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

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

[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])

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Rubi [A]  time = 0.0146343, antiderivative size = 76, normalized size of antiderivative = 1., 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 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^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*}

Mathematica [C]  time = 0.0025907, 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]

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

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Maple [C]  time = 0.014, size = 84, normalized size = 1.1 \begin{align*} -{\frac{1}{3\,{x}^{3}}\sqrt{{x}^{4}+1}}-{\frac{x}{2}{\frac{1}{\sqrt{{x}^{4}+1}}}}-{\frac{5\,{\it EllipticF} \left ( x \left ( 1/2\,\sqrt{2}+i/2\sqrt{2} \right ) ,i \right ) }{3\,\sqrt{2}+3\,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^4/(x^4+1)^(3/2),x)

[Out]

-1/3*(x^4+1)^(1/2)/x^3-1/2*x/(x^4+1)^(1/2)-5/6/(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^{4}}\,{d x} \end{align*}

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

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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Sympy [C]  time = 0.964751, size = 32, normalized size = 0.42 \begin{align*} \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 )} \end{align*}

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

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)