3.58.57 \(\int \frac {-2 x+e^{-3-x^2} (-4 x-2 x^3) \log (2+x^2)}{(2+x^2) \log (2+x^2)} \, dx\) [5757]

Optimal. Leaf size=20 \[ 5+e^{-3-x^2}-\log \left (\log \left (2+x^2\right )\right ) \]

[Out]

exp(-x^2-3)+5-ln(ln(x^2+2))

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Rubi [A]
time = 0.23, antiderivative size = 19, normalized size of antiderivative = 0.95, number of steps used = 7, number of rules used = 6, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {6820, 2240, 2525, 2437, 2339, 29} \begin {gather*} e^{-x^2-3}-\log \left (\log \left (x^2+2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(-3 - x^2) - Log[Log[2 + x^2]]

Rule 29

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

Rule 2240

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^n*(
F^(a + b*(c + d*x)^n)/(b*f*n*(c + d*x)^n*Log[F])), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 2339

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2437

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[(f*(x/d))^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rule 2525

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.)*((f_) + (g_.)*(x_)^(s_))^(r_.),
x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(f + g*x^(s/n))^r*(a + b*Log[c*(d + e*x)^p])^q,
x], x, x^n], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, q, r, s}, x] && IntegerQ[r] && IntegerQ[s/n] && Intege
rQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0])

Rule 6820

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

Rubi steps

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

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Mathematica [A]
time = 0.05, size = 19, normalized size = 0.95 \begin {gather*} e^{-3-x^2}-\log \left (\log \left (2+x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(-3 - x^2) - Log[Log[2 + x^2]]

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Maple [A]
time = 0.71, size = 19, normalized size = 0.95

method	result	size
default	\({\mathrm e}^{-x^{2}-3}-\ln \left (\ln \left (x^{2}+2\right )\right )\)	\(19\)
norman	\({\mathrm e}^{-x^{2}-3}-\ln \left (\ln \left (x^{2}+2\right )\right )\)	\(19\)
risch	\({\mathrm e}^{-x^{2}-3}-\ln \left (\ln \left (x^{2}+2\right )\right )\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.46, size = 18, normalized size = 0.90 \begin {gather*} e^{\left (-x^{2} - 3\right )} - \log \left (\log \left (x^{2} + 2\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(-x^2 - 3) - log(log(x^2 + 2))

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Fricas [A]
time = 0.38, size = 18, normalized size = 0.90 \begin {gather*} e^{\left (-x^{2} - 3\right )} - \log \left (\log \left (x^{2} + 2\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(-x^2 - 3) - log(log(x^2 + 2))

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Sympy [A]
time = 0.11, size = 15, normalized size = 0.75 \begin {gather*} e^{- x^{2} - 3} - \log {\left (\log {\left (x^{2} + 2 \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(-x**2 - 3) - log(log(x**2 + 2))

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Giac [A]
time = 0.40, size = 18, normalized size = 0.90 \begin {gather*} e^{\left (-x^{2} - 3\right )} - \log \left (\log \left (x^{2} + 2\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(-x^2 - 3) - log(log(x^2 + 2))

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Mupad [B]
time = 3.68, size = 18, normalized size = 0.90 \begin {gather*} {\mathrm {e}}^{-x^2-3}-\ln \left (\ln \left (x^2+2\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(- x^2 - 3) - log(log(x^2 + 2))

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