\(\int \frac {x^9}{\sqrt {1+x^8}} \, dx\) [1525]

Optimal result

Integrand size = 13, antiderivative size = 62 \[ \int \frac {x^9}{\sqrt {1+x^8}} \, dx=\frac {1}{6} x^2 \sqrt {1+x^8}-\frac {\left (1+x^4\right ) \sqrt {\frac {1+x^8}{\left (1+x^4\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (x^2\right ),\frac {1}{2}\right )}{12 \sqrt {1+x^8}} \]

[Out]

1/6*x^2*(x^8+1)^(1/2)-1/12*(x^4+1)*(cos(2*arctan(x^2))^2)^(1/2)/cos(2*arctan(x^2))*EllipticF(sin(2*arctan(x^2)
),1/2*2^(1/2))*((x^8+1)/(x^4+1)^2)^(1/2)/(x^8+1)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {281, 327, 226} \[ \int \frac {x^9}{\sqrt {1+x^8}} \, dx=\frac {1}{6} x^2 \sqrt {x^8+1}-\frac {\left (x^4+1\right ) \sqrt {\frac {x^8+1}{\left (x^4+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (x^2\right ),\frac {1}{2}\right )}{12 \sqrt {x^8+1}} \]

[In]

Int[x^9/Sqrt[1 + x^8],x]

[Out]

(x^2*Sqrt[1 + x^8])/6 - ((1 + x^4)*Sqrt[(1 + x^8)/(1 + x^4)^2]*EllipticF[2*ArcTan[x^2], 1/2])/(12*Sqrt[1 + x^8
])

Rule 226

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

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]

Rule 327

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*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {x^4}{\sqrt {1+x^4}} \, dx,x,x^2\right ) \\ & = \frac {1}{6} x^2 \sqrt {1+x^8}-\frac {1}{6} \text {Subst}\left (\int \frac {1}{\sqrt {1+x^4}} \, dx,x,x^2\right ) \\ & = \frac {1}{6} x^2 \sqrt {1+x^8}-\frac {\left (1+x^4\right ) \sqrt {\frac {1+x^8}{\left (1+x^4\right )^2}} F\left (2 \tan ^{-1}\left (x^2\right )|\frac {1}{2}\right )}{12 \sqrt {1+x^8}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.02 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.55 \[ \int \frac {x^9}{\sqrt {1+x^8}} \, dx=\frac {1}{6} x^2 \left (\sqrt {1+x^8}-\operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-x^8\right )\right ) \]

[In]

Integrate[x^9/Sqrt[1 + x^8],x]

[Out]

(x^2*(Sqrt[1 + x^8] - Hypergeometric2F1[1/4, 1/2, 5/4, -x^8]))/6

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4.

Time = 6.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.27

method	result	size
meijerg	\(\frac {x^{10} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{2},\frac {5}{4};\frac {9}{4};-x^{8}\right )}{10}\)	\(17\)
risch	\(\frac {x^{2} \sqrt {x^{8}+1}}{6}-\frac {x^{2} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-x^{8}\right )}{6}\)	\(30\)

[In]

int(x^9/(x^8+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/10*x^10*hypergeom([1/2,5/4],[9/4],-x^8)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.08 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.45 \[ \int \frac {x^9}{\sqrt {1+x^8}} \, dx=\frac {1}{6} \, \sqrt {x^{8} + 1} x^{2} - \frac {1}{6} i \, \sqrt {i} F(\arcsin \left (\frac {\sqrt {i}}{x^{2}}\right )\,|\,-1) \]

[In]

integrate(x^9/(x^8+1)^(1/2),x, algorithm="fricas")

[Out]

1/6*sqrt(x^8 + 1)*x^2 - 1/6*I*sqrt(I)*elliptic_f(arcsin(sqrt(I)/x^2), -1)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.48 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.47 \[ \int \frac {x^9}{\sqrt {1+x^8}} \, dx=\frac {x^{10} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {x^{8} e^{i \pi }} \right )}}{8 \Gamma \left (\frac {9}{4}\right )} \]

[In]

integrate(x**9/(x**8+1)**(1/2),x)

[Out]

x**10*gamma(5/4)*hyper((1/2, 5/4), (9/4,), x**8*exp_polar(I*pi))/(8*gamma(9/4))

Maxima [F]

\[ \int \frac {x^9}{\sqrt {1+x^8}} \, dx=\int { \frac {x^{9}}{\sqrt {x^{8} + 1}} \,d x } \]

[In]

integrate(x^9/(x^8+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^9/sqrt(x^8 + 1), x)

Giac [F]

\[ \int \frac {x^9}{\sqrt {1+x^8}} \, dx=\int { \frac {x^{9}}{\sqrt {x^{8} + 1}} \,d x } \]

[In]

integrate(x^9/(x^8+1)^(1/2),x, algorithm="giac")

[Out]

integrate(x^9/sqrt(x^8 + 1), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^9}{\sqrt {1+x^8}} \, dx=\int \frac {x^9}{\sqrt {x^8+1}} \,d x \]

[In]

int(x^9/(x^8 + 1)^(1/2),x)

[Out]

int(x^9/(x^8 + 1)^(1/2), x)