\(\int x^2 \sqrt {-2+x^6} \, dx\) [1382]

Optimal result

Integrand size = 13, antiderivative size = 35 \[ \int x^2 \sqrt {-2+x^6} \, dx=\frac {1}{6} x^3 \sqrt {-2+x^6}-\frac {1}{3} \text {arctanh}\left (\frac {x^3}{\sqrt {-2+x^6}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {281, 201, 223, 212} \[ \int x^2 \sqrt {-2+x^6} \, dx=\frac {1}{6} x^3 \sqrt {x^6-2}-\frac {1}{3} \text {arctanh}\left (\frac {x^3}{\sqrt {x^6-2}}\right ) \]

[In]

Int[x^2*Sqrt[-2 + x^6],x]

[Out]

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

Rule 201

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 212

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

Rule 223

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 281

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int x^2 \sqrt {-2+x^6} \, dx=\frac {1}{6} x^3 \sqrt {-2+x^6}-\frac {1}{3} \log \left (x^3+\sqrt {-2+x^6}\right ) \]

[In]

Integrate[x^2*Sqrt[-2 + x^6],x]

[Out]

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

Maple [A] (verified)

Time = 4.76 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.80

method	result	size
pseudoelliptic	\(\frac {x^{3} \sqrt {x^{6}-2}}{6}-\frac {\ln \left (x^{3}+\sqrt {x^{6}-2}\right )}{3}\)	\(28\)
trager	\(\frac {x^{3} \sqrt {x^{6}-2}}{6}+\frac {\ln \left (x^{3}-\sqrt {x^{6}-2}\right )}{3}\)	\(30\)
risch	\(\frac {x^{3} \sqrt {x^{6}-2}}{6}-\frac {\sqrt {-\operatorname {signum}\left (-1+\frac {x^{6}}{2}\right )}\, \arcsin \left (\frac {x^{3} \sqrt {2}}{2}\right )}{3 \sqrt {\operatorname {signum}\left (-1+\frac {x^{6}}{2}\right )}}\)	\(47\)
meijerg	\(\frac {i \sqrt {\operatorname {signum}\left (-1+\frac {x^{6}}{2}\right )}\, \left (-i \sqrt {\pi }\, x^{3} \sqrt {2}\, \sqrt {-\frac {x^{6}}{2}+1}-2 i \sqrt {\pi }\, \arcsin \left (\frac {x^{3} \sqrt {2}}{2}\right )\right )}{6 \sqrt {\pi }\, \sqrt {-\operatorname {signum}\left (-1+\frac {x^{6}}{2}\right )}}\)	\(66\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.83 \[ \int x^2 \sqrt {-2+x^6} \, dx=\frac {1}{6} \, \sqrt {x^{6} - 2} x^{3} + \frac {1}{3} \, \log \left (-x^{3} + \sqrt {x^{6} - 2}\right ) \]

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.92 (sec) , antiderivative size = 88, normalized size of antiderivative = 2.51 \[ \int x^2 \sqrt {-2+x^6} \, dx=\begin {cases} \frac {x^{9}}{6 \sqrt {x^{6} - 2}} - \frac {x^{3}}{3 \sqrt {x^{6} - 2}} - \frac {\operatorname {acosh}{\left (\frac {\sqrt {2} x^{3}}{2} \right )}}{3} & \text {for}\: \left |{x^{6}}\right | > 2 \\- \frac {i x^{9}}{6 \sqrt {2 - x^{6}}} + \frac {i x^{3}}{3 \sqrt {2 - x^{6}}} + \frac {i \operatorname {asin}{\left (\frac {\sqrt {2} x^{3}}{2} \right )}}{3} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((x**9/(6*sqrt(x**6 - 2)) - x**3/(3*sqrt(x**6 - 2)) - acosh(sqrt(2)*x**3/2)/3, Abs(x**6) > 2), (-I*x*
*9/(6*sqrt(2 - x**6)) + I*x**3/(3*sqrt(2 - x**6)) + I*asin(sqrt(2)*x**3/2)/3, True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (27) = 54\).

Time = 0.19 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.66 \[ \int x^2 \sqrt {-2+x^6} \, dx=-\frac {\sqrt {x^{6} - 2}}{3 \, x^{3} {\left (\frac {x^{6} - 2}{x^{6}} - 1\right )}} - \frac {1}{6} \, \log \left (\frac {\sqrt {x^{6} - 2}}{x^{3}} + 1\right ) + \frac {1}{6} \, \log \left (\frac {\sqrt {x^{6} - 2}}{x^{3}} - 1\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.86 \[ \int x^2 \sqrt {-2+x^6} \, dx=\frac {1}{6} \, \sqrt {x^{6} - 2} x^{3} + \frac {1}{3} \, \log \left ({\left | -x^{3} + \sqrt {x^{6} - 2} \right |}\right ) \]

[In]

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

[Out]

1/6*sqrt(x^6 - 2)*x^3 + 1/3*log(abs(-x^3 + sqrt(x^6 - 2)))

Mupad [F(-1)]

Timed out. \[ \int x^2 \sqrt {-2+x^6} \, dx=\int x^2\,\sqrt {x^6-2} \,d x \]

[In]

int(x^2*(x^6 - 2)^(1/2),x)

[Out]

int(x^2*(x^6 - 2)^(1/2), x)