3.1513 \(\int x^3 \sqrt {-2+x^8} \, dx\)

Optimal. Leaf size=35 \[ \frac {1}{8} x^4 \sqrt {x^8-2}-\frac {1}{4} \tanh ^{-1}\left (\frac {x^4}{\sqrt {x^8-2}}\right ) \]

[Out]

-1/4*arctanh(x^4/(x^8-2)^(1/2))+1/8*x^4*(x^8-2)^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {275, 195, 217, 206} \[ \frac {1}{8} x^4 \sqrt {x^8-2}-\frac {1}{4} \tanh ^{-1}\left (\frac {x^4}{\sqrt {x^8-2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^3*Sqrt[-2 + x^8],x]

[Out]

(x^4*Sqrt[-2 + x^8])/8 - ArcTanh[x^4/Sqrt[-2 + x^8]]/4

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rubi steps

\begin {align*} \int x^3 \sqrt {-2+x^8} \, dx &=\frac {1}{4} \operatorname {Subst}\left (\int \sqrt {-2+x^2} \, dx,x,x^4\right )\\ &=\frac {1}{8} x^4 \sqrt {-2+x^8}-\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{\sqrt {-2+x^2}} \, dx,x,x^4\right )\\ &=\frac {1}{8} x^4 \sqrt {-2+x^8}-\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x^4}{\sqrt {-2+x^8}}\right )\\ &=\frac {1}{8} x^4 \sqrt {-2+x^8}-\frac {1}{4} \tanh ^{-1}\left (\frac {x^4}{\sqrt {-2+x^8}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 50, normalized size = 1.43 \[ \frac {\left (x^8-2\right ) \left (2 \sin ^{-1}\left (\frac {x^4}{\sqrt {2}}\right )+\sqrt {2-x^8} x^4\right )}{8 \sqrt {-\left (x^8-2\right )^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*Sqrt[-2 + x^8],x]

[Out]

((-2 + x^8)*(x^4*Sqrt[2 - x^8] + 2*ArcSin[x^4/Sqrt[2]]))/(8*Sqrt[-(-2 + x^8)^2])

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fricas [A]  time = 0.88, size = 29, normalized size = 0.83 \[ \frac {1}{8} \, \sqrt {x^{8} - 2} x^{4} + \frac {1}{4} \, \log \left (-x^{4} + \sqrt {x^{8} - 2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(x^8-2)^(1/2),x, algorithm="fricas")

[Out]

1/8*sqrt(x^8 - 2)*x^4 + 1/4*log(-x^4 + sqrt(x^8 - 2))

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giac [A]  time = 0.16, size = 29, normalized size = 0.83 \[ \frac {1}{8} \, \sqrt {x^{8} - 2} x^{4} + \frac {1}{4} \, \log \left (x^{4} - \sqrt {x^{8} - 2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(x^8-2)^(1/2),x, algorithm="giac")

[Out]

1/8*sqrt(x^8 - 2)*x^4 + 1/4*log(x^4 - sqrt(x^8 - 2))

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maple [C]  time = 0.16, size = 47, normalized size = 1.34 \[ \frac {\sqrt {x^{8}-2}\, x^{4}}{8}-\frac {\sqrt {-\mathrm {signum}\left (\frac {x^{8}}{2}-1\right )}\, \arcsin \left (\frac {\sqrt {2}\, x^{4}}{2}\right )}{4 \sqrt {\mathrm {signum}\left (\frac {x^{8}}{2}-1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(x^8-2)^(1/2),x)

[Out]

1/8*x^4*(x^8-2)^(1/2)-1/4/signum(-1+1/2*x^8)^(1/2)*(-signum(-1+1/2*x^8))^(1/2)*arcsin(1/2*x^4*2^(1/2))

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maxima [B]  time = 1.08, size = 58, normalized size = 1.66 \[ -\frac {\sqrt {x^{8} - 2}}{4 \, x^{4} {\left (\frac {x^{8} - 2}{x^{8}} - 1\right )}} - \frac {1}{8} \, \log \left (\frac {\sqrt {x^{8} - 2}}{x^{4}} + 1\right ) + \frac {1}{8} \, \log \left (\frac {\sqrt {x^{8} - 2}}{x^{4}} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(x^8-2)^(1/2),x, algorithm="maxima")

[Out]

-1/4*sqrt(x^8 - 2)/(x^4*((x^8 - 2)/x^8 - 1)) - 1/8*log(sqrt(x^8 - 2)/x^4 + 1) + 1/8*log(sqrt(x^8 - 2)/x^4 - 1)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int x^3\,\sqrt {x^8-2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(x^8 - 2)^(1/2),x)

[Out]

int(x^3*(x^8 - 2)^(1/2), x)

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sympy [A]  time = 1.77, size = 90, normalized size = 2.57 \[ \begin {cases} \frac {x^{12}}{8 \sqrt {x^{8} - 2}} - \frac {x^{4}}{4 \sqrt {x^{8} - 2}} - \frac {\operatorname {acosh}{\left (\frac {\sqrt {2} x^{4}}{2} \right )}}{4} & \text {for}\: \frac {\left |{x^{8}}\right |}{2} > 1 \\- \frac {i x^{12}}{8 \sqrt {2 - x^{8}}} + \frac {i x^{4}}{4 \sqrt {2 - x^{8}}} + \frac {i \operatorname {asin}{\left (\frac {\sqrt {2} x^{4}}{2} \right )}}{4} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(x**8-2)**(1/2),x)

[Out]

Piecewise((x**12/(8*sqrt(x**8 - 2)) - x**4/(4*sqrt(x**8 - 2)) - acosh(sqrt(2)*x**4/2)/4, Abs(x**8)/2 > 1), (-I
*x**12/(8*sqrt(2 - x**8)) + I*x**4/(4*sqrt(2 - x**8)) + I*asin(sqrt(2)*x**4/2)/4, True))

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