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

Optimal. Leaf size=14 \[ \frac {3}{2} x \sqrt [3]{x^3+x} \]

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Rubi [A]  time = 0.08, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {2056, 449} \begin {gather*} \frac {3}{2} x \sqrt [3]{x^3+x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*x*(x + x^3)^(1/3))/2

Rule 449

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 2056

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

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Mathematica [A]  time = 0.04, size = 14, normalized size = 1.00 \begin {gather*} \frac {3}{2} x \sqrt [3]{x^3+x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*x*(x + x^3)^(1/3))/2

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IntegrateAlgebraic [A]  time = 0.18, size = 14, normalized size = 1.00 \begin {gather*} \frac {3}{2} x \sqrt [3]{x+x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*x*(x + x^3)^(1/3))/2

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fricas [A]  time = 0.45, size = 10, normalized size = 0.71 \begin {gather*} \frac {3}{2} \, {\left (x^{3} + x\right )}^{\frac {1}{3}} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2*(x^3 + x)^(1/3)*x

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giac [A]  time = 0.42, size = 12, normalized size = 0.86 \begin {gather*} \frac {3}{2} \, x^{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((3*x^2+2)*(x^3+x)^(1/3)/(x^2+1),x, algorithm="giac")

[Out]

3/2*x^2*(1/x^2 + 1)^(1/3)

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maple [A]  time = 0.07, size = 11, normalized size = 0.79

method	result	size
gosper	\(\frac {3 x \left (x^{3}+x \right )^{\frac {1}{3}}}{2}\)	\(11\)
trager	\(\frac {3 x \left (x^{3}+x \right )^{\frac {1}{3}}}{2}\)	\(11\)
risch	\(\frac {3 \left (\left (x^{2}+1\right ) x \right )^{\frac {1}{3}} x}{2}\)	\(13\)
meijerg	\(\frac {3 \hypergeom \left (\left [\frac {2}{3}, \frac {2}{3}\right ], \left [\frac {5}{3}\right ], -x^{2}\right ) x^{\frac {4}{3}}}{2}+\frac {9 \hypergeom \left (\left [\frac {2}{3}, \frac {5}{3}\right ], \left [\frac {8}{3}\right ], -x^{2}\right ) x^{\frac {10}{3}}}{10}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x^2+2)*(x^3+x)^(1/3)/(x^2+1),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*} \int \frac {{\left (x^{3} + x\right )}^{\frac {1}{3}} {\left (3 \, x^{2} + 2\right )}}{x^{2} + 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.19, size = 10, normalized size = 0.71 \begin {gather*} \frac {3\,x\,{\left (x^3+x\right )}^{1/3}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*x*(x + x^3)^(1/3))/2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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