\(\int \frac {x^4}{(1+x^4)^{3/2}} \, dx\) [949]

Optimal result

Integrand size = 13, antiderivative size = 58 \[ \int \frac {x^4}{\left (1+x^4\right )^{3/2}} \, dx=-\frac {x}{2 \sqrt {1+x^4}}+\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{4 \sqrt {1+x^4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {294, 226} \[ \int \frac {x^4}{\left (1+x^4\right )^{3/2}} \, dx=\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{4 \sqrt {x^4+1}}-\frac {x}{2 \sqrt {x^4+1}} \]

[In]

Int[x^4/(1 + x^4)^(3/2),x]

[Out]

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

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 294

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*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

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

Mathematica [C] (verified)

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

Time = 4.56 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.55 \[ \int \frac {x^4}{\left (1+x^4\right )^{3/2}} \, dx=\frac {1}{2} x \left (-\frac {1}{\sqrt {1+x^4}}+\operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-x^4\right )\right ) \]

[In]

Integrate[x^4/(1 + x^4)^(3/2),x]

[Out]

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

Maple [C] (verified)

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

Time = 4.35 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.29

method	result	size
meijerg	\(\frac {x^{5} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {5}{4},\frac {3}{2};\frac {9}{4};-x^{4}\right )}{5}\)	\(17\)
default	\(-\frac {x}{2 \sqrt {x^{4}+1}}+\frac {\sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, F\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{2 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\)	\(72\)
risch	\(-\frac {x}{2 \sqrt {x^{4}+1}}+\frac {\sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, F\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{2 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\)	\(72\)
elliptic	\(-\frac {x}{2 \sqrt {x^{4}+1}}+\frac {\sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, F\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{2 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\)	\(72\)

[In]

int(x^4/(x^4+1)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/5*x^5*hypergeom([5/4,3/2],[9/4],-x^4)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.08 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.67 \[ \int \frac {x^4}{\left (1+x^4\right )^{3/2}} \, dx=\frac {\sqrt {i} {\left (-i \, x^{4} - i\right )} F(\arcsin \left (\sqrt {i} x\right )\,|\,-1) - \sqrt {x^{4} + 1} x}{2 \, {\left (x^{4} + 1\right )}} \]

[In]

integrate(x^4/(x^4+1)^(3/2),x, algorithm="fricas")

[Out]

1/2*(sqrt(I)*(-I*x^4 - I)*elliptic_f(arcsin(sqrt(I)*x), -1) - sqrt(x^4 + 1)*x)/(x^4 + 1)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.37 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.50 \[ \int \frac {x^4}{\left (1+x^4\right )^{3/2}} \, dx=\frac {x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {3}{2} \\ \frac {9}{4} \end {matrix}\middle | {x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {9}{4}\right )} \]

[In]

integrate(x**4/(x**4+1)**(3/2),x)

[Out]

x**5*gamma(5/4)*hyper((5/4, 3/2), (9/4,), x**4*exp_polar(I*pi))/(4*gamma(9/4))

Maxima [F]

\[ \int \frac {x^4}{\left (1+x^4\right )^{3/2}} \, dx=\int { \frac {x^{4}}{{\left (x^{4} + 1\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(x^4/(x^4+1)^(3/2),x, algorithm="maxima")

[Out]

integrate(x^4/(x^4 + 1)^(3/2), x)

Giac [F]

\[ \int \frac {x^4}{\left (1+x^4\right )^{3/2}} \, dx=\int { \frac {x^{4}}{{\left (x^{4} + 1\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(x^4/(x^4+1)^(3/2),x, algorithm="giac")

[Out]

integrate(x^4/(x^4 + 1)^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4}{\left (1+x^4\right )^{3/2}} \, dx=\int \frac {x^4}{{\left (x^4+1\right )}^{3/2}} \,d x \]

[In]

int(x^4/(x^4 + 1)^(3/2),x)

[Out]

int(x^4/(x^4 + 1)^(3/2), x)