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

Optimal. Leaf size=25 \[ -\tan ^{-1}\left (\sqrt [4]{x^2+1}\right )-\tanh ^{-1}\left (\sqrt [4]{x^2+1}\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 25, 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 = {266, 63, 212, 206, 203} \begin {gather*} -\tan ^{-1}\left (\sqrt [4]{x^2+1}\right )-\tanh ^{-1}\left (\sqrt [4]{x^2+1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*(1 + x^2)^(3/4)),x]

[Out]

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

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 203

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

Rule 206

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

Rule 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/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^2\right )^{3/4}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x (1+x)^{3/4}} \, dx,x,x^2\right )\\ &=2 \operatorname {Subst}\left (\int \frac {1}{-1+x^4} \, dx,x,\sqrt [4]{1+x^2}\right )\\ &=-\operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt [4]{1+x^2}\right )-\operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt [4]{1+x^2}\right )\\ &=-\tan ^{-1}\left (\sqrt [4]{1+x^2}\right )-\tanh ^{-1}\left (\sqrt [4]{1+x^2}\right )\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 25, normalized size = 1.00 \begin {gather*} -\tan ^{-1}\left (\sqrt [4]{x^2+1}\right )-\tanh ^{-1}\left (\sqrt [4]{x^2+1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(1 + x^2)^(3/4)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.02, size = 25, normalized size = 1.00 \begin {gather*} -\tan ^{-1}\left (\sqrt [4]{1+x^2}\right )-\tanh ^{-1}\left (\sqrt [4]{1+x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x*(1 + x^2)^(3/4)),x]

[Out]

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

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fricas [A]  time = 0.47, size = 35, normalized size = 1.40 \begin {gather*} -\arctan \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{2} \, \log \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{2} \, \log \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.44, size = 35, normalized size = 1.40 \begin {gather*} -\arctan \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{2} \, \log \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{2} \, \log \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 1.32, size = 43, normalized size = 1.72

method	result	size
meijerg	\(\frac {\left (-3 \ln \relax (2)+\frac {\pi }{2}+2 \ln \relax (x )\right ) \Gamma \left (\frac {3}{4}\right )-\frac {3 \hypergeom \left (\left [1, 1, \frac {7}{4}\right ], \left [2, 2\right ], -x^{2}\right ) \Gamma \left (\frac {3}{4}\right ) x^{2}}{4}}{2 \Gamma \left (\frac {3}{4}\right )}\)	\(43\)
trager	\(\frac {\ln \left (-\frac {2 \left (x^{2}+1\right )^{\frac {3}{4}}-2 \sqrt {x^{2}+1}-x^{2}+2 \left (x^{2}+1\right )^{\frac {1}{4}}-2}{x^{2}}\right )}{2}+\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {2 \RootOf \left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{2}+1}-\RootOf \left (\textit {\_Z}^{2}+1\right ) x^{2}-2 \left (x^{2}+1\right )^{\frac {3}{4}}-2 \RootOf \left (\textit {\_Z}^{2}+1\right )+2 \left (x^{2}+1\right )^{\frac {1}{4}}}{x^{2}}\right )}{2}\)	\(110\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/GAMMA(3/4)*((-3*ln(2)+1/2*Pi+2*ln(x))*GAMMA(3/4)-3/4*hypergeom([1,1,7/4],[2,2],-x^2)*GAMMA(3/4)*x^2)

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maxima [A]  time = 0.53, size = 35, normalized size = 1.40 \begin {gather*} -\arctan \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{2} \, \log \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{2} \, \log \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.23, size = 21, normalized size = 0.84 \begin {gather*} -\mathrm {atan}\left ({\left (x^2+1\right )}^{1/4}\right )-\mathrm {atanh}\left ({\left (x^2+1\right )}^{1/4}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(x^2 + 1)^(3/4)),x)

[Out]

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

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sympy [C]  time = 0.85, size = 32, normalized size = 1.28 \begin {gather*} - \frac {\Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{2}}} \right )}}{2 x^{\frac {3}{2}} \Gamma \left (\frac {7}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-gamma(3/4)*hyper((3/4, 3/4), (7/4,), exp_polar(I*pi)/x**2)/(2*x**(3/2)*gamma(7/4))

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