\(\int \frac {1}{x^6 (1+x^4)^{3/2}} \, dx\) [958]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (verified)
   Maple [C] (verified)
   Fricas [C] (verification not implemented)
   Sympy [C] (verification not implemented)
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 13, antiderivative size = 156 \[ \int \frac {1}{x^6 \left (1+x^4\right )^{3/2}} \, dx=\frac {1}{2 x^5 \sqrt {1+x^4}}-\frac {7 \sqrt {1+x^4}}{10 x^5}+\frac {21 \sqrt {1+x^4}}{10 x}-\frac {21 x \sqrt {1+x^4}}{10 \left (1+x^2\right )}+\frac {21 \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} E\left (2 \arctan (x)\left |\frac {1}{2}\right .\right )}{10 \sqrt {1+x^4}}-\frac {21 \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{20 \sqrt {1+x^4}} \]

[Out]

1/2/x^5/(x^4+1)^(1/2)-7/10*(x^4+1)^(1/2)/x^5+21/10*(x^4+1)^(1/2)/x-21/10*x*(x^4+1)^(1/2)/(x^2+1)+21/10*(x^2+1)
*(cos(2*arctan(x))^2)^(1/2)/cos(2*arctan(x))*EllipticE(sin(2*arctan(x)),1/2*2^(1/2))*((x^4+1)/(x^2+1)^2)^(1/2)
/(x^4+1)^(1/2)-21/20*(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 = 0.03 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {296, 331, 311, 226, 1210} \[ \int \frac {1}{x^6 \left (1+x^4\right )^{3/2}} \, dx=-\frac {21 \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{20 \sqrt {x^4+1}}+\frac {21 \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} E\left (2 \arctan (x)\left |\frac {1}{2}\right .\right )}{10 \sqrt {x^4+1}}+\frac {21 \sqrt {x^4+1}}{10 x}-\frac {7 \sqrt {x^4+1}}{10 x^5}+\frac {1}{2 \sqrt {x^4+1} x^5}-\frac {21 \sqrt {x^4+1} x}{10 \left (x^2+1\right )} \]

[In]

Int[1/(x^6*(1 + x^4)^(3/2)),x]

[Out]

1/(2*x^5*Sqrt[1 + x^4]) - (7*Sqrt[1 + x^4])/(10*x^5) + (21*Sqrt[1 + x^4])/(10*x) - (21*x*Sqrt[1 + x^4])/(10*(1
 + x^2)) + (21*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticE[2*ArcTan[x], 1/2])/(10*Sqrt[1 + x^4]) - (21*(1
+ x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/2])/(20*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 311

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

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]

Rule 1210

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a +
 c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d*(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4
]))*EllipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2 x^5 \sqrt {1+x^4}}+\frac {7}{2} \int \frac {1}{x^6 \sqrt {1+x^4}} \, dx \\ & = \frac {1}{2 x^5 \sqrt {1+x^4}}-\frac {7 \sqrt {1+x^4}}{10 x^5}-\frac {21}{10} \int \frac {1}{x^2 \sqrt {1+x^4}} \, dx \\ & = \frac {1}{2 x^5 \sqrt {1+x^4}}-\frac {7 \sqrt {1+x^4}}{10 x^5}+\frac {21 \sqrt {1+x^4}}{10 x}-\frac {21}{10} \int \frac {x^2}{\sqrt {1+x^4}} \, dx \\ & = \frac {1}{2 x^5 \sqrt {1+x^4}}-\frac {7 \sqrt {1+x^4}}{10 x^5}+\frac {21 \sqrt {1+x^4}}{10 x}-\frac {21}{10} \int \frac {1}{\sqrt {1+x^4}} \, dx+\frac {21}{10} \int \frac {1-x^2}{\sqrt {1+x^4}} \, dx \\ & = \frac {1}{2 x^5 \sqrt {1+x^4}}-\frac {7 \sqrt {1+x^4}}{10 x^5}+\frac {21 \sqrt {1+x^4}}{10 x}-\frac {21 x \sqrt {1+x^4}}{10 \left (1+x^2\right )}+\frac {21 \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} E\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{10 \sqrt {1+x^4}}-\frac {21 \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 )}{20 \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 = 0.14 \[ \int \frac {1}{x^6 \left (1+x^4\right )^{3/2}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {3}{2},-\frac {1}{4},-x^4\right )}{5 x^5} \]

[In]

Integrate[1/(x^6*(1 + x^4)^(3/2)),x]

[Out]

-1/5*Hypergeometric2F1[-5/4, 3/2, -1/4, -x^4]/x^5

Maple [C] (verified)

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

Time = 4.32 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.11

method result size
meijerg \(-\frac {{}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (-\frac {5}{4},\frac {3}{2};-\frac {1}{4};-x^{4}\right )}{5 x^{5}}\) \(17\)
risch \(\frac {21 x^{8}+14 x^{4}-2}{10 x^{5} \sqrt {x^{4}+1}}-\frac {21 i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \left (F\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )-E\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )\right )}{10 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\) \(107\)
default \(\frac {x^{3}}{2 \sqrt {x^{4}+1}}-\frac {\sqrt {x^{4}+1}}{5 x^{5}}+\frac {8 \sqrt {x^{4}+1}}{5 x}-\frac {21 i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \left (F\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )-E\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )\right )}{10 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\) \(119\)
elliptic \(\frac {x^{3}}{2 \sqrt {x^{4}+1}}-\frac {\sqrt {x^{4}+1}}{5 x^{5}}+\frac {8 \sqrt {x^{4}+1}}{5 x}-\frac {21 i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \left (F\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )-E\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )\right )}{10 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\) \(119\)

[In]

int(1/x^6/(x^4+1)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.52 \[ \int \frac {1}{x^6 \left (1+x^4\right )^{3/2}} \, dx=-\frac {21 \, \sqrt {i} {\left (-i \, x^{9} - i \, x^{5}\right )} E(\arcsin \left (\sqrt {i} x\right )\,|\,-1) + 21 \, \sqrt {i} {\left (i \, x^{9} + i \, x^{5}\right )} F(\arcsin \left (\sqrt {i} x\right )\,|\,-1) - {\left (21 \, x^{8} + 14 \, x^{4} - 2\right )} \sqrt {x^{4} + 1}}{10 \, {\left (x^{9} + x^{5}\right )}} \]

[In]

integrate(1/x^6/(x^4+1)^(3/2),x, algorithm="fricas")

[Out]

-1/10*(21*sqrt(I)*(-I*x^9 - I*x^5)*elliptic_e(arcsin(sqrt(I)*x), -1) + 21*sqrt(I)*(I*x^9 + I*x^5)*elliptic_f(a
rcsin(sqrt(I)*x), -1) - (21*x^8 + 14*x^4 - 2)*sqrt(x^4 + 1))/(x^9 + x^5)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.54 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.23 \[ \int \frac {1}{x^6 \left (1+x^4\right )^{3/2}} \, dx=\frac {\Gamma \left (- \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, \frac {3}{2} \\ - \frac {1}{4} \end {matrix}\middle | {x^{4} e^{i \pi }} \right )}}{4 x^{5} \Gamma \left (- \frac {1}{4}\right )} \]

[In]

integrate(1/x**6/(x**4+1)**(3/2),x)

[Out]

gamma(-5/4)*hyper((-5/4, 3/2), (-1/4,), x**4*exp_polar(I*pi))/(4*x**5*gamma(-1/4))

Maxima [F]

\[ \int \frac {1}{x^6 \left (1+x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (x^{4} + 1\right )}^{\frac {3}{2}} x^{6}} \,d x } \]

[In]

integrate(1/x^6/(x^4+1)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((x^4 + 1)^(3/2)*x^6), x)

Giac [F]

\[ \int \frac {1}{x^6 \left (1+x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (x^{4} + 1\right )}^{\frac {3}{2}} x^{6}} \,d x } \]

[In]

integrate(1/x^6/(x^4+1)^(3/2),x, algorithm="giac")

[Out]

integrate(1/((x^4 + 1)^(3/2)*x^6), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^6 \left (1+x^4\right )^{3/2}} \, dx=\int \frac {1}{x^6\,{\left (x^4+1\right )}^{3/2}} \,d x \]

[In]

int(1/(x^6*(x^4 + 1)^(3/2)),x)

[Out]

int(1/(x^6*(x^4 + 1)^(3/2)), x)