3.477 \(\int \frac{1}{(1-2 x^2)^{7/2}} \, dx\)

Optimal. Leaf size=49 \[ \frac{8 x}{15 \sqrt{1-2 x^2}}+\frac{4 x}{15 \left (1-2 x^2\right )^{3/2}}+\frac{x}{5 \left (1-2 x^2\right )^{5/2}} \]

[Out]

x/(5*(1 - 2*x^2)^(5/2)) + (4*x)/(15*(1 - 2*x^2)^(3/2)) + (8*x)/(15*Sqrt[1 - 2*x^2])

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Rubi [A]  time = 0.0074578, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {192, 191} \[ \frac{8 x}{15 \sqrt{1-2 x^2}}+\frac{4 x}{15 \left (1-2 x^2\right )^{3/2}}+\frac{x}{5 \left (1-2 x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(1 - 2*x^2)^(-7/2),x]

[Out]

x/(5*(1 - 2*x^2)^(5/2)) + (4*x)/(15*(1 - 2*x^2)^(3/2)) + (8*x)/(15*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 )^{7/2}} \, dx &=\frac{x}{5 \left (1-2 x^2\right )^{5/2}}+\frac{4}{5} \int \frac{1}{\left (1-2 x^2\right )^{5/2}} \, dx\\ &=\frac{x}{5 \left (1-2 x^2\right )^{5/2}}+\frac{4 x}{15 \left (1-2 x^2\right )^{3/2}}+\frac{8}{15} \int \frac{1}{\left (1-2 x^2\right )^{3/2}} \, dx\\ &=\frac{x}{5 \left (1-2 x^2\right )^{5/2}}+\frac{4 x}{15 \left (1-2 x^2\right )^{3/2}}+\frac{8 x}{15 \sqrt{1-2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0081629, size = 28, normalized size = 0.57 \[ \frac{x \left (32 x^4-40 x^2+15\right )}{15 \left (1-2 x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - 2*x^2)^(-7/2),x]

[Out]

(x*(15 - 40*x^2 + 32*x^4))/(15*(1 - 2*x^2)^(5/2))

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Maple [A]  time = 0.002, size = 25, normalized size = 0.5 \begin{align*}{\frac{x \left ( 32\,{x}^{4}-40\,{x}^{2}+15 \right ) }{15} \left ( -2\,{x}^{2}+1 \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-2*x^2+1)^(7/2),x)

[Out]

1/15*x*(32*x^4-40*x^2+15)/(-2*x^2+1)^(5/2)

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Maxima [A]  time = 0.924798, size = 50, normalized size = 1.02 \begin{align*} \frac{8 \, x}{15 \, \sqrt{-2 \, x^{2} + 1}} + \frac{4 \, x}{15 \,{\left (-2 \, x^{2} + 1\right )}^{\frac{3}{2}}} + \frac{x}{5 \,{\left (-2 \, x^{2} + 1\right )}^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-2*x^2+1)^(7/2),x, algorithm="maxima")

[Out]

8/15*x/sqrt(-2*x^2 + 1) + 4/15*x/(-2*x^2 + 1)^(3/2) + 1/5*x/(-2*x^2 + 1)^(5/2)

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Fricas [A]  time = 2.10818, size = 105, normalized size = 2.14 \begin{align*} -\frac{{\left (32 \, x^{5} - 40 \, x^{3} + 15 \, x\right )} \sqrt{-2 \, x^{2} + 1}}{15 \,{\left (8 \, x^{6} - 12 \, x^{4} + 6 \, x^{2} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-2*x^2+1)^(7/2),x, algorithm="fricas")

[Out]

-1/15*(32*x^5 - 40*x^3 + 15*x)*sqrt(-2*x^2 + 1)/(8*x^6 - 12*x^4 + 6*x^2 - 1)

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Sympy [B]  time = 85.6648, size = 291, normalized size = 5.94 \begin{align*} \begin{cases} - \frac{32 i x^{5}}{60 x^{4} \sqrt{2 x^{2} - 1} - 60 x^{2} \sqrt{2 x^{2} - 1} + 15 \sqrt{2 x^{2} - 1}} + \frac{40 i x^{3}}{60 x^{4} \sqrt{2 x^{2} - 1} - 60 x^{2} \sqrt{2 x^{2} - 1} + 15 \sqrt{2 x^{2} - 1}} - \frac{15 i x}{60 x^{4} \sqrt{2 x^{2} - 1} - 60 x^{2} \sqrt{2 x^{2} - 1} + 15 \sqrt{2 x^{2} - 1}} & \text{for}\: 2 \left |{x^{2}}\right | > 1 \\\frac{32 x^{5}}{60 x^{4} \sqrt{1 - 2 x^{2}} - 60 x^{2} \sqrt{1 - 2 x^{2}} + 15 \sqrt{1 - 2 x^{2}}} - \frac{40 x^{3}}{60 x^{4} \sqrt{1 - 2 x^{2}} - 60 x^{2} \sqrt{1 - 2 x^{2}} + 15 \sqrt{1 - 2 x^{2}}} + \frac{15 x}{60 x^{4} \sqrt{1 - 2 x^{2}} - 60 x^{2} \sqrt{1 - 2 x^{2}} + 15 \sqrt{1 - 2 x^{2}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-2*x**2+1)**(7/2),x)

[Out]

Piecewise((-32*I*x**5/(60*x**4*sqrt(2*x**2 - 1) - 60*x**2*sqrt(2*x**2 - 1) + 15*sqrt(2*x**2 - 1)) + 40*I*x**3/
(60*x**4*sqrt(2*x**2 - 1) - 60*x**2*sqrt(2*x**2 - 1) + 15*sqrt(2*x**2 - 1)) - 15*I*x/(60*x**4*sqrt(2*x**2 - 1)
- 60*x**2*sqrt(2*x**2 - 1) + 15*sqrt(2*x**2 - 1)), 2*Abs(x**2) > 1), (32*x**5/(60*x**4*sqrt(1 - 2*x**2) - 60*
x**2*sqrt(1 - 2*x**2) + 15*sqrt(1 - 2*x**2)) - 40*x**3/(60*x**4*sqrt(1 - 2*x**2) - 60*x**2*sqrt(1 - 2*x**2) +
15*sqrt(1 - 2*x**2)) + 15*x/(60*x**4*sqrt(1 - 2*x**2) - 60*x**2*sqrt(1 - 2*x**2) + 15*sqrt(1 - 2*x**2)), True)
)

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Giac [A]  time = 1.08415, size = 47, normalized size = 0.96 \begin{align*} -\frac{{\left (8 \,{\left (4 \, x^{2} - 5\right )} x^{2} + 15\right )} \sqrt{-2 \, x^{2} + 1} x}{15 \,{\left (2 \, x^{2} - 1\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-2*x^2+1)^(7/2),x, algorithm="giac")

[Out]

-1/15*(8*(4*x^2 - 5)*x^2 + 15)*sqrt(-2*x^2 + 1)*x/(2*x^2 - 1)^3