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

Optimal. Leaf size=76 \[ -\frac {5 \sqrt {x^4+1}}{6 x^3}+\frac {1}{2 x^3 \sqrt {x^4+1}}-\frac {5 \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{12 \sqrt {x^4+1}} \]

[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)

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Rubi [A]  time = 0.01, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {290, 325, 220} \[ -\frac {5 \sqrt {x^4+1}}{6 x^3}+\frac {1}{2 x^3 \sqrt {x^4+1}}-\frac {5 \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{12 \sqrt {x^4+1}} \]

Antiderivative was successfully verified.

[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 220

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)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 290

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] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 325

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

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Mathematica [C]  time = 0.00, size = 22, normalized size = 0.29 \[ -\frac {\, _2F_1\left (-\frac {3}{4},\frac {3}{2};\frac {1}{4};-x^4\right )}{3 x^3} \]

Antiderivative was successfully verified.

[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

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fricas [F]  time = 0.92, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {x^{4} + 1}}{x^{12} + 2 \, x^{8} + x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(x^4 + 1)/(x^12 + 2*x^8 + x^4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (x^{4} + 1\right )}^{\frac {3}{2}} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((x^4 + 1)^(3/2)*x^4), x)

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maple [C]  time = 0.01, size = 84, normalized size = 1.11 \[ -\frac {x}{2 \sqrt {x^{4}+1}}-\frac {5 \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticF \left (\left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) x , i\right )}{6 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}-\frac {\sqrt {x^{4}+1}}{3 x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(x^4+1)^(1/2)/x^3-1/2/(x^4+1)^(1/2)*x-5/6/(1/2*2^(1/2)+1/2*I*2^(1/2))*(-I*x^2+1)^(1/2)*(I*x^2+1)^(1/2)/(x
^4+1)^(1/2)*EllipticF((1/2*2^(1/2)+1/2*I*2^(1/2))*x,I)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (x^{4} + 1\right )}^{\frac {3}{2}} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((x^4 + 1)^(3/2)*x^4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x^4\,{\left (x^4+1\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*(x^4 + 1)^(3/2)),x)

[Out]

int(1/(x^4*(x^4 + 1)^(3/2)), x)

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sympy [C]  time = 1.39, size = 32, normalized size = 0.42 \[ \frac {\Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {3}{2} \\ \frac {1}{4} \end {matrix}\middle | {x^{4} e^{i \pi }} \right )}}{4 x^{3} \Gamma \left (\frac {1}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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