3.468 \(\int \frac{1}{(1+2 x^2)^{5/2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0042804, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {192, 191} \[ \frac{2 x}{3 \sqrt{2 x^2+1}}+\frac{x}{3 \left (2 x^2+1\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(1 + 2*x^2)^(-5/2),x]

[Out]

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

Rule 192

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p + 1
], 0] && NeQ[p, -1]

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rubi steps

\begin{align*} \int \frac{1}{\left (1+2 x^2\right )^{5/2}} \, dx &=\frac{x}{3 \left (1+2 x^2\right )^{3/2}}+\frac{2}{3} \int \frac{1}{\left (1+2 x^2\right )^{3/2}} \, dx\\ &=\frac{x}{3 \left (1+2 x^2\right )^{3/2}}+\frac{2 x}{3 \sqrt{1+2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0053973, size = 23, normalized size = 0.7 \[ \frac{x \left (4 x^2+3\right )}{3 \left (2 x^2+1\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + 2*x^2)^(-5/2),x]

[Out]

(x*(3 + 4*x^2))/(3*(1 + 2*x^2)^(3/2))

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Maple [A]  time = 0.001, size = 20, normalized size = 0.6 \begin{align*}{\frac{x \left ( 4\,{x}^{2}+3 \right ) }{3} \left ( 2\,{x}^{2}+1 \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*x^2+1)^(5/2),x)

[Out]

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

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Maxima [A]  time = 0.937497, size = 34, normalized size = 1.03 \begin{align*} \frac{2 \, x}{3 \, \sqrt{2 \, x^{2} + 1}} + \frac{x}{3 \,{\left (2 \, x^{2} + 1\right )}^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*x^2+1)^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.07408, size = 74, normalized size = 2.24 \begin{align*} \frac{{\left (4 \, x^{3} + 3 \, x\right )} \sqrt{2 \, x^{2} + 1}}{3 \,{\left (4 \, x^{4} + 4 \, x^{2} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*x^2+1)^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 6.52519, size = 61, normalized size = 1.85 \begin{align*} \frac{4 x^{3}}{6 x^{2} \sqrt{2 x^{2} + 1} + 3 \sqrt{2 x^{2} + 1}} + \frac{3 x}{6 x^{2} \sqrt{2 x^{2} + 1} + 3 \sqrt{2 x^{2} + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*x**2+1)**(5/2),x)

[Out]

4*x**3/(6*x**2*sqrt(2*x**2 + 1) + 3*sqrt(2*x**2 + 1)) + 3*x/(6*x**2*sqrt(2*x**2 + 1) + 3*sqrt(2*x**2 + 1))

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Giac [A]  time = 1.07411, size = 26, normalized size = 0.79 \begin{align*} \frac{{\left (4 \, x^{2} + 3\right )} x}{3 \,{\left (2 \, x^{2} + 1\right )}^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*x^2+1)^(5/2),x, algorithm="giac")

[Out]

1/3*(4*x^2 + 3)*x/(2*x^2 + 1)^(3/2)