3.2.68 \(\int \frac {1}{(1+x^2) \sqrt [3]{x+x^3}} \, dx\) [168]

Optimal. Leaf size=20 \[ \frac {3 \left (x+x^3\right )^{2/3}}{2 \left (1+x^2\right )} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 14, normalized size of antiderivative = 0.70, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2081, 270} \begin {gather*} \frac {3 x}{2 \sqrt [3]{x^3+x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((1 + x^2)*(x + x^3)^(1/3)),x]

[Out]

(3*x)/(2*(x + x^3)^(1/3))

Rule 270

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

Rubi steps

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

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Mathematica [A]
time = 0.54, size = 14, normalized size = 0.70 \begin {gather*} \frac {3 x}{2 \sqrt [3]{x+x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*x)/(2*(x + x^3)^(1/3))

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Maple [A]
time = 0.40, size = 11, normalized size = 0.55

method	result	size
gosper	\(\frac {3 x}{2 \left (x^{3}+x \right )^{\frac {1}{3}}}\)	\(11\)
meijerg	\(\frac {3 x^{\frac {2}{3}}}{2 \left (x^{2}+1\right )^{\frac {1}{3}}}\)	\(13\)
risch	\(\frac {3 x}{2 \left (\left (x^{2}+1\right ) x \right )^{\frac {1}{3}}}\)	\(13\)
trager	\(\frac {3 \left (x^{3}+x \right )^{\frac {2}{3}}}{2 \left (x^{2}+1\right )}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

-3/4*(x^3 + x)/((x^(7/3) + x^(1/3))*(x^2 + 1)^(1/3)) + integrate(3/2*(x^2 + 1)^(2/3)/(x^(13/3) + 2*x^(7/3) + x
^(1/3)), x)

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Fricas [A]
time = 0.37, size = 16, normalized size = 0.80 \begin {gather*} \frac {3 \, {\left (x^{3} + x\right )}^{\frac {2}{3}}}{2 \, {\left (x^{2} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2*(x^3 + x)^(2/3)/(x^2 + 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.40, size = 9, normalized size = 0.45 \begin {gather*} \frac {3}{2 \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2/(1/x^2 + 1)^(1/3)

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Mupad [B]
time = 0.15, size = 18, normalized size = 0.90 \begin {gather*} \frac {3\,{\left (x^3+x\right )}^{2/3}}{2\,\left (x^2+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((x^2 + 1)*(x + x^3)^(1/3)),x)

[Out]

(3*(x + x^3)^(2/3))/(2*(x^2 + 1))

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