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

Optimal. Leaf size=21 \[ \frac {e^x}{x \log \left (3 x \left (e^x+x+x^2\right )\right )} \]

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Rubi [F]  time = 6.39, 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^{2 x} (-1-x)+e^x \left (-2 x-3 x^2\right )+\left (e^{2 x} (-1+x)+e^x \left (-x+x^3\right )\right ) \log \left (3 e^x x+3 x^2+3 x^3\right )}{\left (e^x x^2+x^3+x^4\right ) \log ^2\left (3 e^x x+3 x^2+3 x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 1.48, size = 21, normalized size = 1.00 \begin {gather*} \frac {e^x}{x \log \left (3 x \left (e^x+x+x^2\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^x/(x*Log[3*x*(E^x + x + x^2)])

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fricas [A]  time = 0.65, size = 25, normalized size = 1.19 \begin {gather*} \frac {e^{x}}{x \log \left (3 \, x^{3} + 3 \, x^{2} + 3 \, x e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x-1)*exp(x)^2+(x^3-x)*exp(x))*log(3*exp(x)*x+3*x^3+3*x^2)+(-x-1)*exp(x)^2+(-3*x^2-2*x)*exp(x))/(e
xp(x)*x^2+x^4+x^3)/log(3*exp(x)*x+3*x^3+3*x^2)^2,x, algorithm="fricas")

[Out]

e^x/(x*log(3*x^3 + 3*x^2 + 3*x*e^x))

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giac [A]  time = 0.43, size = 25, normalized size = 1.19 \begin {gather*} \frac {e^{x}}{x \log \left (3 \, x^{3} + 3 \, x^{2} + 3 \, x e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x-1)*exp(x)^2+(x^3-x)*exp(x))*log(3*exp(x)*x+3*x^3+3*x^2)+(-x-1)*exp(x)^2+(-3*x^2-2*x)*exp(x))/(e
xp(x)*x^2+x^4+x^3)/log(3*exp(x)*x+3*x^3+3*x^2)^2,x, algorithm="giac")

[Out]

e^x/(x*log(3*x^3 + 3*x^2 + 3*x*e^x))

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maple [C]  time = 0.11, size = 129, normalized size = 6.14

method	result	size
risch	\(\frac {2 i {\mathrm e}^{x}}{x \left (\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{x}+x^{2}+x \right )\right ) \mathrm {csgn}\left (i x \left ({\mathrm e}^{x}+x^{2}+x \right )\right )-\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \left ({\mathrm e}^{x}+x^{2}+x \right )\right )^{2}-\pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{x}+x^{2}+x \right )\right ) \mathrm {csgn}\left (i x \left ({\mathrm e}^{x}+x^{2}+x \right )\right )^{2}+\pi \mathrm {csgn}\left (i x \left ({\mathrm e}^{x}+x^{2}+x \right )\right )^{3}+2 i \ln \relax (3)+2 i \ln \relax (x )+2 i \ln \left ({\mathrm e}^{x}+x^{2}+x \right )\right )}\)	\(129\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.59, size = 24, normalized size = 1.14 \begin {gather*} \frac {e^{x}}{x \log \relax (3) + x \log \left (x^{2} + x + e^{x}\right ) + x \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x-1)*exp(x)^2+(x^3-x)*exp(x))*log(3*exp(x)*x+3*x^3+3*x^2)+(-x-1)*exp(x)^2+(-3*x^2-2*x)*exp(x))/(e
xp(x)*x^2+x^4+x^3)/log(3*exp(x)*x+3*x^3+3*x^2)^2,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int -\frac {\ln \left (3\,x\,{\mathrm {e}}^x+3\,x^2+3\,x^3\right )\,\left ({\mathrm {e}}^x\,\left (x-x^3\right )-{\mathrm {e}}^{2\,x}\,\left (x-1\right )\right )+{\mathrm {e}}^{2\,x}\,\left (x+1\right )+{\mathrm {e}}^x\,\left (3\,x^2+2\,x\right )}{{\ln \left (3\,x\,{\mathrm {e}}^x+3\,x^2+3\,x^3\right )}^2\,\left (x^2\,{\mathrm {e}}^x+x^3+x^4\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.21, size = 22, normalized size = 1.05 \begin {gather*} \frac {e^{x}}{x \log {\left (3 x^{3} + 3 x^{2} + 3 x e^{x} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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