\(\int \frac {e^{x^2} (x+2 x^3) \log (x)+e^{x^2} (-6-12 x^2+(1+2 x^2) \log (5)) \log ^2(x)+e^{\log ^2(\frac {x+(-6+\log (5)) \log (x)}{\log (x)})} (-10+10 \log (x)) \log (\frac {x+(-6+\log (5)) \log (x)}{\log (x)})}{5 x \log (x)+(-30+5 \log (5)) \log ^2(x)} \, dx\) [5577]

Optimal result

Integrand size = 104, antiderivative size = 26 \[ \int \frac {e^{x^2} \left (x+2 x^3\right ) \log (x)+e^{x^2} \left (-6-12 x^2+\left (1+2 x^2\right ) \log (5)\right ) \log ^2(x)+e^{\log ^2\left (\frac {x+(-6+\log (5)) \log (x)}{\log (x)}\right )} (-10+10 \log (x)) \log \left (\frac {x+(-6+\log (5)) \log (x)}{\log (x)}\right )}{5 x \log (x)+(-30+5 \log (5)) \log ^2(x)} \, dx=e^{\log ^2\left (-6+\log (5)+\frac {x}{\log (x)}\right )}+\frac {e^{x^2} x}{5} \]

[Out]

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

Rubi [A] (verified)

Time = 0.99 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {6873, 12, 6874, 2258, 2235, 2243, 6820, 6838} \[ \int \frac {e^{x^2} \left (x+2 x^3\right ) \log (x)+e^{x^2} \left (-6-12 x^2+\left (1+2 x^2\right ) \log (5)\right ) \log ^2(x)+e^{\log ^2\left (\frac {x+(-6+\log (5)) \log (x)}{\log (x)}\right )} (-10+10 \log (x)) \log \left (\frac {x+(-6+\log (5)) \log (x)}{\log (x)}\right )}{5 x \log (x)+(-30+5 \log (5)) \log ^2(x)} \, dx=\frac {e^{x^2} x}{5}+e^{\log ^2\left (\frac {x}{\log (x)}-6+\log (5)\right )} \]

[In]

Int[(E^x^2*(x + 2*x^3)*Log[x] + E^x^2*(-6 - 12*x^2 + (1 + 2*x^2)*Log[5])*Log[x]^2 + E^Log[(x + (-6 + Log[5])*L
og[x])/Log[x]]^2*(-10 + 10*Log[x])*Log[(x + (-6 + Log[5])*Log[x])/Log[x]])/(5*x*Log[x] + (-30 + 5*Log[5])*Log[
x]^2),x]

[Out]

E^Log[-6 + Log[5] + x/Log[x]]^2 + (E^x^2*x)/5

Rule 12

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

Rule 2235

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

Rule 2243

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^(m
- n + 1)*(F^(a + b*(c + d*x)^n)/(b*d*n*Log[F])), x] - Dist[(m - n + 1)/(b*n*Log[F]), Int[(c + d*x)^(m - n)*F^(
a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2*((m + 1)/n)] && LtQ[0, (m + 1)/n, 5] &&
IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n, 0])

Rule 2258

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*(u_), x_Symbol] :> Int[ExpandLinearProduct[F^(a + b*(c + d*
x)^n), u, c, d, x], x] /; FreeQ[{F, a, b, c, d, n}, x] && PolynomialQ[u, x]

Rule 6820

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

Rule 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rule 6873

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

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{x^2} \left (x+2 x^3\right ) \log (x)+e^{x^2} \left (-6-12 x^2+\left (1+2 x^2\right ) \log (5)\right ) \log ^2(x)+e^{\log ^2\left (\frac {x+(-6+\log (5)) \log (x)}{\log (x)}\right )} (-10+10 \log (x)) \log \left (\frac {x+(-6+\log (5)) \log (x)}{\log (x)}\right )}{5 \log (x) \left (x-6 \left (1-\frac {\log (5)}{6}\right ) \log (x)\right )} \, dx \\ & = \frac {1}{5} \int \frac {e^{x^2} \left (x+2 x^3\right ) \log (x)+e^{x^2} \left (-6-12 x^2+\left (1+2 x^2\right ) \log (5)\right ) \log ^2(x)+e^{\log ^2\left (\frac {x+(-6+\log (5)) \log (x)}{\log (x)}\right )} (-10+10 \log (x)) \log \left (\frac {x+(-6+\log (5)) \log (x)}{\log (x)}\right )}{\log (x) \left (x-6 \left (1-\frac {\log (5)}{6}\right ) \log (x)\right )} \, dx \\ & = \frac {1}{5} \int \left (e^{x^2} \left (1+2 x^2\right )+\frac {10 e^{\log ^2\left (-6 \left (1-\frac {\log (5)}{6}\right )+\frac {x}{\log (x)}\right )} (-1+\log (x)) \log \left (-6 \left (1-\frac {\log (5)}{6}\right )+\frac {x}{\log (x)}\right )}{\log (x) \left (x-6 \left (1-\frac {\log (5)}{6}\right ) \log (x)\right )}\right ) \, dx \\ & = \frac {1}{5} \int e^{x^2} \left (1+2 x^2\right ) \, dx+2 \int \frac {e^{\log ^2\left (-6 \left (1-\frac {\log (5)}{6}\right )+\frac {x}{\log (x)}\right )} (-1+\log (x)) \log \left (-6 \left (1-\frac {\log (5)}{6}\right )+\frac {x}{\log (x)}\right )}{\log (x) \left (x-6 \left (1-\frac {\log (5)}{6}\right ) \log (x)\right )} \, dx \\ & = \frac {1}{5} \int \left (e^{x^2}+2 e^{x^2} x^2\right ) \, dx+2 \int \frac {e^{\log ^2\left (-6 \left (1-\frac {\log (5)}{6}\right )+\frac {x}{\log (x)}\right )} (-1+\log (x)) \log \left (-6 \left (1-\frac {\log (5)}{6}\right )+\frac {x}{\log (x)}\right )}{\log (x) (x+(-6+\log (5)) \log (x))} \, dx \\ & = e^{\log ^2\left (-6+\log (5)+\frac {x}{\log (x)}\right )}+\frac {1}{5} \int e^{x^2} \, dx+\frac {2}{5} \int e^{x^2} x^2 \, dx \\ & = e^{\log ^2\left (-6+\log (5)+\frac {x}{\log (x)}\right )}+\frac {e^{x^2} x}{5}+\frac {1}{10} \sqrt {\pi } \text {erfi}(x)-\frac {1}{5} \int e^{x^2} \, dx \\ & = e^{\log ^2\left (-6+\log (5)+\frac {x}{\log (x)}\right )}+\frac {e^{x^2} x}{5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {e^{x^2} \left (x+2 x^3\right ) \log (x)+e^{x^2} \left (-6-12 x^2+\left (1+2 x^2\right ) \log (5)\right ) \log ^2(x)+e^{\log ^2\left (\frac {x+(-6+\log (5)) \log (x)}{\log (x)}\right )} (-10+10 \log (x)) \log \left (\frac {x+(-6+\log (5)) \log (x)}{\log (x)}\right )}{5 x \log (x)+(-30+5 \log (5)) \log ^2(x)} \, dx=\frac {1}{5} \left (5 e^{\log ^2\left (-6+\log (5)+\frac {x}{\log (x)}\right )}+e^{x^2} x\right ) \]

[In]

Integrate[(E^x^2*(x + 2*x^3)*Log[x] + E^x^2*(-6 - 12*x^2 + (1 + 2*x^2)*Log[5])*Log[x]^2 + E^Log[(x + (-6 + Log
[5])*Log[x])/Log[x]]^2*(-10 + 10*Log[x])*Log[(x + (-6 + Log[5])*Log[x])/Log[x]])/(5*x*Log[x] + (-30 + 5*Log[5]
)*Log[x]^2),x]

[Out]

(5*E^Log[-6 + Log[5] + x/Log[x]]^2 + E^x^2*x)/5

Maple [A] (verified)

Time = 101.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04

method	result	size
parallelrisch	\(\frac {{\mathrm e}^{x^{2}} x}{5}+{\mathrm e}^{\ln \left (\frac {\left (\ln \left (5\right )-6\right ) \ln \left (x \right )+x}{\ln \left (x \right )}\right )^{2}}\)	\(27\)
risch	\(\frac {{\mathrm e}^{x^{2}} x}{5}+{\mathrm e}^{\frac {\left (i \operatorname {csgn}\left (\frac {i \left (\ln \left (5\right ) \ln \left (x \right )-6 \ln \left (x \right )+x \right )}{\ln \left (x \right )}\right )^{3} \pi -i \operatorname {csgn}\left (\frac {i \left (\ln \left (5\right ) \ln \left (x \right )-6 \ln \left (x \right )+x \right )}{\ln \left (x \right )}\right )^{2} \operatorname {csgn}\left (\frac {i}{\ln \left (x \right )}\right ) \pi -i \operatorname {csgn}\left (\frac {i \left (\ln \left (5\right ) \ln \left (x \right )-6 \ln \left (x \right )+x \right )}{\ln \left (x \right )}\right )^{2} \operatorname {csgn}\left (i \left (\ln \left (5\right ) \ln \left (x \right )-6 \ln \left (x \right )+x \right )\right ) \pi +i \operatorname {csgn}\left (\frac {i \left (\ln \left (5\right ) \ln \left (x \right )-6 \ln \left (x \right )+x \right )}{\ln \left (x \right )}\right ) \operatorname {csgn}\left (\frac {i}{\ln \left (x \right )}\right ) \operatorname {csgn}\left (i \left (\ln \left (5\right ) \ln \left (x \right )-6 \ln \left (x \right )+x \right )\right ) \pi -2 \ln \left (\ln \left (5\right ) \ln \left (x \right )-6 \ln \left (x \right )+x \right )+2 \ln \left (\ln \left (x \right )\right )\right )^{2}}{4}}\)	\(178\)

[In]

int(((10*ln(x)-10)*ln(((ln(5)-6)*ln(x)+x)/ln(x))*exp(ln(((ln(5)-6)*ln(x)+x)/ln(x))^2)+((2*x^2+1)*ln(5)-12*x^2-
6)*exp(x^2)*ln(x)^2+(2*x^3+x)*exp(x^2)*ln(x))/((5*ln(5)-30)*ln(x)^2+5*x*ln(x)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^{x^2} \left (x+2 x^3\right ) \log (x)+e^{x^2} \left (-6-12 x^2+\left (1+2 x^2\right ) \log (5)\right ) \log ^2(x)+e^{\log ^2\left (\frac {x+(-6+\log (5)) \log (x)}{\log (x)}\right )} (-10+10 \log (x)) \log \left (\frac {x+(-6+\log (5)) \log (x)}{\log (x)}\right )}{5 x \log (x)+(-30+5 \log (5)) \log ^2(x)} \, dx=\frac {1}{5} \, x e^{\left (x^{2}\right )} + e^{\left (\log \left (\frac {{\left (\log \left (5\right ) - 6\right )} \log \left (x\right ) + x}{\log \left (x\right )}\right )^{2}\right )} \]

[In]

integrate(((10*log(x)-10)*log(((log(5)-6)*log(x)+x)/log(x))*exp(log(((log(5)-6)*log(x)+x)/log(x))^2)+((2*x^2+1
)*log(5)-12*x^2-6)*exp(x^2)*log(x)^2+(2*x^3+x)*exp(x^2)*log(x))/((5*log(5)-30)*log(x)^2+5*x*log(x)),x, algorit
hm="fricas")

[Out]

1/5*x*e^(x^2) + e^(log(((log(5) - 6)*log(x) + x)/log(x))^2)

Sympy [A] (verification not implemented)

Time = 29.03 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^{x^2} \left (x+2 x^3\right ) \log (x)+e^{x^2} \left (-6-12 x^2+\left (1+2 x^2\right ) \log (5)\right ) \log ^2(x)+e^{\log ^2\left (\frac {x+(-6+\log (5)) \log (x)}{\log (x)}\right )} (-10+10 \log (x)) \log \left (\frac {x+(-6+\log (5)) \log (x)}{\log (x)}\right )}{5 x \log (x)+(-30+5 \log (5)) \log ^2(x)} \, dx=\frac {x e^{x^{2}}}{5} + e^{\log {\left (\frac {x + \left (-6 + \log {\left (5 \right )}\right ) \log {\left (x \right )}}{\log {\left (x \right )}} \right )}^{2}} \]

[In]

integrate(((10*ln(x)-10)*ln(((ln(5)-6)*ln(x)+x)/ln(x))*exp(ln(((ln(5)-6)*ln(x)+x)/ln(x))**2)+((2*x**2+1)*ln(5)
-12*x**2-6)*exp(x**2)*ln(x)**2+(2*x**3+x)*exp(x**2)*ln(x))/((5*ln(5)-30)*ln(x)**2+5*x*ln(x)),x)

[Out]

x*exp(x**2)/5 + exp(log((x + (-6 + log(5))*log(x))/log(x))**2)

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62 \[ \int \frac {e^{x^2} \left (x+2 x^3\right ) \log (x)+e^{x^2} \left (-6-12 x^2+\left (1+2 x^2\right ) \log (5)\right ) \log ^2(x)+e^{\log ^2\left (\frac {x+(-6+\log (5)) \log (x)}{\log (x)}\right )} (-10+10 \log (x)) \log \left (\frac {x+(-6+\log (5)) \log (x)}{\log (x)}\right )}{5 x \log (x)+(-30+5 \log (5)) \log ^2(x)} \, dx=\frac {1}{5} \, x e^{\left (x^{2}\right )} + e^{\left (\log \left ({\left (\log \left (5\right ) - 6\right )} \log \left (x\right ) + x\right )^{2} - 2 \, \log \left ({\left (\log \left (5\right ) - 6\right )} \log \left (x\right ) + x\right ) \log \left (\log \left (x\right )\right ) + \log \left (\log \left (x\right )\right )^{2}\right )} \]

[In]

integrate(((10*log(x)-10)*log(((log(5)-6)*log(x)+x)/log(x))*exp(log(((log(5)-6)*log(x)+x)/log(x))^2)+((2*x^2+1
)*log(5)-12*x^2-6)*exp(x^2)*log(x)^2+(2*x^3+x)*exp(x^2)*log(x))/((5*log(5)-30)*log(x)^2+5*x*log(x)),x, algorit
hm="maxima")

[Out]

1/5*x*e^(x^2) + e^(log((log(5) - 6)*log(x) + x)^2 - 2*log((log(5) - 6)*log(x) + x)*log(log(x)) + log(log(x))^2
)

Giac [F]

\[ \int \frac {e^{x^2} \left (x+2 x^3\right ) \log (x)+e^{x^2} \left (-6-12 x^2+\left (1+2 x^2\right ) \log (5)\right ) \log ^2(x)+e^{\log ^2\left (\frac {x+(-6+\log (5)) \log (x)}{\log (x)}\right )} (-10+10 \log (x)) \log \left (\frac {x+(-6+\log (5)) \log (x)}{\log (x)}\right )}{5 x \log (x)+(-30+5 \log (5)) \log ^2(x)} \, dx=\int { -\frac {{\left (12 \, x^{2} - {\left (2 \, x^{2} + 1\right )} \log \left (5\right ) + 6\right )} e^{\left (x^{2}\right )} \log \left (x\right )^{2} - {\left (2 \, x^{3} + x\right )} e^{\left (x^{2}\right )} \log \left (x\right ) - 10 \, {\left (\log \left (x\right ) - 1\right )} e^{\left (\log \left (\frac {{\left (\log \left (5\right ) - 6\right )} \log \left (x\right ) + x}{\log \left (x\right )}\right )^{2}\right )} \log \left (\frac {{\left (\log \left (5\right ) - 6\right )} \log \left (x\right ) + x}{\log \left (x\right )}\right )}{5 \, {\left ({\left (\log \left (5\right ) - 6\right )} \log \left (x\right )^{2} + x \log \left (x\right )\right )}} \,d x } \]

[In]

integrate(((10*log(x)-10)*log(((log(5)-6)*log(x)+x)/log(x))*exp(log(((log(5)-6)*log(x)+x)/log(x))^2)+((2*x^2+1
)*log(5)-12*x^2-6)*exp(x^2)*log(x)^2+(2*x^3+x)*exp(x^2)*log(x))/((5*log(5)-30)*log(x)^2+5*x*log(x)),x, algorit
hm="giac")

[Out]

undef

Mupad [B] (verification not implemented)

Time = 11.97 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {e^{x^2} \left (x+2 x^3\right ) \log (x)+e^{x^2} \left (-6-12 x^2+\left (1+2 x^2\right ) \log (5)\right ) \log ^2(x)+e^{\log ^2\left (\frac {x+(-6+\log (5)) \log (x)}{\log (x)}\right )} (-10+10 \log (x)) \log \left (\frac {x+(-6+\log (5)) \log (x)}{\log (x)}\right )}{5 x \log (x)+(-30+5 \log (5)) \log ^2(x)} \, dx={\mathrm {e}}^{{\ln \left (\frac {x-6\,\ln \left (x\right )+\ln \left (5\right )\,\ln \left (x\right )}{\ln \left (x\right )}\right )}^2}+\frac {x\,{\mathrm {e}}^{x^2}}{5} \]

[In]

int((exp(x^2)*log(x)*(x + 2*x^3) + exp(log((x + log(x)*(log(5) - 6))/log(x))^2)*log((x + log(x)*(log(5) - 6))/
log(x))*(10*log(x) - 10) - exp(x^2)*log(x)^2*(12*x^2 - log(5)*(2*x^2 + 1) + 6))/(5*x*log(x) + log(x)^2*(5*log(
5) - 30)),x)

[Out]

exp(log((x - 6*log(x) + log(5)*log(x))/log(x))^2) + (x*exp(x^2))/5