3.24.38 \(\int \frac {-3+e^x (3 x+6 x^2+3 x^3)}{(2 e^{e^x (1+x^2)}-x) \log (e^{-e^x (1+x^2)} (-12 e^{e^x (1+x^2)}+6 x))} \, dx\)

Optimal. Leaf size=22 \[ 3 \log \left (\log \left (6 \left (-2+e^{-e^x \left (1+x^2\right )} x\right )\right )\right ) \]

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Rubi [F]  time = 2.68, 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 {-3+e^x \left (3 x+6 x^2+3 x^3\right )}{\left (2 e^{e^x \left (1+x^2\right )}-x\right ) \log \left (e^{-e^x \left (1+x^2\right )} \left (-12 e^{e^x \left (1+x^2\right )}+6 x\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-3 + E^x*(3*x + 6*x^2 + 3*x^3))/((2*E^(E^x*(1 + x^2)) - x)*Log[(-12*E^(E^x*(1 + x^2)) + 6*x)/E^(E^x*(1 +
x^2))]),x]

[Out]

-3*Defer[Int][1/((2*E^(E^x*(1 + x^2)) - x)*Log[-12 + (6*x)/E^(E^x*(1 + x^2))]), x] + 3*Defer[Int][(E^x*x)/((2*
E^(E^x*(1 + x^2)) - x)*Log[-12 + (6*x)/E^(E^x*(1 + x^2))]), x] + 6*Defer[Int][(E^x*x^2)/((2*E^(E^x*(1 + x^2))
- x)*Log[-12 + (6*x)/E^(E^x*(1 + x^2))]), x] + 3*Defer[Int][(E^x*x^3)/((2*E^(E^x*(1 + x^2)) - x)*Log[-12 + (6*
x)/E^(E^x*(1 + x^2))]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {3 \left (-1+e^x x (1+x)^2\right )}{\left (2 e^{e^x \left (1+x^2\right )}-x\right ) \log \left (-12+6 e^{-e^x \left (1+x^2\right )} x\right )} \, dx\\ &=3 \int \frac {-1+e^x x (1+x)^2}{\left (2 e^{e^x \left (1+x^2\right )}-x\right ) \log \left (-12+6 e^{-e^x \left (1+x^2\right )} x\right )} \, dx\\ &=3 \int \left (-\frac {1}{\left (2 e^{e^x \left (1+x^2\right )}-x\right ) \log \left (-12+6 e^{-e^x \left (1+x^2\right )} x\right )}+\frac {e^x x}{\left (2 e^{e^x \left (1+x^2\right )}-x\right ) \log \left (-12+6 e^{-e^x \left (1+x^2\right )} x\right )}+\frac {2 e^x x^2}{\left (2 e^{e^x \left (1+x^2\right )}-x\right ) \log \left (-12+6 e^{-e^x \left (1+x^2\right )} x\right )}+\frac {e^x x^3}{\left (2 e^{e^x \left (1+x^2\right )}-x\right ) \log \left (-12+6 e^{-e^x \left (1+x^2\right )} x\right )}\right ) \, dx\\ &=-\left (3 \int \frac {1}{\left (2 e^{e^x \left (1+x^2\right )}-x\right ) \log \left (-12+6 e^{-e^x \left (1+x^2\right )} x\right )} \, dx\right )+3 \int \frac {e^x x}{\left (2 e^{e^x \left (1+x^2\right )}-x\right ) \log \left (-12+6 e^{-e^x \left (1+x^2\right )} x\right )} \, dx+3 \int \frac {e^x x^3}{\left (2 e^{e^x \left (1+x^2\right )}-x\right ) \log \left (-12+6 e^{-e^x \left (1+x^2\right )} x\right )} \, dx+6 \int \frac {e^x x^2}{\left (2 e^{e^x \left (1+x^2\right )}-x\right ) \log \left (-12+6 e^{-e^x \left (1+x^2\right )} x\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.43, size = 21, normalized size = 0.95 \begin {gather*} 3 \log \left (\log \left (-12+6 e^{-e^x \left (1+x^2\right )} x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3 + E^x*(3*x + 6*x^2 + 3*x^3))/((2*E^(E^x*(1 + x^2)) - x)*Log[(-12*E^(E^x*(1 + x^2)) + 6*x)/E^(E^x
*(1 + x^2))]),x]

[Out]

3*Log[Log[-12 + (6*x)/E^(E^x*(1 + x^2))]]

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fricas [A]  time = 0.89, size = 29, normalized size = 1.32 \begin {gather*} 3 \, \log \left (\log \left (6 \, {\left (x - 2 \, e^{\left ({\left (x^{2} + 1\right )} e^{x}\right )}\right )} e^{\left (-{\left (x^{2} + 1\right )} e^{x}\right )}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^3+6*x^2+3*x)*exp(x)-3)/(2*exp((x^2+1)*exp(x))-x)/log((-12*exp((x^2+1)*exp(x))+6*x)/exp((x^2+1)
*exp(x))),x, algorithm="fricas")

[Out]

3*log(log(6*(x - 2*e^((x^2 + 1)*e^x))*e^(-(x^2 + 1)*e^x)))

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giac [A]  time = 0.40, size = 33, normalized size = 1.50 \begin {gather*} 3 \, \log \left (\log \left (6 \, {\left (x - 2 \, e^{\left (x^{2} e^{x} + e^{x}\right )}\right )} e^{\left (-x^{2} e^{x} - e^{x}\right )}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^3+6*x^2+3*x)*exp(x)-3)/(2*exp((x^2+1)*exp(x))-x)/log((-12*exp((x^2+1)*exp(x))+6*x)/exp((x^2+1)
*exp(x))),x, algorithm="giac")

[Out]

3*log(log(6*(x - 2*e^(x^2*e^x + e^x))*e^(-x^2*e^x - e^x)))

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maple [C]  time = 0.55, size = 232, normalized size = 10.55

method	result	size
risch	\(3 \ln \left (\ln \left ({\mathrm e}^{\left (x^{2}+1\right ) {\mathrm e}^{x}}\right )+\frac {i \left (\pi \,\mathrm {csgn}\left (i \left (-2 \,{\mathrm e}^{\left (x^{2}+1\right ) {\mathrm e}^{x}}+x \right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{-\left (x^{2}+1\right ) {\mathrm e}^{x}}\right ) \mathrm {csgn}\left (i {\mathrm e}^{-\left (x^{2}+1\right ) {\mathrm e}^{x}} \left (-2 \,{\mathrm e}^{\left (x^{2}+1\right ) {\mathrm e}^{x}}+x \right )\right )-\pi \,\mathrm {csgn}\left (i \left (-2 \,{\mathrm e}^{\left (x^{2}+1\right ) {\mathrm e}^{x}}+x \right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{-\left (x^{2}+1\right ) {\mathrm e}^{x}} \left (-2 \,{\mathrm e}^{\left (x^{2}+1\right ) {\mathrm e}^{x}}+x \right )\right )^{2}-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{-\left (x^{2}+1\right ) {\mathrm e}^{x}}\right ) \mathrm {csgn}\left (i {\mathrm e}^{-\left (x^{2}+1\right ) {\mathrm e}^{x}} \left (-2 \,{\mathrm e}^{\left (x^{2}+1\right ) {\mathrm e}^{x}}+x \right )\right )^{2}+\pi \mathrm {csgn}\left (i {\mathrm e}^{-\left (x^{2}+1\right ) {\mathrm e}^{x}} \left (-2 \,{\mathrm e}^{\left (x^{2}+1\right ) {\mathrm e}^{x}}+x \right )\right )^{3}+2 i \ln \relax (2)+2 i \ln \relax (3)+2 i \ln \left (-2 \,{\mathrm e}^{\left (x^{2}+1\right ) {\mathrm e}^{x}}+x \right )\right )}{2}\right )\)	\(232\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((3*x^3+6*x^2+3*x)*exp(x)-3)/(2*exp((x^2+1)*exp(x))-x)/ln((-12*exp((x^2+1)*exp(x))+6*x)/exp((x^2+1)*exp(x)
)),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [C]  time = 0.66, size = 37, normalized size = 1.68 \begin {gather*} 3 \, \log \left (i \, \pi - {\left (x^{2} + 1\right )} e^{x} + \log \relax (3) + \log \relax (2) + \log \left (-x + 2 \, e^{\left (x^{2} e^{x} + e^{x}\right )}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^3+6*x^2+3*x)*exp(x)-3)/(2*exp((x^2+1)*exp(x))-x)/log((-12*exp((x^2+1)*exp(x))+6*x)/exp((x^2+1)
*exp(x))),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.57, size = 22, normalized size = 1.00 \begin {gather*} 3\,\ln \left (\ln \left (6\,x\,{\mathrm {e}}^{-x^2\,{\mathrm {e}}^x}\,{\mathrm {e}}^{-{\mathrm {e}}^x}-12\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x)*(3*x + 6*x^2 + 3*x^3) - 3)/(log(exp(-exp(x)*(x^2 + 1))*(6*x - 12*exp(exp(x)*(x^2 + 1))))*(x - 2*e
xp(exp(x)*(x^2 + 1)))),x)

[Out]

3*log(log(6*x*exp(-x^2*exp(x))*exp(-exp(x)) - 12))

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sympy [A]  time = 0.88, size = 29, normalized size = 1.32 \begin {gather*} 3 \log {\left (\log {\left (\left (6 x - 12 e^{\left (x^{2} + 1\right ) e^{x}}\right ) e^{- \left (x^{2} + 1\right ) e^{x}} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x**3+6*x**2+3*x)*exp(x)-3)/(2*exp((x**2+1)*exp(x))-x)/ln((-12*exp((x**2+1)*exp(x))+6*x)/exp((x**
2+1)*exp(x))),x)

[Out]

3*log(log((6*x - 12*exp((x**2 + 1)*exp(x)))*exp(-(x**2 + 1)*exp(x))))

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