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

Optimal. Leaf size=28 \[ 3 \left (5+e^x-x \left (2-\log \left (\frac {x}{1-2 e^{x^2}}\right )\right )\right ) \]

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Rubi [A]  time = 0.42, antiderivative size = 26, normalized size of antiderivative = 0.93, number of steps used = 7, number of rules used = 3, integrand size = 62, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {6742, 2194, 2548} \begin {gather*} 3 x \log \left (\frac {x}{1-2 e^{x^2}}\right )-6 x+3 e^x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2548

Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/u, x], x] /; InverseFunctionFr
eeQ[u, x]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [A]  time = 0.22, size = 26, normalized size = 0.93 \begin {gather*} 3 e^x-6 x+3 x \log \left (\frac {x}{1-2 e^{x^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.76, size = 25, normalized size = 0.89 \begin {gather*} 3 \, x \log \left (-\frac {x}{2 \, e^{\left (x^{2}\right )} - 1}\right ) - 6 \, x + 3 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.22, size = 25, normalized size = 0.89 \begin {gather*} 3 \, x \log \left (-\frac {x}{2 \, e^{\left (x^{2}\right )} - 1}\right ) - 6 \, x + 3 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.14, size = 26, normalized size = 0.93

method	result	size
default	\(-6 x +3 x \ln \left (-\frac {x}{2 \,{\mathrm e}^{x^{2}}-1}\right )+3 \,{\mathrm e}^{x}\)	\(26\)
risch	\(-3 x \ln \left ({\mathrm e}^{x^{2}}-\frac {1}{2}\right )+3 x \ln \relax (x )-\frac {3 i \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{x^{2}}-\frac {1}{2}}\right ) \mathrm {csgn}\left (\frac {i x}{{\mathrm e}^{x^{2}}-\frac {1}{2}}\right )}{2}+\frac {3 i \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i x}{{\mathrm e}^{x^{2}}-\frac {1}{2}}\right )^{2}}{2}+\frac {3 i \pi x \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{x^{2}}-\frac {1}{2}}\right ) \mathrm {csgn}\left (\frac {i x}{{\mathrm e}^{x^{2}}-\frac {1}{2}}\right )^{2}}{2}-3 i \pi x \mathrm {csgn}\left (\frac {i x}{{\mathrm e}^{x^{2}}-\frac {1}{2}}\right )^{2}+\frac {3 i \pi x \mathrm {csgn}\left (\frac {i x}{{\mathrm e}^{x^{2}}-\frac {1}{2}}\right )^{3}}{2}+3 i \pi x -3 x \ln \relax (2)-6 x +3 \,{\mathrm e}^{x}\)	\(166\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-6*x+3*x*ln(-x/(2*exp(x^2)-1))+3*exp(x)

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maxima [A]  time = 0.39, size = 25, normalized size = 0.89 \begin {gather*} 3 \, x \log \relax (x) - 3 \, x \log \left (-2 \, e^{\left (x^{2}\right )} + 1\right ) - 6 \, x + 3 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.41, size = 27, normalized size = 0.96 \begin {gather*} 3\,{\mathrm {e}}^x-6\,x+3\,x\,\ln \left (-x\right )-3\,x\,\ln \left (2\,{\mathrm {e}}^{x^2}-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.51, size = 24, normalized size = 0.86 \begin {gather*} 3 x \log {\left (- \frac {x}{2 e^{x^{2}} - 1} \right )} - 6 x + 3 e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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