\(\int \frac {e^{-x^2} (-x+(5+x) \log (5+x)+(5+x-10 x^2-2 x^3) \log (5+x) \log (\frac {12 x}{\log (5+x)}))}{(100+20 x) \log (5+x)} \, dx\) [2619]

Optimal result

Integrand size = 61, antiderivative size = 22 \[ \int \frac {e^{-x^2} \left (-x+(5+x) \log (5+x)+\left (5+x-10 x^2-2 x^3\right ) \log (5+x) \log \left (\frac {12 x}{\log (5+x)}\right )\right )}{(100+20 x) \log (5+x)} \, dx=\frac {1}{20} e^{-x^2} x \log \left (\frac {12 x}{\log (5+x)}\right ) \]

[Out]

1/20*x/exp(x^2)*ln(12*x/ln(5+x))

Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {6874, 6820, 2236, 2258, 2243, 2635, 6823} \[ \int \frac {e^{-x^2} \left (-x+(5+x) \log (5+x)+\left (5+x-10 x^2-2 x^3\right ) \log (5+x) \log \left (\frac {12 x}{\log (5+x)}\right )\right )}{(100+20 x) \log (5+x)} \, dx=\frac {1}{20} e^{-x^2} x \log \left (\frac {12 x}{\log (x+5)}\right ) \]

[In]

Int[(-x + (5 + x)*Log[5 + x] + (5 + x - 10*x^2 - 2*x^3)*Log[5 + x]*Log[(12*x)/Log[5 + x]])/(E^x^2*(100 + 20*x)
*Log[5 + x]),x]

[Out]

(x*Log[(12*x)/Log[5 + x]])/(20*E^x^2)

Rule 2236

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 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 2635

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

Rule 6820

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

Rule 6823

Int[(u_)^(m_.)*((a_.)*(u_)^(n_) + (v_))^(p_.)*(w_), x_Symbol] :> Int[u^(m + n*p)*(a + v/u^n)^p*w, x] /; FreeQ[
{a, m, n}, x] && IntegerQ[p] &&  !GtQ[n, 0] &&  !FreeQ[v, x]

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {e^{-x^2} (-x+5 \log (5+x)+x \log (5+x))}{20 (5+x) \log (5+x)}-\frac {1}{20} e^{-x^2} \left (-1+2 x^2\right ) \log \left (\frac {12 x}{\log (5+x)}\right )\right ) \, dx \\ & = \frac {1}{20} \int \frac {e^{-x^2} (-x+5 \log (5+x)+x \log (5+x))}{(5+x) \log (5+x)} \, dx-\frac {1}{20} \int e^{-x^2} \left (-1+2 x^2\right ) \log \left (\frac {12 x}{\log (5+x)}\right ) \, dx \\ & = \frac {1}{20} e^{-x^2} x \log \left (\frac {12 x}{\log (5+x)}\right )-\frac {1}{20} \int e^{-x^2} x \left (\frac {1}{x}-\frac {1}{(5+x) \log (5+x)}\right ) \, dx+\frac {1}{20} \int e^{-x^2} \left (1-\frac {x}{(5+x) \log (5+x)}\right ) \, dx \\ & = \frac {1}{20} e^{-x^2} x \log \left (\frac {12 x}{\log (5+x)}\right )-\frac {1}{20} \int e^{-x^2} \left (1-\frac {x}{(5+x) \log (5+x)}\right ) \, dx+\frac {1}{20} \int \left (e^{-x^2}-\frac {e^{-x^2} x}{(5+x) \log (5+x)}\right ) \, dx \\ & = \frac {1}{20} e^{-x^2} x \log \left (\frac {12 x}{\log (5+x)}\right )+\frac {1}{20} \int e^{-x^2} \, dx-\frac {1}{20} \int \left (e^{-x^2}-\frac {e^{-x^2} x}{(5+x) \log (5+x)}\right ) \, dx-\frac {1}{20} \int \frac {e^{-x^2} x}{(5+x) \log (5+x)} \, dx \\ & = \frac {1}{40} \sqrt {\pi } \text {erf}(x)+\frac {1}{20} e^{-x^2} x \log \left (\frac {12 x}{\log (5+x)}\right )-\frac {1}{20} \int e^{-x^2} \, dx-\frac {1}{20} \int \left (\frac {e^{-x^2}}{\log (5+x)}-\frac {5 e^{-x^2}}{(5+x) \log (5+x)}\right ) \, dx+\frac {1}{20} \int \frac {e^{-x^2} x}{(5+x) \log (5+x)} \, dx \\ & = \frac {1}{20} e^{-x^2} x \log \left (\frac {12 x}{\log (5+x)}\right )+\frac {1}{20} \int \left (\frac {e^{-x^2}}{\log (5+x)}-\frac {5 e^{-x^2}}{(5+x) \log (5+x)}\right ) \, dx-\frac {1}{20} \int \frac {e^{-x^2}}{\log (5+x)} \, dx+\frac {1}{4} \int \frac {e^{-x^2}}{(5+x) \log (5+x)} \, dx \\ & = \frac {1}{20} e^{-x^2} x \log \left (\frac {12 x}{\log (5+x)}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-x^2} \left (-x+(5+x) \log (5+x)+\left (5+x-10 x^2-2 x^3\right ) \log (5+x) \log \left (\frac {12 x}{\log (5+x)}\right )\right )}{(100+20 x) \log (5+x)} \, dx=\frac {1}{20} e^{-x^2} x \log \left (\frac {12 x}{\log (5+x)}\right ) \]

[In]

Integrate[(-x + (5 + x)*Log[5 + x] + (5 + x - 10*x^2 - 2*x^3)*Log[5 + x]*Log[(12*x)/Log[5 + x]])/(E^x^2*(100 +
20*x)*Log[5 + x]),x]

[Out]

(x*Log[(12*x)/Log[5 + x]])/(20*E^x^2)

Maple [A] (verified)

Time = 1.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91

method	result	size
parallelrisch	\(\frac {x \,{\mathrm e}^{-x^{2}} \ln \left (\frac {12 x}{\ln \left (5+x \right )}\right )}{20}\)	\(20\)
risch	\(-\frac {x \,{\mathrm e}^{-x^{2}} \ln \left (\ln \left (5+x \right )\right )}{20}+\frac {x \left (i \pi \,\operatorname {csgn}\left (\frac {i}{\ln \left (5+x \right )}\right ) \operatorname {csgn}\left (\frac {i x}{\ln \left (5+x \right )}\right )^{2}-i \pi \,\operatorname {csgn}\left (\frac {i}{\ln \left (5+x \right )}\right ) \operatorname {csgn}\left (\frac {i x}{\ln \left (5+x \right )}\right ) \operatorname {csgn}\left (i x \right )+i \pi \operatorname {csgn}\left (\frac {i x}{\ln \left (5+x \right )}\right )^{2} \operatorname {csgn}\left (i x \right )-i \pi \operatorname {csgn}\left (\frac {i x}{\ln \left (5+x \right )}\right )^{3}+2 \ln \left (3\right )+4 \ln \left (2\right )+2 \ln \left (x \right )\right ) {\mathrm e}^{-x^{2}}}{40}\)	\(134\)

[In]

int(((-2*x^3-10*x^2+x+5)*ln(5+x)*ln(12*x/ln(5+x))+(5+x)*ln(5+x)-x)/(20*x+100)/exp(x^2)/ln(5+x),x,method=_RETUR
NVERBOSE)

[Out]

1/20*x/exp(x^2)*ln(12*x/ln(5+x))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {e^{-x^2} \left (-x+(5+x) \log (5+x)+\left (5+x-10 x^2-2 x^3\right ) \log (5+x) \log \left (\frac {12 x}{\log (5+x)}\right )\right )}{(100+20 x) \log (5+x)} \, dx=\frac {1}{20} \, x e^{\left (-x^{2}\right )} \log \left (\frac {12 \, x}{\log \left (x + 5\right )}\right ) \]

[In]

integrate(((-2*x^3-10*x^2+x+5)*log(5+x)*log(12*x/log(5+x))+(5+x)*log(5+x)-x)/(20*x+100)/exp(x^2)/log(5+x),x, a
lgorithm="fricas")

[Out]

1/20*x*e^(-x^2)*log(12*x/log(x + 5))

Sympy [A] (verification not implemented)

Time = 5.15 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {e^{-x^2} \left (-x+(5+x) \log (5+x)+\left (5+x-10 x^2-2 x^3\right ) \log (5+x) \log \left (\frac {12 x}{\log (5+x)}\right )\right )}{(100+20 x) \log (5+x)} \, dx=\frac {x e^{- x^{2}} \log {\left (\frac {12 x}{\log {\left (x + 5 \right )}} \right )}}{20} \]

[In]

integrate(((-2*x**3-10*x**2+x+5)*ln(5+x)*ln(12*x/ln(5+x))+(5+x)*ln(5+x)-x)/(20*x+100)/exp(x**2)/ln(5+x),x)

[Out]

x*exp(-x**2)*log(12*x/log(x + 5))/20

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.68 \[ \int \frac {e^{-x^2} \left (-x+(5+x) \log (5+x)+\left (5+x-10 x^2-2 x^3\right ) \log (5+x) \log \left (\frac {12 x}{\log (5+x)}\right )\right )}{(100+20 x) \log (5+x)} \, dx=-\frac {1}{20} \, x e^{\left (-x^{2}\right )} \log \left (\log \left (x + 5\right )\right ) + \frac {1}{20} \, {\left (x {\left (\log \left (3\right ) + 2 \, \log \left (2\right )\right )} + x \log \left (x\right )\right )} e^{\left (-x^{2}\right )} \]

[In]

integrate(((-2*x^3-10*x^2+x+5)*log(5+x)*log(12*x/log(5+x))+(5+x)*log(5+x)-x)/(20*x+100)/exp(x^2)/log(5+x),x, a
lgorithm="maxima")

[Out]

-1/20*x*e^(-x^2)*log(log(x + 5)) + 1/20*(x*(log(3) + 2*log(2)) + x*log(x))*e^(-x^2)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27 \[ \int \frac {e^{-x^2} \left (-x+(5+x) \log (5+x)+\left (5+x-10 x^2-2 x^3\right ) \log (5+x) \log \left (\frac {12 x}{\log (5+x)}\right )\right )}{(100+20 x) \log (5+x)} \, dx=\frac {1}{20} \, x e^{\left (-x^{2}\right )} \log \left (12 \, x\right ) - \frac {1}{20} \, x e^{\left (-x^{2}\right )} \log \left (\log \left (x + 5\right )\right ) \]

[In]

integrate(((-2*x^3-10*x^2+x+5)*log(5+x)*log(12*x/log(5+x))+(5+x)*log(5+x)-x)/(20*x+100)/exp(x^2)/log(5+x),x, a
lgorithm="giac")

[Out]

1/20*x*e^(-x^2)*log(12*x) - 1/20*x*e^(-x^2)*log(log(x + 5))

Mupad [B] (verification not implemented)

Time = 8.63 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {e^{-x^2} \left (-x+(5+x) \log (5+x)+\left (5+x-10 x^2-2 x^3\right ) \log (5+x) \log \left (\frac {12 x}{\log (5+x)}\right )\right )}{(100+20 x) \log (5+x)} \, dx=\frac {x\,{\mathrm {e}}^{-x^2}\,\ln \left (\frac {12\,x}{\ln \left (x+5\right )}\right )}{20} \]

[In]

int((exp(-x^2)*(log(x + 5)*(x + 5) - x + log(x + 5)*log((12*x)/log(x + 5))*(x - 10*x^2 - 2*x^3 + 5)))/(log(x +
5)*(20*x + 100)),x)

[Out]

(x*exp(-x^2)*log((12*x)/log(x + 5)))/20