3.72.8 \(\int \frac {e^{-4 e^{3 e^9}} (10 x+6 x^2+(50+50 x+16 x^2) \log (x)+(20 x+8 x^2) \log ^2(x))}{x} \, dx\)

Optimal. Leaf size=24 \[ e^{-4 e^{3 e^9}} (x+(5+2 x) \log (x))^2 \]

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Rubi [B]  time = 0.11, antiderivative size = 142, normalized size of antiderivative = 5.92, number of steps used = 14, number of rules used = 9, integrand size = 51, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {12, 14, 2357, 2295, 2301, 2304, 2330, 2296, 2305} \begin {gather*} -2 e^{-4 e^{3 e^9}} x^2+4 e^{-4 e^{3 e^9}} x^2 \log ^2(x)+4 e^{-4 e^{3 e^9}} x^2 \log (x)-10 e^{-4 e^{3 e^9}} x+\frac {1}{3} e^{-4 e^{3 e^9}} (3 x+5)^2+20 e^{-4 e^{3 e^9}} x \log ^2(x)+25 e^{-4 e^{3 e^9}} \log ^2(x)+10 e^{-4 e^{3 e^9}} x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(10*x + 6*x^2 + (50 + 50*x + 16*x^2)*Log[x] + (20*x + 8*x^2)*Log[x]^2)/(E^(4*E^(3*E^9))*x),x]

[Out]

(-10*x)/E^(4*E^(3*E^9)) - (2*x^2)/E^(4*E^(3*E^9)) + (5 + 3*x)^2/(3*E^(4*E^(3*E^9))) + (10*x*Log[x])/E^(4*E^(3*
E^9)) + (4*x^2*Log[x])/E^(4*E^(3*E^9)) + (25*Log[x]^2)/E^(4*E^(3*E^9)) + (20*x*Log[x]^2)/E^(4*E^(3*E^9)) + (4*
x^2*Log[x]^2)/E^(4*E^(3*E^9))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2296

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2304

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

Rule 2305

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Lo
g[c*x^n])^p)/(d*(m + 1)), x] - Dist[(b*n*p)/(m + 1), Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{
a, b, c, d, m, n}, x] && NeQ[m, -1] && GtQ[p, 0]

Rule 2330

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = Expand
Integrand[(a + b*Log[c*x^n])^p, (d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n, p, q, r}
, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[r]))

Rule 2357

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=e^{-4 e^{3 e^9}} \int \frac {10 x+6 x^2+\left (50+50 x+16 x^2\right ) \log (x)+\left (20 x+8 x^2\right ) \log ^2(x)}{x} \, dx\\ &=e^{-4 e^{3 e^9}} \int \left (2 (5+3 x)+\frac {2 \left (25+25 x+8 x^2\right ) \log (x)}{x}+4 (5+2 x) \log ^2(x)\right ) \, dx\\ &=\frac {1}{3} e^{-4 e^{3 e^9}} (5+3 x)^2+\left (2 e^{-4 e^{3 e^9}}\right ) \int \frac {\left (25+25 x+8 x^2\right ) \log (x)}{x} \, dx+\left (4 e^{-4 e^{3 e^9}}\right ) \int (5+2 x) \log ^2(x) \, dx\\ &=\frac {1}{3} e^{-4 e^{3 e^9}} (5+3 x)^2+\left (2 e^{-4 e^{3 e^9}}\right ) \int \left (25 \log (x)+\frac {25 \log (x)}{x}+8 x \log (x)\right ) \, dx+\left (4 e^{-4 e^{3 e^9}}\right ) \int \left (5 \log ^2(x)+2 x \log ^2(x)\right ) \, dx\\ &=\frac {1}{3} e^{-4 e^{3 e^9}} (5+3 x)^2+\left (8 e^{-4 e^{3 e^9}}\right ) \int x \log ^2(x) \, dx+\left (16 e^{-4 e^{3 e^9}}\right ) \int x \log (x) \, dx+\left (20 e^{-4 e^{3 e^9}}\right ) \int \log ^2(x) \, dx+\left (50 e^{-4 e^{3 e^9}}\right ) \int \log (x) \, dx+\left (50 e^{-4 e^{3 e^9}}\right ) \int \frac {\log (x)}{x} \, dx\\ &=-50 e^{-4 e^{3 e^9}} x-4 e^{-4 e^{3 e^9}} x^2+\frac {1}{3} e^{-4 e^{3 e^9}} (5+3 x)^2+50 e^{-4 e^{3 e^9}} x \log (x)+8 e^{-4 e^{3 e^9}} x^2 \log (x)+25 e^{-4 e^{3 e^9}} \log ^2(x)+20 e^{-4 e^{3 e^9}} x \log ^2(x)+4 e^{-4 e^{3 e^9}} x^2 \log ^2(x)-\left (8 e^{-4 e^{3 e^9}}\right ) \int x \log (x) \, dx-\left (40 e^{-4 e^{3 e^9}}\right ) \int \log (x) \, dx\\ &=-10 e^{-4 e^{3 e^9}} x-2 e^{-4 e^{3 e^9}} x^2+\frac {1}{3} e^{-4 e^{3 e^9}} (5+3 x)^2+10 e^{-4 e^{3 e^9}} x \log (x)+4 e^{-4 e^{3 e^9}} x^2 \log (x)+25 e^{-4 e^{3 e^9}} \log ^2(x)+20 e^{-4 e^{3 e^9}} x \log ^2(x)+4 e^{-4 e^{3 e^9}} x^2 \log ^2(x)\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.03, size = 57, normalized size = 2.38 \begin {gather*} 2 e^{-4 e^{3 e^9}} \left (\frac {x^2}{2}+5 x \log (x)+2 x^2 \log (x)+\frac {25 \log ^2(x)}{2}+10 x \log ^2(x)+2 x^2 \log ^2(x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10*x + 6*x^2 + (50 + 50*x + 16*x^2)*Log[x] + (20*x + 8*x^2)*Log[x]^2)/(E^(4*E^(3*E^9))*x),x]

[Out]

(2*(x^2/2 + 5*x*Log[x] + 2*x^2*Log[x] + (25*Log[x]^2)/2 + 10*x*Log[x]^2 + 2*x^2*Log[x]^2))/E^(4*E^(3*E^9))

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fricas [A]  time = 0.51, size = 41, normalized size = 1.71 \begin {gather*} {\left ({\left (4 \, x^{2} + 20 \, x + 25\right )} \log \relax (x)^{2} + x^{2} + 2 \, {\left (2 \, x^{2} + 5 \, x\right )} \log \relax (x)\right )} e^{\left (-4 \, e^{\left (3 \, e^{9}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^2+20*x)*log(x)^2+(16*x^2+50*x+50)*log(x)+6*x^2+10*x)/x/exp(2*exp(3*exp(9)))^2,x, algorithm="fr
icas")

[Out]

((4*x^2 + 20*x + 25)*log(x)^2 + x^2 + 2*(2*x^2 + 5*x)*log(x))*e^(-4*e^(3*e^9))

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giac [A]  time = 0.16, size = 41, normalized size = 1.71 \begin {gather*} {\left ({\left (4 \, x^{2} + 20 \, x + 25\right )} \log \relax (x)^{2} + x^{2} + 2 \, {\left (2 \, x^{2} + 5 \, x\right )} \log \relax (x)\right )} e^{\left (-4 \, e^{\left (3 \, e^{9}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^2+20*x)*log(x)^2+(16*x^2+50*x+50)*log(x)+6*x^2+10*x)/x/exp(2*exp(3*exp(9)))^2,x, algorithm="gi
ac")

[Out]

((4*x^2 + 20*x + 25)*log(x)^2 + x^2 + 2*(2*x^2 + 5*x)*log(x))*e^(-4*e^(3*e^9))

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maple [B]  time = 0.05, size = 50, normalized size = 2.08

method	result	size
default	\({\mathrm e}^{-4 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}} \left (4 x^{2} \ln \relax (x )^{2}+4 x^{2} \ln \relax (x )+x^{2}+20 x \ln \relax (x )^{2}+10 x \ln \relax (x )+25 \ln \relax (x )^{2}\right )\)	\(50\)
risch	\({\mathrm e}^{-4 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}} \left (4 x^{2}+20 x +25\right ) \ln \relax (x )^{2}+{\mathrm e}^{-4 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}} \left (4 x^{2}+10 x \right ) \ln \relax (x )+{\mathrm e}^{-4 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}} x^{2}\)	\(57\)
norman	\(\left ({\mathrm e}^{-2 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}} x^{2}+25 \,{\mathrm e}^{-2 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}} \ln \relax (x )^{2}+10 \,{\mathrm e}^{-2 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}} x \ln \relax (x )+20 \,{\mathrm e}^{-2 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}} x \ln \relax (x )^{2}+4 \,{\mathrm e}^{-2 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}} x^{2} \ln \relax (x )+4 \,{\mathrm e}^{-2 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}} x^{2} \ln \relax (x )^{2}\right ) {\mathrm e}^{-2 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}}\)	\(99\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((8*x^2+20*x)*ln(x)^2+(16*x^2+50*x+50)*ln(x)+6*x^2+10*x)/x/exp(2*exp(3*exp(9)))^2,x,method=_RETURNVERBOSE)

[Out]

1/exp(2*exp(3*exp(9)))^2*(4*x^2*ln(x)^2+4*x^2*ln(x)+x^2+20*x*ln(x)^2+10*x*ln(x)+25*ln(x)^2)

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maxima [B]  time = 0.37, size = 66, normalized size = 2.75 \begin {gather*} {\left (2 \, {\left (2 \, \log \relax (x)^{2} - 2 \, \log \relax (x) + 1\right )} x^{2} + 8 \, x^{2} \log \relax (x) + 20 \, {\left (\log \relax (x)^{2} - 2 \, \log \relax (x) + 2\right )} x - x^{2} + 50 \, x \log \relax (x) + 25 \, \log \relax (x)^{2} - 40 \, x\right )} e^{\left (-4 \, e^{\left (3 \, e^{9}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^2+20*x)*log(x)^2+(16*x^2+50*x+50)*log(x)+6*x^2+10*x)/x/exp(2*exp(3*exp(9)))^2,x, algorithm="ma
xima")

[Out]

(2*(2*log(x)^2 - 2*log(x) + 1)*x^2 + 8*x^2*log(x) + 20*(log(x)^2 - 2*log(x) + 2)*x - x^2 + 50*x*log(x) + 25*lo
g(x)^2 - 40*x)*e^(-4*e^(3*e^9))

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mupad [B]  time = 4.37, size = 22, normalized size = 0.92 \begin {gather*} {\mathrm {e}}^{-4\,{\mathrm {e}}^{3\,{\mathrm {e}}^9}}\,{\left (x+5\,\ln \relax (x)+2\,x\,\ln \relax (x)\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-4*exp(3*exp(9)))*(10*x + log(x)^2*(20*x + 8*x^2) + log(x)*(50*x + 16*x^2 + 50) + 6*x^2))/x,x)

[Out]

exp(-4*exp(3*exp(9)))*(x + 5*log(x) + 2*x*log(x))^2

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sympy [B]  time = 0.24, size = 60, normalized size = 2.50 \begin {gather*} \frac {x^{2}}{e^{4 e^{3 e^{9}}}} + \frac {\left (4 x^{2} + 10 x\right ) \log {\relax (x )}}{e^{4 e^{3 e^{9}}}} + \frac {\left (4 x^{2} + 20 x + 25\right ) \log {\relax (x )}^{2}}{e^{4 e^{3 e^{9}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x**2+20*x)*ln(x)**2+(16*x**2+50*x+50)*ln(x)+6*x**2+10*x)/x/exp(2*exp(3*exp(9)))**2,x)

[Out]

x**2*exp(-4*exp(3*exp(9))) + (4*x**2 + 10*x)*exp(-4*exp(3*exp(9)))*log(x) + (4*x**2 + 20*x + 25)*exp(-4*exp(3*
exp(9)))*log(x)**2

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