\(\int \frac {x^2}{\sqrt {1+x^6}} \, dx\) [153]

Optimal result

Integrand size = 13, antiderivative size = 18 \[ \int \frac {x^2}{\sqrt {1+x^6}} \, dx=\frac {1}{3} \log \left (x^3+\sqrt {1+x^6}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.44, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {281, 221} \[ \int \frac {x^2}{\sqrt {1+x^6}} \, dx=\frac {\text {arcsinh}\left (x^3\right )}{3} \]

[In]

Int[x^2/Sqrt[1 + x^6],x]

[Out]

ArcSinh[x^3]/3

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt[a])]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

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 \frac {1}{\sqrt {1+x^2}} \, dx,x,x^3\right ) \\ & = \frac {\text {arcsinh}\left (x^3\right )}{3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {x^2}{\sqrt {1+x^6}} \, dx=\frac {1}{3} \log \left (x^3+\sqrt {1+x^6}\right ) \]

[In]

Integrate[x^2/Sqrt[1 + x^6],x]

[Out]

Log[x^3 + Sqrt[1 + x^6]]/3

Maple [A] (verified)

Time = 0.80 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.39

method	result	size
meijerg	\(\frac {\operatorname {arcsinh}\left (x^{3}\right )}{3}\)	\(7\)
pseudoelliptic	\(\frac {\operatorname {arcsinh}\left (x^{3}\right )}{3}\)	\(7\)
trager	\(-\frac {\ln \left (x^{3}-\sqrt {x^{6}+1}\right )}{3}\)	\(17\)

[In]

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

[Out]

1/3*arcsinh(x^3)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {x^2}{\sqrt {1+x^6}} \, dx=-\frac {1}{3} \, \log \left (-x^{3} + \sqrt {x^{6} + 1}\right ) \]

[In]

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

[Out]

-1/3*log(-x^3 + sqrt(x^6 + 1))

Sympy [A] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 5, normalized size of antiderivative = 0.28 \[ \int \frac {x^2}{\sqrt {1+x^6}} \, dx=\frac {\operatorname {asinh}{\left (x^{3} \right )}}{3} \]

[In]

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

[Out]

asinh(x**3)/3

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 33 vs. \(2 (14) = 28\).

Time = 0.18 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.83 \[ \int \frac {x^2}{\sqrt {1+x^6}} \, dx=\frac {1}{6} \, \log \left (\frac {\sqrt {x^{6} + 1}}{x^{3}} + 1\right ) - \frac {1}{6} \, \log \left (\frac {\sqrt {x^{6} + 1}}{x^{3}} - 1\right ) \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 29 vs. \(2 (14) = 28\).

Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.61 \[ \int \frac {x^2}{\sqrt {1+x^6}} \, dx=\frac {1}{6} \, \sqrt {x^{6} + 1} x^{3} - \frac {1}{6} \, \log \left (-x^{3} + \sqrt {x^{6} + 1}\right ) \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2}{\sqrt {1+x^6}} \, dx=\int \frac {x^2}{\sqrt {x^6+1}} \,d x \]

[In]

int(x^2/(x^6 + 1)^(1/2),x)

[Out]

int(x^2/(x^6 + 1)^(1/2), x)