3.9.21 \(\int \frac {\sqrt {x}-\sqrt {x^3}}{x-x^3} \, dx\) [821]

Optimal. Leaf size=52 \[ \tan ^{-1}\left (\sqrt {x}\right )+\frac {\sqrt {x^3} \tan ^{-1}\left (\sqrt {x}\right )}{x^{3/2}}+\tanh ^{-1}\left (\sqrt {x}\right )-\frac {\sqrt {x^3} \tanh ^{-1}\left (\sqrt {x}\right )}{x^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {1607, 6857, 335, 218, 212, 209, 15, 304} \begin {gather*} \frac {\sqrt {x^3} \text {ArcTan}\left (\sqrt {x}\right )}{x^{3/2}}+\text {ArcTan}\left (\sqrt {x}\right )-\frac {\sqrt {x^3} \tanh ^{-1}\left (\sqrt {x}\right )}{x^{3/2}}+\tanh ^{-1}\left (\sqrt {x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sqrt[x] - Sqrt[x^3])/(x - x^3),x]

[Out]

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

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 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 1607

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

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Mathematica [A]
time = 1.63, size = 39, normalized size = 0.75 \begin {gather*} \tan ^{-1}\left (\sqrt {x}\right )+\tan ^{-1}\left (\frac {\sqrt {x^3}}{x}\right )+\tanh ^{-1}\left (\sqrt {x}\right )-\tanh ^{-1}\left (\frac {\sqrt {x^3}}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[x] - Sqrt[x^3])/(x - x^3),x]

[Out]

ArcTan[Sqrt[x]] + ArcTan[Sqrt[x^3]/x] + ArcTanh[Sqrt[x]] - ArcTanh[Sqrt[x^3]/x]

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Maple [A]
time = 0.49, size = 41, normalized size = 0.79

method	result	size
default	\(\arctan \left (\sqrt {x}\right )+\arctanh \left (\sqrt {x}\right )+\frac {\sqrt {x^{3}}\, \left (\ln \left (-1+\sqrt {x}\right )-\ln \left (1+\sqrt {x}\right )+2 \arctan \left (\sqrt {x}\right )\right )}{2 x^{\frac {3}{2}}}\)	\(41\)
meijerg	\(-\frac {\sqrt {x}\, \left (\ln \left (1-\left (x^{2}\right )^{\frac {1}{4}}\right )-\ln \left (1+\left (x^{2}\right )^{\frac {1}{4}}\right )-2 \arctan \left (\left (x^{2}\right )^{\frac {1}{4}}\right )\right )}{2 \left (x^{2}\right )^{\frac {1}{4}}}+\frac {\sqrt {x^{3}}\, \left (\ln \left (1-\left (x^{2}\right )^{\frac {1}{4}}\right )-\ln \left (1+\left (x^{2}\right )^{\frac {1}{4}}\right )+2 \arctan \left (\left (x^{2}\right )^{\frac {1}{4}}\right )\right )}{2 \left (x^{2}\right )^{\frac {3}{4}}}\)	\(82\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(1/2)-(x^3)^(1/2))/(-x^3+x),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^(1/2)-(x^3)^(1/2))/(-x^3+x),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.34, size = 6, normalized size = 0.12 \begin {gather*} 2 \, \arctan \left (\sqrt {x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^(1/2)-(x^3)^(1/2))/(-x^3+x),x, algorithm="fricas")

[Out]

2*arctan(sqrt(x))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {\sqrt {x}}{x^{3} - x}\, dx - \int \left (- \frac {\sqrt {x^{3}}}{x^{3} - x}\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**(1/2)-(x**3)**(1/2))/(-x**3+x),x)

[Out]

-Integral(sqrt(x)/(x**3 - x), x) - Integral(-sqrt(x**3)/(x**3 - x), x)

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Giac [A]
time = 2.10, size = 6, normalized size = 0.12 \begin {gather*} 2 \, \arctan \left (\sqrt {x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^(1/2)-(x^3)^(1/2))/(-x^3+x),x, algorithm="giac")

[Out]

2*arctan(sqrt(x))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} -\int \frac {\sqrt {x^3}-\sqrt {x}}{x-x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((x^3)^(1/2) - x^(1/2))/(x - x^3),x)

[Out]

-int(((x^3)^(1/2) - x^(1/2))/(x - x^3), x)

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