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

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

[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])

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Rubi [A]  time = 0.0212835, antiderivative size = 92, normalized size of antiderivative = 1., 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 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]

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]

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*}

Mathematica [C]  time = 0.002711, 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]

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

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Maple [C]  time = 0.047, size = 96, normalized size = 1. \begin{align*} -{\frac{1}{7\,{x}^{7}}\sqrt{{x}^{4}+1}}+{\frac{4}{7\,{x}^{3}}\sqrt{{x}^{4}+1}}+{\frac{x}{2}{\frac{1}{\sqrt{{x}^{4}+1}}}}+{\frac{15\,{\it EllipticF} \left ( x \left ( 1/2\,\sqrt{2}+i/2\sqrt{2} \right ) ,i \right ) }{7\,\sqrt{2}+7\,i\sqrt{2}}\sqrt{1-i{x}^{2}}\sqrt{1+i{x}^{2}}{\frac{1}{\sqrt{{x}^{4}+1}}}} \end{align*}

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/(x^4+1)^(1/2)+15/14/(1/2*2^(1/2)+1/2*I*2^(1/2))*(1-I*x^2)^(
1/2)*(1+I*x^2)^(1/2)/(x^4+1)^(1/2)*EllipticF(x*(1/2*2^(1/2)+1/2*I*2^(1/2)),I)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (x^{4} + 1\right )}^{\frac{3}{2}} x^{8}}\,{d x} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{x^{4} + 1}}{x^{16} + 2 \, x^{12} + x^{8}}, x\right ) \end{align*}

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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Sympy [C]  time = 1.78596, size = 36, normalized size = 0.39 \begin{align*} \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 )} \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (x^{4} + 1\right )}^{\frac{3}{2}} x^{8}}\,{d x} \end{align*}

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)