3.4.76 \(\int \frac {\sqrt {a x^3}}{x-x^3} \, dx\) [376]

Optimal. Leaf size=44 \[ -\frac {\sqrt {a x^3} \tan ^{-1}\left (\sqrt {x}\right )}{x^{3/2}}+\frac {\sqrt {a x^3} \tanh ^{-1}\left (\sqrt {x}\right )}{x^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {15, 1598, 335, 304, 209, 212} \begin {gather*} \frac {\sqrt {a x^3} \tanh ^{-1}\left (\sqrt {x}\right )}{x^{3/2}}-\frac {\sqrt {a x^3} \text {ArcTan}\left (\sqrt {x}\right )}{x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a*x^3]/(x - x^3),x]

[Out]

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

Rubi steps

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

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Mathematica [A]
time = 0.02, size = 30, normalized size = 0.68 \begin {gather*} \frac {\sqrt {a x^3} \left (-\tan ^{-1}\left (\sqrt {x}\right )+\tanh ^{-1}\left (\sqrt {x}\right )\right )}{x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a*x^3]/(x - x^3),x]

[Out]

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

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Maple [A]
time = 0.20, size = 44, normalized size = 1.00

method	result	size
default	\(-\frac {\sqrt {a \,x^{3}}\, \sqrt {a}\, \left (\arctan \left (\frac {\sqrt {a x}}{\sqrt {a}}\right )-\arctanh \left (\frac {\sqrt {a x}}{\sqrt {a}}\right )\right )}{x \sqrt {a x}}\)	\(44\)
meijerg	\(-\frac {\sqrt {a \,x^{3}}\, \left (\ln \left (1-\left (x^{2}\right )^{\frac {1}{4}}\right )-\ln \left (1+\left (x^{2}\right )^{\frac {1}{4}}\right )+2 \arctan \left (\left (x^{2}\right )^{\frac {1}{4}}\right )\right )}{2 \left (x^{2}\right )^{\frac {3}{4}}}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.63, size = 32, normalized size = 0.73 \begin {gather*} -\sqrt {a} \arctan \left (\sqrt {x}\right ) + \frac {1}{2} \, \sqrt {a} \log \left (\sqrt {x} + 1\right ) - \frac {1}{2} \, \sqrt {a} \log \left (\sqrt {x} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A] Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (32) = 64\).
time = 0.34, size = 127, normalized size = 2.89 \begin {gather*} \left [-\sqrt {a} \arctan \left (\frac {\sqrt {a x^{3}}}{\sqrt {a} x}\right ) + \frac {1}{2} \, \sqrt {a} \log \left (\frac {a x^{2} + a x + 2 \, \sqrt {a x^{3}} \sqrt {a}}{x^{2} - x}\right ), -\sqrt {-a} \arctan \left (\frac {\sqrt {a x^{3}} \sqrt {-a}}{a x}\right ) + \frac {1}{2} \, \sqrt {-a} \log \left (\frac {a x^{2} - a x - 2 \, \sqrt {a x^{3}} \sqrt {-a}}{x^{2} + x}\right )\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-sqrt(a)*arctan(sqrt(a*x^3)/(sqrt(a)*x)) + 1/2*sqrt(a)*log((a*x^2 + a*x + 2*sqrt(a*x^3)*sqrt(a))/(x^2 - x)),
-sqrt(-a)*arctan(sqrt(a*x^3)*sqrt(-a)/(a*x)) + 1/2*sqrt(-a)*log((a*x^2 - a*x - 2*sqrt(a*x^3)*sqrt(-a))/(x^2 +
x))]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(sqrt(a*x**3)/(x**3 - x), x)

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Giac [A]
time = 3.74, size = 43, normalized size = 0.98 \begin {gather*} -\frac {{\left (\frac {a^{2} \arctan \left (\frac {\sqrt {a x}}{\sqrt {-a}}\right )}{\sqrt {-a}} + a^{\frac {3}{2}} \arctan \left (\frac {\sqrt {a x}}{\sqrt {a}}\right )\right )} \mathrm {sgn}\left (x\right )}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a^2*arctan(sqrt(a*x)/sqrt(-a))/sqrt(-a) + a^(3/2)*arctan(sqrt(a*x)/sqrt(a)))*sgn(x)/a

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^3)^(1/2)/(x - x^3),x)

[Out]

int((a*x^3)^(1/2)/(x - x^3), x)

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