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

Optimal. Leaf size=22 \[ \frac {1}{20} e^{-x^2} x \log \left (\frac {12 x}{\log (5+x)}\right ) \]

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Rubi [A]  time = 1.40, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 7, integrand size = 61, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {6742, 6688, 2205, 2226, 2212, 2555, 6691} \begin {gather*} \frac {1}{20} e^{-x^2} x \log \left (\frac {12 x}{\log (x+5)}\right ) \end {gather*}

Antiderivative was successfully verified.

[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 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 2212

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 2226

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 2555

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 6688

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

Rule 6691

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 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 22, normalized size = 1.00 \begin {gather*} \frac {1}{20} e^{-x^2} x \log \left (\frac {12 x}{\log (5+x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.60, size = 19, normalized size = 0.86 \begin {gather*} \frac {1}{20} \, x e^{\left (-x^{2}\right )} \log \left (\frac {12 \, x}{\log \left (x + 5\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [A]  time = 0.34, size = 28, normalized size = 1.27 \begin {gather*} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maple [C]  time = 0.49, size = 134, normalized size = 6.09

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

Verification of antiderivative is not currently implemented for this CAS.

[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(ln(5+x))+1/40*x*(-I*Pi*csgn(I*x)*csgn(I/ln(5+x))*csgn(I*x/ln(5+x))+I*Pi*csgn(I*x)*csgn(I*
x/ln(5+x))^2+I*Pi*csgn(I/ln(5+x))*csgn(I*x/ln(5+x))^2-I*Pi*csgn(I*x/ln(5+x))^3+4*ln(2)+2*ln(3)+2*ln(x))*exp(-x
^2)

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maxima [A]  time = 0.93, size = 37, normalized size = 1.68 \begin {gather*} -\frac {1}{20} \, x e^{\left (-x^{2}\right )} \log \left (\log \left (x + 5\right )\right ) + \frac {1}{20} \, {\left (x {\left (\log \relax (3) + 2 \, \log \relax (2)\right )} + x \log \relax (x)\right )} e^{\left (-x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 1.80, size = 19, normalized size = 0.86 \begin {gather*} \frac {x\,{\mathrm {e}}^{-x^2}\,\ln \left (\frac {12\,x}{\ln \left (x+5\right )}\right )}{20} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 10.14, size = 17, normalized size = 0.77 \begin {gather*} \frac {x e^{- x^{2}} \log {\left (\frac {12 x}{\log {\left (x + 5 \right )}} \right )}}{20} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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