3.66.86 \(\int (1-e^8+e^4 (-10-4 x)-8 x-3 x^2+(40+8 e^4+16 x) \log (\frac {2}{x})-16 \log ^2(\frac {2}{x})+(-10-2 e^4-4 x+8 \log (\frac {2}{x})) \log (x)-\log ^2(x)) \, dx\)

Optimal. Leaf size=30 \[ x \left (1+x-\left (-e^4-x+4 \log \left (\frac {2}{x}\right )-\log (x)\right )^2\right ) \]

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Rubi [B]  time = 0.16, antiderivative size = 134, normalized size of antiderivative = 4.47, number of steps used = 16, number of rules used = 9, integrand size = 78, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {2313, 9, 2296, 2295, 6741, 12, 6742, 2304, 2361} \begin {gather*} -x^3-3 x^2-2 x^2 \log (x)+8 \left (x^2+\left (5+e^4\right ) x\right ) \log \left (\frac {2}{x}\right )+\left (1-e^8\right ) x+2 \left (5+e^4\right ) x-50 x+4 \left (x+e^4+5\right )^2-\frac {1}{2} e^4 (2 x+5)^2-16 x \log ^2\left (\frac {2}{x}\right )-x \log ^2(x)-40 x \log \left (\frac {2}{x}\right )+8 x \log \left (\frac {2}{x}\right ) \log (x)-2 \left (5+e^4\right ) x \log (x)+10 x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1 - E^8 + E^4*(-10 - 4*x) - 8*x - 3*x^2 + (40 + 8*E^4 + 16*x)*Log[2/x] - 16*Log[2/x]^2 + (-10 - 2*E^4 - 4*
x + 8*Log[2/x])*Log[x] - Log[x]^2,x]

[Out]

-50*x + 2*(5 + E^4)*x + (1 - E^8)*x - 3*x^2 - x^3 + 4*(5 + E^4 + x)^2 - (E^4*(5 + 2*x)^2)/2 - 40*x*Log[2/x] +
8*((5 + E^4)*x + x^2)*Log[2/x] - 16*x*Log[2/x]^2 + 10*x*Log[x] - 2*(5 + E^4)*x*Log[x] - 2*x^2*Log[x] + 8*x*Log
[2/x]*Log[x] - x*Log[x]^2

Rule 9

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

Rule 12

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

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 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 2313

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = IntHide[(d +
e*x^r)^q, x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]] /; FreeQ[{a,
b, c, d, e, n, r}, x] && IGtQ[q, 0]

Rule 2361

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(f_.)*(x_)^(r_.)]*(e_.)), x_Symbol] :> With[{u =
IntHide[(a + b*Log[c*x^n])^p, x]}, Dist[d + e*Log[f*x^r], u, x] - Dist[e*r, Int[SimplifyIntegrand[u/x, x], x],
x]] /; FreeQ[{a, b, c, d, e, f, n, p, r}, x]

Rule 6741

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\left (1-e^8\right ) x-4 x^2-x^3-\frac {1}{2} e^4 (5+2 x)^2-16 \int \log ^2\left (\frac {2}{x}\right ) \, dx+\int \left (40+8 e^4+16 x\right ) \log \left (\frac {2}{x}\right ) \, dx+\int \left (-10-2 e^4-4 x+8 \log \left (\frac {2}{x}\right )\right ) \log (x) \, dx-\int \log ^2(x) \, dx\\ &=\left (1-e^8\right ) x-4 x^2-x^3-\frac {1}{2} e^4 (5+2 x)^2+8 \left (\left (5+e^4\right ) x+x^2\right ) \log \left (\frac {2}{x}\right )-16 x \log ^2\left (\frac {2}{x}\right )-x \log ^2(x)+2 \int \log (x) \, dx-32 \int \log \left (\frac {2}{x}\right ) \, dx+\int 8 \left (5+e^4+x\right ) \, dx+\int 2 \left (-5 \left (1+\frac {e^4}{5}\right )-2 x+4 \log \left (\frac {2}{x}\right )\right ) \log (x) \, dx\\ &=-34 x+\left (1-e^8\right ) x-4 x^2-x^3+4 \left (5+e^4+x\right )^2-\frac {1}{2} e^4 (5+2 x)^2-32 x \log \left (\frac {2}{x}\right )+8 \left (\left (5+e^4\right ) x+x^2\right ) \log \left (\frac {2}{x}\right )-16 x \log ^2\left (\frac {2}{x}\right )+2 x \log (x)-x \log ^2(x)+2 \int \left (-5 \left (1+\frac {e^4}{5}\right )-2 x+4 \log \left (\frac {2}{x}\right )\right ) \log (x) \, dx\\ &=-34 x+\left (1-e^8\right ) x-4 x^2-x^3+4 \left (5+e^4+x\right )^2-\frac {1}{2} e^4 (5+2 x)^2-32 x \log \left (\frac {2}{x}\right )+8 \left (\left (5+e^4\right ) x+x^2\right ) \log \left (\frac {2}{x}\right )-16 x \log ^2\left (\frac {2}{x}\right )+2 x \log (x)-x \log ^2(x)+2 \int \left (-\left (\left (5+e^4\right ) \log (x)\right )-2 x \log (x)+4 \log \left (\frac {2}{x}\right ) \log (x)\right ) \, dx\\ &=-34 x+\left (1-e^8\right ) x-4 x^2-x^3+4 \left (5+e^4+x\right )^2-\frac {1}{2} e^4 (5+2 x)^2-32 x \log \left (\frac {2}{x}\right )+8 \left (\left (5+e^4\right ) x+x^2\right ) \log \left (\frac {2}{x}\right )-16 x \log ^2\left (\frac {2}{x}\right )+2 x \log (x)-x \log ^2(x)-4 \int x \log (x) \, dx+8 \int \log \left (\frac {2}{x}\right ) \log (x) \, dx-\left (2 \left (5+e^4\right )\right ) \int \log (x) \, dx\\ &=-34 x+2 \left (5+e^4\right ) x+\left (1-e^8\right ) x-3 x^2-x^3+4 \left (5+e^4+x\right )^2-\frac {1}{2} e^4 (5+2 x)^2-32 x \log \left (\frac {2}{x}\right )+8 \left (\left (5+e^4\right ) x+x^2\right ) \log \left (\frac {2}{x}\right )-16 x \log ^2\left (\frac {2}{x}\right )+10 x \log (x)-2 \left (5+e^4\right ) x \log (x)-2 x^2 \log (x)+8 x \log \left (\frac {2}{x}\right ) \log (x)-x \log ^2(x)-8 \int \left (1+\log \left (\frac {2}{x}\right )\right ) \, dx\\ &=-42 x+2 \left (5+e^4\right ) x+\left (1-e^8\right ) x-3 x^2-x^3+4 \left (5+e^4+x\right )^2-\frac {1}{2} e^4 (5+2 x)^2-32 x \log \left (\frac {2}{x}\right )+8 \left (\left (5+e^4\right ) x+x^2\right ) \log \left (\frac {2}{x}\right )-16 x \log ^2\left (\frac {2}{x}\right )+10 x \log (x)-2 \left (5+e^4\right ) x \log (x)-2 x^2 \log (x)+8 x \log \left (\frac {2}{x}\right ) \log (x)-x \log ^2(x)-8 \int \log \left (\frac {2}{x}\right ) \, dx\\ &=-50 x+2 \left (5+e^4\right ) x+\left (1-e^8\right ) x-3 x^2-x^3+4 \left (5+e^4+x\right )^2-\frac {1}{2} e^4 (5+2 x)^2-40 x \log \left (\frac {2}{x}\right )+8 \left (\left (5+e^4\right ) x+x^2\right ) \log \left (\frac {2}{x}\right )-16 x \log ^2\left (\frac {2}{x}\right )+10 x \log (x)-2 \left (5+e^4\right ) x \log (x)-2 x^2 \log (x)+8 x \log \left (\frac {2}{x}\right ) \log (x)-x \log ^2(x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 58, normalized size = 1.93 \begin {gather*} -x \left (-1+e^8-x+2 e^4 x+x^2+16 \log ^2\left (\frac {2}{x}\right )+2 \left (e^4+x\right ) \log (x)+\log ^2(x)-8 \log \left (\frac {2}{x}\right ) \left (e^4+x+\log (x)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1 - E^8 + E^4*(-10 - 4*x) - 8*x - 3*x^2 + (40 + 8*E^4 + 16*x)*Log[2/x] - 16*Log[2/x]^2 + (-10 - 2*E^
4 - 4*x + 8*Log[2/x])*Log[x] - Log[x]^2,x]

[Out]

-(x*(-1 + E^8 - x + 2*E^4*x + x^2 + 16*Log[2/x]^2 + 2*(E^4 + x)*Log[x] + Log[x]^2 - 8*Log[2/x]*(E^4 + x + Log[
x])))

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fricas [B]  time = 0.57, size = 72, normalized size = 2.40 \begin {gather*} -x^{3} - 2 \, x^{2} e^{4} - x \log \relax (2)^{2} - 25 \, x \log \left (\frac {2}{x}\right )^{2} + x^{2} - x e^{8} - 2 \, {\left (x^{2} + x e^{4}\right )} \log \relax (2) + 10 \, {\left (x^{2} + x e^{4} + x \log \relax (2)\right )} \log \left (\frac {2}{x}\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-log(x)^2+(8*log(2/x)-2*exp(4)-4*x-10)*log(x)-16*log(2/x)^2+(8*exp(4)+16*x+40)*log(2/x)-exp(4)^2+(-4
*x-10)*exp(4)-3*x^2-8*x+1,x, algorithm="fricas")

[Out]

-x^3 - 2*x^2*e^4 - x*log(2)^2 - 25*x*log(2/x)^2 + x^2 - x*e^8 - 2*(x^2 + x*e^4)*log(2) + 10*(x^2 + x*e^4 + x*l
og(2))*log(2/x) + x

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giac [B]  time = 0.22, size = 131, normalized size = 4.37 \begin {gather*} -x^{3} + 4 \, x^{2} {\left (\frac {2 \, e^{4}}{x} + \frac {10}{x} + 1\right )} - 2 \, x^{2} \log \relax (x) - 2 \, x e^{4} \log \relax (x) + 8 \, x \log \relax (2) \log \relax (x) - 9 \, x \log \relax (x)^{2} - 16 \, x \log \left (\frac {2}{x}\right )^{2} - 3 \, x^{2} - x e^{8} - 2 \, {\left (x^{2} + 5 \, x\right )} e^{4} + 2 \, x e^{4} - 8 \, x \log \relax (2) + 8 \, x \log \relax (x) + 8 \, {\left (x^{2} + x e^{4} + 5 \, x\right )} \log \left (\frac {2}{x}\right ) - 32 \, x \log \left (\frac {2}{x}\right ) - 39 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-log(x)^2+(8*log(2/x)-2*exp(4)-4*x-10)*log(x)-16*log(2/x)^2+(8*exp(4)+16*x+40)*log(2/x)-exp(4)^2+(-4
*x-10)*exp(4)-3*x^2-8*x+1,x, algorithm="giac")

[Out]

-x^3 + 4*x^2*(2*e^4/x + 10/x + 1) - 2*x^2*log(x) - 2*x*e^4*log(x) + 8*x*log(2)*log(x) - 9*x*log(x)^2 - 16*x*lo
g(2/x)^2 - 3*x^2 - x*e^8 - 2*(x^2 + 5*x)*e^4 + 2*x*e^4 - 8*x*log(2) + 8*x*log(x) + 8*(x^2 + x*e^4 + 5*x)*log(2
/x) - 32*x*log(2/x) - 39*x

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maple [B]  time = 0.16, size = 103, normalized size = 3.43

method	result	size
risch	\(-9 x \ln \relax (x )^{2}+2 x \ln \relax (x )+\left (-2 x \,{\mathrm e}^{4}+8 x \ln \relax (2)-2 x^{2}+6 x \right ) \ln \relax (x )-8 x \ln \relax (2)+x^{2}-16 x \ln \left (\frac {2}{x}\right )^{2}-32 x \ln \left (\frac {2}{x}\right )+\left (8 x \,{\mathrm e}^{4}+8 x^{2}+40 x \right ) \ln \left (\frac {2}{x}\right )-x \,{\mathrm e}^{8}-2 x^{2} {\mathrm e}^{4}-x^{3}+x\)	\(103\)
default	\(x +{\mathrm e}^{4} \left (-2 x^{2}-10 x \right )+8 \,{\mathrm e}^{4} x \ln \left (\frac {2}{x}\right )+10 x \,{\mathrm e}^{4}+8 x \ln \left (\frac {2}{x}\right )+8 x^{2} \ln \left (\frac {2}{x}\right )+x^{2}-2 x^{2} \ln \relax (x )+8 x \ln \relax (2) \ln \relax (x )-8 x \ln \relax (2)-2 x \,{\mathrm e}^{4} \ln \relax (x )-8 x \ln \left (\frac {1}{x}\right )+8 \ln \relax (x ) \ln \left (\frac {1}{x}\right ) x -x^{3}-x \,{\mathrm e}^{8}-x \ln \relax (x )^{2}-16 x \ln \left (\frac {2}{x}\right )^{2}\)	\(126\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-ln(x)^2+(8*ln(2/x)-2*exp(4)-4*x-10)*ln(x)-16*ln(2/x)^2+(8*exp(4)+16*x+40)*ln(2/x)-exp(4)^2+(-4*x-10)*exp(
4)-3*x^2-8*x+1,x,method=_RETURNVERBOSE)

[Out]

-9*x*ln(x)^2+2*x*ln(x)+(-2*x*exp(4)+8*x*ln(2)-2*x^2+6*x)*ln(x)-8*x*ln(2)+x^2-16*x*ln(2/x)^2-32*x*ln(2/x)+(8*x*
exp(4)+8*x^2+40*x)*ln(2/x)-x*exp(8)-2*x^2*exp(4)-x^3+x

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maxima [B]  time = 0.50, size = 125, normalized size = 4.17 \begin {gather*} -x^{3} - 16 \, x \log \left (\frac {2}{x}\right )^{2} - {\left (\log \relax (x)^{2} - 2 \, \log \relax (x) + 2\right )} x + x^{2} + 2 \, x {\left (e^{4} - 4 \, \log \relax (2) - 3\right )} + 8 \, x {\left (e^{4} + 5\right )} - x e^{8} - 2 \, {\left (x^{2} + 5 \, x\right )} e^{4} - 2 \, {\left (x^{2} + x e^{4} - 4 \, x \log \left (\frac {2}{x}\right ) + x\right )} \log \relax (x) + 8 \, x \log \relax (x) + 8 \, {\left (x^{2} + x e^{4} + 5 \, x\right )} \log \left (\frac {2}{x}\right ) - 32 \, x \log \left (\frac {2}{x}\right ) - 31 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-log(x)^2+(8*log(2/x)-2*exp(4)-4*x-10)*log(x)-16*log(2/x)^2+(8*exp(4)+16*x+40)*log(2/x)-exp(4)^2+(-4
*x-10)*exp(4)-3*x^2-8*x+1,x, algorithm="maxima")

[Out]

-x^3 - 16*x*log(2/x)^2 - (log(x)^2 - 2*log(x) + 2)*x + x^2 + 2*x*(e^4 - 4*log(2) - 3) + 8*x*(e^4 + 5) - x*e^8
- 2*(x^2 + 5*x)*e^4 - 2*(x^2 + x*e^4 - 4*x*log(2/x) + x)*log(x) + 8*x*log(x) + 8*(x^2 + x*e^4 + 5*x)*log(2/x)
- 32*x*log(2/x) - 31*x

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mupad [B]  time = 4.30, size = 72, normalized size = 2.40 \begin {gather*} -x\,\left ({\mathrm {e}}^8-x+16\,{\ln \left (\frac {2}{x}\right )}^2+2\,x\,{\mathrm {e}}^4+{\ln \relax (x)}^2-8\,{\mathrm {e}}^4\,\ln \left (\frac {2}{x}\right )-8\,x\,\ln \left (\frac {2}{x}\right )+2\,{\mathrm {e}}^4\,\ln \relax (x)+2\,x\,\ln \relax (x)+x^2-8\,\ln \left (\frac {2}{x}\right )\,\ln \relax (x)-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(2/x)*(16*x + 8*exp(4) + 40) - exp(8) - 16*log(2/x)^2 - log(x)*(4*x + 2*exp(4) - 8*log(2/x) + 10) - log
(x)^2 - 3*x^2 - exp(4)*(4*x + 10) - 8*x + 1,x)

[Out]

-x*(exp(8) - x + 16*log(2/x)^2 + 2*x*exp(4) + log(x)^2 - 8*exp(4)*log(2/x) - 8*x*log(2/x) + 2*exp(4)*log(x) +
2*x*log(x) + x^2 - 8*log(2/x)*log(x) - 1)

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sympy [B]  time = 0.29, size = 70, normalized size = 2.33 \begin {gather*} - x^{3} + x^{2} \left (- 2 e^{4} + 1 + 8 \log {\relax (2 )}\right ) - 25 x \log {\relax (x )}^{2} + x \left (- e^{8} - 16 \log {\relax (2 )}^{2} + 1 + 8 e^{4} \log {\relax (2 )}\right ) + \left (- 10 x^{2} - 10 x e^{4} + 40 x \log {\relax (2 )}\right ) \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-ln(x)**2+(8*ln(2/x)-2*exp(4)-4*x-10)*ln(x)-16*ln(2/x)**2+(8*exp(4)+16*x+40)*ln(2/x)-exp(4)**2+(-4*x
-10)*exp(4)-3*x**2-8*x+1,x)

[Out]

-x**3 + x**2*(-2*exp(4) + 1 + 8*log(2)) - 25*x*log(x)**2 + x*(-exp(8) - 16*log(2)**2 + 1 + 8*exp(4)*log(2)) +
(-10*x**2 - 10*x*exp(4) + 40*x*log(2))*log(x)

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