\(\int \frac {-1+2 x^2}{(-1+4 x) (1+x^2)} \, dx\) [262]

Optimal result

Integrand size = 22, antiderivative size = 27 \[ \int \frac {-1+2 x^2}{(-1+4 x) \left (1+x^2\right )} \, dx=\frac {3 \arctan (x)}{17}-\frac {7}{34} \log (1-4 x)+\frac {6}{17} \log \left (1+x^2\right ) \]

[Out]

3/17*arctan(x)-7/34*ln(1-4*x)+6/17*ln(x^2+1)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1643, 649, 209, 266} \[ \int \frac {-1+2 x^2}{(-1+4 x) \left (1+x^2\right )} \, dx=\frac {3 \arctan (x)}{17}+\frac {6}{17} \log \left (x^2+1\right )-\frac {7}{34} \log (1-4 x) \]

[In]

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

[Out]

(3*ArcTan[x])/17 - (7*Log[1 - 4*x])/34 + (6*Log[1 + x^2])/17

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 1643

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*
Pq*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {14}{17 (-1+4 x)}+\frac {3 (1+4 x)}{17 \left (1+x^2\right )}\right ) \, dx \\ & = -\frac {7}{34} \log (1-4 x)+\frac {3}{17} \int \frac {1+4 x}{1+x^2} \, dx \\ & = -\frac {7}{34} \log (1-4 x)+\frac {3}{17} \int \frac {1}{1+x^2} \, dx+\frac {12}{17} \int \frac {x}{1+x^2} \, dx \\ & = \frac {3}{17} \tan ^{-1}(x)-\frac {7}{34} \log (1-4 x)+\frac {6}{17} \log \left (1+x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.41 \[ \int \frac {-1+2 x^2}{(-1+4 x) \left (1+x^2\right )} \, dx=\frac {3 \arctan (x)}{17}-\frac {7}{34} \log (-1+4 x)+\frac {6}{17} \log \left (17+2 (-1+4 x)+(-1+4 x)^2\right ) \]

[In]

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

[Out]

(3*ArcTan[x])/17 - (7*Log[-1 + 4*x])/34 + (6*Log[17 + 2*(-1 + 4*x) + (-1 + 4*x)^2])/17

Maple [A] (verified)

Time = 0.81 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81

method	result	size
default	\(\frac {6 \ln \left (x^{2}+1\right )}{17}+\frac {3 \arctan \left (x \right )}{17}-\frac {7 \ln \left (-1+4 x \right )}{34}\)	\(22\)
risch	\(\frac {6 \ln \left (x^{2}+1\right )}{17}+\frac {3 \arctan \left (x \right )}{17}-\frac {7 \ln \left (-1+4 x \right )}{34}\)	\(22\)
parallelrisch	\(-\frac {7 \ln \left (x -\frac {1}{4}\right )}{34}+\frac {6 \ln \left (x -i\right )}{17}-\frac {3 i \ln \left (x -i\right )}{34}+\frac {6 \ln \left (x +i\right )}{17}+\frac {3 i \ln \left (x +i\right )}{34}\)	\(38\)

[In]

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

[Out]

6/17*ln(x^2+1)+3/17*arctan(x)-7/34*ln(-1+4*x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {-1+2 x^2}{(-1+4 x) \left (1+x^2\right )} \, dx=\frac {3}{17} \, \arctan \left (x\right ) + \frac {6}{17} \, \log \left (x^{2} + 1\right ) - \frac {7}{34} \, \log \left (4 \, x - 1\right ) \]

[In]

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

[Out]

3/17*arctan(x) + 6/17*log(x^2 + 1) - 7/34*log(4*x - 1)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {-1+2 x^2}{(-1+4 x) \left (1+x^2\right )} \, dx=- \frac {7 \log {\left (x - \frac {1}{4} \right )}}{34} + \frac {6 \log {\left (x^{2} + 1 \right )}}{17} + \frac {3 \operatorname {atan}{\left (x \right )}}{17} \]

[In]

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

[Out]

-7*log(x - 1/4)/34 + 6*log(x**2 + 1)/17 + 3*atan(x)/17

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {-1+2 x^2}{(-1+4 x) \left (1+x^2\right )} \, dx=\frac {3}{17} \, \arctan \left (x\right ) + \frac {6}{17} \, \log \left (x^{2} + 1\right ) - \frac {7}{34} \, \log \left (4 \, x - 1\right ) \]

[In]

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

[Out]

3/17*arctan(x) + 6/17*log(x^2 + 1) - 7/34*log(4*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {-1+2 x^2}{(-1+4 x) \left (1+x^2\right )} \, dx=\frac {3}{17} \, \arctan \left (x\right ) + \frac {6}{17} \, \log \left (x^{2} + 1\right ) - \frac {7}{34} \, \log \left ({\left | 4 \, x - 1 \right |}\right ) \]

[In]

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

[Out]

3/17*arctan(x) + 6/17*log(x^2 + 1) - 7/34*log(abs(4*x - 1))

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {-1+2 x^2}{(-1+4 x) \left (1+x^2\right )} \, dx=-\frac {7\,\ln \left (x-\frac {1}{4}\right )}{34}+\ln \left (x-\mathrm {i}\right )\,\left (\frac {6}{17}-\frac {3}{34}{}\mathrm {i}\right )+\ln \left (x+1{}\mathrm {i}\right )\,\left (\frac {6}{17}+\frac {3}{34}{}\mathrm {i}\right ) \]

[In]

int((2*x^2 - 1)/((4*x - 1)*(x^2 + 1)),x)

[Out]

log(x - 1i)*(6/17 - 3i/34) - (7*log(x - 1/4))/34 + log(x + 1i)*(6/17 + 3i/34)