\(\int x^3 \sqrt {-2+x^8} \, dx\) [1513]

Optimal result

Integrand size = 13, antiderivative size = 35 \[ \int x^3 \sqrt {-2+x^8} \, dx=\frac {1}{8} x^4 \sqrt {-2+x^8}-\frac {1}{4} \text {arctanh}\left (\frac {x^4}{\sqrt {-2+x^8}}\right ) \]

[Out]

-1/4*arctanh(x^4/(x^8-2)^(1/2))+1/8*x^4*(x^8-2)^(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} \[ \int x^3 \sqrt {-2+x^8} \, dx=\frac {1}{8} x^4 \sqrt {x^8-2}-\frac {1}{4} \text {arctanh}\left (\frac {x^4}{\sqrt {x^8-2}}\right ) \]

[In]

Int[x^3*Sqrt[-2 + x^8],x]

[Out]

(x^4*Sqrt[-2 + x^8])/8 - ArcTanh[x^4/Sqrt[-2 + x^8]]/4

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}{4} \text {Subst}\left (\int \sqrt {-2+x^2} \, dx,x,x^4\right ) \\ & = \frac {1}{8} x^4 \sqrt {-2+x^8}-\frac {1}{4} \text {Subst}\left (\int \frac {1}{\sqrt {-2+x^2}} \, dx,x,x^4\right ) \\ & = \frac {1}{8} x^4 \sqrt {-2+x^8}-\frac {1}{4} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x^4}{\sqrt {-2+x^8}}\right ) \\ & = \frac {1}{8} x^4 \sqrt {-2+x^8}-\frac {1}{4} \tanh ^{-1}\left (\frac {x^4}{\sqrt {-2+x^8}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int x^3 \sqrt {-2+x^8} \, dx=\frac {1}{8} x^4 \sqrt {-2+x^8}-\frac {1}{4} \log \left (x^4+\sqrt {-2+x^8}\right ) \]

[In]

Integrate[x^3*Sqrt[-2 + x^8],x]

[Out]

(x^4*Sqrt[-2 + x^8])/8 - Log[x^4 + Sqrt[-2 + x^8]]/4

Maple [A] (verified)

Time = 3.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.80

method	result	size
trager	\(\frac {x^{4} \sqrt {x^{8}-2}}{8}-\frac {\ln \left (x^{4}+\sqrt {x^{8}-2}\right )}{4}\)	\(28\)
pseudoelliptic	\(\frac {x^{4} \sqrt {x^{8}-2}}{8}-\frac {\ln \left (x^{4}+\sqrt {x^{8}-2}\right )}{4}\)	\(28\)
risch	\(\frac {x^{4} \sqrt {x^{8}-2}}{8}-\frac {\sqrt {-\operatorname {signum}\left (-1+\frac {x^{8}}{2}\right )}\, \arcsin \left (\frac {x^{4} \sqrt {2}}{2}\right )}{4 \sqrt {\operatorname {signum}\left (-1+\frac {x^{8}}{2}\right )}}\)	\(47\)
meijerg	\(\frac {i \sqrt {\operatorname {signum}\left (-1+\frac {x^{8}}{2}\right )}\, \left (-i \sqrt {\pi }\, x^{4} \sqrt {2}\, \sqrt {-\frac {x^{8}}{2}+1}-2 i \sqrt {\pi }\, \arcsin \left (\frac {x^{4} \sqrt {2}}{2}\right )\right )}{8 \sqrt {\pi }\, \sqrt {-\operatorname {signum}\left (-1+\frac {x^{8}}{2}\right )}}\)	\(66\)

[In]

int(x^3*(x^8-2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/8*x^4*(x^8-2)^(1/2)-1/4*ln(x^4+(x^8-2)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.83 \[ \int x^3 \sqrt {-2+x^8} \, dx=\frac {1}{8} \, \sqrt {x^{8} - 2} x^{4} + \frac {1}{4} \, \log \left (-x^{4} + \sqrt {x^{8} - 2}\right ) \]

[In]

integrate(x^3*(x^8-2)^(1/2),x, algorithm="fricas")

[Out]

1/8*sqrt(x^8 - 2)*x^4 + 1/4*log(-x^4 + sqrt(x^8 - 2))

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.90 (sec) , antiderivative size = 88, normalized size of antiderivative = 2.51 \[ \int x^3 \sqrt {-2+x^8} \, dx=\begin {cases} \frac {x^{12}}{8 \sqrt {x^{8} - 2}} - \frac {x^{4}}{4 \sqrt {x^{8} - 2}} - \frac {\operatorname {acosh}{\left (\frac {\sqrt {2} x^{4}}{2} \right )}}{4} & \text {for}\: \left |{x^{8}}\right | > 2 \\- \frac {i x^{12}}{8 \sqrt {2 - x^{8}}} + \frac {i x^{4}}{4 \sqrt {2 - x^{8}}} + \frac {i \operatorname {asin}{\left (\frac {\sqrt {2} x^{4}}{2} \right )}}{4} & \text {otherwise} \end {cases} \]

[In]

integrate(x**3*(x**8-2)**(1/2),x)

[Out]

Piecewise((x**12/(8*sqrt(x**8 - 2)) - x**4/(4*sqrt(x**8 - 2)) - acosh(sqrt(2)*x**4/2)/4, Abs(x**8) > 2), (-I*x
**12/(8*sqrt(2 - x**8)) + I*x**4/(4*sqrt(2 - x**8)) + I*asin(sqrt(2)*x**4/2)/4, 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 \[ \int x^3 \sqrt {-2+x^8} \, dx=-\frac {\sqrt {x^{8} - 2}}{4 \, x^{4} {\left (\frac {x^{8} - 2}{x^{8}} - 1\right )}} - \frac {1}{8} \, \log \left (\frac {\sqrt {x^{8} - 2}}{x^{4}} + 1\right ) + \frac {1}{8} \, \log \left (\frac {\sqrt {x^{8} - 2}}{x^{4}} - 1\right ) \]

[In]

integrate(x^3*(x^8-2)^(1/2),x, algorithm="maxima")

[Out]

-1/4*sqrt(x^8 - 2)/(x^4*((x^8 - 2)/x^8 - 1)) - 1/8*log(sqrt(x^8 - 2)/x^4 + 1) + 1/8*log(sqrt(x^8 - 2)/x^4 - 1)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.83 \[ \int x^3 \sqrt {-2+x^8} \, dx=\frac {1}{8} \, \sqrt {x^{8} - 2} x^{4} + \frac {1}{4} \, \log \left (x^{4} - \sqrt {x^{8} - 2}\right ) \]

[In]

integrate(x^3*(x^8-2)^(1/2),x, algorithm="giac")

[Out]

1/8*sqrt(x^8 - 2)*x^4 + 1/4*log(x^4 - sqrt(x^8 - 2))

Mupad [F(-1)]

Timed out. \[ \int x^3 \sqrt {-2+x^8} \, dx=\int x^3\,\sqrt {x^8-2} \,d x \]

[In]

int(x^3*(x^8 - 2)^(1/2),x)

[Out]

int(x^3*(x^8 - 2)^(1/2), x)