\(\int \frac {\sqrt {x}}{(-2+x^2)^{3/4}} \, dx\) [426]

Optimal result

Integrand size = 15, antiderivative size = 35 \[ \int \frac {\sqrt {x}}{\left (-2+x^2\right )^{3/4}} \, dx=-\arctan \left (\frac {\sqrt {x}}{\sqrt [4]{-2+x^2}}\right )+\text {arctanh}\left (\frac {\sqrt {x}}{\sqrt [4]{-2+x^2}}\right ) \]

[Out]

-arctan(x^(1/2)/(x^2-2)^(1/4))+arctanh(x^(1/2)/(x^2-2)^(1/4))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {335, 338, 304, 209, 212} \[ \int \frac {\sqrt {x}}{\left (-2+x^2\right )^{3/4}} \, dx=\text {arctanh}\left (\frac {\sqrt {x}}{\sqrt [4]{x^2-2}}\right )-\arctan \left (\frac {\sqrt {x}}{\sqrt [4]{x^2-2}}\right ) \]

[In]

Int[Sqrt[x]/(-2 + x^2)^(3/4),x]

[Out]

-ArcTan[Sqrt[x]/(-2 + x^2)^(1/4)] + ArcTanh[Sqrt[x]/(-2 + x^2)^(1/4)]

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 338

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (m + 1)/n]

Rubi steps \begin{align*} \text {integral}& = 2 \text {Subst}\left (\int \frac {x^2}{\left (-2+x^4\right )^{3/4}} \, dx,x,\sqrt {x}\right ) \\ & = 2 \text {Subst}\left (\int \frac {x^2}{1-x^4} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-2+x^2}}\right ) \\ & = \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-2+x^2}}\right )-\text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-2+x^2}}\right ) \\ & = -\arctan \left (\frac {\sqrt {x}}{\sqrt [4]{-2+x^2}}\right )+\text {arctanh}\left (\frac {\sqrt {x}}{\sqrt [4]{-2+x^2}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {x}}{\left (-2+x^2\right )^{3/4}} \, dx=-\arctan \left (\frac {\sqrt {x}}{\sqrt [4]{-2+x^2}}\right )+\text {arctanh}\left (\frac {\sqrt {x}}{\sqrt [4]{-2+x^2}}\right ) \]

[In]

Integrate[Sqrt[x]/(-2 + x^2)^(3/4),x]

[Out]

-ArcTan[Sqrt[x]/(-2 + x^2)^(1/4)] + ArcTanh[Sqrt[x]/(-2 + x^2)^(1/4)]

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.86 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.20

method	result	size
meijerg	\(\frac {2^{\frac {1}{4}} {\left (-\operatorname {signum}\left (-1+\frac {x^{2}}{2}\right )\right )}^{\frac {3}{4}} x^{\frac {3}{2}} \operatorname {hypergeom}\left (\left [\frac {3}{4}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], \frac {x^{2}}{2}\right )}{3 \operatorname {signum}\left (-1+\frac {x^{2}}{2}\right )^{\frac {3}{4}}}\)	\(42\)

[In]

int(x^(1/2)/(x^2-2)^(3/4),x,method=_RETURNVERBOSE)

[Out]

1/3*2^(1/4)/signum(-1+1/2*x^2)^(3/4)*(-signum(-1+1/2*x^2))^(3/4)*x^(3/2)*hypergeom([3/4,3/4],[7/4],1/2*x^2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 83 vs. \(2 (27) = 54\).

Time = 0.50 (sec) , antiderivative size = 83, normalized size of antiderivative = 2.37 \[ \int \frac {\sqrt {x}}{\left (-2+x^2\right )^{3/4}} \, dx=-\frac {1}{2} \, \arctan \left (\frac {{\left (x^{2} - 2\right )}^{\frac {3}{4}} x^{\frac {3}{2}} - {\left (x^{2} - 2\right )}^{\frac {5}{4}} \sqrt {x}}{2 \, {\left (x^{3} - 2 \, x\right )}}\right ) + \frac {1}{2} \, \log \left (-x^{2} - {\left (x^{2} - 2\right )}^{\frac {1}{4}} x^{\frac {3}{2}} - \sqrt {x^{2} - 2} x - {\left (x^{2} - 2\right )}^{\frac {3}{4}} \sqrt {x} + 1\right ) \]

[In]

integrate(x^(1/2)/(x^2-2)^(3/4),x, algorithm="fricas")

[Out]

-1/2*arctan(1/2*((x^2 - 2)^(3/4)*x^(3/2) - (x^2 - 2)^(5/4)*sqrt(x))/(x^3 - 2*x)) + 1/2*log(-x^2 - (x^2 - 2)^(1
/4)*x^(3/2) - sqrt(x^2 - 2)*x - (x^2 - 2)^(3/4)*sqrt(x) + 1)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.49 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.17 \[ \int \frac {\sqrt {x}}{\left (-2+x^2\right )^{3/4}} \, dx=\frac {\sqrt [4]{2} x^{\frac {3}{2}} e^{- \frac {3 i \pi }{4}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {x^{2}}{2}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} \]

[In]

integrate(x**(1/2)/(x**2-2)**(3/4),x)

[Out]

2**(1/4)*x**(3/2)*exp(-3*I*pi/4)*gamma(3/4)*hyper((3/4, 3/4), (7/4,), x**2/2)/(4*gamma(7/4))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.29 \[ \int \frac {\sqrt {x}}{\left (-2+x^2\right )^{3/4}} \, dx=\arctan \left (\frac {{\left (x^{2} - 2\right )}^{\frac {1}{4}}}{\sqrt {x}}\right ) + \frac {1}{2} \, \log \left (\frac {{\left (x^{2} - 2\right )}^{\frac {1}{4}}}{\sqrt {x}} + 1\right ) - \frac {1}{2} \, \log \left (\frac {{\left (x^{2} - 2\right )}^{\frac {1}{4}}}{\sqrt {x}} - 1\right ) \]

[In]

integrate(x^(1/2)/(x^2-2)^(3/4),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {\sqrt {x}}{\left (-2+x^2\right )^{3/4}} \, dx=\int { \frac {\sqrt {x}}{{\left (x^{2} - 2\right )}^{\frac {3}{4}}} \,d x } \]

[In]

integrate(x^(1/2)/(x^2-2)^(3/4),x, algorithm="giac")

[Out]

integrate(sqrt(x)/(x^2 - 2)^(3/4), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {x}}{\left (-2+x^2\right )^{3/4}} \, dx=\int \frac {\sqrt {x}}{{\left (x^2-2\right )}^{3/4}} \,d x \]

[In]

int(x^(1/2)/(x^2 - 2)^(3/4),x)

[Out]

int(x^(1/2)/(x^2 - 2)^(3/4), x)