3.2.20 \(\int \frac {x^3}{(-1+x^2)^{3/4}} \, dx\)

Optimal. Leaf size=18 \[ \frac {2}{5} \sqrt [4]{x^2-1} \left (x^2+4\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 25, normalized size of antiderivative = 1.39, number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {266, 43} \begin {gather*} \frac {2}{5} \left (x^2-1\right )^{5/4}+2 \sqrt [4]{x^2-1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3/(-1 + x^2)^(3/4),x]

[Out]

2*(-1 + x^2)^(1/4) + (2*(-1 + x^2)^(5/4))/5

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

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

Rubi steps

\begin {align*} \int \frac {x^3}{\left (-1+x^2\right )^{3/4}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x}{(-1+x)^{3/4}} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {1}{(-1+x)^{3/4}}+\sqrt [4]{-1+x}\right ) \, dx,x,x^2\right )\\ &=2 \sqrt [4]{-1+x^2}+\frac {2}{5} \left (-1+x^2\right )^{5/4}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 18, normalized size = 1.00 \begin {gather*} \frac {2}{5} \sqrt [4]{x^2-1} \left (x^2+4\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3/(-1 + x^2)^(3/4),x]

[Out]

(2*(-1 + x^2)^(1/4)*(4 + x^2))/5

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IntegrateAlgebraic [A]  time = 0.02, size = 18, normalized size = 1.00 \begin {gather*} \frac {2}{5} \sqrt [4]{-1+x^2} \left (4+x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^3/(-1 + x^2)^(3/4),x]

[Out]

(2*(-1 + x^2)^(1/4)*(4 + x^2))/5

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fricas [A]  time = 0.44, size = 14, normalized size = 0.78 \begin {gather*} \frac {2}{5} \, {\left (x^{2} + 4\right )} {\left (x^{2} - 1\right )}^{\frac {1}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(x^2 + 4)*(x^2 - 1)^(1/4)

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giac [A]  time = 0.29, size = 19, normalized size = 1.06 \begin {gather*} \frac {2}{5} \, {\left (x^{2} - 1\right )}^{\frac {5}{4}} + 2 \, {\left (x^{2} - 1\right )}^{\frac {1}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(x^2 - 1)^(5/4) + 2*(x^2 - 1)^(1/4)

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maple [A]  time = 0.07, size = 15, normalized size = 0.83

method	result	size
risch	\(\frac {2 \left (x^{2}-1\right )^{\frac {1}{4}} \left (x^{2}+4\right )}{5}\)	\(15\)
trager	\(\left (\frac {2 x^{2}}{5}+\frac {8}{5}\right ) \left (x^{2}-1\right )^{\frac {1}{4}}\)	\(16\)
gosper	\(\frac {2 \left (-1+x \right ) \left (1+x \right ) \left (x^{2}+4\right )}{5 \left (x^{2}-1\right )^{\frac {3}{4}}}\)	\(21\)
meijerg	\(\frac {\left (-\mathrm {signum}\left (x^{2}-1\right )\right )^{\frac {3}{4}} \hypergeom \left (\left [\frac {3}{4}, 2\right ], \relax [3], x^{2}\right ) x^{4}}{4 \mathrm {signum}\left (x^{2}-1\right )^{\frac {3}{4}}}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(x^2-1)^(1/4)*(x^2+4)

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maxima [A]  time = 0.34, size = 19, normalized size = 1.06 \begin {gather*} \frac {2}{5} \, {\left (x^{2} - 1\right )}^{\frac {5}{4}} + 2 \, {\left (x^{2} - 1\right )}^{\frac {1}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(x^2 - 1)^(5/4) + 2*(x^2 - 1)^(1/4)

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mupad [B]  time = 0.21, size = 14, normalized size = 0.78 \begin {gather*} \frac {2\,{\left (x^2-1\right )}^{1/4}\,\left (x^2+4\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(x^2 - 1)^(3/4),x)

[Out]

(2*(x^2 - 1)^(1/4)*(x^2 + 4))/5

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sympy [A]  time = 0.83, size = 75, normalized size = 4.17 \begin {gather*} \begin {cases} \frac {2 x^{2} \sqrt [4]{x^{2} - 1}}{5} + \frac {8 \sqrt [4]{x^{2} - 1}}{5} & \text {for}\: \left |{x^{2}}\right | > 1 \\- \frac {2 x^{2} \sqrt [4]{1 - x^{2}} e^{- \frac {3 i \pi }{4}}}{5} - \frac {8 \sqrt [4]{1 - x^{2}} e^{- \frac {3 i \pi }{4}}}{5} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((2*x**2*(x**2 - 1)**(1/4)/5 + 8*(x**2 - 1)**(1/4)/5, Abs(x**2) > 1), (-2*x**2*(1 - x**2)**(1/4)*exp(
-3*I*pi/4)/5 - 8*(1 - x**2)**(1/4)*exp(-3*I*pi/4)/5, True))

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