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

Optimal. Leaf size=23 \[ \left (x^2+\log \left (\frac {3 e^5 (-x+\log (x))}{2 x}\right )\right )^2 \]

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Rubi [A]  time = 0.32, antiderivative size = 21, normalized size of antiderivative = 0.91, number of steps used = 4, number of rules used = 4, integrand size = 78, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {2561, 6688, 12, 6686} \begin {gather*} \left (x^2+\log \left (-\frac {3}{2} \left (1-\frac {\log (x)}{x}\right )\right )+5\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5 + x^2 + Log[(-3*(1 - Log[x]/x))/2])^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 2561

Int[(u_.)*((a_.)*(x_)^(m_.) + Log[(c_.)*(x_)^(n_.)]^(q_.)*(b_.)*(x_)^(r_.))^(p_.), x_Symbol] :> Int[u*x^(p*r)*
(a*x^(m - r) + b*Log[c*x^n]^q)^p, x] /; FreeQ[{a, b, c, m, n, p, q, r}, x] && IntegerQ[p]

Rule 6686

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

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

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Mathematica [A]  time = 0.03, size = 20, normalized size = 0.87 \begin {gather*} \left (5+x^2+\log \left (\frac {3}{2} \left (-1+\frac {\log (x)}{x}\right )\right )\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5 + x^2 + Log[(3*(-1 + Log[x]/x))/2])^2

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fricas [B]  time = 0.73, size = 45, normalized size = 1.96 \begin {gather*} x^{4} + 2 \, x^{2} \log \left (-\frac {3 \, {\left (x e^{5} - e^{5} \log \relax (x)\right )}}{2 \, x}\right ) + \log \left (-\frac {3 \, {\left (x e^{5} - e^{5} \log \relax (x)\right )}}{2 \, x}\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x^2-2)*log(x)-4*x^3+2)*log(1/2*(3*exp(5)*log(x)-3*x*exp(5))/x)+(4*x^4-2*x^2)*log(x)-4*x^5+2*x^2
)/(x*log(x)-x^2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.37, size = 75, normalized size = 3.26 \begin {gather*} x^{4} - 2 \, x^{2} {\left (\log \relax (2) - 5\right )} - 2 \, x^{2} \log \relax (x) + 2 \, {\left (\log \relax (2) - 5\right )} \log \relax (x) + \log \relax (x)^{2} - 2 \, {\left (\log \relax (2) - 5\right )} \log \left (-x + \log \relax (x)\right ) + 2 \, {\left (x^{2} - \log \relax (x)\right )} \log \left (-3 \, x + 3 \, \log \relax (x)\right ) + \log \left (-3 \, x + 3 \, \log \relax (x)\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x^2-2)*log(x)-4*x^3+2)*log(1/2*(3*exp(5)*log(x)-3*x*exp(5))/x)+(4*x^4-2*x^2)*log(x)-4*x^5+2*x^2
)/(x*log(x)-x^2),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.19, size = 574, normalized size = 24.96

method	result	size
risch	\(-2 x^{2} \ln \relax (x )+2 x^{2} \ln \relax (3)-10 \ln \relax (x )+\ln \relax (x )^{2}+x^{4}+10 x^{2}+2 \ln \relax (2) \ln \relax (x )-2 \ln \relax (3) \ln \relax (x )-2 x^{2} \ln \relax (2)-i \pi \,x^{2} \mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left (\ln \relax (x )-x \right )\right ) \mathrm {csgn}\left (\frac {i \left (\ln \relax (x )-x \right )}{x}\right )-i \pi \ln \left (\ln \relax (x )-x \right ) \mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left (\ln \relax (x )-x \right )\right ) \mathrm {csgn}\left (\frac {i \left (\ln \relax (x )-x \right )}{x}\right )+i \ln \relax (x ) \pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left (\ln \relax (x )-x \right )\right ) \mathrm {csgn}\left (\frac {i \left (\ln \relax (x )-x \right )}{x}\right )+\left (2 x^{2}-2 \ln \relax (x )\right ) \ln \left (x -\ln \relax (x )\right )-2 \ln \relax (2) \ln \left (\ln \relax (x )-x \right )+2 \ln \relax (3) \ln \left (\ln \relax (x )-x \right )+10 \ln \left (\ln \relax (x )-x \right )+i \ln \relax (x ) \pi \mathrm {csgn}\left (\frac {i \left (\ln \relax (x )-x \right )}{x}\right )^{3}-2 i \pi \,x^{2} \mathrm {csgn}\left (\frac {i \left (\ln \relax (x )-x \right )}{x}\right )^{2}-2 i \pi \ln \left (\ln \relax (x )-x \right ) \mathrm {csgn}\left (\frac {i \left (\ln \relax (x )-x \right )}{x}\right )^{2}-i \pi \,x^{2} \mathrm {csgn}\left (\frac {i \left (\ln \relax (x )-x \right )}{x}\right )^{3}-i \pi \ln \left (\ln \relax (x )-x \right ) \mathrm {csgn}\left (\frac {i \left (\ln \relax (x )-x \right )}{x}\right )^{3}+2 i \ln \relax (x ) \pi \mathrm {csgn}\left (\frac {i \left (\ln \relax (x )-x \right )}{x}\right )^{2}+i \pi \ln \left (\ln \relax (x )-x \right ) \mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left (\ln \relax (x )-x \right )}{x}\right )^{2}+2 i \pi \,x^{2}-2 i \ln \relax (x ) \pi +2 i \pi \ln \left (\ln \relax (x )-x \right )+\ln \left (x -\ln \relax (x )\right )^{2}-i \ln \relax (x ) \pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left (\ln \relax (x )-x \right )}{x}\right )^{2}+i \pi \,x^{2} \mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left (\ln \relax (x )-x \right )}{x}\right )^{2}+i \ln \relax (x ) \pi \,\mathrm {csgn}\left (i \left (\ln \relax (x )-x \right )\right ) \mathrm {csgn}\left (\frac {i \left (\ln \relax (x )-x \right )}{x}\right )^{2}-i \pi \,x^{2} \mathrm {csgn}\left (i \left (\ln \relax (x )-x \right )\right ) \mathrm {csgn}\left (\frac {i \left (\ln \relax (x )-x \right )}{x}\right )^{2}-i \pi \ln \left (\ln \relax (x )-x \right ) \mathrm {csgn}\left (i \left (\ln \relax (x )-x \right )\right ) \mathrm {csgn}\left (\frac {i \left (\ln \relax (x )-x \right )}{x}\right )^{2}\)	\(574\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((4*x^2-2)*ln(x)-4*x^3+2)*ln(1/2*(3*exp(5)*ln(x)-3*x*exp(5))/x)+(4*x^4-2*x^2)*ln(x)-4*x^5+2*x^2)/(x*ln(x)
-x^2),x,method=_RETURNVERBOSE)

[Out]

-2*x^2*ln(x)+2*x^2*ln(3)-10*ln(x)+ln(x)^2+x^4+10*x^2+2*ln(2)*ln(x)-2*ln(3)*ln(x)-2*x^2*ln(2)+I*Pi*ln(ln(x)-x)*
csgn(I/x)*csgn(I*(ln(x)-x)/x)^2-I*ln(x)*Pi*csgn(I/x)*csgn(I*(ln(x)-x)/x)^2+I*ln(x)*Pi*csgn(I*(ln(x)-x)/x)^3+I*
Pi*x^2*csgn(I/x)*csgn(I*(ln(x)-x)/x)^2+(2*x^2-2*ln(x))*ln(x-ln(x))-2*ln(2)*ln(ln(x)-x)+2*ln(3)*ln(ln(x)-x)+10*
ln(ln(x)-x)-2*I*Pi*x^2*csgn(I*(ln(x)-x)/x)^2-2*I*Pi*ln(ln(x)-x)*csgn(I*(ln(x)-x)/x)^2-I*Pi*x^2*csgn(I*(ln(x)-x
)/x)^3-I*Pi*ln(ln(x)-x)*csgn(I*(ln(x)-x)/x)^3+2*I*ln(x)*Pi*csgn(I*(ln(x)-x)/x)^2+2*I*Pi*x^2-2*I*ln(x)*Pi+2*I*P
i*ln(ln(x)-x)+I*ln(x)*Pi*csgn(I*(ln(x)-x))*csgn(I*(ln(x)-x)/x)^2-I*Pi*x^2*csgn(I*(ln(x)-x))*csgn(I*(ln(x)-x)/x
)^2-I*Pi*ln(ln(x)-x)*csgn(I*(ln(x)-x))*csgn(I*(ln(x)-x)/x)^2+ln(x-ln(x))^2+I*ln(x)*Pi*csgn(I/x)*csgn(I*(ln(x)-
x))*csgn(I*(ln(x)-x)/x)-I*Pi*x^2*csgn(I/x)*csgn(I*(ln(x)-x))*csgn(I*(ln(x)-x)/x)-I*Pi*ln(ln(x)-x)*csgn(I/x)*cs
gn(I*(ln(x)-x))*csgn(I*(ln(x)-x)/x)

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maxima [B]  time = 0.57, size = 69, normalized size = 3.00 \begin {gather*} x^{4} + 2 \, x^{2} {\left (\log \relax (3) - \log \relax (2) + 5\right )} - 2 \, {\left (x^{2} + \log \relax (3) - \log \relax (2) + 5\right )} \log \relax (x) + \log \relax (x)^{2} + 2 \, {\left (x^{2} + \log \relax (3) - \log \relax (2) - \log \relax (x) + 5\right )} \log \left (-x + \log \relax (x)\right ) + \log \left (-x + \log \relax (x)\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x^2-2)*log(x)-4*x^3+2)*log(1/2*(3*exp(5)*log(x)-3*x*exp(5))/x)+(4*x^4-2*x^2)*log(x)-4*x^5+2*x^2
)/(x*log(x)-x^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.50, size = 24, normalized size = 1.04 \begin {gather*} {\left (\ln \left (-\frac {3\,x\,{\mathrm {e}}^5-3\,{\mathrm {e}}^5\,\ln \relax (x)}{2\,x}\right )+x^2\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 0.39, size = 53, normalized size = 2.30 \begin {gather*} x^{4} + 2 x^{2} \log {\left (\frac {- \frac {3 x e^{5}}{2} + \frac {3 e^{5} \log {\relax (x )}}{2}}{x} \right )} + \log {\left (\frac {- \frac {3 x e^{5}}{2} + \frac {3 e^{5} \log {\relax (x )}}{2}}{x} \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x**2-2)*ln(x)-4*x**3+2)*ln(1/2*(3*exp(5)*ln(x)-3*x*exp(5))/x)+(4*x**4-2*x**2)*ln(x)-4*x**5+2*x*
*2)/(x*ln(x)-x**2),x)

[Out]

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

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