3.35.72 \(\int \frac {e^{-2 x^2} (6 x-24 x^3-4 x^4+e^{2 x^2} (36 x^2+12 x^3+x^4)+e^{2 x^2} (-36-12 x-x^2) \log (\frac {55}{4} e^{\frac {e^{-2 x^2} x}{6+x}}))}{36 x^2+12 x^3+x^4} \, dx\)

Optimal. Leaf size=27 \[ x+\frac {\log \left (\frac {55}{4} e^{\frac {e^{-2 x^2} x}{6+x}}\right )}{x} \]

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Rubi [A]  time = 2.00, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 9, integrand size = 99, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1594, 27, 6688, 14, 6742, 2205, 2210, 2220, 2551} \begin {gather*} \frac {\log \left (\frac {55}{4} e^{\frac {e^{-2 x^2} x}{x+6}}\right )}{x}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(6*x - 24*x^3 - 4*x^4 + E^(2*x^2)*(36*x^2 + 12*x^3 + x^4) + E^(2*x^2)*(-36 - 12*x - x^2)*Log[(55*E^(x/(E^(
2*x^2)*(6 + x))))/4])/(E^(2*x^2)*(36*x^2 + 12*x^3 + x^4)),x]

[Out]

x + Log[(55*E^(x/(E^(2*x^2)*(6 + x))))/4]/x

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

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 2210

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

Rule 2220

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2)*((e_.) + (f_.)*(x_))^(m_), x_Symbol] :> Simp[(f*(e + f*x)^(m +
1)*F^(a + b*(c + d*x)^2))/((m + 1)*f^2), x] + (-Dist[(2*b*d^2*Log[F])/(f^2*(m + 1)), Int[(e + f*x)^(m + 2)*F^
(a + b*(c + d*x)^2), x], x] + Dist[(2*b*d*(d*e - c*f)*Log[F])/(f^2*(m + 1)), Int[(e + f*x)^(m + 1)*F^(a + b*(c
+ d*x)^2), x], x]) /; FreeQ[{F, a, b, c, d, e, f}, x] && NeQ[d*e - c*f, 0] && LtQ[m, -1]

Rule 2551

Int[Log[u_]*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Log[u])/(b*(m + 1)), x] - Dist[1/
(b*(m + 1)), Int[SimplifyIntegrand[((a + b*x)^(m + 1)*D[u, x])/u, x], x], x] /; FreeQ[{a, b, m}, x] && Inverse
FunctionFreeQ[u, x] && NeQ[m, -1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, 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 \frac {e^{-2 x^2} \left (6 x-24 x^3-4 x^4+e^{2 x^2} \left (36 x^2+12 x^3+x^4\right )+e^{2 x^2} \left (-36-12 x-x^2\right ) \log \left (\frac {55}{4} e^{\frac {e^{-2 x^2} x}{6+x}}\right )\right )}{x^2 \left (36+12 x+x^2\right )} \, dx\\ &=\int \frac {e^{-2 x^2} \left (6 x-24 x^3-4 x^4+e^{2 x^2} \left (36 x^2+12 x^3+x^4\right )+e^{2 x^2} \left (-36-12 x-x^2\right ) \log \left (\frac {55}{4} e^{\frac {e^{-2 x^2} x}{6+x}}\right )\right )}{x^2 (6+x)^2} \, dx\\ &=\int \frac {\frac {e^{-2 x^2} x \left (6-24 x^2-4 x^3+e^{2 x^2} x (6+x)^2\right )}{(6+x)^2}-\log \left (\frac {55}{4} e^{\frac {e^{-2 x^2} x}{6+x}}\right )}{x^2} \, dx\\ &=\int \left (-\frac {2 e^{-2 x^2} \left (-3+12 x^2+2 x^3\right )}{x (6+x)^2}+\frac {x^2-\log \left (\frac {55}{4} e^{\frac {e^{-2 x^2} x}{6+x}}\right )}{x^2}\right ) \, dx\\ &=-\left (2 \int \frac {e^{-2 x^2} \left (-3+12 x^2+2 x^3\right )}{x (6+x)^2} \, dx\right )+\int \frac {x^2-\log \left (\frac {55}{4} e^{\frac {e^{-2 x^2} x}{6+x}}\right )}{x^2} \, dx\\ &=-\left (2 \int \left (2 e^{-2 x^2}-\frac {e^{-2 x^2}}{12 x}+\frac {e^{-2 x^2}}{2 (6+x)^2}-\frac {143 e^{-2 x^2}}{12 (6+x)}\right ) \, dx\right )+\int \left (1-\frac {\log \left (\frac {55}{4} e^{\frac {e^{-2 x^2} x}{6+x}}\right )}{x^2}\right ) \, dx\\ &=x+\frac {1}{6} \int \frac {e^{-2 x^2}}{x} \, dx-4 \int e^{-2 x^2} \, dx+\frac {143}{6} \int \frac {e^{-2 x^2}}{6+x} \, dx-\int \frac {e^{-2 x^2}}{(6+x)^2} \, dx-\int \frac {\log \left (\frac {55}{4} e^{\frac {e^{-2 x^2} x}{6+x}}\right )}{x^2} \, dx\\ &=x+\frac {e^{-2 x^2}}{6+x}-\sqrt {2 \pi } \text {erf}\left (\sqrt {2} x\right )+\frac {\text {Ei}\left (-2 x^2\right )}{12}+\frac {\log \left (\frac {55}{4} e^{\frac {e^{-2 x^2} x}{6+x}}\right )}{x}+4 \int e^{-2 x^2} \, dx+\frac {143}{6} \int \frac {e^{-2 x^2}}{6+x} \, dx-24 \int \frac {e^{-2 x^2}}{6+x} \, dx-\int \frac {e^{-2 x^2} \left (6-24 x^2-4 x^3\right )}{x (6+x)^2} \, dx\\ &=x+\frac {e^{-2 x^2}}{6+x}+\frac {\text {Ei}\left (-2 x^2\right )}{12}+\frac {\log \left (\frac {55}{4} e^{\frac {e^{-2 x^2} x}{6+x}}\right )}{x}+\frac {143}{6} \int \frac {e^{-2 x^2}}{6+x} \, dx-24 \int \frac {e^{-2 x^2}}{6+x} \, dx-\int \left (-4 e^{-2 x^2}+\frac {e^{-2 x^2}}{6 x}-\frac {e^{-2 x^2}}{(6+x)^2}+\frac {143 e^{-2 x^2}}{6 (6+x)}\right ) \, dx\\ &=x+\frac {e^{-2 x^2}}{6+x}+\frac {\text {Ei}\left (-2 x^2\right )}{12}+\frac {\log \left (\frac {55}{4} e^{\frac {e^{-2 x^2} x}{6+x}}\right )}{x}-\frac {1}{6} \int \frac {e^{-2 x^2}}{x} \, dx+4 \int e^{-2 x^2} \, dx-24 \int \frac {e^{-2 x^2}}{6+x} \, dx+\int \frac {e^{-2 x^2}}{(6+x)^2} \, dx\\ &=x+\sqrt {2 \pi } \text {erf}\left (\sqrt {2} x\right )+\frac {\log \left (\frac {55}{4} e^{\frac {e^{-2 x^2} x}{6+x}}\right )}{x}-4 \int e^{-2 x^2} \, dx\\ &=x+\frac {\log \left (\frac {55}{4} e^{\frac {e^{-2 x^2} x}{6+x}}\right )}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.19, size = 27, normalized size = 1.00 \begin {gather*} x+\frac {\log \left (\frac {55}{4} e^{\frac {e^{-2 x^2} x}{6+x}}\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(6*x - 24*x^3 - 4*x^4 + E^(2*x^2)*(36*x^2 + 12*x^3 + x^4) + E^(2*x^2)*(-36 - 12*x - x^2)*Log[(55*E^(
x/(E^(2*x^2)*(6 + x))))/4])/(E^(2*x^2)*(36*x^2 + 12*x^3 + x^4)),x]

[Out]

x + Log[(55*E^(x/(E^(2*x^2)*(6 + x))))/4]/x

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fricas [A]  time = 0.69, size = 40, normalized size = 1.48 \begin {gather*} \frac {{\left ({\left (x^{3} + 6 \, x^{2} + {\left (x + 6\right )} \log \left (\frac {55}{4}\right )\right )} e^{\left (2 \, x^{2}\right )} + x\right )} e^{\left (-2 \, x^{2}\right )}}{x^{2} + 6 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2-12*x-36)*exp(x^2)^2*log(55/4*exp(x/(x+6)/exp(x^2)^2))+(x^4+12*x^3+36*x^2)*exp(x^2)^2-4*x^4-24
*x^3+6*x)/(x^4+12*x^3+36*x^2)/exp(x^2)^2,x, algorithm="fricas")

[Out]

((x^3 + 6*x^2 + (x + 6)*log(55/4))*e^(2*x^2) + x)*e^(-2*x^2)/(x^2 + 6*x)

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giac [A]  time = 0.32, size = 44, normalized size = 1.63 \begin {gather*} \frac {x^{3} + 6 \, x^{2} + x e^{\left (-2 \, x^{2}\right )} + x \log \left (55\right ) - 2 \, x \log \relax (2) + 6 \, \log \left (55\right ) - 12 \, \log \relax (2)}{x^{2} + 6 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2-12*x-36)*exp(x^2)^2*log(55/4*exp(x/(x+6)/exp(x^2)^2))+(x^4+12*x^3+36*x^2)*exp(x^2)^2-4*x^4-24
*x^3+6*x)/(x^4+12*x^3+36*x^2)/exp(x^2)^2,x, algorithm="giac")

[Out]

(x^3 + 6*x^2 + x*e^(-2*x^2) + x*log(55) - 2*x*log(2) + 6*log(55) - 12*log(2))/(x^2 + 6*x)

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maple [A]  time = 0.37, size = 44, normalized size = 1.63

method	result	size
risch	\(\frac {\ln \left ({\mathrm e}^{\frac {x \,{\mathrm e}^{-2 x^{2}}}{x +6}}\right )}{x}-\frac {-2 x^{2}+4 \ln \relax (2)-2 \ln \relax (5)-2 \ln \left (11\right )}{2 x}\)	\(44\)
default	\(\frac {\left (x^{4} {\mathrm e}^{4 x^{2}}+x^{2} {\mathrm e}^{2 x^{2}}+\left (-432+12 \ln \left (\frac {55 \,{\mathrm e}^{\frac {x \,{\mathrm e}^{-2 x^{2}}}{x +6}}}{4}\right )-\frac {12 x \,{\mathrm e}^{-2 x^{2}}}{x +6}\right ) x \,{\mathrm e}^{4 x^{2}}+\left (-108+\ln \left (\frac {55 \,{\mathrm e}^{\frac {x \,{\mathrm e}^{-2 x^{2}}}{x +6}}}{4}\right )-\frac {x \,{\mathrm e}^{-2 x^{2}}}{x +6}\right ) x^{2} {\mathrm e}^{4 x^{2}}+36 \left (\ln \left (\frac {55 \,{\mathrm e}^{\frac {x \,{\mathrm e}^{-2 x^{2}}}{x +6}}}{4}\right )-\frac {x \,{\mathrm e}^{-2 x^{2}}}{x +6}\right ) {\mathrm e}^{4 x^{2}}+6 x \,{\mathrm e}^{2 x^{2}}\right ) {\mathrm e}^{-4 x^{2}}}{x \left (x +6\right )^{2}}\)	\(172\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^2-12*x-36)*exp(x^2)^2*ln(55/4*exp(x/(x+6)/exp(x^2)^2))+(x^4+12*x^3+36*x^2)*exp(x^2)^2-4*x^4-24*x^3+6*
x)/(x^4+12*x^3+36*x^2)/exp(x^2)^2,x,method=_RETURNVERBOSE)

[Out]

1/x*ln(exp(x/(x+6)*exp(-2*x^2)))-1/2*(-2*x^2+4*ln(2)-2*ln(5)-2*ln(11))/x

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maxima [B]  time = 0.69, size = 50, normalized size = 1.85 \begin {gather*} \frac {x^{3} + 6 \, x^{2} + x {\left (\log \left (11\right ) + \log \relax (5) - 2 \, \log \relax (2)\right )} + x e^{\left (-2 \, x^{2}\right )} + 6 \, \log \left (11\right ) + 6 \, \log \relax (5) - 12 \, \log \relax (2)}{x^{2} + 6 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2-12*x-36)*exp(x^2)^2*log(55/4*exp(x/(x+6)/exp(x^2)^2))+(x^4+12*x^3+36*x^2)*exp(x^2)^2-4*x^4-24
*x^3+6*x)/(x^4+12*x^3+36*x^2)/exp(x^2)^2,x, algorithm="maxima")

[Out]

(x^3 + 6*x^2 + x*(log(11) + log(5) - 2*log(2)) + x*e^(-2*x^2) + 6*log(11) + 6*log(5) - 12*log(2))/(x^2 + 6*x)

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mupad [B]  time = 2.25, size = 34, normalized size = 1.26 \begin {gather*} x+\frac {1}{6\,{\mathrm {e}}^{2\,x^2}+x\,{\mathrm {e}}^{2\,x^2}}-\frac {2\,\ln \relax (2)}{x}+\frac {\ln \left (55\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-2*x^2)*(24*x^3 - exp(2*x^2)*(36*x^2 + 12*x^3 + x^4) - 6*x + 4*x^4 + exp(2*x^2)*log((55*exp((x*exp(-
2*x^2))/(x + 6)))/4)*(12*x + x^2 + 36)))/(36*x^2 + 12*x^3 + x^4),x)

[Out]

x + 1/(6*exp(2*x^2) + x*exp(2*x^2)) - (2*log(2))/x + log(55)/x

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sympy [A]  time = 0.57, size = 22, normalized size = 0.81 \begin {gather*} x + \frac {e^{- 2 x^{2}}}{x + 6} + \frac {- 2 \log {\relax (2 )} + \log {\left (55 \right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**2-12*x-36)*exp(x**2)**2*ln(55/4*exp(x/(x+6)/exp(x**2)**2))+(x**4+12*x**3+36*x**2)*exp(x**2)**2
-4*x**4-24*x**3+6*x)/(x**4+12*x**3+36*x**2)/exp(x**2)**2,x)

[Out]

x + exp(-2*x**2)/(x + 6) + (-2*log(2) + log(55))/x

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