\(\int \frac {1}{x^4 (1+x^4)^{3/2}} \, dx\) [951]

Optimal result

Integrand size = 13, antiderivative size = 76 \[ \int \frac {1}{x^4 \left (1+x^4\right )^{3/2}} \, dx=\frac {1}{2 x^3 \sqrt {1+x^4}}-\frac {5 \sqrt {1+x^4}}{6 x^3}-\frac {5 \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{12 \sqrt {1+x^4}} \]

[Out]

1/2/x^3/(x^4+1)^(1/2)-5/6*(x^4+1)^(1/2)/x^3-5/12*(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 = 76, 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 = {296, 331, 226} \[ \int \frac {1}{x^4 \left (1+x^4\right )^{3/2}} \, dx=-\frac {5 \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{12 \sqrt {x^4+1}}-\frac {5 \sqrt {x^4+1}}{6 x^3}+\frac {1}{2 x^3 \sqrt {x^4+1}} \]

[In]

Int[1/(x^4*(1 + x^4)^(3/2)),x]

[Out]

1/(2*x^3*Sqrt[1 + x^4]) - (5*Sqrt[1 + x^4])/(6*x^3) - (5*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*Arc
Tan[x], 1/2])/(12*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 296

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-(c*x)^(m + 1))*((a + b*x^n)^(p + 1)/
(a*c*n*(p + 1))), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; Free
Q[{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2 x^3 \sqrt {1+x^4}}+\frac {5}{2} \int \frac {1}{x^4 \sqrt {1+x^4}} \, dx \\ & = \frac {1}{2 x^3 \sqrt {1+x^4}}-\frac {5 \sqrt {1+x^4}}{6 x^3}-\frac {5}{6} \int \frac {1}{\sqrt {1+x^4}} \, dx \\ & = \frac {1}{2 x^3 \sqrt {1+x^4}}-\frac {5 \sqrt {1+x^4}}{6 x^3}-\frac {5 \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 )}{12 \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 = 10.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.29 \[ \int \frac {1}{x^4 \left (1+x^4\right )^{3/2}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {3}{2},\frac {1}{4},-x^4\right )}{3 x^3} \]

[In]

Integrate[1/(x^4*(1 + x^4)^(3/2)),x]

[Out]

-1/3*Hypergeometric2F1[-3/4, 3/2, 1/4, -x^4]/x^3

Maple [C] (verified)

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

Time = 4.19 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.22

method	result	size
meijerg	\(-\frac {{}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (-\frac {3}{4},\frac {3}{2};\frac {1}{4};-x^{4}\right )}{3 x^{3}}\)	\(17\)
risch	\(-\frac {5 x^{4}+2}{6 x^{3} \sqrt {x^{4}+1}}-\frac {5 \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, F\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{6 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\)	\(81\)
default	\(-\frac {x}{2 \sqrt {x^{4}+1}}-\frac {\sqrt {x^{4}+1}}{3 x^{3}}-\frac {5 \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, F\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{6 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\)	\(84\)
elliptic	\(-\frac {x}{2 \sqrt {x^{4}+1}}-\frac {\sqrt {x^{4}+1}}{3 x^{3}}-\frac {5 \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, F\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{6 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\)	\(84\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.08 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.67 \[ \int \frac {1}{x^4 \left (1+x^4\right )^{3/2}} \, dx=-\frac {5 \, \sqrt {i} {\left (-i \, x^{7} - i \, x^{3}\right )} F(\arcsin \left (\sqrt {i} x\right )\,|\,-1) + {\left (5 \, x^{4} + 2\right )} \sqrt {x^{4} + 1}}{6 \, {\left (x^{7} + x^{3}\right )}} \]

[In]

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

[Out]

-1/6*(5*sqrt(I)*(-I*x^7 - I*x^3)*elliptic_f(arcsin(sqrt(I)*x), -1) + (5*x^4 + 2)*sqrt(x^4 + 1))/(x^7 + x^3)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.47 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.42 \[ \int \frac {1}{x^4 \left (1+x^4\right )^{3/2}} \, dx=\frac {\Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {3}{2} \\ \frac {1}{4} \end {matrix}\middle | {x^{4} e^{i \pi }} \right )}}{4 x^{3} \Gamma \left (\frac {1}{4}\right )} \]

[In]

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

[Out]

gamma(-3/4)*hyper((-3/4, 3/2), (1/4,), x**4*exp_polar(I*pi))/(4*x**3*gamma(1/4))

Maxima [F]

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

[In]

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

[Out]

integrate(1/((x^4 + 1)^(3/2)*x^4), x)

Giac [F]

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

[In]

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

[Out]

integrate(1/((x^4 + 1)^(3/2)*x^4), x)

Mupad [F(-1)]

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

[In]

int(1/(x^4*(x^4 + 1)^(3/2)),x)

[Out]

int(1/(x^4*(x^4 + 1)^(3/2)), x)