3.473 \(\int \frac{(\frac{2+x^2}{x^2})^{7/9}}{(2+x^2)^{3/2}} \, dx\)

Optimal. Leaf size=25 \[ -\frac{9 \left (\frac{2}{x^2}+1\right )^{7/9} x}{10 \sqrt{x^2+2}} \]

[Out]

(-9*(1 + 2/x^2)^(7/9)*x)/(10*Sqrt[2 + x^2])

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Rubi [A]  time = 0.0569161, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {1991, 435, 264} \[ -\frac{9 \left (\frac{2}{x^2}+1\right )^{7/9} x}{10 \sqrt{x^2+2}} \]

Antiderivative was successfully verified.

[In]

Int[((2 + x^2)/x^2)^(7/9)/(2 + x^2)^(3/2),x]

[Out]

(-9*(1 + 2/x^2)^(7/9)*x)/(10*Sqrt[2 + x^2])

Rule 1991

Int[(u_)^(q_.)*(v_)^(p_.), x_Symbol] :> Int[ExpandToSum[u, x]^q*ExpandToSum[v, x]^p, x] /; FreeQ[{p, q}, x] &&
BinomialQ[u, x] && BinomialQ[v, x] &&  !(BinomialMatchQ[u, x] && BinomialMatchQ[v, x])

Rule 435

Int[((c_) + (d_.)*(x_)^(mn_.))^(q_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(x^(n*FracPart[q])*(c +
d/x^n)^FracPart[q])/(d + c*x^n)^FracPart[q], Int[((a + b*x^n)^p*(d + c*x^n)^q)/x^(n*q), x], x] /; FreeQ[{a, b,
c, d, n, p, q}, x] && EqQ[mn, -n] &&  !IntegerQ[q] &&  !IntegerQ[p]

Rule 264

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]

Rubi steps

\begin{align*} \int \frac{\left (\frac{2+x^2}{x^2}\right )^{7/9}}{\left (2+x^2\right )^{3/2}} \, dx &=\int \frac{\left (1+\frac{2}{x^2}\right )^{7/9}}{\left (2+x^2\right )^{3/2}} \, dx\\ &=\frac{\left (\left (1+\frac{2}{x^2}\right )^{7/9} x^{14/9}\right ) \int \frac{1}{x^{14/9} \left (2+x^2\right )^{13/18}} \, dx}{\left (2+x^2\right )^{7/9}}\\ &=-\frac{9 \left (1+\frac{2}{x^2}\right )^{7/9} x}{10 \sqrt{2+x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0085718, size = 25, normalized size = 1. \[ -\frac{9 \left (\frac{2}{x^2}+1\right )^{7/9} x}{10 \sqrt{x^2+2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((2 + x^2)/x^2)^(7/9)/(2 + x^2)^(3/2),x]

[Out]

(-9*(1 + 2/x^2)^(7/9)*x)/(10*Sqrt[2 + x^2])

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Maple [A]  time = 0.004, size = 22, normalized size = 0.9 \begin{align*} -{\frac{9\,x}{10} \left ({\frac{{x}^{2}+2}{{x}^{2}}} \right ) ^{{\frac{7}{9}}}{\frac{1}{\sqrt{{x}^{2}+2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^2+2)/x^2)^(7/9)/(x^2+2)^(3/2),x)

[Out]

-9/10*x/(x^2+2)^(1/2)*((x^2+2)/x^2)^(7/9)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\frac{x^{2} + 2}{x^{2}}\right )^{\frac{7}{9}}}{{\left (x^{2} + 2\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+2)/x^2)^(7/9)/(x^2+2)^(3/2),x, algorithm="maxima")

[Out]

integrate(((x^2 + 2)/x^2)^(7/9)/(x^2 + 2)^(3/2), x)

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Fricas [A]  time = 4.55282, size = 61, normalized size = 2.44 \begin{align*} -\frac{9 \, x \left (\frac{x^{2} + 2}{x^{2}}\right )^{\frac{7}{9}}}{10 \, \sqrt{x^{2} + 2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+2)/x^2)^(7/9)/(x^2+2)^(3/2),x, algorithm="fricas")

[Out]

-9/10*x*((x^2 + 2)/x^2)^(7/9)/sqrt(x^2 + 2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**2+2)/x**2)**(7/9)/(x**2+2)**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\frac{x^{2} + 2}{x^{2}}\right )^{\frac{7}{9}}}{{\left (x^{2} + 2\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+2)/x^2)^(7/9)/(x^2+2)^(3/2),x, algorithm="giac")

[Out]

integrate(((x^2 + 2)/x^2)^(7/9)/(x^2 + 2)^(3/2), x)