\(\int \frac {1}{x^5 (1-x^4)^{3/2}} \, dx\) [898]

Optimal result

Integrand size = 15, antiderivative size = 50 \[ \int \frac {1}{x^5 \left (1-x^4\right )^{3/2}} \, dx=\frac {3}{4 \sqrt {1-x^4}}-\frac {1}{4 x^4 \sqrt {1-x^4}}-\frac {3}{4} \text {arctanh}\left (\sqrt {1-x^4}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {272, 44, 53, 65, 212} \[ \int \frac {1}{x^5 \left (1-x^4\right )^{3/2}} \, dx=-\frac {3}{4} \text {arctanh}\left (\sqrt {1-x^4}\right )-\frac {1}{4 x^4 \sqrt {1-x^4}}+\frac {3}{4 \sqrt {1-x^4}} \]

[In]

Int[1/(x^5*(1 - x^4)^(3/2)),x]

[Out]

3/(4*Sqrt[1 - x^4]) - 1/(4*x^4*Sqrt[1 - x^4]) - (3*ArcTanh[Sqrt[1 - x^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 53

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.82 \[ \int \frac {1}{x^5 \left (1-x^4\right )^{3/2}} \, dx=\frac {1}{4} \left (\frac {-1+3 x^4}{x^4 \sqrt {1-x^4}}-3 \text {arctanh}\left (\sqrt {1-x^4}\right )\right ) \]

[In]

Integrate[1/(x^5*(1 - x^4)^(3/2)),x]

[Out]

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

Maple [A] (verified)

Time = 4.16 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.70

method	result	size
risch	\(\frac {3 x^{4}-1}{4 x^{4} \sqrt {-x^{4}+1}}-\frac {3 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-x^{4}+1}}\right )}{4}\)	\(35\)
pseudoelliptic	\(-\frac {3 \left (\operatorname {arctanh}\left (\frac {1}{\sqrt {-x^{4}+1}}\right ) x^{4} \sqrt {-x^{4}+1}-x^{4}+\frac {1}{3}\right )}{4 \sqrt {-x^{4}+1}\, x^{4}}\)	\(45\)
trager	\(-\frac {\left (3 x^{4}-1\right ) \sqrt {-x^{4}+1}}{4 \left (x^{4}-1\right ) x^{4}}+\frac {3 \ln \left (\frac {-1+\sqrt {-x^{4}+1}}{x^{2}}\right )}{4}\)	\(48\)
default	\(-\frac {\sqrt {-x^{4}+1}}{4 x^{4}}-\frac {3 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-x^{4}+1}}\right )}{4}-\frac {\sqrt {-\left (x^{2}-1\right )^{2}+2-2 x^{2}}}{4 \left (x^{2}-1\right )}+\frac {\sqrt {-\left (x^{2}+1\right )^{2}+2+2 x^{2}}}{4 x^{2}+4}\)	\(82\)
elliptic	\(-\frac {\sqrt {-x^{4}+1}}{4 x^{4}}-\frac {3 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-x^{4}+1}}\right )}{4}-\frac {\sqrt {-\left (x^{2}-1\right )^{2}+2-2 x^{2}}}{4 \left (x^{2}-1\right )}+\frac {\sqrt {-\left (x^{2}+1\right )^{2}+2+2 x^{2}}}{4 x^{2}+4}\)	\(82\)
meijerg	\(-\frac {\frac {\sqrt {\pi }}{2 x^{4}}-\frac {3 \left (\frac {5}{3}-2 \ln \left (2\right )+4 \ln \left (x \right )+i \pi \right ) \sqrt {\pi }}{4}-\frac {\sqrt {\pi }\, \left (-20 x^{4}+8\right )}{16 x^{4}}+\frac {\sqrt {\pi }\, \left (-24 x^{4}+8\right )}{16 x^{4} \sqrt {-x^{4}+1}}+\frac {3 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-x^{4}+1}}{2}\right )}{2}}{2 \sqrt {\pi }}\)	\(92\)

[In]

int(1/x^5/(-x^4+1)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 78 vs. \(2 (38) = 76\).

Time = 0.29 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.56 \[ \int \frac {1}{x^5 \left (1-x^4\right )^{3/2}} \, dx=-\frac {3 \, {\left (x^{8} - x^{4}\right )} \log \left (\sqrt {-x^{4} + 1} + 1\right ) - 3 \, {\left (x^{8} - x^{4}\right )} \log \left (\sqrt {-x^{4} + 1} - 1\right ) + 2 \, {\left (3 \, x^{4} - 1\right )} \sqrt {-x^{4} + 1}}{8 \, {\left (x^{8} - x^{4}\right )}} \]

[In]

integrate(1/x^5/(-x^4+1)^(3/2),x, algorithm="fricas")

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.48 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.90 \[ \int \frac {1}{x^5 \left (1-x^4\right )^{3/2}} \, dx=\begin {cases} - \frac {3 \operatorname {acosh}{\left (\frac {1}{x^{2}} \right )}}{4} + \frac {3}{4 x^{2} \sqrt {-1 + \frac {1}{x^{4}}}} - \frac {1}{4 x^{6} \sqrt {-1 + \frac {1}{x^{4}}}} & \text {for}\: \frac {1}{\left |{x^{4}}\right |} > 1 \\\frac {3 i \operatorname {asin}{\left (\frac {1}{x^{2}} \right )}}{4} - \frac {3 i}{4 x^{2} \sqrt {1 - \frac {1}{x^{4}}}} + \frac {i}{4 x^{6} \sqrt {1 - \frac {1}{x^{4}}}} & \text {otherwise} \end {cases} \]

[In]

integrate(1/x**5/(-x**4+1)**(3/2),x)

[Out]

Piecewise((-3*acosh(x**(-2))/4 + 3/(4*x**2*sqrt(-1 + x**(-4))) - 1/(4*x**6*sqrt(-1 + x**(-4))), 1/Abs(x**4) >
1), (3*I*asin(x**(-2))/4 - 3*I/(4*x**2*sqrt(1 - 1/x**4)) + I/(4*x**6*sqrt(1 - 1/x**4)), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.22 \[ \int \frac {1}{x^5 \left (1-x^4\right )^{3/2}} \, dx=-\frac {3 \, x^{4} - 1}{4 \, {\left ({\left (-x^{4} + 1\right )}^{\frac {3}{2}} - \sqrt {-x^{4} + 1}\right )}} - \frac {3}{8} \, \log \left (\sqrt {-x^{4} + 1} + 1\right ) + \frac {3}{8} \, \log \left (\sqrt {-x^{4} + 1} - 1\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.26 \[ \int \frac {1}{x^5 \left (1-x^4\right )^{3/2}} \, dx=-\frac {3 \, x^{4} - 1}{4 \, {\left ({\left (-x^{4} + 1\right )}^{\frac {3}{2}} - \sqrt {-x^{4} + 1}\right )}} - \frac {3}{8} \, \log \left (\sqrt {-x^{4} + 1} + 1\right ) + \frac {3}{8} \, \log \left (-\sqrt {-x^{4} + 1} + 1\right ) \]

[In]

integrate(1/x^5/(-x^4+1)^(3/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.89 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.76 \[ \int \frac {1}{x^5 \left (1-x^4\right )^{3/2}} \, dx=\frac {3}{4\,\sqrt {1-x^4}}-\frac {1}{4\,x^4\,\sqrt {1-x^4}}-\frac {3\,\mathrm {atanh}\left (\sqrt {1-x^4}\right )}{4} \]

[In]

int(1/(x^5*(1 - x^4)^(3/2)),x)

[Out]

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