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]

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

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Rubi [A]  time = 0.0120813, antiderivative size = 35, normalized size of antiderivative = 1., 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 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]

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 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 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])

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*}

Mathematica [A]  time = 0.017889, size = 50, normalized size = 1.43 \[ \frac{\left (x^8-2\right ) \left (\sqrt{2-x^8} x^4+2 \sin ^{-1}\left (\frac{x^4}{\sqrt{2}}\right )\right )}{8 \sqrt{-\left (x^8-2\right )^2}} \]

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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Maple [C]  time = 0.046, size = 47, normalized size = 1.3 \begin{align*}{\frac{{x}^{4}}{8}\sqrt{{x}^{8}-2}}-{\frac{1}{4}\sqrt{-{\it signum} \left ( -1+{\frac{{x}^{8}}{2}} \right ) }\arcsin \left ({\frac{{x}^{4}\sqrt{2}}{2}} \right ){\frac{1}{\sqrt{{\it signum} \left ( -1+{\frac{{x}^{8}}{2}} \right ) }}}} \end{align*}

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 = 0.966931, size = 78, normalized size = 2.23 \begin{align*} -\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 ) \end{align*}

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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Fricas [A]  time = 1.28885, size = 74, normalized size = 2.11 \begin{align*} \frac{1}{8} \, \sqrt{x^{8} - 2} x^{4} + \frac{1}{4} \, \log \left (-x^{4} + \sqrt{x^{8} - 2}\right ) \end{align*}

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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Sympy [A]  time = 1.5826, size = 90, normalized size = 2.57 \begin{align*} \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} \end{align*}

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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Giac [A]  time = 1.16387, size = 39, normalized size = 1.11 \begin{align*} \frac{1}{8} \, \sqrt{x^{8} - 2} x^{4} + \frac{1}{4} \, \log \left (x^{4} - \sqrt{x^{8} - 2}\right ) \end{align*}

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))