3.952 \(\int \frac {1}{x^8 (1+x^4)^{3/2}} \, dx\)

Optimal. Leaf size=92 \[ -\frac {9 \sqrt {x^4+1}}{14 x^7}+\frac {1}{2 x^7 \sqrt {x^4+1}}+\frac {15 \sqrt {x^4+1}}{14 x^3}+\frac {15 \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{28 \sqrt {x^4+1}} \]

[Out]

1/2/x^7/(x^4+1)^(1/2)-9/14*(x^4+1)^(1/2)/x^7+15/14*(x^4+1)^(1/2)/x^3+15/28*(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 = 0.02, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {290, 325, 220} \[ \frac {15 \sqrt {x^4+1}}{14 x^3}-\frac {9 \sqrt {x^4+1}}{14 x^7}+\frac {1}{2 x^7 \sqrt {x^4+1}}+\frac {15 \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{28 \sqrt {x^4+1}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^8*(1 + x^4)^(3/2)),x]

[Out]

1/(2*x^7*Sqrt[1 + x^4]) - (9*Sqrt[1 + x^4])/(14*x^7) + (15*Sqrt[1 + x^4])/(14*x^3) + (15*(1 + x^2)*Sqrt[(1 + x
^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/2])/(28*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^8 \left (1+x^4\right )^{3/2}} \, dx &=\frac {1}{2 x^7 \sqrt {1+x^4}}+\frac {9}{2} \int \frac {1}{x^8 \sqrt {1+x^4}} \, dx\\ &=\frac {1}{2 x^7 \sqrt {1+x^4}}-\frac {9 \sqrt {1+x^4}}{14 x^7}-\frac {45}{14} \int \frac {1}{x^4 \sqrt {1+x^4}} \, dx\\ &=\frac {1}{2 x^7 \sqrt {1+x^4}}-\frac {9 \sqrt {1+x^4}}{14 x^7}+\frac {15 \sqrt {1+x^4}}{14 x^3}+\frac {15}{14} \int \frac {1}{\sqrt {1+x^4}} \, dx\\ &=\frac {1}{2 x^7 \sqrt {1+x^4}}-\frac {9 \sqrt {1+x^4}}{14 x^7}+\frac {15 \sqrt {1+x^4}}{14 x^3}+\frac {15 \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 )}{28 \sqrt {1+x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.00, size = 22, normalized size = 0.24 \[ -\frac {\, _2F_1\left (-\frac {7}{4},\frac {3}{2};-\frac {3}{4};-x^4\right )}{7 x^7} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^8*(1 + x^4)^(3/2)),x]

[Out]

-1/7*Hypergeometric2F1[-7/4, 3/2, -3/4, -x^4]/x^7

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fricas [F]  time = 0.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {x^{4} + 1}}{x^{16} + 2 \, x^{12} + x^{8}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^8/(x^4+1)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(x^4 + 1)/(x^16 + 2*x^12 + x^8), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (x^{4} + 1\right )}^{\frac {3}{2}} x^{8}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^8/(x^4+1)^(3/2),x, algorithm="giac")

[Out]

integrate(1/((x^4 + 1)^(3/2)*x^8), x)

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maple [C]  time = 0.02, size = 96, normalized size = 1.04 \[ \frac {x}{2 \sqrt {x^{4}+1}}+\frac {15 \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticF \left (\left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) x , i\right )}{14 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}+\frac {4 \sqrt {x^{4}+1}}{7 x^{3}}-\frac {\sqrt {x^{4}+1}}{7 x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^8/(x^4+1)^(3/2),x)

[Out]

-1/7*(x^4+1)^(1/2)/x^7+4/7*(x^4+1)^(1/2)/x^3+1/2/(x^4+1)^(1/2)*x+15/14/(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 \[ \int \frac {1}{{\left (x^{4} + 1\right )}^{\frac {3}{2}} x^{8}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^8/(x^4+1)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((x^4 + 1)^(3/2)*x^8), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x^8\,{\left (x^4+1\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^8*(x^4 + 1)^(3/2)),x)

[Out]

int(1/(x^8*(x^4 + 1)^(3/2)), x)

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sympy [C]  time = 2.06, size = 36, normalized size = 0.39 \[ \frac {\Gamma \left (- \frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {7}{4}, \frac {3}{2} \\ - \frac {3}{4} \end {matrix}\middle | {x^{4} e^{i \pi }} \right )}}{4 x^{7} \Gamma \left (- \frac {3}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**8/(x**4+1)**(3/2),x)

[Out]

gamma(-7/4)*hyper((-7/4, 3/2), (-3/4,), x**4*exp_polar(I*pi))/(4*x**7*gamma(-3/4))

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