3.363 \(\int \frac{x^4}{\sqrt{-2+x^{10}}} \, dx\)

Optimal. Leaf size=18 \[ \frac{1}{5} \tanh ^{-1}\left (\frac{x^5}{\sqrt{x^{10}-2}}\right ) \]

[Out]

ArcTanh[x^5/Sqrt[-2 + x^10]]/5

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Rubi [A]  time = 0.007273, antiderivative size = 18, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {275, 217, 206} \[ \frac{1}{5} \tanh ^{-1}\left (\frac{x^5}{\sqrt{x^{10}-2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^4/Sqrt[-2 + x^10],x]

[Out]

ArcTanh[x^5/Sqrt[-2 + x^10]]/5

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 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 \frac{x^4}{\sqrt{-2+x^{10}}} \, dx &=\frac{1}{5} \operatorname{Subst}\left (\int \frac{1}{\sqrt{-2+x^2}} \, dx,x,x^5\right )\\ &=\frac{1}{5} \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\frac{x^5}{\sqrt{-2+x^{10}}}\right )\\ &=\frac{1}{5} \tanh ^{-1}\left (\frac{x^5}{\sqrt{-2+x^{10}}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0030268, size = 18, normalized size = 1. \[ \frac{1}{5} \tanh ^{-1}\left (\frac{x^5}{\sqrt{x^{10}-2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/Sqrt[-2 + x^10],x]

[Out]

ArcTanh[x^5/Sqrt[-2 + x^10]]/5

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Maple [C]  time = 0.043, size = 34, normalized size = 1.9 \begin{align*}{\frac{1}{5}\sqrt{-{\it signum} \left ( -1+{\frac{{x}^{10}}{2}} \right ) }\arcsin \left ({\frac{{x}^{5}\sqrt{2}}{2}} \right ){\frac{1}{\sqrt{{\it signum} \left ( -1+{\frac{{x}^{10}}{2}} \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(x^10-2)^(1/2),x)

[Out]

1/5/signum(-1+1/2*x^10)^(1/2)*(-signum(-1+1/2*x^10))^(1/2)*arcsin(1/2*x^5*2^(1/2))

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Maxima [B]  time = 0.931524, size = 45, normalized size = 2.5 \begin{align*} \frac{1}{10} \, \log \left (\frac{\sqrt{x^{10} - 2}}{x^{5}} + 1\right ) - \frac{1}{10} \, \log \left (\frac{\sqrt{x^{10} - 2}}{x^{5}} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(x^10-2)^(1/2),x, algorithm="maxima")

[Out]

1/10*log(sqrt(x^10 - 2)/x^5 + 1) - 1/10*log(sqrt(x^10 - 2)/x^5 - 1)

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Fricas [A]  time = 1.81069, size = 45, normalized size = 2.5 \begin{align*} -\frac{1}{5} \, \log \left (-x^{5} + \sqrt{x^{10} - 2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(x^10-2)^(1/2),x, algorithm="fricas")

[Out]

-1/5*log(-x^5 + sqrt(x^10 - 2))

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Sympy [A]  time = 1.01159, size = 34, normalized size = 1.89 \begin{align*} \begin{cases} \frac{\operatorname{acosh}{\left (\frac{\sqrt{2} x^{5}}{2} \right )}}{5} & \text{for}\: \frac{\left |{x^{10}}\right |}{2} > 1 \\- \frac{i \operatorname{asin}{\left (\frac{\sqrt{2} x^{5}}{2} \right )}}{5} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(x**10-2)**(1/2),x)

[Out]

Piecewise((acosh(sqrt(2)*x**5/2)/5, Abs(x**10)/2 > 1), (-I*asin(sqrt(2)*x**5/2)/5, True))

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Giac [A]  time = 1.07437, size = 23, normalized size = 1.28 \begin{align*} -\frac{1}{5} \, \log \left ({\left | -x^{5} + \sqrt{x^{10} - 2} \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(x^10-2)^(1/2),x, algorithm="giac")

[Out]

-1/5*log(abs(-x^5 + sqrt(x^10 - 2)))