3.23.7 \(\int \frac {e^{-\log ^2(x)} (e^{\log ^2(x)} (-2 x^2-4 e^{8+2 x^2} x^2+e^{4+x^2} (-2 x-4 x^3))+(-20-4 e^{8+2 x^2}-8 e^{4+x^2} x-4 x^2) \log (x))}{5 x+e^{8+2 x^2} x+2 e^{4+x^2} x^2+x^3} \, dx\)

Optimal. Leaf size=28 \[ 4+2 e^{-\log ^2(x)}-\log \left (5+\left (e^{4+x^2}+x\right )^2\right ) \]

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Rubi [A]  time = 0.64, antiderivative size = 38, normalized size of antiderivative = 1.36, number of steps used = 5, number of rules used = 5, integrand size = 117, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {6688, 6684, 2276, 2205, 2209} \begin {gather*} 2 e^{-\log ^2(x)}-\log \left (x^2+2 e^{x^2+4} x+e^{2 x^2+8}+5\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^Log[x]^2*(-2*x^2 - 4*E^(8 + 2*x^2)*x^2 + E^(4 + x^2)*(-2*x - 4*x^3)) + (-20 - 4*E^(8 + 2*x^2) - 8*E^(4
+ x^2)*x - 4*x^2)*Log[x])/(E^Log[x]^2*(5*x + E^(8 + 2*x^2)*x + 2*E^(4 + x^2)*x^2 + x^3)),x]

[Out]

2/E^Log[x]^2 - Log[5 + E^(8 + 2*x^2) + 2*E^(4 + x^2)*x + x^2]

Rule 2205

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erf[(c + d*x)*Rt[-(b*Log[F]),
2]])/(2*d*Rt[-(b*Log[F]), 2]), x] /; FreeQ[{F, a, b, c, d}, x] && NegQ[b]

Rule 2209

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 2276

Int[(F_)^(((a_.) + Log[(c_.)*(x_)^(n_.)]^2*(b_.))*(d_.))*((e_.)*(x_))^(m_.), x_Symbol] :> Dist[(e*x)^(m + 1)/(
e*n*(c*x^n)^((m + 1)/n)), Subst[Int[E^(a*d*Log[F] + ((m + 1)*x)/n + b*d*Log[F]*x^2), x], x, Log[c*x^n]], x] /;
FreeQ[{F, a, b, c, d, e, m, n}, 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]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {2 \left (x+2 e^{8+2 x^2} x+e^{4+x^2} \left (1+2 x^2\right )\right )}{5+e^{8+2 x^2}+2 e^{4+x^2} x+x^2}-\frac {4 e^{-\log ^2(x)} \log (x)}{x}\right ) \, dx\\ &=-\left (2 \int \frac {x+2 e^{8+2 x^2} x+e^{4+x^2} \left (1+2 x^2\right )}{5+e^{8+2 x^2}+2 e^{4+x^2} x+x^2} \, dx\right )-4 \int \frac {e^{-\log ^2(x)} \log (x)}{x} \, dx\\ &=-\log \left (5+e^{8+2 x^2}+2 e^{4+x^2} x+x^2\right )-4 \operatorname {Subst}\left (\int e^{-x^2} x \, dx,x,\log (x)\right )\\ &=2 e^{-\log ^2(x)}-\log \left (5+e^{8+2 x^2}+2 e^{4+x^2} x+x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.34, size = 38, normalized size = 1.36 \begin {gather*} 2 e^{-\log ^2(x)}-\log \left (5+e^{8+2 x^2}+2 e^{4+x^2} x+x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^Log[x]^2*(-2*x^2 - 4*E^(8 + 2*x^2)*x^2 + E^(4 + x^2)*(-2*x - 4*x^3)) + (-20 - 4*E^(8 + 2*x^2) - 8
*E^(4 + x^2)*x - 4*x^2)*Log[x])/(E^Log[x]^2*(5*x + E^(8 + 2*x^2)*x + 2*E^(4 + x^2)*x^2 + x^3)),x]

[Out]

2/E^Log[x]^2 - Log[5 + E^(8 + 2*x^2) + 2*E^(4 + x^2)*x + x^2]

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fricas [A]  time = 1.00, size = 40, normalized size = 1.43 \begin {gather*} -{\left (e^{\left (\log \relax (x)^{2}\right )} \log \left (x^{2} + 2 \, x e^{\left (x^{2} + 4\right )} + e^{\left (2 \, x^{2} + 8\right )} + 5\right ) - 2\right )} e^{\left (-\log \relax (x)^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^2*exp(x^2+4)^2+(-4*x^3-2*x)*exp(x^2+4)-2*x^2)*exp(log(x)^2)+(-4*exp(x^2+4)^2-8*x*exp(x^2+4)-4
*x^2-20)*log(x))/(x*exp(x^2+4)^2+2*x^2*exp(x^2+4)+x^3+5*x)/exp(log(x)^2),x, algorithm="fricas")

[Out]

-(e^(log(x)^2)*log(x^2 + 2*x*e^(x^2 + 4) + e^(2*x^2 + 8) + 5) - 2)*e^(-log(x)^2)

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giac [A]  time = 1.02, size = 40, normalized size = 1.43 \begin {gather*} -{\left (e^{\left (\log \relax (x)^{2}\right )} \log \left (x^{2} + 2 \, x e^{\left (x^{2} + 4\right )} + e^{\left (2 \, x^{2} + 8\right )} + 5\right ) - 2\right )} e^{\left (-\log \relax (x)^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^2*exp(x^2+4)^2+(-4*x^3-2*x)*exp(x^2+4)-2*x^2)*exp(log(x)^2)+(-4*exp(x^2+4)^2-8*x*exp(x^2+4)-4
*x^2-20)*log(x))/(x*exp(x^2+4)^2+2*x^2*exp(x^2+4)+x^3+5*x)/exp(log(x)^2),x, algorithm="giac")

[Out]

-(e^(log(x)^2)*log(x^2 + 2*x*e^(x^2 + 4) + e^(2*x^2 + 8) + 5) - 2)*e^(-log(x)^2)

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maple [A]  time = 0.18, size = 37, normalized size = 1.32

method	result	size
risch	\(8-\ln \left (2 x \,{\mathrm e}^{x^{2}+4}+x^{2}+{\mathrm e}^{2 x^{2}+8}+5\right )+2 \,{\mathrm e}^{-\ln \relax (x )^{2}}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x^2*exp(x^2+4)^2+(-4*x^3-2*x)*exp(x^2+4)-2*x^2)*exp(ln(x)^2)+(-4*exp(x^2+4)^2-8*x*exp(x^2+4)-4*x^2-20
)*ln(x))/(x*exp(x^2+4)^2+2*x^2*exp(x^2+4)+x^3+5*x)/exp(ln(x)^2),x,method=_RETURNVERBOSE)

[Out]

8-ln(2*x*exp(x^2+4)+x^2+exp(2*x^2+8)+5)+2*exp(-ln(x)^2)

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maxima [A]  time = 0.58, size = 38, normalized size = 1.36 \begin {gather*} 2 \, e^{\left (-\log \relax (x)^{2}\right )} - \log \left ({\left (x^{2} + 2 \, x e^{\left (x^{2} + 4\right )} + e^{\left (2 \, x^{2} + 8\right )} + 5\right )} e^{\left (-8\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^2*exp(x^2+4)^2+(-4*x^3-2*x)*exp(x^2+4)-2*x^2)*exp(log(x)^2)+(-4*exp(x^2+4)^2-8*x*exp(x^2+4)-4
*x^2-20)*log(x))/(x*exp(x^2+4)^2+2*x^2*exp(x^2+4)+x^3+5*x)/exp(log(x)^2),x, algorithm="maxima")

[Out]

2*e^(-log(x)^2) - log((x^2 + 2*x*e^(x^2 + 4) + e^(2*x^2 + 8) + 5)*e^(-8))

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mupad [B]  time = 1.74, size = 36, normalized size = 1.29 \begin {gather*} 2\,{\mathrm {e}}^{-{\ln \relax (x)}^2}-\ln \left ({\mathrm {e}}^8\,{\mathrm {e}}^{2\,x^2}+x^2+2\,x\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^4+5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-log(x)^2)*(exp(log(x)^2)*(4*x^2*exp(2*x^2 + 8) + exp(x^2 + 4)*(2*x + 4*x^3) + 2*x^2) + log(x)*(4*ex
p(2*x^2 + 8) + 8*x*exp(x^2 + 4) + 4*x^2 + 20)))/(5*x + x*exp(2*x^2 + 8) + 2*x^2*exp(x^2 + 4) + x^3),x)

[Out]

2*exp(-log(x)^2) - log(exp(8)*exp(2*x^2) + x^2 + 2*x*exp(x^2)*exp(4) + 5)

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sympy [A]  time = 0.45, size = 32, normalized size = 1.14 \begin {gather*} - \log {\left (x^{2} + 2 x e^{x^{2} + 4} + e^{2 x^{2} + 8} + 5 \right )} + 2 e^{- \log {\relax (x )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x**2*exp(x**2+4)**2+(-4*x**3-2*x)*exp(x**2+4)-2*x**2)*exp(ln(x)**2)+(-4*exp(x**2+4)**2-8*x*exp(
x**2+4)-4*x**2-20)*ln(x))/(x*exp(x**2+4)**2+2*x**2*exp(x**2+4)+x**3+5*x)/exp(ln(x)**2),x)

[Out]

-log(x**2 + 2*x*exp(x**2 + 4) + exp(2*x**2 + 8) + 5) + 2*exp(-log(x)**2)

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