\(\int \frac {1}{-1+4 x-4 x^2+16 x^3} \, dx\) [22]

Optimal result

Integrand size = 17, antiderivative size = 31 \[ \int \frac {1}{-1+4 x-4 x^2+16 x^3} \, dx=-\frac {1}{10} \arctan (2 x)+\frac {1}{5} \log (1-4 x)-\frac {1}{10} \log \left (1+4 x^2\right ) \]

[Out]

-1/10*arctan(2*x)+1/5*ln(1-4*x)-1/10*ln(4*x^2+1)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {2083, 649, 209, 266} \[ \int \frac {1}{-1+4 x-4 x^2+16 x^3} \, dx=-\frac {1}{10} \arctan (2 x)-\frac {1}{10} \log \left (4 x^2+1\right )+\frac {1}{5} \log (1-4 x) \]

[In]

Int[(-1 + 4*x - 4*x^2 + 16*x^3)^(-1),x]

[Out]

-1/10*ArcTan[2*x] + Log[1 - 4*x]/5 - Log[1 + 4*x^2]/10

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 266

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

Rule 649

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[(-a)*c]

Rule 2083

Int[(P_)^(p_), x_Symbol] :> With[{u = Factor[P]}, Int[ExpandIntegrand[u^p, x], x] /;  !SumQ[NonfreeFactors[u,
x]]] /; PolyQ[P, x] && ILtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {4}{5 (-1+4 x)}+\frac {-1-4 x}{5 \left (1+4 x^2\right )}\right ) \, dx \\ & = \frac {1}{5} \log (1-4 x)+\frac {1}{5} \int \frac {-1-4 x}{1+4 x^2} \, dx \\ & = \frac {1}{5} \log (1-4 x)-\frac {1}{5} \int \frac {1}{1+4 x^2} \, dx-\frac {4}{5} \int \frac {x}{1+4 x^2} \, dx \\ & = -\frac {1}{10} \tan ^{-1}(2 x)+\frac {1}{5} \log (1-4 x)-\frac {1}{10} \log \left (1+4 x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {1}{-1+4 x-4 x^2+16 x^3} \, dx=-\frac {1}{10} \arctan (2 x)+\frac {1}{5} \log (1-4 x)-\frac {1}{10} \log \left (1+4 x^2\right ) \]

[In]

Integrate[(-1 + 4*x - 4*x^2 + 16*x^3)^(-1),x]

[Out]

-1/10*ArcTan[2*x] + Log[1 - 4*x]/5 - Log[1 + 4*x^2]/10

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84

method	result	size
default	\(-\frac {\ln \left (4 x^{2}+1\right )}{10}-\frac {\arctan \left (2 x \right )}{10}+\frac {\ln \left (-1+4 x \right )}{5}\)	\(26\)
risch	\(-\frac {\ln \left (4 x^{2}+1\right )}{10}-\frac {\arctan \left (2 x \right )}{10}+\frac {\ln \left (-1+4 x \right )}{5}\)	\(26\)
parallelrisch	\(-\frac {\ln \left (x -\frac {i}{2}\right )}{10}+\frac {i \ln \left (x -\frac {i}{2}\right )}{20}-\frac {\ln \left (x +\frac {i}{2}\right )}{10}-\frac {i \ln \left (x +\frac {i}{2}\right )}{20}+\frac {\ln \left (x -\frac {1}{4}\right )}{5}\)	\(38\)

[In]

int(1/(16*x^3-4*x^2+4*x-1),x,method=_RETURNVERBOSE)

[Out]

-1/10*ln(4*x^2+1)-1/10*arctan(2*x)+1/5*ln(-1+4*x)

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \frac {1}{-1+4 x-4 x^2+16 x^3} \, dx=-\frac {1}{10} \, \arctan \left (2 \, x\right ) - \frac {1}{10} \, \log \left (4 \, x^{2} + 1\right ) + \frac {1}{5} \, \log \left (4 \, x - 1\right ) \]

[In]

integrate(1/(16*x^3-4*x^2+4*x-1),x, algorithm="fricas")

[Out]

-1/10*arctan(2*x) - 1/10*log(4*x^2 + 1) + 1/5*log(4*x - 1)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {1}{-1+4 x-4 x^2+16 x^3} \, dx=\frac {\log {\left (x - \frac {1}{4} \right )}}{5} - \frac {\log {\left (x^{2} + \frac {1}{4} \right )}}{10} - \frac {\operatorname {atan}{\left (2 x \right )}}{10} \]

[In]

integrate(1/(16*x**3-4*x**2+4*x-1),x)

[Out]

log(x - 1/4)/5 - log(x**2 + 1/4)/10 - atan(2*x)/10

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \frac {1}{-1+4 x-4 x^2+16 x^3} \, dx=-\frac {1}{10} \, \arctan \left (2 \, x\right ) - \frac {1}{10} \, \log \left (4 \, x^{2} + 1\right ) + \frac {1}{5} \, \log \left (4 \, x - 1\right ) \]

[In]

integrate(1/(16*x^3-4*x^2+4*x-1),x, algorithm="maxima")

[Out]

-1/10*arctan(2*x) - 1/10*log(4*x^2 + 1) + 1/5*log(4*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {1}{-1+4 x-4 x^2+16 x^3} \, dx=-\frac {1}{10} \, \arctan \left (2 \, x\right ) - \frac {1}{10} \, \log \left (4 \, x^{2} + 1\right ) + \frac {1}{5} \, \log \left ({\left | 4 \, x - 1 \right |}\right ) \]

[In]

integrate(1/(16*x^3-4*x^2+4*x-1),x, algorithm="giac")

[Out]

-1/10*arctan(2*x) - 1/10*log(4*x^2 + 1) + 1/5*log(abs(4*x - 1))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \frac {1}{-1+4 x-4 x^2+16 x^3} \, dx=\frac {\ln \left (x-\frac {1}{4}\right )}{5}+\ln \left (x-\frac {1}{2}{}\mathrm {i}\right )\,\left (-\frac {1}{10}+\frac {1}{20}{}\mathrm {i}\right )+\ln \left (x+\frac {1}{2}{}\mathrm {i}\right )\,\left (-\frac {1}{10}-\frac {1}{20}{}\mathrm {i}\right ) \]

[In]

int(1/(4*x - 4*x^2 + 16*x^3 - 1),x)

[Out]

log(x - 1/4)/5 - log(x - 1i/2)*(1/10 - 1i/20) - log(x + 1i/2)*(1/10 + 1i/20)