3.37.97 \(\int \frac {(-32 x^2+16 x^3+e^2 (8 x-6 x^2)+(-16 x^2+4 x^3+e^2 (8 x-2 x^2)) \log (32 x^2-16 x^3+2 x^4+e^2 (-16 x+8 x^2-x^3))) \log (\frac {5}{x \log (32 x^2-16 x^3+2 x^4+e^2 (-16 x+8 x^2-x^3))})+(16 x^2-4 x^3+e^2 (-8 x+2 x^2)) \log (32 x^2-16 x^3+2 x^4+e^2 (-16 x+8 x^2-x^3)) \log ^2(\frac {5}{x \log (32 x^2-16 x^3+2 x^4+e^2 (-16 x+8 x^2-x^3))})}{(e^2 (-4+x)+8 x-2 x^2) \log (32 x^2-16 x^3+2 x^4+e^2 (-16 x+8 x^2-x^3))} \, dx\)

Optimal. Leaf size=33 \[ x^2 \log ^2\left (\frac {5}{x \log \left ((4-x)^2 x \left (-e^2+2 x\right )\right )}\right ) \]

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Rubi [A]  time = 3.79, antiderivative size = 32, normalized size of antiderivative = 0.97, number of steps used = 3, number of rules used = 7, integrand size = 290, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {6688, 12, 14, 6742, 30, 2555, 6687} \begin {gather*} x^2 \log ^2\left (\frac {5}{x \log \left (-\left (\left (e^2-2 x\right ) (4-x)^2 x\right )\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-32*x^2 + 16*x^3 + E^2*(8*x - 6*x^2) + (-16*x^2 + 4*x^3 + E^2*(8*x - 2*x^2))*Log[32*x^2 - 16*x^3 + 2*x^4
+ E^2*(-16*x + 8*x^2 - x^3)])*Log[5/(x*Log[32*x^2 - 16*x^3 + 2*x^4 + E^2*(-16*x + 8*x^2 - x^3)])] + (16*x^2 -
4*x^3 + E^2*(-8*x + 2*x^2))*Log[32*x^2 - 16*x^3 + 2*x^4 + E^2*(-16*x + 8*x^2 - x^3)]*Log[5/(x*Log[32*x^2 - 16
*x^3 + 2*x^4 + E^2*(-16*x + 8*x^2 - x^3)])]^2)/((E^2*(-4 + x) + 8*x - 2*x^2)*Log[32*x^2 - 16*x^3 + 2*x^4 + E^2
*(-16*x + 8*x^2 - x^3)]),x]

[Out]

x^2*Log[5/(x*Log[-((E^2 - 2*x)*(4 - x)^2*x)])]^2

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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 6687

Int[(u_)*(y_)^(m_.)*(z_)^(n_.), x_Symbol] :> With[{q = DerivativeDivides[y*z, u*z^(n - m), x]}, Simp[(q*y^(m +
1)*z^(m + 1))/(m + 1), x] /;  !FalseQ[q]] /; FreeQ[{m, n}, x] && NeQ[m, -1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, 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 2 x \log \left (\frac {5}{x \log \left (-\left (\left (e^2-2 x\right ) (-4+x)^2 x\right )\right )}\right ) \left (\frac {e^2 (4-3 x)+8 (-2+x) x-\left (e^2-2 x\right ) (-4+x) \log \left (-\left (\left (e^2-2 x\right ) (-4+x)^2 x\right )\right )}{\left (e^2-2 x\right ) (-4+x) \log \left (-\left (\left (e^2-2 x\right ) (-4+x)^2 x\right )\right )}+\log \left (\frac {5}{x \log \left (-\left (\left (e^2-2 x\right ) (-4+x)^2 x\right )\right )}\right )\right ) \, dx\\ &=2 \int x \log \left (\frac {5}{x \log \left (-\left (\left (e^2-2 x\right ) (-4+x)^2 x\right )\right )}\right ) \left (\frac {e^2 (4-3 x)+8 (-2+x) x-\left (e^2-2 x\right ) (-4+x) \log \left (-\left (\left (e^2-2 x\right ) (-4+x)^2 x\right )\right )}{\left (e^2-2 x\right ) (-4+x) \log \left (-\left (\left (e^2-2 x\right ) (-4+x)^2 x\right )\right )}+\log \left (\frac {5}{x \log \left (-\left (\left (e^2-2 x\right ) (-4+x)^2 x\right )\right )}\right )\right ) \, dx\\ &=x^2 \log ^2\left (\frac {5}{x \log \left (-\left (\left (e^2-2 x\right ) (4-x)^2 x\right )\right )}\right )\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 0.54, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (-32 x^2+16 x^3+e^2 \left (8 x-6 x^2\right )+\left (-16 x^2+4 x^3+e^2 \left (8 x-2 x^2\right )\right ) \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right )\right ) \log \left (\frac {5}{x \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right )}\right )+\left (16 x^2-4 x^3+e^2 \left (-8 x+2 x^2\right )\right ) \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right ) \log ^2\left (\frac {5}{x \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right )}\right )}{\left (e^2 (-4+x)+8 x-2 x^2\right ) \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[((-32*x^2 + 16*x^3 + E^2*(8*x - 6*x^2) + (-16*x^2 + 4*x^3 + E^2*(8*x - 2*x^2))*Log[32*x^2 - 16*x^3 +
2*x^4 + E^2*(-16*x + 8*x^2 - x^3)])*Log[5/(x*Log[32*x^2 - 16*x^3 + 2*x^4 + E^2*(-16*x + 8*x^2 - x^3)])] + (16
*x^2 - 4*x^3 + E^2*(-8*x + 2*x^2))*Log[32*x^2 - 16*x^3 + 2*x^4 + E^2*(-16*x + 8*x^2 - x^3)]*Log[5/(x*Log[32*x^
2 - 16*x^3 + 2*x^4 + E^2*(-16*x + 8*x^2 - x^3)])]^2)/((E^2*(-4 + x) + 8*x - 2*x^2)*Log[32*x^2 - 16*x^3 + 2*x^4
+ E^2*(-16*x + 8*x^2 - x^3)]),x]

[Out]

Integrate[((-32*x^2 + 16*x^3 + E^2*(8*x - 6*x^2) + (-16*x^2 + 4*x^3 + E^2*(8*x - 2*x^2))*Log[32*x^2 - 16*x^3 +
2*x^4 + E^2*(-16*x + 8*x^2 - x^3)])*Log[5/(x*Log[32*x^2 - 16*x^3 + 2*x^4 + E^2*(-16*x + 8*x^2 - x^3)])] + (16
*x^2 - 4*x^3 + E^2*(-8*x + 2*x^2))*Log[32*x^2 - 16*x^3 + 2*x^4 + E^2*(-16*x + 8*x^2 - x^3)]*Log[5/(x*Log[32*x^
2 - 16*x^3 + 2*x^4 + E^2*(-16*x + 8*x^2 - x^3)])]^2)/((E^2*(-4 + x) + 8*x - 2*x^2)*Log[32*x^2 - 16*x^3 + 2*x^4
+ E^2*(-16*x + 8*x^2 - x^3)]), x]

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fricas [A]  time = 0.74, size = 47, normalized size = 1.42 \begin {gather*} x^{2} \log \left (\frac {5}{x \log \left (2 \, x^{4} - 16 \, x^{3} + 32 \, x^{2} - {\left (x^{3} - 8 \, x^{2} + 16 \, x\right )} e^{2}\right )}\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2-8*x)*exp(2)-4*x^3+16*x^2)*log((-x^3+8*x^2-16*x)*exp(2)+2*x^4-16*x^3+32*x^2)*log(5/x/log((-x
^3+8*x^2-16*x)*exp(2)+2*x^4-16*x^3+32*x^2))^2+(((-2*x^2+8*x)*exp(2)+4*x^3-16*x^2)*log((-x^3+8*x^2-16*x)*exp(2)
+2*x^4-16*x^3+32*x^2)+(-6*x^2+8*x)*exp(2)+16*x^3-32*x^2)*log(5/x/log((-x^3+8*x^2-16*x)*exp(2)+2*x^4-16*x^3+32*
x^2)))/((x-4)*exp(2)-2*x^2+8*x)/log((-x^3+8*x^2-16*x)*exp(2)+2*x^4-16*x^3+32*x^2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, {\left ({\left (2 \, x^{3} - 8 \, x^{2} - {\left (x^{2} - 4 \, x\right )} e^{2}\right )} \log \left (2 \, x^{4} - 16 \, x^{3} + 32 \, x^{2} - {\left (x^{3} - 8 \, x^{2} + 16 \, x\right )} e^{2}\right ) \log \left (\frac {5}{x \log \left (2 \, x^{4} - 16 \, x^{3} + 32 \, x^{2} - {\left (x^{3} - 8 \, x^{2} + 16 \, x\right )} e^{2}\right )}\right )^{2} - {\left (8 \, x^{3} - 16 \, x^{2} - {\left (3 \, x^{2} - 4 \, x\right )} e^{2} + {\left (2 \, x^{3} - 8 \, x^{2} - {\left (x^{2} - 4 \, x\right )} e^{2}\right )} \log \left (2 \, x^{4} - 16 \, x^{3} + 32 \, x^{2} - {\left (x^{3} - 8 \, x^{2} + 16 \, x\right )} e^{2}\right )\right )} \log \left (\frac {5}{x \log \left (2 \, x^{4} - 16 \, x^{3} + 32 \, x^{2} - {\left (x^{3} - 8 \, x^{2} + 16 \, x\right )} e^{2}\right )}\right )\right )}}{{\left (2 \, x^{2} - {\left (x - 4\right )} e^{2} - 8 \, x\right )} \log \left (2 \, x^{4} - 16 \, x^{3} + 32 \, x^{2} - {\left (x^{3} - 8 \, x^{2} + 16 \, x\right )} e^{2}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2-8*x)*exp(2)-4*x^3+16*x^2)*log((-x^3+8*x^2-16*x)*exp(2)+2*x^4-16*x^3+32*x^2)*log(5/x/log((-x
^3+8*x^2-16*x)*exp(2)+2*x^4-16*x^3+32*x^2))^2+(((-2*x^2+8*x)*exp(2)+4*x^3-16*x^2)*log((-x^3+8*x^2-16*x)*exp(2)
+2*x^4-16*x^3+32*x^2)+(-6*x^2+8*x)*exp(2)+16*x^3-32*x^2)*log(5/x/log((-x^3+8*x^2-16*x)*exp(2)+2*x^4-16*x^3+32*
x^2)))/((x-4)*exp(2)-2*x^2+8*x)/log((-x^3+8*x^2-16*x)*exp(2)+2*x^4-16*x^3+32*x^2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (2 x^{2}-8 x \right ) {\mathrm e}^{2}-4 x^{3}+16 x^{2}\right ) \ln \left (\left (-x^{3}+8 x^{2}-16 x \right ) {\mathrm e}^{2}+2 x^{4}-16 x^{3}+32 x^{2}\right ) \ln \left (\frac {5}{x \ln \left (\left (-x^{3}+8 x^{2}-16 x \right ) {\mathrm e}^{2}+2 x^{4}-16 x^{3}+32 x^{2}\right )}\right )^{2}+\left (\left (\left (-2 x^{2}+8 x \right ) {\mathrm e}^{2}+4 x^{3}-16 x^{2}\right ) \ln \left (\left (-x^{3}+8 x^{2}-16 x \right ) {\mathrm e}^{2}+2 x^{4}-16 x^{3}+32 x^{2}\right )+\left (-6 x^{2}+8 x \right ) {\mathrm e}^{2}+16 x^{3}-32 x^{2}\right ) \ln \left (\frac {5}{x \ln \left (\left (-x^{3}+8 x^{2}-16 x \right ) {\mathrm e}^{2}+2 x^{4}-16 x^{3}+32 x^{2}\right )}\right )}{\left (\left (x -4\right ) {\mathrm e}^{2}-2 x^{2}+8 x \right ) \ln \left (\left (-x^{3}+8 x^{2}-16 x \right ) {\mathrm e}^{2}+2 x^{4}-16 x^{3}+32 x^{2}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x^2-8*x)*exp(2)-4*x^3+16*x^2)*ln((-x^3+8*x^2-16*x)*exp(2)+2*x^4-16*x^3+32*x^2)*ln(5/x/ln((-x^3+8*x^2-
16*x)*exp(2)+2*x^4-16*x^3+32*x^2))^2+(((-2*x^2+8*x)*exp(2)+4*x^3-16*x^2)*ln((-x^3+8*x^2-16*x)*exp(2)+2*x^4-16*
x^3+32*x^2)+(-6*x^2+8*x)*exp(2)+16*x^3-32*x^2)*ln(5/x/ln((-x^3+8*x^2-16*x)*exp(2)+2*x^4-16*x^3+32*x^2)))/((x-4
)*exp(2)-2*x^2+8*x)/ln((-x^3+8*x^2-16*x)*exp(2)+2*x^4-16*x^3+32*x^2),x)

[Out]

int((((2*x^2-8*x)*exp(2)-4*x^3+16*x^2)*ln((-x^3+8*x^2-16*x)*exp(2)+2*x^4-16*x^3+32*x^2)*ln(5/x/ln((-x^3+8*x^2-
16*x)*exp(2)+2*x^4-16*x^3+32*x^2))^2+(((-2*x^2+8*x)*exp(2)+4*x^3-16*x^2)*ln((-x^3+8*x^2-16*x)*exp(2)+2*x^4-16*
x^3+32*x^2)+(-6*x^2+8*x)*exp(2)+16*x^3-32*x^2)*ln(5/x/ln((-x^3+8*x^2-16*x)*exp(2)+2*x^4-16*x^3+32*x^2)))/((x-4
)*exp(2)-2*x^2+8*x)/ln((-x^3+8*x^2-16*x)*exp(2)+2*x^4-16*x^3+32*x^2),x)

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maxima [B]  time = 0.57, size = 86, normalized size = 2.61 \begin {gather*} x^{2} \log \relax (5)^{2} - 2 \, x^{2} \log \relax (5) \log \relax (x) + x^{2} \log \relax (x)^{2} + x^{2} \log \left (\log \left (2 \, x - e^{2}\right ) + 2 \, \log \left (x - 4\right ) + \log \relax (x)\right )^{2} - 2 \, {\left (x^{2} \log \relax (5) - x^{2} \log \relax (x)\right )} \log \left (\log \left (2 \, x - e^{2}\right ) + 2 \, \log \left (x - 4\right ) + \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2-8*x)*exp(2)-4*x^3+16*x^2)*log((-x^3+8*x^2-16*x)*exp(2)+2*x^4-16*x^3+32*x^2)*log(5/x/log((-x
^3+8*x^2-16*x)*exp(2)+2*x^4-16*x^3+32*x^2))^2+(((-2*x^2+8*x)*exp(2)+4*x^3-16*x^2)*log((-x^3+8*x^2-16*x)*exp(2)
+2*x^4-16*x^3+32*x^2)+(-6*x^2+8*x)*exp(2)+16*x^3-32*x^2)*log(5/x/log((-x^3+8*x^2-16*x)*exp(2)+2*x^4-16*x^3+32*
x^2)))/((x-4)*exp(2)-2*x^2+8*x)/log((-x^3+8*x^2-16*x)*exp(2)+2*x^4-16*x^3+32*x^2),x, algorithm="maxima")

[Out]

x^2*log(5)^2 - 2*x^2*log(5)*log(x) + x^2*log(x)^2 + x^2*log(log(2*x - e^2) + 2*log(x - 4) + log(x))^2 - 2*(x^2
*log(5) - x^2*log(x))*log(log(2*x - e^2) + 2*log(x - 4) + log(x))

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mupad [B]  time = 3.96, size = 47, normalized size = 1.42 \begin {gather*} x^2\,{\ln \left (\frac {5}{x\,\ln \left (32\,x^2-{\mathrm {e}}^2\,\left (x^3-8\,x^2+16\,x\right )-16\,x^3+2\,x^4\right )}\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(5/(x*log(32*x^2 - exp(2)*(16*x - 8*x^2 + x^3) - 16*x^3 + 2*x^4)))*(exp(2)*(8*x - 6*x^2) - 32*x^2 + 16
*x^3 + log(32*x^2 - exp(2)*(16*x - 8*x^2 + x^3) - 16*x^3 + 2*x^4)*(exp(2)*(8*x - 2*x^2) - 16*x^2 + 4*x^3)) - l
og(32*x^2 - exp(2)*(16*x - 8*x^2 + x^3) - 16*x^3 + 2*x^4)*log(5/(x*log(32*x^2 - exp(2)*(16*x - 8*x^2 + x^3) -
16*x^3 + 2*x^4)))^2*(exp(2)*(8*x - 2*x^2) - 16*x^2 + 4*x^3))/(log(32*x^2 - exp(2)*(16*x - 8*x^2 + x^3) - 16*x^
3 + 2*x^4)*(8*x + exp(2)*(x - 4) - 2*x^2)),x)

[Out]

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

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: CoercionFailed} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x**2-8*x)*exp(2)-4*x**3+16*x**2)*ln((-x**3+8*x**2-16*x)*exp(2)+2*x**4-16*x**3+32*x**2)*ln(5/x/l
n((-x**3+8*x**2-16*x)*exp(2)+2*x**4-16*x**3+32*x**2))**2+(((-2*x**2+8*x)*exp(2)+4*x**3-16*x**2)*ln((-x**3+8*x*
*2-16*x)*exp(2)+2*x**4-16*x**3+32*x**2)+(-6*x**2+8*x)*exp(2)+16*x**3-32*x**2)*ln(5/x/ln((-x**3+8*x**2-16*x)*ex
p(2)+2*x**4-16*x**3+32*x**2)))/((x-4)*exp(2)-2*x**2+8*x)/ln((-x**3+8*x**2-16*x)*exp(2)+2*x**4-16*x**3+32*x**2)
,x)

[Out]

Exception raised: CoercionFailed

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