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

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

[Out]

exp(x^2)*ln(x)-ln(x)/(x+ln(x)^2)+ln(x+ln(x)^2)

________________________________________________________________________________________

Rubi [A]  time = 0.21, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {2210, 2209, 2554, 12, 2547, 6742, 2538} \[ e^{x^2} \log (x)-\frac {\log (x)}{x+\log ^2(x)}+\log \left (x+\log ^2(x)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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 2538

Int[Log[(c_.)*(x_)^(n_.)]^(r_.)/((x_)*(Log[(c_.)*(x_)^(n_.)]^(q_)*(b_.) + (a_.)*(x_)^(m_.))), x_Symbol] :> Sim
p[Log[a*x^m + b*Log[c*x^n]^q]/(b*n*q), x] - Dist[(a*m)/(b*n*q), Int[x^(m - 1)/(a*x^m + b*Log[c*x^n]^q), x], x]
/; FreeQ[{a, b, c, m, n, q, r}, x] && EqQ[r, q - 1]

Rule 2547

Int[(Log[(c_.)*(x_)^(n_.)]*(e_.) + (d_))/(Log[(c_.)*(x_)^(n_.)]^(q_)*(b_.) + (a_.)*(x_))^2, x_Symbol] :> -Simp
[(e*Log[c*x^n])/(a*(a*x + b*Log[c*x^n]^q)), x] + Dist[(d + e*n)/a, Int[1/(x*(a*x + b*Log[c*x^n]^q)), x], x] /;
FreeQ[{a, b, c, d, e, n, q}, x] && EqQ[d + e*n*q, 0]

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 6742

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

Rubi steps

\begin {align*} \int \left (\frac {e^{x^2}}{x}+2 e^{x^2} x \log (x)+\frac {-2+\log (x)}{\left (x+\log ^2(x)\right )^2}+\frac {1+\frac {1}{x}+\frac {2 \log (x)}{x}}{x+\log ^2(x)}\right ) \, dx &=2 \int e^{x^2} x \log (x) \, dx+\int \frac {e^{x^2}}{x} \, dx+\int \frac {-2+\log (x)}{\left (x+\log ^2(x)\right )^2} \, dx+\int \frac {1+\frac {1}{x}+\frac {2 \log (x)}{x}}{x+\log ^2(x)} \, dx\\ &=\frac {\text {Ei}\left (x^2\right )}{2}+e^{x^2} \log (x)-\frac {\log (x)}{x+\log ^2(x)}-2 \int \frac {e^{x^2}}{2 x} \, dx-\int \frac {1}{x \left (x+\log ^2(x)\right )} \, dx+\int \left (\frac {1}{x+\log ^2(x)}+\frac {1}{x \left (x+\log ^2(x)\right )}+\frac {2 \log (x)}{x \left (x+\log ^2(x)\right )}\right ) \, dx\\ &=\frac {\text {Ei}\left (x^2\right )}{2}+e^{x^2} \log (x)-\frac {\log (x)}{x+\log ^2(x)}+2 \int \frac {\log (x)}{x \left (x+\log ^2(x)\right )} \, dx-\int \frac {e^{x^2}}{x} \, dx+\int \frac {1}{x+\log ^2(x)} \, dx\\ &=e^{x^2} \log (x)-\frac {\log (x)}{x+\log ^2(x)}+\log \left (x+\log ^2(x)\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.17, size = 28, normalized size = 1.00 \[ e^{x^2} \log (x)-\frac {\log (x)}{x+\log ^2(x)}+\log \left (x+\log ^2(x)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.43, size = 44, normalized size = 1.57 \[ \frac {e^{\left (x^{2}\right )} \log \relax (x)^{3} + {\left (\log \relax (x)^{2} + x\right )} \log \left (\log \relax (x)^{2} + x\right ) + {\left (x e^{\left (x^{2}\right )} - 1\right )} \log \relax (x)}{\log \relax (x)^{2} + x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x^2)/x+2*exp(x^2)*x*log(x)+(-2+log(x))/(x+log(x)^2)^2+(1+1/x+2*log(x)/x)/(x+log(x)^2),x, algorit
hm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 1.11, size = 27, normalized size = 0.96 \[ e^{\left (x^{2}\right )} \log \relax (x) - \frac {3 \, \log \relax (x)}{\log \relax (x)^{2} + x} + \log \left (\log \relax (x)^{2} + x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x^2)/x+2*exp(x^2)*x*log(x)+(-2+log(x))/(x+log(x)^2)^2+(1+1/x+2*log(x)/x)/(x+log(x)^2),x, algorit
hm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.04, size = 28, normalized size = 1.00 \[ {\mathrm e}^{x^{2}} \ln \relax (x )-\frac {\ln \relax (x )}{\ln \relax (x )^{2}+x}+\ln \left (\ln \relax (x )^{2}+x \right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(x^2)*ln(x)-ln(x)/(x+ln(x)^2)+ln(x+ln(x)^2)

________________________________________________________________________________________

maxima [A]  time = 0.56, size = 27, normalized size = 0.96 \[ e^{\left (x^{2}\right )} \log \relax (x) - \frac {\log \relax (x)}{\log \relax (x)^{2} + x} + \log \left (\log \relax (x)^{2} + x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x^2)/x+2*exp(x^2)*x*log(x)+(-2+log(x))/(x+log(x)^2)^2+(1+1/x+2*log(x)/x)/(x+log(x)^2),x, algorit
hm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.34, size = 27, normalized size = 0.96 \[ \ln \left ({\ln \relax (x)}^2+x\right )+{\mathrm {e}}^{x^2}\,\ln \relax (x)-\frac {\ln \relax (x)}{{\ln \relax (x)}^2+x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x + log(x)^2) + exp(x^2)*log(x) - log(x)/(x + log(x)^2)

________________________________________________________________________________________

sympy [A]  time = 0.47, size = 26, normalized size = 0.93 \[ e^{x^{2}} \log {\relax (x )} + \log {\left (x + \log {\relax (x )}^{2} \right )} - \frac {\log {\relax (x )}}{x + \log {\relax (x )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________