3.820 \(\int \frac{x}{x+\sqrt{x^6}} \, dx\)

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]

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

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Rubi [A]  time = 0.133012, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.615, Rules used = {6729, 1584, 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/(x + Sqrt[x^6]),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 6729

Int[(u_.)/((a_.)*(x_)^(m_.) + (b_.)*Sqrt[(c_.)*(x_)^(n_)]), x_Symbol] :> Int[(u*(a*x^m - b*Sqrt[c*x^n]))/(a^2*
x^(2*m) - b^2*c*x^n), x] /; FreeQ[{a, b, c, m, n}, x]

Rule 1584

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, 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]

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 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 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 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 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]

Rubi steps

\begin{align*} \int \frac{x}{x+\sqrt{x^6}} \, dx &=\int \frac{x \left (x-\sqrt{x^6}\right )}{x^2-x^6} \, 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*}

Mathematica [A]  time = 0.0341553, size = 27, normalized size = 0.6 \[ \frac{1}{2} \left (\frac{\sqrt{x^6} \left (\tan ^{-1}(x)-\tanh ^{-1}(x)\right )}{x^3}+\tan ^{-1}(x)+\tanh ^{-1}(x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x/(x + Sqrt[x^6]),x]

[Out]

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

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Maple [A]  time = 0.007, size = 27, normalized size = 0.6 \begin{align*}{\arctan \left ( \sqrt{{\frac{1}{{x}^{3}}\sqrt{{x}^{6}}}}x \right ){\frac{1}{\sqrt{{\frac{1}{{x}^{3}}\sqrt{{x}^{6}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/((x^6)^(1/2)/x^3)^(1/2)*arctan(((x^6)^(1/2)/x^3)^(1/2)*x)

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Maxima [A]  time = 1.70077, size = 3, normalized size = 0.07 \begin{align*} \arctan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan(x)

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Fricas [A]  time = 1.68473, size = 15, normalized size = 0.33 \begin{align*} \arctan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan(x)

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Sympy [A]  time = 0.091828, size = 2, normalized size = 0.04 \begin{align*} \operatorname{atan}{\left (x \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

atan(x)

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Giac [A]  time = 1.11389, size = 16, normalized size = 0.36 \begin{align*} \frac{\arctan \left (x \sqrt{\mathrm{sgn}\left (x\right )}\right )}{\sqrt{\mathrm{sgn}\left (x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan(x*sqrt(sgn(x)))/sqrt(sgn(x))