\(\int \frac {4 x-6 x^2+24 x^3+(-1+4 x) \log (\frac {1}{2} (-x+4 x^2))}{-1+4 x} \, dx\) [1117]

Optimal result

Integrand size = 42, antiderivative size = 23 \[ \int \frac {4 x-6 x^2+24 x^3+(-1+4 x) \log \left (\frac {1}{2} \left (-x+4 x^2\right )\right )}{-1+4 x} \, dx=-x+x \left (2 x^2+\log \left (\frac {1}{2} x (-1+4 x)\right )\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.119, Rules used = {6874, 785, 2579, 31, 8} \[ \int \frac {4 x-6 x^2+24 x^3+(-1+4 x) \log \left (\frac {1}{2} \left (-x+4 x^2\right )\right )}{-1+4 x} \, dx=2 x^3-x+x \log \left (-\frac {1}{2} (1-4 x) x\right ) \]

[In]

Int[(4*x - 6*x^2 + 24*x^3 + (-1 + 4*x)*Log[(-x + 4*x^2)/2])/(-1 + 4*x),x]

[Out]

-x + 2*x^3 + x*Log[-1/2*((1 - 4*x)*x)]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 31

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

Rule 785

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(d + e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && N
eQ[b^2 - 4*a*c, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rule 2579

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]^(s_.), x_Symbol] :> Simp[(a
+ b*x)*(Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^s/b), x] + (Dist[q*r*s*((b*c - a*d)/b), Int[Log[e*(f*(a + b*x)^p
*(c + d*x)^q)^r]^(s - 1)/(c + d*x), x], x] - Dist[r*s*(p + q), Int[Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^(s - 1
), x], x]) /; FreeQ[{a, b, c, d, e, f, p, q, r, s}, x] && NeQ[b*c - a*d, 0] && NeQ[p + q, 0] && IGtQ[s, 0] &&
LtQ[s, 4]

Rule 6874

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

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.96 \[ \int \frac {4 x-6 x^2+24 x^3+(-1+4 x) \log \left (\frac {1}{2} \left (-x+4 x^2\right )\right )}{-1+4 x} \, dx=-\frac {9}{32}-x+2 x^3-\frac {1}{4} \log (1-4 x)+\frac {1}{4} \log (-1+4 x)+x \log \left (\frac {1}{2} x (-1+4 x)\right ) \]

[In]

Integrate[(4*x - 6*x^2 + 24*x^3 + (-1 + 4*x)*Log[(-x + 4*x^2)/2])/(-1 + 4*x),x]

[Out]

-9/32 - x + 2*x^3 - Log[1 - 4*x]/4 + Log[-1 + 4*x]/4 + x*Log[(x*(-1 + 4*x))/2]

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96

method	result	size
norman	\(\ln \left (2 x^{2}-\frac {1}{2} x \right ) x -x +2 x^{3}\)	\(22\)
risch	\(\ln \left (2 x^{2}-\frac {1}{2} x \right ) x -x +2 x^{3}\)	\(22\)
parallelrisch	\(2 x^{3}-\frac {1}{2}+\ln \left (2 x^{2}-\frac {1}{2} x \right ) x -x\)	\(23\)
default	\(2 x^{3}-x -x \ln \left (2\right )+x \ln \left (4 x^{2}-x \right )\)	\(27\)
parts	\(2 x^{3}-x -x \ln \left (2\right )+x \ln \left (4 x^{2}-x \right )\)	\(27\)

[In]

int(((-1+4*x)*ln(2*x^2-1/2*x)+24*x^3-6*x^2+4*x)/(-1+4*x),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {4 x-6 x^2+24 x^3+(-1+4 x) \log \left (\frac {1}{2} \left (-x+4 x^2\right )\right )}{-1+4 x} \, dx=2 \, x^{3} + x \log \left (2 \, x^{2} - \frac {1}{2} \, x\right ) - x \]

[In]

integrate(((-1+4*x)*log(2*x^2-1/2*x)+24*x^3-6*x^2+4*x)/(-1+4*x),x, algorithm="fricas")

[Out]

2*x^3 + x*log(2*x^2 - 1/2*x) - x

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {4 x-6 x^2+24 x^3+(-1+4 x) \log \left (\frac {1}{2} \left (-x+4 x^2\right )\right )}{-1+4 x} \, dx=2 x^{3} + x \log {\left (2 x^{2} - \frac {x}{2} \right )} - x \]

[In]

integrate(((-1+4*x)*ln(2*x**2-1/2*x)+24*x**3-6*x**2+4*x)/(-1+4*x),x)

[Out]

2*x**3 + x*log(2*x**2 - x/2) - x

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.70 \[ \int \frac {4 x-6 x^2+24 x^3+(-1+4 x) \log \left (\frac {1}{2} \left (-x+4 x^2\right )\right )}{-1+4 x} \, dx=2 \, x^{3} - x {\left (\log \left (2\right ) + 2\right )} + \frac {1}{4} \, {\left (4 \, x - 1\right )} \log \left (4 \, x - 1\right ) + x \log \left (x\right ) + x + \frac {1}{4} \, \log \left (4 \, x - 1\right ) \]

[In]

integrate(((-1+4*x)*log(2*x^2-1/2*x)+24*x^3-6*x^2+4*x)/(-1+4*x),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {4 x-6 x^2+24 x^3+(-1+4 x) \log \left (\frac {1}{2} \left (-x+4 x^2\right )\right )}{-1+4 x} \, dx=2 \, x^{3} + x \log \left (2 \, x^{2} - \frac {1}{2} \, x\right ) - x \]

[In]

integrate(((-1+4*x)*log(2*x^2-1/2*x)+24*x^3-6*x^2+4*x)/(-1+4*x),x, algorithm="giac")

[Out]

2*x^3 + x*log(2*x^2 - 1/2*x) - x

Mupad [B] (verification not implemented)

Time = 9.15 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {4 x-6 x^2+24 x^3+(-1+4 x) \log \left (\frac {1}{2} \left (-x+4 x^2\right )\right )}{-1+4 x} \, dx=x\,\left (\ln \left (2\,x^2-\frac {x}{2}\right )-1\right )+2\,x^3 \]

[In]

int((4*x + log(2*x^2 - x/2)*(4*x - 1) - 6*x^2 + 24*x^3)/(4*x - 1),x)

[Out]

x*(log(2*x^2 - x/2) - 1) + 2*x^3