3.6.66 \(\int \frac {1}{\sqrt {x}-x^{5/2}} \, dx\) [566]

Optimal. Leaf size=13 \[ \tan ^{-1}\left (\sqrt {x}\right )+\tanh ^{-1}\left (\sqrt {x}\right ) \]

[Out]

arctan(x^(1/2))+arctanh(x^(1/2))

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Rubi [A]
time = 0.01, antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1607, 335, 218, 212, 209} \begin {gather*} \text {ArcTan}\left (\sqrt {x}\right )+\tanh ^{-1}\left (\sqrt {x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sqrt[x] - x^(5/2))^(-1),x]

[Out]

ArcTan[Sqrt[x]] + ArcTanh[Sqrt[x]]

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 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]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {x}-x^{5/2}} \, dx &=\int \frac {1}{\sqrt {x} \left (1-x^2\right )} \, dx\\ &=2 \text {Subst}\left (\int \frac {1}{1-x^4} \, dx,x,\sqrt {x}\right )\\ &=\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 )\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 13, normalized size = 1.00 \begin {gather*} \tan ^{-1}\left (\sqrt {x}\right )+\tanh ^{-1}\left (\sqrt {x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[x] - x^(5/2))^(-1),x]

[Out]

ArcTan[Sqrt[x]] + ArcTanh[Sqrt[x]]

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Maple [A]
time = 0.20, size = 10, normalized size = 0.77

method	result	size
derivativedivides	\(\arctan \left (\sqrt {x}\right )+\arctanh \left (\sqrt {x}\right )\)	\(10\)
default	\(\arctan \left (\sqrt {x}\right )+\arctanh \left (\sqrt {x}\right )\)	\(10\)
meijerg	\(-\frac {\sqrt {x}\, \left (\ln \left (1-\left (x^{2}\right )^{\frac {1}{4}}\right )-\ln \left (1+\left (x^{2}\right )^{\frac {1}{4}}\right )-2 \arctan \left (\left (x^{2}\right )^{\frac {1}{4}}\right )\right )}{2 \left (x^{2}\right )^{\frac {1}{4}}}\)	\(40\)
trager	\(-\frac {\ln \left (\frac {-1-x +2 \sqrt {x}}{-1+x}\right )}{2}-\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {x \RootOf \left (\textit {\_Z}^{2}+1\right )-\RootOf \left (\textit {\_Z}^{2}+1\right )+2 \sqrt {x}}{1+x}\right )}{2}\)	\(58\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-x^(5/2)+x^(1/2)),x,method=_RETURNVERBOSE)

[Out]

arctan(x^(1/2))+arctanh(x^(1/2))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 21 vs. \(2 (9) = 18\).
time = 0.49, size = 21, normalized size = 1.62 \begin {gather*} \arctan \left (\sqrt {x}\right ) + \frac {1}{2} \, \log \left (\sqrt {x} + 1\right ) - \frac {1}{2} \, \log \left (\sqrt {x} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 21 vs. \(2 (9) = 18\).
time = 0.36, size = 21, normalized size = 1.62 \begin {gather*} \arctan \left (\sqrt {x}\right ) + \frac {1}{2} \, \log \left (\sqrt {x} + 1\right ) - \frac {1}{2} \, \log \left (\sqrt {x} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 26 vs. \(2 (12) = 24\).
time = 0.12, size = 26, normalized size = 2.00 \begin {gather*} - \frac {\log {\left (\sqrt {x} - 1 \right )}}{2} + \frac {\log {\left (\sqrt {x} + 1 \right )}}{2} + \operatorname {atan}{\left (\sqrt {x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(sqrt(x) - 1)/2 + log(sqrt(x) + 1)/2 + atan(sqrt(x))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 22 vs. \(2 (9) = 18\).
time = 4.14, size = 22, normalized size = 1.69 \begin {gather*} \arctan \left (\sqrt {x}\right ) + \frac {1}{2} \, \log \left (\sqrt {x} + 1\right ) - \frac {1}{2} \, \log \left ({\left | \sqrt {x} - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 3.08, size = 9, normalized size = 0.69 \begin {gather*} \mathrm {atan}\left (\sqrt {x}\right )+\mathrm {atanh}\left (\sqrt {x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(1/2) - x^(5/2)),x)

[Out]

atan(x^(1/2)) + atanh(x^(1/2))

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