3.10.57 \(\int \frac {1}{x^2 (1+x^4)^{3/2}} \, dx\) [957]

Optimal. Leaf size=140 \[ \frac {1}{2 x \sqrt {1+x^4}}-\frac {3 \sqrt {1+x^4}}{2 x}+\frac {3 x \sqrt {1+x^4}}{2 \left (1+x^2\right )}-\frac {3 \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} E\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{2 \sqrt {1+x^4}}+\frac {3 \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{4 \sqrt {1+x^4}} \]

[Out]

1/2/x/(x^4+1)^(1/2)-3/2*(x^4+1)^(1/2)/x+3/2*x*(x^4+1)^(1/2)/(x^2+1)-3/2*(x^2+1)*(cos(2*arctan(x))^2)^(1/2)/cos
(2*arctan(x))*EllipticE(sin(2*arctan(x)),1/2*2^(1/2))*((x^4+1)/(x^2+1)^2)^(1/2)/(x^4+1)^(1/2)+3/4*(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 = 140, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {296, 331, 311, 226, 1210} \begin {gather*} \frac {3 \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} F\left (2 \text {ArcTan}(x)\left |\frac {1}{2}\right .\right )}{4 \sqrt {x^4+1}}-\frac {3 \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} E\left (2 \text {ArcTan}(x)\left |\frac {1}{2}\right .\right )}{2 \sqrt {x^4+1}}-\frac {3 \sqrt {x^4+1}}{2 x}+\frac {1}{2 \sqrt {x^4+1} x}+\frac {3 \sqrt {x^4+1} x}{2 \left (x^2+1\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(1 + x^4)^(3/2)),x]

[Out]

1/(2*x*Sqrt[1 + x^4]) - (3*Sqrt[1 + x^4])/(2*x) + (3*x*Sqrt[1 + x^4])/(2*(1 + x^2)) - (3*(1 + x^2)*Sqrt[(1 + x
^4)/(1 + x^2)^2]*EllipticE[2*ArcTan[x], 1/2])/(2*Sqrt[1 + x^4]) + (3*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*Ell
ipticF[2*ArcTan[x], 1/2])/(4*Sqrt[1 + x^4])

Rule 226

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

Rule 296

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

Rule 311

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 331

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 1210

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a +
c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d*(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4
]))*EllipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin {align*} \int \frac {1}{x^2 \left (1+x^4\right )^{3/2}} \, dx &=\frac {1}{2 x \sqrt {1+x^4}}+\frac {3}{2} \int \frac {1}{x^2 \sqrt {1+x^4}} \, dx\\ &=\frac {1}{2 x \sqrt {1+x^4}}-\frac {3 \sqrt {1+x^4}}{2 x}+\frac {3}{2} \int \frac {x^2}{\sqrt {1+x^4}} \, dx\\ &=\frac {1}{2 x \sqrt {1+x^4}}-\frac {3 \sqrt {1+x^4}}{2 x}+\frac {3}{2} \int \frac {1}{\sqrt {1+x^4}} \, dx-\frac {3}{2} \int \frac {1-x^2}{\sqrt {1+x^4}} \, dx\\ &=\frac {1}{2 x \sqrt {1+x^4}}-\frac {3 \sqrt {1+x^4}}{2 x}+\frac {3 x \sqrt {1+x^4}}{2 \left (1+x^2\right )}-\frac {3 \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} E\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{2 \sqrt {1+x^4}}+\frac {3 \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 )}{4 \sqrt {1+x^4}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 8.46, size = 20, normalized size = 0.14 \begin {gather*} -\frac {\, _2F_1\left (-\frac {1}{4},\frac {3}{2};\frac {3}{4};-x^4\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(1 + x^4)^(3/2)),x]

[Out]

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

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Maple [C] Result contains complex when optimal does not.
time = 0.15, size = 107, normalized size = 0.76

method	result	size
meijerg	\(-\frac {\hypergeom \left (\left [-\frac {1}{4}, \frac {3}{2}\right ], \left [\frac {3}{4}\right ], -x^{4}\right )}{x}\)	\(17\)
risch	\(-\frac {3 x^{4}+2}{2 x \sqrt {x^{4}+1}}+\frac {3 i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \left (\EllipticF \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )-\EllipticE \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )\right )}{2 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\)	\(102\)
default	\(-\frac {x^{3}}{2 \sqrt {x^{4}+1}}-\frac {\sqrt {x^{4}+1}}{x}+\frac {3 i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \left (\EllipticF \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )-\EllipticE \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )\right )}{2 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\)	\(107\)
elliptic	\(-\frac {x^{3}}{2 \sqrt {x^{4}+1}}-\frac {\sqrt {x^{4}+1}}{x}+\frac {3 i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \left (\EllipticF \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )-\EllipticE \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )\right )}{2 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\)	\(107\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*x^3/(x^4+1)^(1/2)-(x^4+1)^(1/2)/x+3/2*I/(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)-EllipticE(x*(1/2*2^(1/2)+1/2*I*2^(1/2)),I))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((x^4 + 1)^(3/2)*x^2), x)

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Fricas [C] Result contains complex when optimal does not.
time = 0.09, size = 69, normalized size = 0.49 \begin {gather*} -\frac {3 \, \sqrt {i} {\left (i \, x^{5} + i \, x\right )} E(\arcsin \left (\sqrt {i} x\right )\,|\,-1) + 3 \, \sqrt {i} {\left (-i \, x^{5} - i \, x\right )} F(\arcsin \left (\sqrt {i} x\right )\,|\,-1) + {\left (3 \, x^{4} + 2\right )} \sqrt {x^{4} + 1}}{2 \, {\left (x^{5} + x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(3*sqrt(I)*(I*x^5 + I*x)*elliptic_e(arcsin(sqrt(I)*x), -1) + 3*sqrt(I)*(-I*x^5 - I*x)*elliptic_f(arcsin(s
qrt(I)*x), -1) + (3*x^4 + 2)*sqrt(x^4 + 1))/(x^5 + x)

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Sympy [C] Result contains complex when optimal does not.
time = 0.38, size = 31, normalized size = 0.22 \begin {gather*} \frac {\Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {3}{2} \\ \frac {3}{4} \end {matrix}\middle | {x^{4} e^{i \pi }} \right )}}{4 x \Gamma \left (\frac {3}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

gamma(-1/4)*hyper((-1/4, 3/2), (3/4,), x**4*exp_polar(I*pi))/(4*x*gamma(3/4))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((x^4 + 1)^(3/2)*x^2), x)

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Mupad [B]
time = 1.14, size = 15, normalized size = 0.11 \begin {gather*} -\frac {{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {3}{2};\ \frac {3}{4};\ -x^4\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(x^4 + 1)^(3/2)),x)

[Out]

-hypergeom([-1/4, 3/2], 3/4, -x^4)/x

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