3.60.25 \(\int \frac {-4 x^3+4 x^4-x^5+(8 x^3-8 x^4+2 x^5) \log (x)+(40-40 x) \log ^2(x)}{(4 x^2-4 x^3+x^4) \log ^2(x)} \, dx\)

Optimal. Leaf size=26 \[ 9+e^3-\frac {20}{2 x-x^2}+\frac {x^2}{\log (x)} \]

________________________________________________________________________________________

Rubi [A]  time = 0.35, antiderivative size = 21, normalized size of antiderivative = 0.81, number of steps used = 10, number of rules used = 7, integrand size = 66, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.106, Rules used = {1594, 27, 6688, 74, 2306, 2309, 2178} \begin {gather*} \frac {x^2}{\log (x)}-\frac {20}{(2-x) x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4*x^3 + 4*x^4 - x^5 + (8*x^3 - 8*x^4 + 2*x^5)*Log[x] + (40 - 40*x)*Log[x]^2)/((4*x^2 - 4*x^3 + x^4)*Log[
x]^2),x]

[Out]

-20/((2 - x)*x) + x^2/Log[x]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 74

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] &
& EqQ[a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 0]

Rule 1594

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

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2306

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log
[c*x^n])^(p + 1))/(b*d*n*(p + 1)), x] - Dist[(m + 1)/(b*n*(p + 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] && LtQ[p, -1]

Rule 2309

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[E^((m + 1)*x)*(a
+ b*x)^p, x], x, Log[c*x]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[m]

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 x^3+4 x^4-x^5+\left (8 x^3-8 x^4+2 x^5\right ) \log (x)+(40-40 x) \log ^2(x)}{x^2 \left (4-4 x+x^2\right ) \log ^2(x)} \, dx\\ &=\int \frac {-4 x^3+4 x^4-x^5+\left (8 x^3-8 x^4+2 x^5\right ) \log (x)+(40-40 x) \log ^2(x)}{(-2+x)^2 x^2 \log ^2(x)} \, dx\\ &=\int \left (-\frac {40 (-1+x)}{(-2+x)^2 x^2}-\frac {x}{\log ^2(x)}+\frac {2 x}{\log (x)}\right ) \, dx\\ &=2 \int \frac {x}{\log (x)} \, dx-40 \int \frac {-1+x}{(-2+x)^2 x^2} \, dx-\int \frac {x}{\log ^2(x)} \, dx\\ &=-\frac {20}{(2-x) x}+\frac {x^2}{\log (x)}-2 \int \frac {x}{\log (x)} \, dx+2 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )\\ &=-\frac {20}{(2-x) x}+2 \text {Ei}(2 \log (x))+\frac {x^2}{\log (x)}-2 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )\\ &=-\frac {20}{(2-x) x}+\frac {x^2}{\log (x)}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 21, normalized size = 0.81 \begin {gather*} \frac {10}{-2+x}-\frac {10}{x}+\frac {x^2}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4*x^3 + 4*x^4 - x^5 + (8*x^3 - 8*x^4 + 2*x^5)*Log[x] + (40 - 40*x)*Log[x]^2)/((4*x^2 - 4*x^3 + x^4
)*Log[x]^2),x]

[Out]

10/(-2 + x) - 10/x + x^2/Log[x]

________________________________________________________________________________________

fricas [A]  time = 0.66, size = 27, normalized size = 1.04 \begin {gather*} \frac {x^{4} - 2 \, x^{3} + 20 \, \log \relax (x)}{{\left (x^{2} - 2 \, x\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-40*x+40)*log(x)^2+(2*x^5-8*x^4+8*x^3)*log(x)-x^5+4*x^4-4*x^3)/(x^4-4*x^3+4*x^2)/log(x)^2,x, algor
ithm="fricas")

[Out]

(x^4 - 2*x^3 + 20*log(x))/((x^2 - 2*x)*log(x))

________________________________________________________________________________________

giac [A]  time = 0.19, size = 21, normalized size = 0.81 \begin {gather*} \frac {x^{2}}{\log \relax (x)} + \frac {10}{x - 2} - \frac {10}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-40*x+40)*log(x)^2+(2*x^5-8*x^4+8*x^3)*log(x)-x^5+4*x^4-4*x^3)/(x^4-4*x^3+4*x^2)/log(x)^2,x, algor
ithm="giac")

[Out]

x^2/log(x) + 10/(x - 2) - 10/x

________________________________________________________________________________________

maple [A]  time = 0.05, size = 20, normalized size = 0.77

method	result	size
risch	\(\frac {20}{\left (x -2\right ) x}+\frac {x^{2}}{\ln \relax (x )}\)	\(20\)
default	\(\frac {x^{2}}{\ln \relax (x )}+\frac {10}{x -2}-\frac {10}{x}\)	\(22\)
norman	\(\frac {x^{4}-2 x^{3}+20 \ln \relax (x )}{x \left (x -2\right ) \ln \relax (x )}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-40*x+40)*ln(x)^2+(2*x^5-8*x^4+8*x^3)*ln(x)-x^5+4*x^4-4*x^3)/(x^4-4*x^3+4*x^2)/ln(x)^2,x,method=_RETURNV
ERBOSE)

[Out]

20/(x-2)/x+x^2/ln(x)

________________________________________________________________________________________

maxima [A]  time = 0.38, size = 27, normalized size = 1.04 \begin {gather*} \frac {x^{4} - 2 \, x^{3} + 20 \, \log \relax (x)}{{\left (x^{2} - 2 \, x\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-40*x+40)*log(x)^2+(2*x^5-8*x^4+8*x^3)*log(x)-x^5+4*x^4-4*x^3)/(x^4-4*x^3+4*x^2)/log(x)^2,x, algor
ithm="maxima")

[Out]

(x^4 - 2*x^3 + 20*log(x))/((x^2 - 2*x)*log(x))

________________________________________________________________________________________

mupad [B]  time = 4.37, size = 19, normalized size = 0.73 \begin {gather*} \frac {20}{x\,\left (x-2\right )}+\frac {x^2}{\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*x^3 - log(x)*(8*x^3 - 8*x^4 + 2*x^5) - 4*x^4 + x^5 + log(x)^2*(40*x - 40))/(log(x)^2*(4*x^2 - 4*x^3 +
x^4)),x)

[Out]

20/(x*(x - 2)) + x^2/log(x)

________________________________________________________________________________________

sympy [A]  time = 0.13, size = 14, normalized size = 0.54 \begin {gather*} \frac {x^{2}}{\log {\relax (x )}} + \frac {20}{x^{2} - 2 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-40*x+40)*ln(x)**2+(2*x**5-8*x**4+8*x**3)*ln(x)-x**5+4*x**4-4*x**3)/(x**4-4*x**3+4*x**2)/ln(x)**2,
x)

[Out]

x**2/log(x) + 20/(x**2 - 2*x)

________________________________________________________________________________________