\(\int \frac {1}{x^5 \sqrt {1+x^4}} \, dx\) [921]

Optimal result

Integrand size = 13, antiderivative size = 31 \[ \int \frac {1}{x^5 \sqrt {1+x^4}} \, dx=-\frac {\sqrt {1+x^4}}{4 x^4}+\frac {1}{4} \text {arctanh}\left (\sqrt {1+x^4}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {272, 44, 65, 213} \[ \int \frac {1}{x^5 \sqrt {1+x^4}} \, dx=\frac {1}{4} \text {arctanh}\left (\sqrt {x^4+1}\right )-\frac {\sqrt {x^4+1}}{4 x^4} \]

[In]

Int[1/(x^5*Sqrt[1 + x^4]),x]

[Out]

-1/4*Sqrt[1 + x^4]/x^4 + 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 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 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[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}{x^2 \sqrt {1+x}} \, dx,x,x^4\right ) \\ & = -\frac {\sqrt {1+x^4}}{4 x^4}-\frac {1}{8} \text {Subst}\left (\int \frac {1}{x \sqrt {1+x}} \, dx,x,x^4\right ) \\ & = -\frac {\sqrt {1+x^4}}{4 x^4}-\frac {1}{4} \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sqrt {1+x^4}\right ) \\ & = -\frac {\sqrt {1+x^4}}{4 x^4}+\frac {1}{4} \tanh ^{-1}\left (\sqrt {1+x^4}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^5 \sqrt {1+x^4}} \, dx=-\frac {\sqrt {1+x^4}}{4 x^4}+\frac {1}{4} \text {arctanh}\left (\sqrt {1+x^4}\right ) \]

[In]

Integrate[1/(x^5*Sqrt[1 + x^4]),x]

[Out]

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

Maple [A] (verified)

Time = 4.13 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77

method	result	size
default	\(-\frac {\sqrt {x^{4}+1}}{4 x^{4}}+\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {x^{4}+1}}\right )}{4}\)	\(24\)
risch	\(-\frac {\sqrt {x^{4}+1}}{4 x^{4}}+\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {x^{4}+1}}\right )}{4}\)	\(24\)
elliptic	\(-\frac {\sqrt {x^{4}+1}}{4 x^{4}}+\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {x^{4}+1}}\right )}{4}\)	\(24\)
pseudoelliptic	\(\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {x^{4}+1}}\right ) x^{4}-\sqrt {x^{4}+1}}{4 x^{4}}\)	\(28\)
trager	\(-\frac {\sqrt {x^{4}+1}}{4 x^{4}}-\frac {\ln \left (\frac {-1+\sqrt {x^{4}+1}}{x^{2}}\right )}{4}\)	\(30\)
meijerg	\(\frac {-\frac {\sqrt {\pi }}{x^{4}}-\frac {\left (1-2 \ln \left (2\right )+4 \ln \left (x \right )\right ) \sqrt {\pi }}{2}+\frac {\sqrt {\pi }\, \left (4 x^{4}+8\right )}{8 x^{4}}-\frac {\sqrt {\pi }\, \sqrt {x^{4}+1}}{x^{4}}+\sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {x^{4}+1}}{2}\right )}{4 \sqrt {\pi }}\)	\(76\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.42 \[ \int \frac {1}{x^5 \sqrt {1+x^4}} \, dx=\frac {x^{4} \log \left (\sqrt {x^{4} + 1} + 1\right ) - x^{4} \log \left (\sqrt {x^{4} + 1} - 1\right ) - 2 \, \sqrt {x^{4} + 1}}{8 \, x^{4}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 1.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.71 \[ \int \frac {1}{x^5 \sqrt {1+x^4}} \, dx=\frac {\operatorname {asinh}{\left (\frac {1}{x^{2}} \right )}}{4} - \frac {\sqrt {1 + \frac {1}{x^{4}}}}{4 x^{2}} \]

[In]

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

[Out]

asinh(x**(-2))/4 - sqrt(1 + x**(-4))/(4*x**2)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.19 \[ \int \frac {1}{x^5 \sqrt {1+x^4}} \, dx=-\frac {\sqrt {x^{4} + 1}}{4 \, x^{4}} + \frac {1}{8} \, \log \left (\sqrt {x^{4} + 1} + 1\right ) - \frac {1}{8} \, \log \left (\sqrt {x^{4} + 1} - 1\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.19 \[ \int \frac {1}{x^5 \sqrt {1+x^4}} \, dx=-\frac {\sqrt {x^{4} + 1}}{4 \, x^{4}} + \frac {1}{8} \, \log \left (\sqrt {x^{4} + 1} + 1\right ) - \frac {1}{8} \, \log \left (\sqrt {x^{4} + 1} - 1\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.59 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.74 \[ \int \frac {1}{x^5 \sqrt {1+x^4}} \, dx=\frac {\mathrm {atanh}\left (\sqrt {x^4+1}\right )}{4}-\frac {\sqrt {x^4+1}}{4\,x^4} \]

[In]

int(1/(x^5*(x^4 + 1)^(1/2)),x)

[Out]

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