\(\int \frac {1-\frac {\sqrt {x^6}}{x}}{1-x^4} \, dx\) [818]

Optimal result

Integrand size = 24, antiderivative size = 45 \[ \int \frac {1-\frac {\sqrt {x^6}}{x}}{1-x^4} \, dx=\frac {\arctan (x)}{2}+\frac {\sqrt {x^6} \arctan (x)}{2 x^3}+\frac {\text {arctanh}(x)}{2}-\frac {\sqrt {x^6} \text {arctanh}(x)}{2 x^3} \]

[Out]

1/2*arctan(x)+1/2*arctanh(x)+1/2*arctan(x)*(x^6)^(1/2)/x^3-1/2*arctanh(x)*(x^6)^(1/2)/x^3

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6857, 218, 212, 209, 15, 304} \[ \int \frac {1-\frac {\sqrt {x^6}}{x}}{1-x^4} \, dx=\frac {\sqrt {x^6} \arctan (x)}{2 x^3}+\frac {\arctan (x)}{2}-\frac {\sqrt {x^6} \text {arctanh}(x)}{2 x^3}+\frac {\text {arctanh}(x)}{2} \]

[In]

Int[(1 - Sqrt[x^6]/x)/(1 - x^4),x]

[Out]

ArcTan[x]/2 + (Sqrt[x^6]*ArcTan[x])/(2*x^3) + ArcTanh[x]/2 - (Sqrt[x^6]*ArcTanh[x])/(2*x^3)

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[a^IntPart[m]*((a*x^n)^FracPart[m]/x^(n*FracPart[m])), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

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 212

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

Rule 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[a/b, 0]

Rule 304

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}
, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !
GtQ[a/b, 0]

Rule 6857

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{1-x^4}+\frac {\sqrt {x^6}}{x \left (-1+x^4\right )}\right ) \, dx \\ & = \int \frac {1}{1-x^4} \, dx+\int \frac {\sqrt {x^6}}{x \left (-1+x^4\right )} \, dx \\ & = \frac {1}{2} \int \frac {1}{1-x^2} \, dx+\frac {1}{2} \int \frac {1}{1+x^2} \, dx+\frac {\sqrt {x^6} \int \frac {x^2}{-1+x^4} \, dx}{x^3} \\ & = \frac {1}{2} \tan ^{-1}(x)+\frac {1}{2} \tanh ^{-1}(x)-\frac {\sqrt {x^6} \int \frac {1}{1-x^2} \, dx}{2 x^3}+\frac {\sqrt {x^6} \int \frac {1}{1+x^2} \, dx}{2 x^3} \\ & = \frac {1}{2} \tan ^{-1}(x)+\frac {\sqrt {x^6} \tan ^{-1}(x)}{2 x^3}+\frac {1}{2} \tanh ^{-1}(x)-\frac {\sqrt {x^6} \tanh ^{-1}(x)}{2 x^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.27 \[ \int \frac {1-\frac {\sqrt {x^6}}{x}}{1-x^4} \, dx=\frac {\arctan (x)}{2}-\frac {1}{2} \arctan \left (\frac {\sqrt {x^6}}{x^4}\right )-\frac {1}{2} \text {arctanh}\left (\frac {\sqrt {x^6}}{x^4}\right )-\frac {1}{4} \log (1-x)+\frac {1}{4} \log (1+x) \]

[In]

Integrate[(1 - Sqrt[x^6]/x)/(1 - x^4),x]

[Out]

ArcTan[x]/2 - ArcTan[Sqrt[x^6]/x^4]/2 - ArcTanh[Sqrt[x^6]/x^4]/2 - Log[1 - x]/4 + Log[1 + x]/4

Maple [A] (verified)

Time = 1.06 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.78

method	result	size
default	\(\frac {\sqrt {x^{6}}\, \left (\ln \left (x -1\right )-\ln \left (x +1\right )+2 \arctan \left (x \right )\right )}{4 x^{3}}+\frac {\arctan \left (x \right )}{2}+\frac {\operatorname {arctanh}\left (x \right )}{2}\)	\(35\)
meijerg	\(-\frac {x \left (\ln \left (1-\left (x^{4}\right )^{\frac {1}{4}}\right )-\ln \left (1+\left (x^{4}\right )^{\frac {1}{4}}\right )-2 \arctan \left (\left (x^{4}\right )^{\frac {1}{4}}\right )\right )}{4 \left (x^{4}\right )^{\frac {1}{4}}}+\frac {\sqrt {x^{6}}\, \left (\ln \left (1-\left (x^{4}\right )^{\frac {1}{4}}\right )-\ln \left (1+\left (x^{4}\right )^{\frac {1}{4}}\right )+2 \arctan \left (\left (x^{4}\right )^{\frac {1}{4}}\right )\right )}{4 \left (x^{4}\right )^{\frac {3}{4}}}\)	\(80\)

[In]

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

[Out]

1/4*(x^6)^(1/2)*(ln(x-1)-ln(x+1)+2*arctan(x))/x^3+1/2*arctan(x)+1/2*arctanh(x)

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 2, normalized size of antiderivative = 0.04 \[ \int \frac {1-\frac {\sqrt {x^6}}{x}}{1-x^4} \, dx=\arctan \left (x\right ) \]

[In]

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

[Out]

arctan(x)

Sympy [F]

\[ \int \frac {1-\frac {\sqrt {x^6}}{x}}{1-x^4} \, dx=- \int \frac {x}{x^{5} - x}\, dx - \int \left (- \frac {\sqrt {x^{6}}}{x^{5} - x}\right )\, dx \]

[In]

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

[Out]

-Integral(x/(x**5 - x), x) - Integral(-sqrt(x**6)/(x**5 - x), x)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 2, normalized size of antiderivative = 0.04 \[ \int \frac {1-\frac {\sqrt {x^6}}{x}}{1-x^4} \, dx=\arctan \left (x\right ) \]

[In]

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

[Out]

arctan(x)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.69 \[ \int \frac {1-\frac {\sqrt {x^6}}{x}}{1-x^4} \, dx=\frac {1}{2} \, {\left (\mathrm {sgn}\left (x\right ) + 1\right )} \arctan \left (x\right ) - \frac {1}{4} \, {\left (\mathrm {sgn}\left (x\right ) - 1\right )} \log \left ({\left | x + 1 \right |}\right ) + \frac {1}{4} \, {\left (\mathrm {sgn}\left (x\right ) - 1\right )} \log \left ({\left | x - 1 \right |}\right ) \]

[In]

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

[Out]

1/2*(sgn(x) + 1)*arctan(x) - 1/4*(sgn(x) - 1)*log(abs(x + 1)) + 1/4*(sgn(x) - 1)*log(abs(x - 1))

Mupad [F(-1)]

Timed out. \[ \int \frac {1-\frac {\sqrt {x^6}}{x}}{1-x^4} \, dx=\int \frac {\frac {\sqrt {x^6}}{x}-1}{x^4-1} \,d x \]

[In]

int(((x^6)^(1/2)/x - 1)/(x^4 - 1),x)

[Out]

int(((x^6)^(1/2)/x - 1)/(x^4 - 1), x)