\(\int \frac {1}{x^3 (1+x^2)^{3/4}} \, dx\) [577]

Optimal result

Integrand size = 13, antiderivative size = 45 \[ \int \frac {1}{x^3 \left (1+x^2\right )^{3/4}} \, dx=-\frac {\sqrt [4]{1+x^2}}{2 x^2}+\frac {3}{4} \arctan \left (\sqrt [4]{1+x^2}\right )+\frac {3}{4} \text {arctanh}\left (\sqrt [4]{1+x^2}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {272, 44, 65, 218, 212, 209} \[ \int \frac {1}{x^3 \left (1+x^2\right )^{3/4}} \, dx=\frac {3}{4} \arctan \left (\sqrt [4]{x^2+1}\right )+\frac {3}{4} \text {arctanh}\left (\sqrt [4]{x^2+1}\right )-\frac {\sqrt [4]{x^2+1}}{2 x^2} \]

[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 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

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 209

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

Rule 212

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

Rule 218

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] &&  !Gt
Q[a/b, 0]

Rule 272

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*} \text {integral}& = \frac {1}{2} \text {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} \text {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} \text {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} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt [4]{1+x^2}\right )+\frac {3}{4} \text {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} \arctan \left (\sqrt [4]{1+x^2}\right )+\frac {3}{4} \text {arctanh}\left (\sqrt [4]{1+x^2}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^3 \left (1+x^2\right )^{3/4}} \, dx=-\frac {\sqrt [4]{1+x^2}}{2 x^2}+\frac {3}{4} \arctan \left (\sqrt [4]{1+x^2}\right )+\frac {3}{4} \text {arctanh}\left (\sqrt [4]{1+x^2}\right ) \]

[In]

Integrate[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

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3.

Time = 1.46 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.16

method	result	size
meijerg	\(\frac {\frac {21 \Gamma \left (\frac {3}{4}\right ) x^{2} \operatorname {hypergeom}\left (\left [1, 1, \frac {11}{4}\right ], \left [2, 3\right ], -x^{2}\right )}{32}-\frac {3 \left (\frac {1}{3}-3 \ln \left (2\right )+\frac {\pi }{2}+2 \ln \left (x \right )\right ) \Gamma \left (\frac {3}{4}\right )}{4}-\frac {\Gamma \left (\frac {3}{4}\right )}{x^{2}}}{2 \Gamma \left (\frac {3}{4}\right )}\)	\(52\)
risch	\(-\frac {\left (x^{2}+1\right )^{\frac {1}{4}}}{2 x^{2}}-\frac {3 \left (-\frac {3 \Gamma \left (\frac {3}{4}\right ) x^{2} \operatorname {hypergeom}\left (\left [1, 1, \frac {7}{4}\right ], \left [2, 2\right ], -x^{2}\right )}{4}+\left (-3 \ln \left (2\right )+\frac {\pi }{2}+2 \ln \left (x \right )\right ) \Gamma \left (\frac {3}{4}\right )\right )}{8 \Gamma \left (\frac {3}{4}\right )}\)	\(56\)
pseudoelliptic	\(\frac {-3 \ln \left (\left (x^{2}+1\right )^{\frac {1}{4}}-1\right ) x^{2}+6 \arctan \left (\left (x^{2}+1\right )^{\frac {1}{4}}\right ) x^{2}+3 \ln \left (\left (x^{2}+1\right )^{\frac {1}{4}}+1\right ) x^{2}-4 \left (x^{2}+1\right )^{\frac {1}{4}}}{8 x^{2}}\)	\(59\)
trager	\(-\frac {\left (x^{2}+1\right )^{\frac {1}{4}}}{2 x^{2}}-\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{2}+1}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}-2 \left (x^{2}+1\right )^{\frac {3}{4}}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )+2 \left (x^{2}+1\right )^{\frac {1}{4}}}{x^{2}}\right )}{8}-\frac {3 \ln \left (-\frac {2 \left (x^{2}+1\right )^{\frac {3}{4}}-2 \sqrt {x^{2}+1}-x^{2}+2 \left (x^{2}+1\right )^{\frac {1}{4}}-2}{x^{2}}\right )}{8}\)	\(122\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.29 \[ \int \frac {1}{x^3 \left (1+x^2\right )^{3/4}} \, dx=\frac {6 \, x^{2} \arctan \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}}\right ) + 3 \, x^{2} \log \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}} + 1\right ) - 3 \, x^{2} \log \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}} - 1\right ) - 4 \, {\left (x^{2} + 1\right )}^{\frac {1}{4}}}{8 \, x^{2}} \]

[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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.64 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.71 \[ \int \frac {1}{x^3 \left (1+x^2\right )^{3/4}} \, dx=- \frac {\Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{2}}} \right )}}{2 x^{\frac {7}{2}} \Gamma \left (\frac {11}{4}\right )} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.04 \[ \int \frac {1}{x^3 \left (1+x^2\right )^{3/4}} \, dx=-\frac {{\left (x^{2} + 1\right )}^{\frac {1}{4}}}{2 \, x^{2}} + \frac {3}{4} \, \arctan \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}}\right ) + \frac {3}{8} \, \log \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}} + 1\right ) - \frac {3}{8} \, \log \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}} - 1\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.04 \[ \int \frac {1}{x^3 \left (1+x^2\right )^{3/4}} \, dx=-\frac {{\left (x^{2} + 1\right )}^{\frac {1}{4}}}{2 \, x^{2}} + \frac {3}{4} \, \arctan \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}}\right ) + \frac {3}{8} \, \log \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}} + 1\right ) - \frac {3}{8} \, \log \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}} - 1\right ) \]

[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)

Mupad [B] (verification not implemented)

Time = 5.58 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.73 \[ \int \frac {1}{x^3 \left (1+x^2\right )^{3/4}} \, dx=\frac {3\,\mathrm {atan}\left ({\left (x^2+1\right )}^{1/4}\right )}{4}+\frac {3\,\mathrm {atanh}\left ({\left (x^2+1\right )}^{1/4}\right )}{4}-\frac {{\left (x^2+1\right )}^{1/4}}{2\,x^2} \]

[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)