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

Optimal. Leaf size=18 \[ -\frac {4 \left (x^3+x^2\right )^{3/4}}{3 x^3} \]

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Rubi [A]  time = 0.02, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {1590} \begin {gather*} -\frac {4 \left (x^3+x^2\right )^{3/4}}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 + x)/(x^2*(x^2 + x^3)^(1/4)),x]

[Out]

(-4*(x^2 + x^3)^(3/4))/(3*x^3)

Rule 1590

Int[(Pp_)*(Qq_)^(m_.)*(Rr_)^(n_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x], r = Expon[Rr, x]}, S
imp[(Coeff[Pp, x, p]*x^(p - q - r + 1)*Qq^(m + 1)*Rr^(n + 1))/((p + m*q + n*r + 1)*Coeff[Qq, x, q]*Coeff[Rr, x
, r]), x] /; NeQ[p + m*q + n*r + 1, 0] && EqQ[(p + m*q + n*r + 1)*Coeff[Qq, x, q]*Coeff[Rr, x, r]*Pp, Coeff[Pp
, x, p]*x^(p - q - r)*((p - q - r + 1)*Qq*Rr + (m + 1)*x*Rr*D[Qq, x] + (n + 1)*x*Qq*D[Rr, x])]] /; FreeQ[{m, n
}, x] && PolyQ[Pp, x] && PolyQ[Qq, x] && PolyQ[Rr, x] && NeQ[m, -1] && NeQ[n, -1]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 18, normalized size = 1.00 \begin {gather*} -\frac {4 \left (x^2 (x+1)\right )^{3/4}}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + x)/(x^2*(x^2 + x^3)^(1/4)),x]

[Out]

(-4*(x^2*(1 + x))^(3/4))/(3*x^3)

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IntegrateAlgebraic [A]  time = 0.17, size = 18, normalized size = 1.00 \begin {gather*} -\frac {4 \left (x^3+x^2\right )^{3/4}}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(x^2 + x^3)^(3/4))/(3*x^3)

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fricas [A]  time = 0.40, size = 14, normalized size = 0.78 \begin {gather*} -\frac {4 \, {\left (x^{3} + x^{2}\right )}^{\frac {3}{4}}}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/3*(x^3 + x^2)^(3/4)/x^3

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giac [A]  time = 0.41, size = 11, normalized size = 0.61 \begin {gather*} -\frac {4}{3} \, {\left (\frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {3}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/3*(1/x + 1/x^2)^(3/4)

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maple [A]  time = 0.00, size = 18, normalized size = 1.00 \begin {gather*} -\frac {4 \left (1+x \right )}{3 x \left (x^{3}+x^{2}\right )^{\frac {1}{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2+x)/x^2/(x^3+x^2)^(1/4),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x + 2}{{\left (x^{3} + x^{2}\right )}^{\frac {1}{4}} x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.23, size = 14, normalized size = 0.78 \begin {gather*} -\frac {4\,{\left (x^3+x^2\right )}^{3/4}}{3\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + 2)/(x^2*(x^2 + x^3)^(1/4)),x)

[Out]

-(4*(x^2 + x^3)^(3/4))/(3*x^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x + 2}{x^{2} \sqrt [4]{x^{2} \left (x + 1\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((x + 2)/(x**2*(x**2*(x + 1))**(1/4)), x)

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