3.5.77 \(\int \frac {1}{(1-2 x^2)^{7/2}} \, dx\) [477]

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

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 49, normalized size of antiderivative = 1.00, 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 = {198, 197} \begin {gather*} \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}} \end {gather*}

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 197

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]

Rule 198

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]

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*}

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Mathematica [A]
time = 0.18, size = 28, normalized size = 0.57 \begin {gather*} \frac {x \left (15-40 x^2+32 x^4\right )}{15 \left (1-2 x^2\right )^{5/2}} \end {gather*}

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 = 0.08, size = 38, normalized size = 0.78

method	result	size
gosper	\(\frac {x \left (32 x^{4}-40 x^{2}+15\right )}{15 \left (-2 x^{2}+1\right )^{\frac {5}{2}}}\)	\(25\)
meijerg	\(\frac {x \left (32 x^{4}-40 x^{2}+15\right )}{15 \left (-2 x^{2}+1\right )^{\frac {5}{2}}}\)	\(25\)
trager	\(-\frac {\left (32 x^{4}-40 x^{2}+15\right ) x \sqrt {-2 x^{2}+1}}{15 \left (2 x^{2}-1\right )^{3}}\)	\(34\)
risch	\(\frac {x \left (32 x^{4}-40 x^{2}+15\right )}{15 \left (2 x^{2}-1\right )^{2} \sqrt {-2 x^{2}+1}}\)	\(34\)
default	\(\frac {x}{5 \left (-2 x^{2}+1\right )^{\frac {5}{2}}}+\frac {4 x}{15 \left (-2 x^{2}+1\right )^{\frac {3}{2}}}+\frac {8 x}{15 \sqrt {-2 x^{2}+1}}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.79, size = 37, normalized size = 0.76 \begin {gather*} \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 {gather*}

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 = 0.83, size = 44, normalized size = 0.90 \begin {gather*} -\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 {gather*}

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 [C] Result contains complex when optimal does not.
time = 3.84, size = 291, normalized size = 5.94 \begin {gather*} \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}\: \left |{x^{2}}\right | > \frac {1}{2} \\\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 {gather*}

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)), Abs(x**2) > 1/2), (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 = 0.58, size = 35, normalized size = 0.71 \begin {gather*} -\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 {gather*}

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

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Mupad [B]
time = 0.28, size = 179, normalized size = 3.65 \begin {gather*} \frac {19\,\sqrt {\frac {1}{2}-x^2}}{480\,\left (x^2-\sqrt {2}\,x+\frac {1}{2}\right )}-\frac {19\,\sqrt {\frac {1}{2}-x^2}}{480\,\left (x^2+\sqrt {2}\,x+\frac {1}{2}\right )}-\frac {\sqrt {2}\,\sqrt {\frac {1}{2}-x^2}}{160\,\left (x^3-\frac {3\,\sqrt {2}\,x^2}{2}+\frac {3\,x}{2}-\frac {\sqrt {2}}{4}\right )}-\frac {\sqrt {2}\,\sqrt {\frac {1}{2}-x^2}}{160\,\left (x^3+\frac {3\,\sqrt {2}\,x^2}{2}+\frac {3\,x}{2}+\frac {\sqrt {2}}{4}\right )}-\frac {2\,\sqrt {2}\,\sqrt {\frac {1}{2}-x^2}}{15\,\left (x-\frac {\sqrt {2}}{2}\right )}-\frac {2\,\sqrt {2}\,\sqrt {\frac {1}{2}-x^2}}{15\,\left (x+\frac {\sqrt {2}}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1 - 2*x^2)^(7/2),x)

[Out]

(19*(1/2 - x^2)^(1/2))/(480*(x^2 - 2^(1/2)*x + 1/2)) - (19*(1/2 - x^2)^(1/2))/(480*(2^(1/2)*x + x^2 + 1/2)) -
(2^(1/2)*(1/2 - x^2)^(1/2))/(160*((3*x)/2 - 2^(1/2)/4 - (3*2^(1/2)*x^2)/2 + x^3)) - (2^(1/2)*(1/2 - x^2)^(1/2)
)/(160*((3*x)/2 + 2^(1/2)/4 + (3*2^(1/2)*x^2)/2 + x^3)) - (2*2^(1/2)*(1/2 - x^2)^(1/2))/(15*(x - 2^(1/2)/2)) -
(2*2^(1/2)*(1/2 - x^2)^(1/2))/(15*(x + 2^(1/2)/2))

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