\(\int \frac {18 x-216 x^2+(72 x-288 x^2) \log (-x+4 x^2)}{-1+4 x} \, dx\) [2914]

Optimal result

Integrand size = 37, antiderivative size = 21 \[ \int \frac {18 x-216 x^2+\left (72 x-288 x^2\right ) \log \left (-x+4 x^2\right )}{-1+4 x} \, dx=2+9 x^2 \left (1-4 \log \left (-x+4 x^2\right )\right ) \]

[Out]

9*x^2*(1-4*ln(4*x^2-x))+2

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95, number of steps used = 10, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.189, Rules used = {6873, 12, 6874, 78, 2581, 30, 45} \[ \int \frac {18 x-216 x^2+\left (72 x-288 x^2\right ) \log \left (-x+4 x^2\right )}{-1+4 x} \, dx=9 x^2-36 x^2 \log (-((1-4 x) x)) \]

[In]

Int[(18*x - 216*x^2 + (72*x - 288*x^2)*Log[-x + 4*x^2])/(-1 + 4*x),x]

[Out]

9*x^2 - 36*x^2*Log[-((1 - 4*x)*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 30

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

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 2581

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 6873

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

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {18 x (-1+12 x-4 \log (x (-1+4 x))+16 x \log (x (-1+4 x)))}{1-4 x} \, dx \\ & = 18 \int \frac {x (-1+12 x-4 \log (x (-1+4 x))+16 x \log (x (-1+4 x)))}{1-4 x} \, dx \\ & = 18 \int \left (\frac {x (-1+12 x)}{1-4 x}-4 x \log (x (-1+4 x))\right ) \, dx \\ & = 18 \int \frac {x (-1+12 x)}{1-4 x} \, dx-72 \int x \log (x (-1+4 x)) \, dx \\ & = -36 x^2 \log (-((1-4 x) x))+18 \int \left (-\frac {1}{2}-3 x-\frac {1}{2 (-1+4 x)}\right ) \, dx+36 \int x \, dx+144 \int \frac {x^2}{-1+4 x} \, dx \\ & = -9 x-9 x^2-\frac {9}{4} \log (1-4 x)-36 x^2 \log (-((1-4 x) x))+144 \int \left (\frac {1}{16}+\frac {x}{4}+\frac {1}{16 (-1+4 x)}\right ) \, dx \\ & = 9 x^2-36 x^2 \log (-((1-4 x) x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {18 x-216 x^2+\left (72 x-288 x^2\right ) \log \left (-x+4 x^2\right )}{-1+4 x} \, dx=-18 \left (-\frac {x^2}{2}+2 x^2 \log (x (-1+4 x))\right ) \]

[In]

Integrate[(18*x - 216*x^2 + (72*x - 288*x^2)*Log[-x + 4*x^2])/(-1 + 4*x),x]

[Out]

-18*(-1/2*x^2 + 2*x^2*Log[x*(-1 + 4*x)])

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05

method	result	size
default	\(-36 \ln \left (4 x^{2}-x \right ) x^{2}+9 x^{2}\)	\(22\)
norman	\(-36 \ln \left (4 x^{2}-x \right ) x^{2}+9 x^{2}\)	\(22\)
risch	\(-36 \ln \left (4 x^{2}-x \right ) x^{2}+9 x^{2}\)	\(22\)
parts	\(-36 \ln \left (4 x^{2}-x \right ) x^{2}+9 x^{2}\)	\(22\)
parallelrisch	\(-36 \ln \left (4 x^{2}-x \right ) x^{2}-\frac {9}{16}+9 x^{2}\)	\(23\)

[In]

int(((-288*x^2+72*x)*ln(4*x^2-x)-216*x^2+18*x)/(-1+4*x),x,method=_RETURNVERBOSE)

[Out]

-36*ln(4*x^2-x)*x^2+9*x^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {18 x-216 x^2+\left (72 x-288 x^2\right ) \log \left (-x+4 x^2\right )}{-1+4 x} \, dx=-36 \, x^{2} \log \left (4 \, x^{2} - x\right ) + 9 \, x^{2} \]

[In]

integrate(((-288*x^2+72*x)*log(4*x^2-x)-216*x^2+18*x)/(-1+4*x),x, algorithm="fricas")

[Out]

-36*x^2*log(4*x^2 - x) + 9*x^2

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {18 x-216 x^2+\left (72 x-288 x^2\right ) \log \left (-x+4 x^2\right )}{-1+4 x} \, dx=- 36 x^{2} \log {\left (4 x^{2} - x \right )} + 9 x^{2} \]

[In]

integrate(((-288*x**2+72*x)*ln(4*x**2-x)-216*x**2+18*x)/(-1+4*x),x)

[Out]

-36*x**2*log(4*x**2 - x) + 9*x**2

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.71 \[ \int \frac {18 x-216 x^2+\left (72 x-288 x^2\right ) \log \left (-x+4 x^2\right )}{-1+4 x} \, dx=-36 \, x^{2} \log \left (x\right ) + 9 \, x^{2} - \frac {9}{4} \, {\left (16 \, x^{2} - 1\right )} \log \left (4 \, x - 1\right ) - \frac {9}{4} \, \log \left (4 \, x - 1\right ) \]

[In]

integrate(((-288*x^2+72*x)*log(4*x^2-x)-216*x^2+18*x)/(-1+4*x),x, algorithm="maxima")

[Out]

-36*x^2*log(x) + 9*x^2 - 9/4*(16*x^2 - 1)*log(4*x - 1) - 9/4*log(4*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {18 x-216 x^2+\left (72 x-288 x^2\right ) \log \left (-x+4 x^2\right )}{-1+4 x} \, dx=-36 \, x^{2} \log \left (4 \, x^{2} - x\right ) + 9 \, x^{2} \]

[In]

integrate(((-288*x^2+72*x)*log(4*x^2-x)-216*x^2+18*x)/(-1+4*x),x, algorithm="giac")

[Out]

-36*x^2*log(4*x^2 - x) + 9*x^2

Mupad [B] (verification not implemented)

Time = 8.88 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {18 x-216 x^2+\left (72 x-288 x^2\right ) \log \left (-x+4 x^2\right )}{-1+4 x} \, dx=-9\,x^2\,\left (4\,\ln \left (4\,x^2-x\right )-1\right ) \]

[In]

int((18*x + log(4*x^2 - x)*(72*x - 288*x^2) - 216*x^2)/(4*x - 1),x)

[Out]

-9*x^2*(4*log(4*x^2 - x) - 1)