3.9.90 \(\int \frac {\sqrt {\frac {(-1+x^2)^2}{x+x^3}}}{1+x^2} \, dx\) [890]

Optimal. Leaf size=33 \[ \frac {2 x \sqrt {\frac {\left (1-x^2\right )^2}{x+x^3}}}{1-x^2} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6851, 2081, 460} \begin {gather*} \frac {2 x \sqrt {\frac {\left (1-x^2\right )^2}{x^3+x}}}{1-x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[(-1 + x^2)^2/(x + x^3)]/(1 + x^2),x]

[Out]

(2*x*Sqrt[(1 - x^2)^2/(x + x^3)])/(1 - x^2)

Rule 460

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[
a*d*(m + 1) - b*c*(m + n*(p + 1) + 1), 0] && NeQ[m, -1]

Rule 2081

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rule 6851

Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a*v^m*w^n)^FracPart[p]/(v^(m*Fr
acPart[p])*w^(n*FracPart[p]))), Int[u*v^(m*p)*w^(n*p), x], x] /; FreeQ[{a, m, n, p}, x] &&  !IntegerQ[p] &&  !
FreeQ[v, x] &&  !FreeQ[w, x]

Rubi steps

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

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Mathematica [A]
time = 0.00, size = 29, normalized size = 0.88 \begin {gather*} -\frac {2 x \sqrt {\frac {\left (-1+x^2\right )^2}{x+x^3}}}{-1+x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[(-1 + x^2)^2/(x + x^3)]/(1 + x^2),x]

[Out]

(-2*x*Sqrt[(-1 + x^2)^2/(x + x^3)])/(-1 + x^2)

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Maple [A]
time = 0.22, size = 31, normalized size = 0.94

method	result	size
default	\(-\frac {2 \sqrt {\frac {\left (x^{2}-1\right )^{2}}{x \left (x^{2}+1\right )}}\, x}{x^{2}-1}\)	\(31\)
risch	\(-\frac {2 \sqrt {\frac {\left (x^{2}-1\right )^{2}}{x \left (x^{2}+1\right )}}\, x}{x^{2}-1}\)	\(31\)
gosper	\(-\frac {2 x \sqrt {\frac {\left (x^{2}-1\right )^{2}}{x \left (x^{2}+1\right )}}}{\left (1+x \right ) \left (-1+x \right )}\)	\(34\)
trager	\(-\frac {2 x \sqrt {-\frac {-x^{4}+2 x^{2}-1}{x^{3}+x}}}{x^{2}-1}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

integrate(sqrt((x^2 - 1)^2/(x^3 + x))/(x^2 + 1), x)

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Fricas [A]
time = 0.33, size = 30, normalized size = 0.91 \begin {gather*} -\frac {2 \, x \sqrt {\frac {x^{4} - 2 \, x^{2} + 1}{x^{3} + x}}}{x^{2} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {\frac {\left (x - 1\right )^{2} \left (x + 1\right )^{2}}{x^{3} + x}}}{x^{2} + 1}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt((x - 1)**2*(x + 1)**2/(x**3 + x))/(x**2 + 1), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt((x^2 - 1)^2/(x^3 + x))/(x^2 + 1), x)

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Mupad [B]
time = 3.49, size = 43, normalized size = 1.30 \begin {gather*} -\frac {\sqrt {\frac {1}{x^3+x}}\,\left (2\,x^3+2\,x\right )\,\sqrt {{\left (x^2-1\right )}^2}}{\left (x^2-1\right )\,\left (x^2+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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