3.959 \(\int \frac {1}{(1+x^4)^{5/2}} \, dx\)

Optimal. Leaf size=72 \[ \frac {5 x}{12 \sqrt {x^4+1}}+\frac {x}{6 \left (x^4+1\right )^{3/2}}+\frac {5 \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{24 \sqrt {x^4+1}} \]

[Out]

1/6*x/(x^4+1)^(3/2)+5/12*x/(x^4+1)^(1/2)+5/24*(x^2+1)*(cos(2*arctan(x))^2)^(1/2)/cos(2*arctan(x))*EllipticF(si
n(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 = 72, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {199, 220} \[ \frac {5 x}{12 \sqrt {x^4+1}}+\frac {x}{6 \left (x^4+1\right )^{3/2}}+\frac {5 \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{24 \sqrt {x^4+1}} \]

Antiderivative was successfully verified.

[In]

Int[(1 + x^4)^(-5/2),x]

[Out]

x/(6*(1 + x^4)^(3/2)) + (5*x)/(12*Sqrt[1 + x^4]) + (5*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan
[x], 1/2])/(24*Sqrt[1 + x^4])

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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}{\left (1+x^4\right )^{5/2}} \, dx &=\frac {x}{6 \left (1+x^4\right )^{3/2}}+\frac {5}{6} \int \frac {1}{\left (1+x^4\right )^{3/2}} \, dx\\ &=\frac {x}{6 \left (1+x^4\right )^{3/2}}+\frac {5 x}{12 \sqrt {1+x^4}}+\frac {5}{12} \int \frac {1}{\sqrt {1+x^4}} \, dx\\ &=\frac {x}{6 \left (1+x^4\right )^{3/2}}+\frac {5 x}{12 \sqrt {1+x^4}}+\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 )}{24 \sqrt {1+x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 49, normalized size = 0.68 \[ \frac {5}{12} x \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-x^4\right )+\frac {5 x}{12 \sqrt {x^4+1}}+\frac {x}{6 \left (x^4+1\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + x^4)^(-5/2),x]

[Out]

x/(6*(1 + x^4)^(3/2)) + (5*x)/(12*Sqrt[1 + x^4]) + (5*x*Hypergeometric2F1[1/4, 1/2, 5/4, -x^4])/12

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fricas [F]  time = 0.89, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {x^{4} + 1}}{x^{12} + 3 \, x^{8} + 3 \, x^{4} + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^4+1)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(x^4 + 1)/(x^12 + 3*x^8 + 3*x^4 + 1), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (x^{4} + 1\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^4+1)^(5/2),x, algorithm="giac")

[Out]

integrate((x^4 + 1)^(-5/2), x)

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maple [C]  time = 0.01, size = 82, normalized size = 1.14 \[ \frac {x}{6 \left (x^{4}+1\right )^{\frac {3}{2}}}+\frac {5 x}{12 \sqrt {x^{4}+1}}+\frac {5 \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticF \left (\left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) x , i\right )}{12 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4+1)^(5/2),x)

[Out]

1/6*x/(x^4+1)^(3/2)+5/12/(x^4+1)^(1/2)*x+5/12/(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 \[ \int \frac {1}{{\left (x^{4} + 1\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^4+1)^(5/2),x, algorithm="maxima")

[Out]

integrate((x^4 + 1)^(-5/2), x)

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mupad [B]  time = 1.07, size = 12, normalized size = 0.17 \[ x\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {5}{2};\ \frac {5}{4};\ -x^4\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4 + 1)^(5/2),x)

[Out]

x*hypergeom([1/4, 5/2], 5/4, -x^4)

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sympy [C]  time = 1.03, size = 27, normalized size = 0.38 \[ \frac {x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{2} \\ \frac {5}{4} \end {matrix}\middle | {x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x**4+1)**(5/2),x)

[Out]

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

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