+2 x^3\right )+e^{x+x^2} \left (1-x-2 x^2\right ) \log \left (x^2\right )\right )}{4 x^2-4 x^3+x^4+\left (4 x^2-2 x^3\right ) \log \left (x^2\right )+x^2 \log ^2\left (x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(Log[E*x]*(E^(x + x^2)*(-4 + 2*x) - 2*E^(x + x^2)*Log[x^2]) + Log[E*x]^2*(E^(x + x^2)*(4 - 4*x - 3*x^2 + 2
*x^3) + E^(x + x^2)*(1 - x - 2*x^2)*Log[x^2]))/(4*x^2 - 4*x^3 + x^4 + (4*x^2 - 2*x^3)*Log[x^2] + x^2*Log[x^2]^
2),x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 5.14, size = 28, normalized size = 1.00 \begin {gather*} \frac {e^{x+x^2} (1+\log (x))^2}{x \left (-2+x-\log \left (x^2\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Log[E*x]*(E^(x + x^2)*(-4 + 2*x) - 2*E^(x + x^2)*Log[x^2]) + Log[E*x]^2*(E^(x + x^2)*(4 - 4*x - 3*x
^2 + 2*x^3) + E^(x + x^2)*(1 - x - 2*x^2)*Log[x^2]))/(4*x^2 - 4*x^3 + x^4 + (4*x^2 - 2*x^3)*Log[x^2] + x^2*Log
[x^2]^2),x]

[Out]

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

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fricas [A]  time = 0.60, size = 52, normalized size = 1.86 \begin {gather*} \frac {e^{\left (x^{2} + x\right )} \log \left (x^{2}\right )^{2} + 4 \, e^{\left (x^{2} + x\right )} \log \left (x^{2}\right ) + 4 \, e^{\left (x^{2} + x\right )}}{4 \, {\left (x^{2} - x \log \left (x^{2}\right ) - 2 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2-x+1)*exp(x^2+x)*log(x^2)+(2*x^3-3*x^2-4*x+4)*exp(x^2+x))*log(x*exp(1))^2+(-2*exp(x^2+x)*lo
g(x^2)+(2*x-4)*exp(x^2+x))*log(x*exp(1)))/(x^2*log(x^2)^2+(-2*x^3+4*x^2)*log(x^2)+x^4-4*x^3+4*x^2),x, algorith
m="fricas")

[Out]

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

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giac [A]  time = 0.47, size = 43, normalized size = 1.54 \begin {gather*} \frac {e^{\left (x^{2} + x\right )} \log \relax (x)^{2} + 2 \, e^{\left (x^{2} + x\right )} \log \relax (x) + e^{\left (x^{2} + x\right )}}{x^{2} - 2 \, x \log \relax (x) - 2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2-x+1)*exp(x^2+x)*log(x^2)+(2*x^3-3*x^2-4*x+4)*exp(x^2+x))*log(x*exp(1))^2+(-2*exp(x^2+x)*lo
g(x^2)+(2*x-4)*exp(x^2+x))*log(x*exp(1)))/(x^2*log(x^2)^2+(-2*x^3+4*x^2)*log(x^2)+x^4-4*x^3+4*x^2),x, algorith
m="giac")

[Out]

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

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maple [C]  time = 0.35, size = 305, normalized size = 10.89

method	result	size
risch	\(-\frac {{\mathrm e}^{\left (x +1\right ) x} \ln \relax (x )}{2 x}-\frac {\left (i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+4+i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+2 x \right ) {\mathrm e}^{\left (x +1\right ) x}}{8 x}+\frac {\left (4 x^{2}-\pi ^{2} \mathrm {csgn}\left (i x^{2}\right )^{6}+4 i \pi x \mathrm {csgn}\left (i x^{2}\right )^{3}-\pi ^{2} \mathrm {csgn}\left (i x \right )^{4} \mathrm {csgn}\left (i x^{2}\right )^{2}+4 \pi ^{2} \mathrm {csgn}\left (i x \right )^{3} \mathrm {csgn}\left (i x^{2}\right )^{3}-6 \pi ^{2} \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )^{4}+4 \pi ^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{5}+4 i \pi x \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-8 i \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}\right ) {\mathrm e}^{\left (x +1\right ) x}}{8 x \left (i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+2 x -4 \ln \relax (x )-4\right )}\)	\(305\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-2*x^2-x+1)*exp(x^2+x)*ln(x^2)+(2*x^3-3*x^2-4*x+4)*exp(x^2+x))*ln(x*exp(1))^2+(-2*exp(x^2+x)*ln(x^2)+(2
*x-4)*exp(x^2+x))*ln(x*exp(1)))/(x^2*ln(x^2)^2+(-2*x^3+4*x^2)*ln(x^2)+x^4-4*x^3+4*x^2),x,method=_RETURNVERBOSE
)

[Out]

-1/2/x*exp((x+1)*x)*ln(x)-1/8*(I*Pi*csgn(I*x^2)*csgn(I*x)^2-2*I*Pi*csgn(I*x^2)^2*csgn(I*x)+4+I*Pi*csgn(I*x^2)^
3+2*x)/x*exp((x+1)*x)+1/8*(4*x^2-Pi^2*csgn(I*x^2)^6+4*I*Pi*x*csgn(I*x^2)^3-Pi^2*csgn(I*x)^4*csgn(I*x^2)^2+4*Pi
^2*csgn(I*x)^3*csgn(I*x^2)^3-6*Pi^2*csgn(I*x)^2*csgn(I*x^2)^4+4*Pi^2*csgn(I*x)*csgn(I*x^2)^5+4*I*Pi*x*csgn(I*x
)^2*csgn(I*x^2)-8*I*Pi*x*csgn(I*x)*csgn(I*x^2)^2)*exp((x+1)*x)/x/(I*Pi*csgn(I*x^2)*csgn(I*x)^2-2*I*Pi*csgn(I*x
^2)^2*csgn(I*x)+I*Pi*csgn(I*x^2)^3+2*x-4*ln(x)-4)

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maxima [A]  time = 0.63, size = 31, normalized size = 1.11 \begin {gather*} \frac {{\left (\log \relax (x)^{2} + 2 \, \log \relax (x) + 1\right )} e^{\left (x^{2} + x\right )}}{x^{2} - 2 \, x \log \relax (x) - 2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2-x+1)*exp(x^2+x)*log(x^2)+(2*x^3-3*x^2-4*x+4)*exp(x^2+x))*log(x*exp(1))^2+(-2*exp(x^2+x)*lo
g(x^2)+(2*x-4)*exp(x^2+x))*log(x*exp(1)))/(x^2*log(x^2)^2+(-2*x^3+4*x^2)*log(x^2)+x^4-4*x^3+4*x^2),x, algorith
m="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {{\ln \left (x\,\mathrm {e}\right )}^2\,\left ({\mathrm {e}}^{x^2+x}\,\left (-2\,x^3+3\,x^2+4\,x-4\right )+\ln \left (x^2\right )\,{\mathrm {e}}^{x^2+x}\,\left (2\,x^2+x-1\right )\right )-\ln \left (x\,\mathrm {e}\right )\,\left ({\mathrm {e}}^{x^2+x}\,\left (2\,x-4\right )-2\,\ln \left (x^2\right )\,{\mathrm {e}}^{x^2+x}\right )}{\ln \left (x^2\right )\,\left (4\,x^2-2\,x^3\right )+4\,x^2-4\,x^3+x^4+x^2\,{\ln \left (x^2\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x*exp(1))^2*(exp(x + x^2)*(4*x + 3*x^2 - 2*x^3 - 4) + log(x^2)*exp(x + x^2)*(x + 2*x^2 - 1)) - log(x
*exp(1))*(exp(x + x^2)*(2*x - 4) - 2*log(x^2)*exp(x + x^2)))/(log(x^2)*(4*x^2 - 2*x^3) + 4*x^2 - 4*x^3 + x^4 +
x^2*log(x^2)^2),x)

[Out]

int(-(log(x*exp(1))^2*(exp(x + x^2)*(4*x + 3*x^2 - 2*x^3 - 4) + log(x^2)*exp(x + x^2)*(x + 2*x^2 - 1)) - log(x
*exp(1))*(exp(x + x^2)*(2*x - 4) - 2*log(x^2)*exp(x + x^2)))/(log(x^2)*(4*x^2 - 2*x^3) + 4*x^2 - 4*x^3 + x^4 +
x^2*log(x^2)^2), x)

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sympy [A]  time = 0.45, size = 37, normalized size = 1.32 \begin {gather*} \frac {\left (\log {\left (x^{2} \right )}^{2} + 4 \log {\left (x^{2} \right )} + 4\right ) e^{x^{2} + x}}{4 x^{2} - 4 x \log {\left (x^{2} \right )} - 8 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x**2-x+1)*exp(x**2+x)*ln(x**2)+(2*x**3-3*x**2-4*x+4)*exp(x**2+x))*ln(x*exp(1))**2+(-2*exp(x**2
+x)*ln(x**2)+(2*x-4)*exp(x**2+x))*ln(x*exp(1)))/(x**2*ln(x**2)**2+(-2*x**3+4*x**2)*ln(x**2)+x**4-4*x**3+4*x**2
),x)

[Out]

(log(x**2)**2 + 4*log(x**2) + 4)*exp(x**2 + x)/(4*x**2 - 4*x*log(x**2) - 8*x)

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