3.100.14 \(\int \frac {2 x-x^2+(8-6 x) \log (\frac {1}{4} (2 x^2-x^3))+(-4+2 x) \log ^2(\frac {1}{4} (2 x^2-x^3))}{-4 x^3+2 x^4} \, dx\)

Optimal. Leaf size=26 \[ \frac {x-\log ^2\left (\frac {1}{4} (2-x) x^2\right )}{2 x^2} \]

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Rubi [A]  time = 0.58, antiderivative size = 30, normalized size of antiderivative = 1.15, number of steps used = 46, number of rules used = 18, integrand size = 69, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {1593, 6688, 12, 14, 2514, 2494, 2390, 2301, 2394, 2315, 2495, 44, 30, 36, 31, 29, 2316, 2498} \begin {gather*} \frac {1}{2 x}-\frac {\log ^2\left (\frac {1}{4} (2-x) x^2\right )}{2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

1/(2*x) - Log[((2 - x)*x^2)/4]^2/(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 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 30

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

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 2301

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

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2316

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[((a + b*Log[-((c*d)/e)])*Log[d + e*
x])/e, x] + Dist[b, Int[Log[-((e*x)/d)]/(d + e*x), x], x] /; FreeQ[{a, b, c, d, e}, x] && GtQ[-((c*d)/e), 0]

Rule 2390

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

Rule 2394

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

Rule 2494

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]/((g_.) + (h_.)*(x_)), x_Sym
bol] :> Simp[(Log[g + h*x]*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r])/h, x] + (-Dist[(b*p*r)/h, Int[Log[g + h*x]/(a
+ b*x), x], x] - Dist[(d*q*r)/h, Int[Log[g + h*x]/(c + d*x), x], x]) /; FreeQ[{a, b, c, d, e, f, g, h, p, q,
r}, x] && NeQ[b*c - a*d, 0]

Rule 2495

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]*((g_.) + (h_.)*(x_))^(m_.),
x_Symbol] :> Simp[((g + h*x)^(m + 1)*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r])/(h*(m + 1)), x] + (-Dist[(b*p*r)/(
h*(m + 1)), Int[(g + h*x)^(m + 1)/(a + b*x), x], x] - Dist[(d*q*r)/(h*(m + 1)), Int[(g + h*x)^(m + 1)/(c + d*x
), x], x]) /; FreeQ[{a, b, c, d, e, f, g, h, m, p, q, r}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1]

Rule 2498

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]^(s_)*((g_.) + (h_.)*(x_))^(
m_.), x_Symbol] :> Simp[((g + h*x)^(m + 1)*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^s)/(h*(m + 1)), x] + (-Dist[(b
*p*r*s)/(h*(m + 1)), Int[((g + h*x)^(m + 1)*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^(s - 1))/(a + b*x), x], x] -
Dist[(d*q*r*s)/(h*(m + 1)), Int[((g + h*x)^(m + 1)*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^(s - 1))/(c + d*x), x]
, x]) /; FreeQ[{a, b, c, d, e, f, g, h, m, p, q, r, s}, x] && NeQ[b*c - a*d, 0] && IGtQ[s, 0] && NeQ[m, -1]

Rule 2514

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]^(s_.)*(RFx_), x_Symbol] :>
With[{u = ExpandIntegrand[Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^s, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a,
b, c, d, e, f, p, q, r, s}, x] && RationalFunctionQ[RFx, x] && IGtQ[s, 0]

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-x^2+(8-6 x) \log \left (\frac {1}{4} \left (2 x^2-x^3\right )\right )+(-4+2 x) \log ^2\left (\frac {1}{4} \left (2 x^2-x^3\right )\right )}{x^3 (-4+2 x)} \, dx\\ &=\int \frac {-x+\frac {(8-6 x) \log \left (-\frac {1}{4} (-2+x) x^2\right )}{-2+x}+2 \log ^2\left (-\frac {1}{4} (-2+x) x^2\right )}{2 x^3} \, dx\\ &=\frac {1}{2} \int \frac {-x+\frac {(8-6 x) \log \left (-\frac {1}{4} (-2+x) x^2\right )}{-2+x}+2 \log ^2\left (-\frac {1}{4} (-2+x) x^2\right )}{x^3} \, dx\\ &=\frac {1}{2} \int \left (-\frac {1}{x^2}-\frac {2 (-4+3 x) \log \left (-\frac {1}{4} (-2+x) x^2\right )}{(-2+x) x^3}+\frac {2 \log ^2\left (-\frac {1}{4} (-2+x) x^2\right )}{x^3}\right ) \, dx\\ &=\frac {1}{2 x}-\int \frac {(-4+3 x) \log \left (-\frac {1}{4} (-2+x) x^2\right )}{(-2+x) x^3} \, dx+\int \frac {\log ^2\left (-\frac {1}{4} (-2+x) x^2\right )}{x^3} \, dx\\ &=\frac {1}{2 x}-\frac {\log ^2\left (\frac {1}{4} (2-x) x^2\right )}{2 x^2}+2 \int \frac {\log \left (-\frac {1}{4} (-2+x) x^2\right )}{x^3} \, dx+\int \frac {\log \left (-\frac {1}{4} (-2+x) x^2\right )}{(-2+x) x^2} \, dx-\int \left (\frac {\log \left (-\frac {1}{4} (-2+x) x^2\right )}{4 (-2+x)}+\frac {2 \log \left (-\frac {1}{4} (-2+x) x^2\right )}{x^3}-\frac {\log \left (-\frac {1}{4} (-2+x) x^2\right )}{2 x^2}-\frac {\log \left (-\frac {1}{4} (-2+x) x^2\right )}{4 x}\right ) \, dx\\ &=\frac {1}{2 x}-\frac {\log \left (\frac {1}{4} (2-x) x^2\right )}{x^2}-\frac {\log ^2\left (\frac {1}{4} (2-x) x^2\right )}{2 x^2}-\frac {1}{4} \int \frac {\log \left (-\frac {1}{4} (-2+x) x^2\right )}{-2+x} \, dx+\frac {1}{4} \int \frac {\log \left (-\frac {1}{4} (-2+x) x^2\right )}{x} \, dx+\frac {1}{2} \int \frac {\log \left (-\frac {1}{4} (-2+x) x^2\right )}{x^2} \, dx+2 \int \frac {1}{x^3} \, dx-2 \int \frac {\log \left (-\frac {1}{4} (-2+x) x^2\right )}{x^3} \, dx+\int \frac {1}{(-2+x) x^2} \, dx+\int \left (\frac {\log \left (-\frac {1}{4} (-2+x) x^2\right )}{4 (-2+x)}-\frac {\log \left (-\frac {1}{4} (-2+x) x^2\right )}{2 x^2}-\frac {\log \left (-\frac {1}{4} (-2+x) x^2\right )}{4 x}\right ) \, dx\\ &=-\frac {1}{x^2}+\frac {1}{2 x}-\frac {\log \left (\frac {1}{4} (2-x) x^2\right )}{2 x}-\frac {1}{4} \log (-2+x) \log \left (\frac {1}{4} (2-x) x^2\right )+\frac {1}{4} \log (x) \log \left (\frac {1}{4} (2-x) x^2\right )-\frac {\log ^2\left (\frac {1}{4} (2-x) x^2\right )}{2 x^2}+\frac {1}{4} \int \frac {\log (-2+x)}{-2+x} \, dx-\frac {1}{4} \int \frac {\log (x)}{-2+x} \, dx+\frac {1}{4} \int \frac {\log \left (-\frac {1}{4} (-2+x) x^2\right )}{-2+x} \, dx-\frac {1}{4} \int \frac {\log \left (-\frac {1}{4} (-2+x) x^2\right )}{x} \, dx+\frac {1}{2} \int \frac {1}{(-2+x) x} \, dx+\frac {1}{2} \int \frac {\log (-2+x)}{x} \, dx-\frac {1}{2} \int \frac {\log (x)}{x} \, dx-\frac {1}{2} \int \frac {\log \left (-\frac {1}{4} (-2+x) x^2\right )}{x^2} \, dx-2 \int \frac {1}{x^3} \, dx+\int \left (\frac {1}{4 (-2+x)}-\frac {1}{2 x^2}-\frac {1}{4 x}\right ) \, dx+\int \frac {1}{x^2} \, dx-\int \frac {1}{(-2+x) x^2} \, dx\\ &=\frac {1}{4} \log (2-x)-\frac {1}{4} \log (2) \log (-2+x)+\frac {1}{2} \log (-2+x) \log \left (\frac {x}{2}\right )-\frac {\log (x)}{4}-\frac {\log ^2(x)}{4}-\frac {\log ^2\left (\frac {1}{4} (2-x) x^2\right )}{2 x^2}+\frac {1}{4} \int \frac {1}{-2+x} \, dx-\frac {1}{4} \int \frac {1}{x} \, dx-\frac {1}{4} \int \frac {\log (-2+x)}{-2+x} \, dx-\frac {1}{4} \int \frac {\log \left (\frac {x}{2}\right )}{-2+x} \, dx+\frac {1}{4} \int \frac {\log (x)}{-2+x} \, dx+\frac {1}{4} \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,-2+x\right )-\frac {1}{2} \int \frac {1}{(-2+x) x} \, dx-\frac {1}{2} \int \frac {\log (-2+x)}{x} \, dx-\frac {1}{2} \int \frac {\log \left (\frac {x}{2}\right )}{-2+x} \, dx+\frac {1}{2} \int \frac {\log (x)}{x} \, dx-\int \left (\frac {1}{4 (-2+x)}-\frac {1}{2 x^2}-\frac {1}{4 x}\right ) \, dx-\int \frac {1}{x^2} \, dx\\ &=\frac {1}{2 x}+\frac {1}{4} \log (2-x)+\frac {1}{8} \log ^2(-2+x)-\frac {\log (x)}{4}-\frac {\log ^2\left (\frac {1}{4} (2-x) x^2\right )}{2 x^2}+\frac {3}{4} \text {Li}_2\left (1-\frac {x}{2}\right )-\frac {1}{4} \int \frac {1}{-2+x} \, dx+\frac {1}{4} \int \frac {1}{x} \, dx+\frac {1}{4} \int \frac {\log \left (\frac {x}{2}\right )}{-2+x} \, dx-\frac {1}{4} \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,-2+x\right )+\frac {1}{2} \int \frac {\log \left (\frac {x}{2}\right )}{-2+x} \, dx\\ &=\frac {1}{2 x}-\frac {\log ^2\left (\frac {1}{4} (2-x) x^2\right )}{2 x^2}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.12, size = 61, normalized size = 2.35 \begin {gather*} \frac {1}{2} \left (\frac {1}{x}+\frac {\log \left (\frac {1}{4} (2-x) x^2\right )}{x}-\frac {\log ^2\left (\frac {1}{4} (2-x) x^2\right )}{x^2}-\frac {\log \left (-\frac {1}{4} (-2+x) x^2\right )}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.54, size = 23, normalized size = 0.88 \begin {gather*} -\frac {\log \left (-\frac {1}{4} \, x^{3} + \frac {1}{2} \, x^{2}\right )^{2} - x}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x-4)*log(-1/4*x^3+1/2*x^2)^2+(-6*x+8)*log(-1/4*x^3+1/2*x^2)-x^2+2*x)/(2*x^4-4*x^3),x, algorithm=
"fricas")

[Out]

-1/2*(log(-1/4*x^3 + 1/2*x^2)^2 - x)/x^2

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giac [A]  time = 0.18, size = 25, normalized size = 0.96 \begin {gather*} -\frac {\log \left (-\frac {1}{4} \, x^{3} + \frac {1}{2} \, x^{2}\right )^{2}}{2 \, x^{2}} + \frac {1}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x-4)*log(-1/4*x^3+1/2*x^2)^2+(-6*x+8)*log(-1/4*x^3+1/2*x^2)-x^2+2*x)/(2*x^4-4*x^3),x, algorithm=
"giac")

[Out]

-1/2*log(-1/4*x^3 + 1/2*x^2)^2/x^2 + 1/2/x

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maple [A]  time = 0.12, size = 25, normalized size = 0.96

method	result	size
norman	\(\frac {\frac {x}{2}-\frac {\ln \left (-\frac {1}{4} x^{3}+\frac {1}{2} x^{2}\right )^{2}}{2}}{x^{2}}\)	\(25\)
risch	\(-\frac {\ln \left (-\frac {1}{4} x^{3}+\frac {1}{2} x^{2}\right )^{2}}{2 x^{2}}+\frac {1}{2 x}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x-4)*ln(-1/4*x^3+1/2*x^2)^2+(-6*x+8)*ln(-1/4*x^3+1/2*x^2)-x^2+2*x)/(2*x^4-4*x^3),x,method=_RETURNVERBO
SE)

[Out]

(1/2*x-1/2*ln(-1/4*x^3+1/2*x^2)^2)/x^2

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maxima [B]  time = 0.49, size = 53, normalized size = 2.04 \begin {gather*} -\frac {4 \, \log \relax (2)^{2} - 8 \, \log \relax (2) \log \relax (x) + 4 \, \log \relax (x)^{2} - 4 \, {\left (\log \relax (2) - \log \relax (x)\right )} \log \left (-x + 2\right ) + \log \left (-x + 2\right )^{2}}{2 \, x^{2}} + \frac {1}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x-4)*log(-1/4*x^3+1/2*x^2)^2+(-6*x+8)*log(-1/4*x^3+1/2*x^2)-x^2+2*x)/(2*x^4-4*x^3),x, algorithm=
"maxima")

[Out]

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

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mupad [B]  time = 5.75, size = 24, normalized size = 0.92 \begin {gather*} \frac {\frac {x}{2}-\frac {{\ln \left (\frac {x^2}{2}-\frac {x^3}{4}\right )}^2}{2}}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x + log(x^2/2 - x^3/4)^2*(2*x - 4) - log(x^2/2 - x^3/4)*(6*x - 8) - x^2)/(4*x^3 - 2*x^4),x)

[Out]

(x/2 - log(x^2/2 - x^3/4)^2/2)/x^2

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sympy [A]  time = 0.17, size = 22, normalized size = 0.85 \begin {gather*} \frac {1}{2 x} - \frac {\log {\left (- \frac {x^{3}}{4} + \frac {x^{2}}{2} \right )}^{2}}{2 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x-4)*ln(-1/4*x**3+1/2*x**2)**2+(-6*x+8)*ln(-1/4*x**3+1/2*x**2)-x**2+2*x)/(2*x**4-4*x**3),x)

[Out]

1/(2*x) - log(-x**3/4 + x**2/2)**2/(2*x**2)

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