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

Optimal. Leaf size=45 \[ -\frac {\sqrt [4]{x^2+1}}{2 x^2}+\frac {3}{4} \tan ^{-1}\left (\sqrt [4]{x^2+1}\right )+\frac {3}{4} \tanh ^{-1}\left (\sqrt [4]{x^2+1}\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {266, 51, 63, 212, 206, 203} \begin {gather*} -\frac {\sqrt [4]{x^2+1}}{2 x^2}+\frac {3}{4} \tan ^{-1}\left (\sqrt [4]{x^2+1}\right )+\frac {3}{4} \tanh ^{-1}\left (\sqrt [4]{x^2+1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(1 + x^2)^(3/4)),x]

[Out]

-1/2*(1 + x^2)^(1/4)/x^2 + (3*ArcTan[(1 + x^2)^(1/4)])/4 + (3*ArcTanh[(1 + x^2)^(1/4)])/4

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 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^3 \left (1+x^2\right )^{3/4}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x^2 (1+x)^{3/4}} \, dx,x,x^2\right )\\ &=-\frac {\sqrt [4]{1+x^2}}{2 x^2}-\frac {3}{8} \operatorname {Subst}\left (\int \frac {1}{x (1+x)^{3/4}} \, dx,x,x^2\right )\\ &=-\frac {\sqrt [4]{1+x^2}}{2 x^2}-\frac {3}{2} \operatorname {Subst}\left (\int \frac {1}{-1+x^4} \, dx,x,\sqrt [4]{1+x^2}\right )\\ &=-\frac {\sqrt [4]{1+x^2}}{2 x^2}+\frac {3}{4} \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt [4]{1+x^2}\right )+\frac {3}{4} \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt [4]{1+x^2}\right )\\ &=-\frac {\sqrt [4]{1+x^2}}{2 x^2}+\frac {3}{4} \tan ^{-1}\left (\sqrt [4]{1+x^2}\right )+\frac {3}{4} \tanh ^{-1}\left (\sqrt [4]{1+x^2}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.00, size = 24, normalized size = 0.53 \begin {gather*} 2 \sqrt [4]{x^2+1} \, _2F_1\left (\frac {1}{4},2;\frac {5}{4};x^2+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(1 + x^2)^(3/4)),x]

[Out]

2*(1 + x^2)^(1/4)*Hypergeometric2F1[1/4, 2, 5/4, 1 + x^2]

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.05, size = 45, normalized size = 1.00 \begin {gather*} -\frac {\sqrt [4]{1+x^2}}{2 x^2}+\frac {3}{4} \tan ^{-1}\left (\sqrt [4]{1+x^2}\right )+\frac {3}{4} \tanh ^{-1}\left (\sqrt [4]{1+x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(1 + x^2)^(1/4)/x^2 + (3*ArcTan[(1 + x^2)^(1/4)])/4 + (3*ArcTanh[(1 + x^2)^(1/4)])/4

________________________________________________________________________________________

fricas [A]  time = 0.50, size = 58, normalized size = 1.29 \begin {gather*} \frac {6 \, x^{2} \arctan \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}}\right ) + 3 \, x^{2} \log \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}} + 1\right ) - 3 \, x^{2} \log \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}} - 1\right ) - 4 \, {\left (x^{2} + 1\right )}^{\frac {1}{4}}}{8 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(6*x^2*arctan((x^2 + 1)^(1/4)) + 3*x^2*log((x^2 + 1)^(1/4) + 1) - 3*x^2*log((x^2 + 1)^(1/4) - 1) - 4*(x^2
+ 1)^(1/4))/x^2

________________________________________________________________________________________

giac [A]  time = 0.34, size = 47, normalized size = 1.04 \begin {gather*} -\frac {{\left (x^{2} + 1\right )}^{\frac {1}{4}}}{2 \, x^{2}} + \frac {3}{4} \, \arctan \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}}\right ) + \frac {3}{8} \, \log \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}} + 1\right ) - \frac {3}{8} \, \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^3/(x^2+1)^(3/4),x, algorithm="giac")

[Out]

-1/2*(x^2 + 1)^(1/4)/x^2 + 3/4*arctan((x^2 + 1)^(1/4)) + 3/8*log((x^2 + 1)^(1/4) + 1) - 3/8*log((x^2 + 1)^(1/4
) - 1)

________________________________________________________________________________________

maple [C]  time = 1.22, size = 52, normalized size = 1.16

method	result	size
meijerg	\(\frac {-\frac {\Gamma \left (\frac {3}{4}\right )}{x^{2}}-\frac {3 \left (\frac {1}{3}-3 \ln \relax (2)+\frac {\pi }{2}+2 \ln \relax (x )\right ) \Gamma \left (\frac {3}{4}\right )}{4}+\frac {21 \hypergeom \left (\left [1, 1, \frac {11}{4}\right ], \left [2, 3\right ], -x^{2}\right ) \Gamma \left (\frac {3}{4}\right ) x^{2}}{32}}{2 \Gamma \left (\frac {3}{4}\right )}\)	\(52\)
risch	\(-\frac {\left (x^{2}+1\right )^{\frac {1}{4}}}{2 x^{2}}-\frac {3 \left (\left (-3 \ln \relax (2)+\frac {\pi }{2}+2 \ln \relax (x )\right ) \Gamma \left (\frac {3}{4}\right )-\frac {3 \hypergeom \left (\left [1, 1, \frac {7}{4}\right ], \left [2, 2\right ], -x^{2}\right ) \Gamma \left (\frac {3}{4}\right ) x^{2}}{4}\right )}{8 \Gamma \left (\frac {3}{4}\right )}\)	\(56\)
trager	\(-\frac {\left (x^{2}+1\right )^{\frac {1}{4}}}{2 x^{2}}+\frac {3 \ln \left (-\frac {2 \left (x^{2}+1\right )^{\frac {3}{4}}+2 \sqrt {x^{2}+1}+x^{2}+2 \left (x^{2}+1\right )^{\frac {1}{4}}+2}{x^{2}}\right )}{8}+\frac {3 \RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {2 \RootOf \left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{2}+1}-\RootOf \left (\textit {\_Z}^{2}+1\right ) x^{2}+2 \left (x^{2}+1\right )^{\frac {3}{4}}-2 \RootOf \left (\textit {\_Z}^{2}+1\right )-2 \left (x^{2}+1\right )^{\frac {1}{4}}}{x^{2}}\right )}{8}\)	\(121\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/GAMMA(3/4)*(-GAMMA(3/4)/x^2-3/4*(1/3-3*ln(2)+1/2*Pi+2*ln(x))*GAMMA(3/4)+21/32*hypergeom([1,1,11/4],[2,3],-
x^2)*GAMMA(3/4)*x^2)

________________________________________________________________________________________

maxima [A]  time = 0.45, size = 47, normalized size = 1.04 \begin {gather*} -\frac {{\left (x^{2} + 1\right )}^{\frac {1}{4}}}{2 \, x^{2}} + \frac {3}{4} \, \arctan \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}}\right ) + \frac {3}{8} \, \log \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}} + 1\right ) - \frac {3}{8} \, \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^3/(x^2+1)^(3/4),x, algorithm="maxima")

[Out]

-1/2*(x^2 + 1)^(1/4)/x^2 + 3/4*arctan((x^2 + 1)^(1/4)) + 3/8*log((x^2 + 1)^(1/4) + 1) - 3/8*log((x^2 + 1)^(1/4
) - 1)

________________________________________________________________________________________

mupad [B]  time = 0.56, size = 33, normalized size = 0.73 \begin {gather*} \frac {3\,\mathrm {atan}\left ({\left (x^2+1\right )}^{1/4}\right )}{4}+\frac {3\,\mathrm {atanh}\left ({\left (x^2+1\right )}^{1/4}\right )}{4}-\frac {{\left (x^2+1\right )}^{1/4}}{2\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(x^2 + 1)^(3/4)),x)

[Out]

(3*atan((x^2 + 1)^(1/4)))/4 + (3*atanh((x^2 + 1)^(1/4)))/4 - (x^2 + 1)^(1/4)/(2*x^2)

________________________________________________________________________________________

sympy [C]  time = 0.92, size = 32, normalized size = 0.71 \begin {gather*} - \frac {\Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{2}}} \right )}}{2 x^{\frac {7}{2}} \Gamma \left (\frac {11}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-gamma(7/4)*hyper((3/4, 7/4), (11/4,), exp_polar(I*pi)/x**2)/(2*x**(7/2)*gamma(11/4))

________________________________________________________________________________________