3.10.36 \(\int \frac {1}{x^6 \sqrt {1+x^4}} \, dx\) [936]

Optimal. Leaf size=140 \[ -\frac {\sqrt {1+x^4}}{5 x^5}+\frac {3 \sqrt {1+x^4}}{5 x}-\frac {3 x \sqrt {1+x^4}}{5 \left (1+x^2\right )}+\frac {3 \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} E\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{5 \sqrt {1+x^4}}-\frac {3 \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{10 \sqrt {1+x^4}} \]

[Out]

-1/5*(x^4+1)^(1/2)/x^5+3/5*(x^4+1)^(1/2)/x-3/5*x*(x^4+1)^(1/2)/(x^2+1)+3/5*(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)-3/10*(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 = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {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 )}{10 \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 )}{5 \sqrt {x^4+1}}+\frac {3 \sqrt {x^4+1}}{5 x}-\frac {\sqrt {x^4+1}}{5 x^5}-\frac {3 \sqrt {x^4+1} x}{5 \left (x^2+1\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^6*Sqrt[1 + x^4]),x]

[Out]

-1/5*Sqrt[1 + x^4]/x^5 + (3*Sqrt[1 + x^4])/(5*x) - (3*x*Sqrt[1 + x^4])/(5*(1 + x^2)) + (3*(1 + x^2)*Sqrt[(1 +
x^4)/(1 + x^2)^2]*EllipticE[2*ArcTan[x], 1/2])/(5*Sqrt[1 + x^4]) - (3*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*El
lipticF[2*ArcTan[x], 1/2])/(10*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 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^6 \sqrt {1+x^4}} \, dx &=-\frac {\sqrt {1+x^4}}{5 x^5}-\frac {3}{5} \int \frac {1}{x^2 \sqrt {1+x^4}} \, dx\\ &=-\frac {\sqrt {1+x^4}}{5 x^5}+\frac {3 \sqrt {1+x^4}}{5 x}-\frac {3}{5} \int \frac {x^2}{\sqrt {1+x^4}} \, dx\\ &=-\frac {\sqrt {1+x^4}}{5 x^5}+\frac {3 \sqrt {1+x^4}}{5 x}-\frac {3}{5} \int \frac {1}{\sqrt {1+x^4}} \, dx+\frac {3}{5} \int \frac {1-x^2}{\sqrt {1+x^4}} \, dx\\ &=-\frac {\sqrt {1+x^4}}{5 x^5}+\frac {3 \sqrt {1+x^4}}{5 x}-\frac {3 x \sqrt {1+x^4}}{5 \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 )}{5 \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 )}{10 \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 = 10.00, size = 22, normalized size = 0.16 \begin {gather*} -\frac {\, _2F_1\left (-\frac {5}{4},\frac {1}{2};-\frac {1}{4};-x^4\right )}{5 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^6*Sqrt[1 + x^4]),x]

[Out]

-1/5*Hypergeometric2F1[-5/4, 1/2, -1/4, -x^4]/x^5

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Maple [C] Result contains complex when optimal does not.
time = 0.14, size = 107, normalized size = 0.76

method	result	size
meijerg	\(-\frac {\hypergeom \left (\left [-\frac {5}{4}, \frac {1}{2}\right ], \left [-\frac {1}{4}\right ], -x^{4}\right )}{5 x^{5}}\)	\(17\)
default	\(-\frac {\sqrt {x^{4}+1}}{5 x^{5}}+\frac {3 \sqrt {x^{4}+1}}{5 x}-\frac {3 i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \left (\EllipticF \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )-\EllipticE \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )\right )}{5 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\)	\(107\)
risch	\(\frac {3 x^{8}+2 x^{4}-1}{5 x^{5} \sqrt {x^{4}+1}}-\frac {3 i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \left (\EllipticF \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )-\EllipticE \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )\right )}{5 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\)	\(107\)
elliptic	\(-\frac {\sqrt {x^{4}+1}}{5 x^{5}}+\frac {3 \sqrt {x^{4}+1}}{5 x}-\frac {3 i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \left (\EllipticF \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )-\EllipticE \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )\right )}{5 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\)	\(107\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^6/(x^4+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/5*(x^4+1)^(1/2)/x^5+3/5*(x^4+1)^(1/2)/x-3/5*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^6/(x^4+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(x^4 + 1)*x^6), x)

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Fricas [C] Result contains complex when optimal does not.
time = 0.08, size = 53, normalized size = 0.38 \begin {gather*} \frac {3 i \, \sqrt {i} x^{5} E(\arcsin \left (\sqrt {i} x\right )\,|\,-1) - 3 i \, \sqrt {i} x^{5} F(\arcsin \left (\sqrt {i} x\right )\,|\,-1) + {\left (3 \, x^{4} - 1\right )} \sqrt {x^{4} + 1}}{5 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(x^4+1)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 0.42, size = 36, normalized size = 0.26 \begin {gather*} \frac {\Gamma \left (- \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, \frac {1}{2} \\ - \frac {1}{4} \end {matrix}\middle | {x^{4} e^{i \pi }} \right )}}{4 x^{5} \Gamma \left (- \frac {1}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**6/(x**4+1)**(1/2),x)

[Out]

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

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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^6/(x^4+1)^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(x^4 + 1)*x^6), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^6\,\sqrt {x^4+1}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^6*(x^4 + 1)^(1/2)),x)

[Out]

int(1/(x^6*(x^4 + 1)^(1/2)), x)

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