3.819 \(\int \frac {x-\sqrt {x^6}}{x-x^5} \, dx\)

Optimal. Leaf size=45 \[ \frac {\sqrt {x^6} \tan ^{-1}(x)}{2 x^3}-\frac {\sqrt {x^6} \tanh ^{-1}(x)}{2 x^3}+\frac {1}{2} \tan ^{-1}(x)+\frac {1}{2} \tanh ^{-1}(x) \]

[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

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Rubi [A]  time = 0.10, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1593, 6725, 212, 206, 203, 15, 298} \[ \frac {\sqrt {x^6} \tan ^{-1}(x)}{2 x^3}-\frac {\sqrt {x^6} \tanh ^{-1}(x)}{2 x^3}+\frac {1}{2} \tan ^{-1}(x)+\frac {1}{2} \tanh ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(x - Sqrt[x^6])/(x - x^5),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 203

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

Rule 206

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

Rule 212

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] &&
!GtQ[a/b, 0]

Rule 298

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 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 6725

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

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Mathematica [A]  time = 0.01, size = 27, normalized size = 0.60 \[ \frac {1}{2} \left (\frac {\sqrt {x^6} \left (\tan ^{-1}(x)-\tanh ^{-1}(x)\right )}{x^3}+\tan ^{-1}(x)+\tanh ^{-1}(x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(x - Sqrt[x^6])/(x - x^5),x]

[Out]

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

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fricas [A]  time = 0.40, size = 2, normalized size = 0.04 \[ \arctan \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x-(x^6)^(1/2))/(-x^5+x),x, algorithm="fricas")

[Out]

arctan(x)

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giac [A]  time = 0.36, size = 31, normalized size = 0.69 \[ \frac {1}{2} \, {\left (\mathrm {sgn}\relax (x) + 1\right )} \arctan \relax (x) - \frac {1}{4} \, {\left (\mathrm {sgn}\relax (x) - 1\right )} \log \left ({\left | x + 1 \right |}\right ) + \frac {1}{4} \, {\left (\mathrm {sgn}\relax (x) - 1\right )} \log \left ({\left | x - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x-(x^6)^(1/2))/(-x^5+x),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))

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maple [A]  time = 0.01, size = 35, normalized size = 0.78 \[ \frac {\arctanh \relax (x )}{2}+\frac {\arctan \relax (x )}{2}+\frac {\sqrt {x^{6}}\, \left (2 \arctan \relax (x )+\ln \left (x -1\right )-\ln \left (x +1\right )\right )}{4 x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x-(x^6)^(1/2))/(-x^5+x),x)

[Out]

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

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maxima [A]  time = 0.98, size = 2, normalized size = 0.04 \[ \arctan \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x-(x^6)^(1/2))/(-x^5+x),x, algorithm="maxima")

[Out]

arctan(x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x-\sqrt {x^6}}{x-x^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x - (x^6)^(1/2))/(x - x^5),x)

[Out]

int((x - (x^6)^(1/2))/(x - x^5), x)

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sympy [A]  time = 0.10, size = 2, normalized size = 0.04 \[ \operatorname {atan}{\relax (x )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x-(x**6)**(1/2))/(-x**5+x),x)

[Out]

atan(x)

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