3.5.40 \(\int \frac {x^8}{\sqrt {-1+x^6}} \, dx\)

Optimal. Leaf size=35 \[ \frac {1}{6} \sqrt {x^6-1} x^3+\frac {1}{6} \log \left (\sqrt {x^6-1}+x^3\right ) \]

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Rubi [A]  time = 0.02, 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, 321, 217, 206} \begin {gather*} \frac {1}{6} \sqrt {x^6-1} x^3+\frac {1}{6} \tanh ^{-1}\left (\frac {x^3}{\sqrt {x^6-1}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^8/Sqrt[-1 + x^6],x]

[Out]

(x^3*Sqrt[-1 + x^6])/6 + ArcTanh[x^3/Sqrt[-1 + x^6]]/6

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]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rubi steps

\begin {align*} \int \frac {x^8}{\sqrt {-1+x^6}} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {-1+x^2}} \, dx,x,x^3\right )\\ &=\frac {1}{6} x^3 \sqrt {-1+x^6}+\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{\sqrt {-1+x^2}} \, dx,x,x^3\right )\\ &=\frac {1}{6} x^3 \sqrt {-1+x^6}+\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x^3}{\sqrt {-1+x^6}}\right )\\ &=\frac {1}{6} x^3 \sqrt {-1+x^6}+\frac {1}{6} \tanh ^{-1}\left (\frac {x^3}{\sqrt {-1+x^6}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 32, normalized size = 0.91 \begin {gather*} \frac {1}{6} \left (\sqrt {x^6-1} x^3+\tanh ^{-1}\left (\frac {x^3}{\sqrt {x^6-1}}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^8/Sqrt[-1 + x^6],x]

[Out]

(x^3*Sqrt[-1 + x^6] + ArcTanh[x^3/Sqrt[-1 + x^6]])/6

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IntegrateAlgebraic [A]  time = 0.16, size = 35, normalized size = 1.00 \begin {gather*} \frac {1}{6} x^3 \sqrt {-1+x^6}+\frac {1}{6} \log \left (x^3+\sqrt {-1+x^6}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^8/Sqrt[-1 + x^6],x]

[Out]

(x^3*Sqrt[-1 + x^6])/6 + Log[x^3 + Sqrt[-1 + x^6]]/6

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fricas [A]  time = 0.64, size = 29, normalized size = 0.83 \begin {gather*} \frac {1}{6} \, \sqrt {x^{6} - 1} x^{3} - \frac {1}{6} \, \log \left (-x^{3} + \sqrt {x^{6} - 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(x^6-1)^(1/2),x, algorithm="fricas")

[Out]

1/6*sqrt(x^6 - 1)*x^3 - 1/6*log(-x^3 + sqrt(x^6 - 1))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{8}}{\sqrt {x^{6} - 1}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(x^6-1)^(1/2),x, algorithm="giac")

[Out]

integrate(x^8/sqrt(x^6 - 1), x)

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maple [A]  time = 0.19, size = 28, normalized size = 0.80

method	result	size
trager	\(\frac {x^{3} \sqrt {x^{6}-1}}{6}+\frac {\ln \left (x^{3}+\sqrt {x^{6}-1}\right )}{6}\)	\(28\)
risch	\(\frac {x^{3} \sqrt {x^{6}-1}}{6}+\frac {\sqrt {-\mathrm {signum}\left (x^{6}-1\right )}\, \arcsin \left (x^{3}\right )}{6 \sqrt {\mathrm {signum}\left (x^{6}-1\right )}}\)	\(38\)
meijerg	\(\frac {i \sqrt {-\mathrm {signum}\left (x^{6}-1\right )}\, \left (i \sqrt {\pi }\, x^{3} \sqrt {-x^{6}+1}-i \sqrt {\pi }\, \arcsin \left (x^{3}\right )\right )}{6 \sqrt {\mathrm {signum}\left (x^{6}-1\right )}\, \sqrt {\pi }}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/(x^6-1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/6*x^3*(x^6-1)^(1/2)+1/6*ln(x^3+(x^6-1)^(1/2))

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maxima [B]  time = 0.34, size = 58, normalized size = 1.66 \begin {gather*} -\frac {\sqrt {x^{6} - 1}}{6 \, x^{3} {\left (\frac {x^{6} - 1}{x^{6}} - 1\right )}} + \frac {1}{12} \, \log \left (\frac {\sqrt {x^{6} - 1}}{x^{3}} + 1\right ) - \frac {1}{12} \, \log \left (\frac {\sqrt {x^{6} - 1}}{x^{3}} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(x^6-1)^(1/2),x, algorithm="maxima")

[Out]

-1/6*sqrt(x^6 - 1)/(x^3*((x^6 - 1)/x^6 - 1)) + 1/12*log(sqrt(x^6 - 1)/x^3 + 1) - 1/12*log(sqrt(x^6 - 1)/x^3 -
1)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {x^8}{\sqrt {x^6-1}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/(x^6 - 1)^(1/2),x)

[Out]

int(x^8/(x^6 - 1)^(1/2), x)

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sympy [A]  time = 1.56, size = 61, normalized size = 1.74 \begin {gather*} \begin {cases} \frac {x^{3} \sqrt {x^{6} - 1}}{6} + \frac {\operatorname {acosh}{\left (x^{3} \right )}}{6} & \text {for}\: \left |{x^{6}}\right | > 1 \\- \frac {i x^{9}}{6 \sqrt {1 - x^{6}}} + \frac {i x^{3}}{6 \sqrt {1 - x^{6}}} - \frac {i \operatorname {asin}{\left (x^{3} \right )}}{6} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8/(x**6-1)**(1/2),x)

[Out]

Piecewise((x**3*sqrt(x**6 - 1)/6 + acosh(x**3)/6, Abs(x**6) > 1), (-I*x**9/(6*sqrt(1 - x**6)) + I*x**3/(6*sqrt
(1 - x**6)) - I*asin(x**3)/6, True))

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