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

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

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Rubi [A]  time = 0.39, antiderivative size = 21, normalized size of antiderivative = 0.95, number of steps used = 10, number of rules used = 8, integrand size = 60, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {6688, 6742, 6715, 2282, 36, 31, 29, 6684} \begin {gather*} -\log \left (3-e^{x^2}\right )+x+\log \left (\log \left (e^x x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x - Log[3 - E^x^2] + Log[Log[E^x*x]]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6688

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

Rule 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), 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 {-3+e^{x^2} (1-2 x)}{-3+e^{x^2}}+\frac {1+\frac {1}{x}}{\log \left (e^x x\right )}\right ) \, dx\\ &=\int \frac {-3+e^{x^2} (1-2 x)}{-3+e^{x^2}} \, dx+\int \frac {1+\frac {1}{x}}{\log \left (e^x x\right )} \, dx\\ &=\log \left (\log \left (e^x x\right )\right )+\int \left (1-2 x-\frac {6 x}{-3+e^{x^2}}\right ) \, dx\\ &=x-x^2+\log \left (\log \left (e^x x\right )\right )-6 \int \frac {x}{-3+e^{x^2}} \, dx\\ &=x-x^2+\log \left (\log \left (e^x x\right )\right )-3 \operatorname {Subst}\left (\int \frac {1}{-3+e^x} \, dx,x,x^2\right )\\ &=x-x^2+\log \left (\log \left (e^x x\right )\right )-3 \operatorname {Subst}\left (\int \frac {1}{(-3+x) x} \, dx,x,e^{x^2}\right )\\ &=x-x^2+\log \left (\log \left (e^x x\right )\right )-\operatorname {Subst}\left (\int \frac {1}{-3+x} \, dx,x,e^{x^2}\right )+\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^{x^2}\right )\\ &=x-\log \left (3-e^{x^2}\right )+\log \left (\log \left (e^x x\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 21, normalized size = 0.95 \begin {gather*} x-\log \left (3-e^{x^2}\right )+\log \left (\log \left (e^x x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x - Log[3 - E^x^2] + Log[Log[E^x*x]]

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fricas [A]  time = 0.64, size = 17, normalized size = 0.77 \begin {gather*} x - \log \left (e^{\left (x^{2}\right )} - 3\right ) + \log \left (\log \left (x e^{x}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2+x)*exp(x^2)-3*x)*log(exp(x)*x)+(x+1)*exp(x^2)-3*x-3)/(exp(x^2)*x-3*x)/log(exp(x)*x),x, alg
orithm="fricas")

[Out]

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

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giac [A]  time = 0.18, size = 16, normalized size = 0.73 \begin {gather*} x + \log \left (x + \log \relax (x)\right ) - \log \left (e^{\left (x^{2}\right )} - 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2+x)*exp(x^2)-3*x)*log(exp(x)*x)+(x+1)*exp(x^2)-3*x-3)/(exp(x^2)*x-3*x)/log(exp(x)*x),x, alg
orithm="giac")

[Out]

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

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maple [C]  time = 0.09, size = 92, normalized size = 4.18

method	result	size
risch	\(x -\ln \left ({\mathrm e}^{x^{2}}-3\right )+\ln \left (\ln \left ({\mathrm e}^{x}\right )-\frac {i \left (\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{x}\right )-\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{x}\right )^{2}-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{x}\right )^{2}+\pi \mathrm {csgn}\left (i x \,{\mathrm e}^{x}\right )^{3}+2 i \ln \relax (x )\right )}{2}\right )\)	\(92\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.40, size = 16, normalized size = 0.73 \begin {gather*} x + \log \left (x + \log \relax (x)\right ) - \log \left (e^{\left (x^{2}\right )} - 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2+x)*exp(x^2)-3*x)*log(exp(x)*x)+(x+1)*exp(x^2)-3*x-3)/(exp(x^2)*x-3*x)/log(exp(x)*x),x, alg
orithm="maxima")

[Out]

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

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mupad [B]  time = 4.32, size = 16, normalized size = 0.73 \begin {gather*} x-\ln \left ({\mathrm {e}}^{x^2}-3\right )+\ln \left (x+\ln \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.19, size = 17, normalized size = 0.77 \begin {gather*} x - \log {\left (e^{x^{2}} - 3 \right )} + \log {\left (\log {\left (x e^{x} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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