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

Optimal. Leaf size=28 \[ \frac {1}{2 \sqrt {x^4+1}}-\frac {1}{2} \tanh ^{-1}\left (\sqrt {x^4+1}\right ) \]

[Out]

-1/2*arctanh((x^4+1)^(1/2))+1/2/(x^4+1)^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {266, 51, 63, 207} \[ \frac {1}{2 \sqrt {x^4+1}}-\frac {1}{2} \tanh ^{-1}\left (\sqrt {x^4+1}\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(1 + x^4)^(3/2)),x]

[Out]

1/(2*Sqrt[1 + x^4]) - ArcTanh[Sqrt[1 + x^4]]/2

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 26, normalized size = 0.93 \[ \frac {\, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};x^4+1\right )}{2 \sqrt {x^4+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(1 + x^4)^(3/2)),x]

[Out]

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

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fricas [B]  time = 0.87, size = 52, normalized size = 1.86 \[ -\frac {{\left (x^{4} + 1\right )} \log \left (\sqrt {x^{4} + 1} + 1\right ) - {\left (x^{4} + 1\right )} \log \left (\sqrt {x^{4} + 1} - 1\right ) - 2 \, \sqrt {x^{4} + 1}}{4 \, {\left (x^{4} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*((x^4 + 1)*log(sqrt(x^4 + 1) + 1) - (x^4 + 1)*log(sqrt(x^4 + 1) - 1) - 2*sqrt(x^4 + 1))/(x^4 + 1)

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giac [A]  time = 0.17, size = 34, normalized size = 1.21 \[ \frac {1}{2 \, \sqrt {x^{4} + 1}} - \frac {1}{4} \, \log \left (\sqrt {x^{4} + 1} + 1\right ) + \frac {1}{4} \, \log \left (\sqrt {x^{4} + 1} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/sqrt(x^4 + 1) - 1/4*log(sqrt(x^4 + 1) + 1) + 1/4*log(sqrt(x^4 + 1) - 1)

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maple [A]  time = 0.01, size = 21, normalized size = 0.75 \[ -\frac {\arctanh \left (\frac {1}{\sqrt {x^{4}+1}}\right )}{2}+\frac {1}{2 \sqrt {x^{4}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/(x^4+1)^(1/2)-1/2*arctanh(1/(x^4+1)^(1/2))

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maxima [A]  time = 1.28, size = 34, normalized size = 1.21 \[ \frac {1}{2 \, \sqrt {x^{4} + 1}} - \frac {1}{4} \, \log \left (\sqrt {x^{4} + 1} + 1\right ) + \frac {1}{4} \, \log \left (\sqrt {x^{4} + 1} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/sqrt(x^4 + 1) - 1/4*log(sqrt(x^4 + 1) + 1) + 1/4*log(sqrt(x^4 + 1) - 1)

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mupad [B]  time = 1.19, size = 20, normalized size = 0.71 \[ \frac {1}{2\,\sqrt {x^4+1}}-\frac {\mathrm {atanh}\left (\sqrt {x^4+1}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(x^4 + 1)^(3/2)),x)

[Out]

1/(2*(x^4 + 1)^(1/2)) - atanh((x^4 + 1)^(1/2))/2

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sympy [B]  time = 1.89, size = 87, normalized size = 3.11 \[ \frac {x^{4} \log {\left (x^{4} \right )}}{4 x^{4} + 4} - \frac {2 x^{4} \log {\left (\sqrt {x^{4} + 1} + 1 \right )}}{4 x^{4} + 4} + \frac {2 \sqrt {x^{4} + 1}}{4 x^{4} + 4} + \frac {\log {\left (x^{4} \right )}}{4 x^{4} + 4} - \frac {2 \log {\left (\sqrt {x^{4} + 1} + 1 \right )}}{4 x^{4} + 4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**4*log(x**4)/(4*x**4 + 4) - 2*x**4*log(sqrt(x**4 + 1) + 1)/(4*x**4 + 4) + 2*sqrt(x**4 + 1)/(4*x**4 + 4) + lo
g(x**4)/(4*x**4 + 4) - 2*log(sqrt(x**4 + 1) + 1)/(4*x**4 + 4)

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