3.9.20 \(\int \frac {x}{x+\sqrt {x^6}} \, dx\) [820]

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

[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 = 0.09, antiderivative size = 45, normalized size of antiderivative = 1.00, 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 = {6861, 1598, 6857, 218, 212, 209, 15, 304} \begin {gather*} \frac {\sqrt {x^6} \text {ArcTan}(x)}{2 x^3}+\frac {\text {ArcTan}(x)}{2}-\frac {\sqrt {x^6} \tanh ^{-1}(x)}{2 x^3}+\frac {1}{2} \tanh ^{-1}(x) \end {gather*}

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 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 1598

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 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]

Rule 6861

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]

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*}

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Mathematica [A]
time = 3.89, size = 57, normalized size = 1.27 \begin {gather*} \frac {1}{2} \tan ^{-1}(x)+\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {x^6}}{x^2}\right )-\frac {1}{2} \tanh ^{-1}\left (\frac {\sqrt {x^6}}{x^2}\right )-\frac {1}{4} \log (1-x)+\frac {1}{4} \log (1+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/(x + Sqrt[x^6]),x]

[Out]

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

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Maple [A]
time = 0.56, size = 27, normalized size = 0.60

method	result	size
meijerg	\(\frac {x^{\frac {3}{2}} \arctan \left (\frac {\left (x^{6}\right )^{\frac {1}{4}}}{\sqrt {x}}\right )}{\left (x^{6}\right )^{\frac {1}{4}}}\)	\(20\)
default	\(\frac {\arctan \left (\sqrt {\frac {\sqrt {x^{6}}}{x^{3}}}\, x \right )}{\sqrt {\frac {\sqrt {x^{6}}}{x^{3}}}}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.50, size = 2, normalized size = 0.04 \begin {gather*} \arctan \left (x\right ) \end {gather*}

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 = 0.33, size = 2, normalized size = 0.04 \begin {gather*} \arctan \left (x\right ) \end {gather*}

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 = 0.03, size = 2, normalized size = 0.04 \begin {gather*} \operatorname {atan}{\left (x \right )} \end {gather*}

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 = 2.50, size = 12, normalized size = 0.27 \begin {gather*} \frac {\arctan \left (x \sqrt {\mathrm {sgn}\left (x\right )}\right )}{\sqrt {\mathrm {sgn}\left (x\right )}} \end {gather*}

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))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x}{x+\sqrt {x^6}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(x + (x^6)^(1/2)),x)

[Out]

int(x/(x + (x^6)^(1/2)), x)

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