3.79.93 \(\int \frac {4 \log (5)+(-2 x+5 x^2-2 x^3+e^2 (x^2-2 x^3)) \log (5) \log ^2(\frac {4-4 x+x^2+e^4 x^2+e^2 (-4 x+2 x^2)}{x^2})}{(-2 x+x^2+e^2 x^2) \log ^2(\frac {4-4 x+x^2+e^4 x^2+e^2 (-4 x+2 x^2)}{x^2})} \, dx\)

Optimal. Leaf size=31 \[ \log (5) \left (x-x^2-\frac {1}{\log \left (\left (e^2-\frac {2-x}{x}\right )^2\right )}\right ) \]

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Rubi [A]  time = 0.50, antiderivative size = 35, normalized size of antiderivative = 1.13, number of steps used = 8, number of rules used = 6, integrand size = 122, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {6, 1593, 6688, 12, 2516, 2505} \begin {gather*} x^2 (-\log (5))-\frac {\log (5)}{\log \left (\frac {\left (2-\left (1+e^2\right ) x\right )^2}{x^2}\right )}+x \log (5) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x*Log[5] - x^2*Log[5] - Log[5]/Log[(2 - (1 + E^2)*x)^2/x^2]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2505

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]^(s_.)*(u_), x_Symbol] :> Wi
th[{h = Simplify[u*(a + b*x)*(c + d*x)]}, Simp[(h*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^(s + 1))/(p*r*(s + 1)*(
b*c - a*d)), x] /; FreeQ[h, x]] /; FreeQ[{a, b, c, d, e, f, p, q, r, s}, x] && NeQ[b*c - a*d, 0] && EqQ[p + q,
0] && NeQ[s, -1]

Rule 2516

Int[Log[(e_.)*((f_.)*(v_)^(p_.)*(w_)^(q_.))^(r_.)]^(s_.)*(u_.), x_Symbol] :> Int[u*Log[e*(f*ExpandToSum[v, x]^
p*ExpandToSum[w, x]^q)^r]^s, x] /; FreeQ[{e, f, p, q, r, s}, x] && LinearQ[{v, w}, x] &&  !LinearMatchQ[{v, w}
, x] && AlgebraicFunctionQ[u, x]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 \log (5)+\left (-2 x+5 x^2-2 x^3+e^2 \left (x^2-2 x^3\right )\right ) \log (5) \log ^2\left (\frac {4-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )}{x^2}\right )}{\left (-2 x+\left (1+e^2\right ) x^2\right ) \log ^2\left (\frac {4-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )}{x^2}\right )} \, dx\\ &=\int \frac {4 \log (5)+\left (-2 x+5 x^2-2 x^3+e^2 \left (x^2-2 x^3\right )\right ) \log (5) \log ^2\left (\frac {4-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )}{x^2}\right )}{x \left (-2+\left (1+e^2\right ) x\right ) \log ^2\left (\frac {4-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )}{x^2}\right )} \, dx\\ &=\int \log (5) \left (1-2 x+\frac {4}{x \left (-2+x+e^2 x\right ) \log ^2\left (\frac {\left (-2+x+e^2 x\right )^2}{x^2}\right )}\right ) \, dx\\ &=\log (5) \int \left (1-2 x+\frac {4}{x \left (-2+x+e^2 x\right ) \log ^2\left (\frac {\left (-2+x+e^2 x\right )^2}{x^2}\right )}\right ) \, dx\\ &=x \log (5)-x^2 \log (5)+(4 \log (5)) \int \frac {1}{x \left (-2+x+e^2 x\right ) \log ^2\left (\frac {\left (-2+x+e^2 x\right )^2}{x^2}\right )} \, dx\\ &=x \log (5)-x^2 \log (5)+(4 \log (5)) \int \frac {1}{x \left (-2+\left (1+e^2\right ) x\right ) \log ^2\left (\frac {\left (-2+x+e^2 x\right )^2}{x^2}\right )} \, dx\\ &=x \log (5)-x^2 \log (5)+(4 \log (5)) \int \frac {1}{x \left (-2+\left (1+e^2\right ) x\right ) \log ^2\left (\frac {\left (-2+\left (1+e^2\right ) x\right )^2}{x^2}\right )} \, dx\\ &=x \log (5)-x^2 \log (5)-\frac {\log (5)}{\log \left (\frac {\left (2-\left (1+e^2\right ) x\right )^2}{x^2}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 29, normalized size = 0.94 \begin {gather*} \log (5) \left (x-x^2-\frac {1}{\log \left (\frac {\left (-2+x+e^2 x\right )^2}{x^2}\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[5]*(x - x^2 - Log[(-2 + x + E^2*x)^2/x^2]^(-1))

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fricas [B]  time = 0.68, size = 77, normalized size = 2.48 \begin {gather*} -\frac {{\left (x^{2} - x\right )} \log \relax (5) \log \left (\frac {x^{2} e^{4} + x^{2} + 2 \, {\left (x^{2} - 2 \, x\right )} e^{2} - 4 \, x + 4}{x^{2}}\right ) + \log \relax (5)}{\log \left (\frac {x^{2} e^{4} + x^{2} + 2 \, {\left (x^{2} - 2 \, x\right )} e^{2} - 4 \, x + 4}{x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 1.21, size = 125, normalized size = 4.03 \begin {gather*} -\frac {x^{2} \log \relax (5) \log \left (x^{2} e^{4} + 2 \, x^{2} e^{2} + x^{2} - 4 \, x e^{2} - 4 \, x + 4\right ) - x^{2} \log \relax (5) \log \left (x^{2}\right ) - x \log \relax (5) \log \left (x^{2} e^{4} + 2 \, x^{2} e^{2} + x^{2} - 4 \, x e^{2} - 4 \, x + 4\right ) + x \log \relax (5) \log \left (x^{2}\right ) + \log \relax (5)}{\log \left (x^{2} e^{4} + 2 \, x^{2} e^{2} + x^{2} - 4 \, x e^{2} - 4 \, x + 4\right ) - \log \left (x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.25, size = 42, normalized size = 1.35

method	result	size
derivativedivides	\(-\ln \relax (5) \left (x^{2}-x +\frac {1}{\ln \left ({\mathrm e}^{4}-\frac {4 \,{\mathrm e}^{2}}{x}+\frac {4}{x^{2}}+2 \,{\mathrm e}^{2}-\frac {4}{x}+1\right )}\right )\)	\(42\)
default	\(-\ln \relax (5) \left (x^{2}-x +\frac {1}{\ln \left ({\mathrm e}^{4}-\frac {4 \,{\mathrm e}^{2}}{x}+\frac {4}{x^{2}}+2 \,{\mathrm e}^{2}-\frac {4}{x}+1\right )}\right )\)	\(42\)
risch	\(-\ln \relax (5) x \left (x -1\right )-\frac {\ln \relax (5)}{\ln \left (\frac {x^{2} {\mathrm e}^{4}+\left (2 x^{2}-4 x \right ) {\mathrm e}^{2}+x^{2}-4 x +4}{x^{2}}\right )}\)	\(47\)
norman	\(\frac {x \ln \relax (5) \ln \left (\frac {x^{2} {\mathrm e}^{4}+\left (2 x^{2}-4 x \right ) {\mathrm e}^{2}+x^{2}-4 x +4}{x^{2}}\right )-x^{2} \ln \relax (5) \ln \left (\frac {x^{2} {\mathrm e}^{4}+\left (2 x^{2}-4 x \right ) {\mathrm e}^{2}+x^{2}-4 x +4}{x^{2}}\right )-\ln \relax (5)}{\ln \left (\frac {x^{2} {\mathrm e}^{4}+\left (2 x^{2}-4 x \right ) {\mathrm e}^{2}+x^{2}-4 x +4}{x^{2}}\right )}\)	\(119\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-2*x^3+x^2)*exp(2)-2*x^3+5*x^2-2*x)*ln(5)*ln((x^2*exp(2)^2+(2*x^2-4*x)*exp(2)+x^2-4*x+4)/x^2)^2+4*ln(5)
)/(x^2*exp(2)+x^2-2*x)/ln((x^2*exp(2)^2+(2*x^2-4*x)*exp(2)+x^2-4*x+4)/x^2)^2,x,method=_RETURNVERBOSE)

[Out]

-ln(5)*(x^2-x+1/ln(exp(2)^2-4*exp(2)/x+4/x^2+2*exp(2)-4/x+1))

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maxima [B]  time = 0.47, size = 217, normalized size = 7.00 \begin {gather*} -{\left (\frac {x^{2} {\left (e^{2} + 1\right )} + 4 \, x}{e^{4} + 2 \, e^{2} + 1} + \frac {8 \, \log \left (x {\left (e^{2} + 1\right )} - 2\right )}{e^{6} + 3 \, e^{4} + 3 \, e^{2} + 1}\right )} e^{2} \log \relax (5) + {\left (\frac {x}{e^{2} + 1} + \frac {2 \, \log \left (x {\left (e^{2} + 1\right )} - 2\right )}{e^{4} + 2 \, e^{2} + 1}\right )} e^{2} \log \relax (5) - {\left (\frac {x^{2} {\left (e^{2} + 1\right )} + 4 \, x}{e^{4} + 2 \, e^{2} + 1} + \frac {8 \, \log \left (x {\left (e^{2} + 1\right )} - 2\right )}{e^{6} + 3 \, e^{4} + 3 \, e^{2} + 1}\right )} \log \relax (5) + 5 \, {\left (\frac {x}{e^{2} + 1} + \frac {2 \, \log \left (x {\left (e^{2} + 1\right )} - 2\right )}{e^{4} + 2 \, e^{2} + 1}\right )} \log \relax (5) - \frac {2 \, \log \relax (5) \log \left (x {\left (e^{2} + 1\right )} - 2\right )}{e^{2} + 1} - \frac {\log \relax (5)}{2 \, {\left (\log \left (x {\left (e^{2} + 1\right )} - 2\right ) - \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((x^2*(e^2 + 1) + 4*x)/(e^4 + 2*e^2 + 1) + 8*log(x*(e^2 + 1) - 2)/(e^6 + 3*e^4 + 3*e^2 + 1))*e^2*log(5) + (x/
(e^2 + 1) + 2*log(x*(e^2 + 1) - 2)/(e^4 + 2*e^2 + 1))*e^2*log(5) - ((x^2*(e^2 + 1) + 4*x)/(e^4 + 2*e^2 + 1) +
8*log(x*(e^2 + 1) - 2)/(e^6 + 3*e^4 + 3*e^2 + 1))*log(5) + 5*(x/(e^2 + 1) + 2*log(x*(e^2 + 1) - 2)/(e^4 + 2*e^
2 + 1))*log(5) - 2*log(5)*log(x*(e^2 + 1) - 2)/(e^2 + 1) - 1/2*log(5)/(log(x*(e^2 + 1) - 2) - log(x))

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mupad [B]  time = 5.82, size = 50, normalized size = 1.61 \begin {gather*} x\,\ln \relax (5)-x^2\,\ln \relax (5)-\frac {\ln \relax (5)}{\ln \left (\frac {x^2\,{\mathrm {e}}^4-{\mathrm {e}}^2\,\left (4\,x-2\,x^2\right )-4\,x+x^2+4}{x^2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*log(5) - log(5)*log((x^2*exp(4) - exp(2)*(4*x - 2*x^2) - 4*x + x^2 + 4)/x^2)^2*(2*x - exp(2)*(x^2 - 2*x
^3) - 5*x^2 + 2*x^3))/(log((x^2*exp(4) - exp(2)*(4*x - 2*x^2) - 4*x + x^2 + 4)/x^2)^2*(x^2*exp(2) - 2*x + x^2)
),x)

[Out]

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

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sympy [B]  time = 0.19, size = 46, normalized size = 1.48 \begin {gather*} - x^{2} \log {\relax (5 )} + x \log {\relax (5 )} - \frac {\log {\relax (5 )}}{\log {\left (\frac {x^{2} + x^{2} e^{4} - 4 x + \left (2 x^{2} - 4 x\right ) e^{2} + 4}{x^{2}} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x**3+x**2)*exp(2)-2*x**3+5*x**2-2*x)*ln(5)*ln((x**2*exp(2)**2+(2*x**2-4*x)*exp(2)+x**2-4*x+4)/
x**2)**2+4*ln(5))/(x**2*exp(2)+x**2-2*x)/ln((x**2*exp(2)**2+(2*x**2-4*x)*exp(2)+x**2-4*x+4)/x**2)**2,x)

[Out]

-x**2*log(5) + x*log(5) - log(5)/log((x**2 + x**2*exp(4) - 4*x + (2*x**2 - 4*x)*exp(2) + 4)/x**2)

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