3.5.42 \(\int x^2 \sqrt {-1+x^6} \, dx\)

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

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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} \begin {gather*} \frac {1}{6} x^3 \sqrt {x^6-1}-\frac {1}{6} \tanh ^{-1}\left (\frac {x^3}{\sqrt {x^6-1}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*Sqrt[-1 + x^6],x]

[Out]

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

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^2 \sqrt {-1+x^6} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \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 = 42, normalized size = 1.20 \begin {gather*} \frac {\left (x^6-1\right ) \left (\sin ^{-1}\left (x^3\right )+\sqrt {1-x^6} x^3\right )}{6 \sqrt {-\left (x^6-1\right )^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*Sqrt[-1 + x^6],x]

[Out]

((-1 + x^6)*(x^3*Sqrt[1 - x^6] + ArcSin[x^3]))/(6*Sqrt[-(-1 + x^6)^2])

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IntegrateAlgebraic [A]  time = 0.11, 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^2*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 = 0.51, 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^2*(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 [A]  time = 0.29, size = 30, normalized size = 0.86 \begin {gather*} \frac {1}{6} \, \sqrt {x^{6} - 1} x^{3} + \frac {1}{6} \, \log \left ({\left | -x^{3} + \sqrt {x^{6} - 1} \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.18, size = 30, normalized size = 0.86

method	result	size
trager	\(\frac {x^{3} \sqrt {x^{6}-1}}{6}+\frac {\ln \left (x^{3}-\sqrt {x^{6}-1}\right )}{6}\)	\(30\)
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	\(\frac {i \sqrt {\mathrm {signum}\left (x^{6}-1\right )}\, \left (-2 i \sqrt {\pi }\, x^{3} \sqrt {-x^{6}+1}-2 i \sqrt {\pi }\, \arcsin \left (x^{3}\right )\right )}{12 \sqrt {-\mathrm {signum}\left (x^{6}-1\right )}\, \sqrt {\pi }}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(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 = 0.40, 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^2*(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 x^2\,\sqrt {x^6-1} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(x^6 - 1)^(1/2),x)

[Out]

int(x^2*(x^6 - 1)^(1/2), x)

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sympy [A]  time = 1.30, size = 60, normalized size = 1.71 \begin {gather*} \begin {cases} \frac {x^{9}}{6 \sqrt {x^{6} - 1}} - \frac {x^{3}}{6 \sqrt {x^{6} - 1}} - \frac {\operatorname {acosh}{\left (x^{3} \right )}}{6} & \text {for}\: \left |{x^{6}}\right | > 1 \\\frac {i x^{3} \sqrt {1 - x^{6}}}{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**2*(x**6-1)**(1/2),x)

[Out]

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

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