3.16.21 \(\int \frac {1}{x^9 \sqrt {1+x^8}} \, dx\) [1521]

Optimal. Leaf size=31 \[ -\frac {\sqrt {1+x^8}}{8 x^8}+\frac {1}{8} \tanh ^{-1}\left (\sqrt {1+x^8}\right ) \]

[Out]

1/8*arctanh((x^8+1)^(1/2))-1/8*(x^8+1)^(1/2)/x^8

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Rubi [A]
time = 0.01, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {272, 44, 65, 213} \begin {gather*} \frac {1}{8} \tanh ^{-1}\left (\sqrt {x^8+1}\right )-\frac {\sqrt {x^8+1}}{8 x^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^9*Sqrt[1 + x^8]),x]

[Out]

-1/8*Sqrt[1 + x^8]/x^8 + ArcTanh[Sqrt[1 + x^8]]/8

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*} \int \frac {1}{x^9 \sqrt {1+x^8}} \, dx &=\frac {1}{8} \text {Subst}\left (\int \frac {1}{x^2 \sqrt {1+x}} \, dx,x,x^8\right )\\ &=-\frac {\sqrt {1+x^8}}{8 x^8}-\frac {1}{16} \text {Subst}\left (\int \frac {1}{x \sqrt {1+x}} \, dx,x,x^8\right )\\ &=-\frac {\sqrt {1+x^8}}{8 x^8}-\frac {1}{8} \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sqrt {1+x^8}\right )\\ &=-\frac {\sqrt {1+x^8}}{8 x^8}+\frac {1}{8} \tanh ^{-1}\left (\sqrt {1+x^8}\right )\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 31, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {1+x^8}}{8 x^8}+\frac {1}{8} \tanh ^{-1}\left (\sqrt {1+x^8}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^9*Sqrt[1 + x^8]),x]

[Out]

-1/8*Sqrt[1 + x^8]/x^8 + ArcTanh[Sqrt[1 + x^8]]/8

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Maple [A]
time = 0.30, size = 30, normalized size = 0.97

method	result	size
trager	\(-\frac {\sqrt {x^{8}+1}}{8 x^{8}}-\frac {\ln \left (\frac {\sqrt {x^{8}+1}-1}{x^{4}}\right )}{8}\)	\(30\)
risch	\(-\frac {\sqrt {x^{8}+1}}{8 x^{8}}-\frac {\ln \left (\frac {\sqrt {x^{8}+1}-1}{\sqrt {x^{8}}}\right )}{8}\)	\(32\)
meijerg	\(\frac {\frac {\sqrt {\pi }\, \left (4 x^{8}+8\right )}{8 x^{8}}-\frac {\sqrt {\pi }\, \sqrt {x^{8}+1}}{x^{8}}+\sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {x^{8}+1}}{2}\right )-\frac {\left (1-2 \ln \left (2\right )+8 \ln \left (x \right )\right ) \sqrt {\pi }}{2}-\frac {\sqrt {\pi }}{x^{8}}}{8 \sqrt {\pi }}\)	\(76\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^9/(x^8+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 37, normalized size = 1.19 \begin {gather*} -\frac {\sqrt {x^{8} + 1}}{8 \, x^{8}} + \frac {1}{16} \, \log \left (\sqrt {x^{8} + 1} + 1\right ) - \frac {1}{16} \, \log \left (\sqrt {x^{8} + 1} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^9/(x^8+1)^(1/2),x, algorithm="maxima")

[Out]

-1/8*sqrt(x^8 + 1)/x^8 + 1/16*log(sqrt(x^8 + 1) + 1) - 1/16*log(sqrt(x^8 + 1) - 1)

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Fricas [A]
time = 0.37, size = 44, normalized size = 1.42 \begin {gather*} \frac {x^{8} \log \left (\sqrt {x^{8} + 1} + 1\right ) - x^{8} \log \left (\sqrt {x^{8} + 1} - 1\right ) - 2 \, \sqrt {x^{8} + 1}}{16 \, x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^9/(x^8+1)^(1/2),x, algorithm="fricas")

[Out]

1/16*(x^8*log(sqrt(x^8 + 1) + 1) - x^8*log(sqrt(x^8 + 1) - 1) - 2*sqrt(x^8 + 1))/x^8

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Sympy [A]
time = 1.05, size = 22, normalized size = 0.71 \begin {gather*} \frac {\operatorname {asinh}{\left (\frac {1}{x^{4}} \right )}}{8} - \frac {\sqrt {1 + \frac {1}{x^{8}}}}{8 x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**9/(x**8+1)**(1/2),x)

[Out]

asinh(x**(-4))/8 - sqrt(1 + x**(-8))/(8*x**4)

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Giac [A]
time = 2.01, size = 37, normalized size = 1.19 \begin {gather*} -\frac {\sqrt {x^{8} + 1}}{8 \, x^{8}} + \frac {1}{16} \, \log \left (\sqrt {x^{8} + 1} + 1\right ) - \frac {1}{16} \, \log \left (\sqrt {x^{8} + 1} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^9/(x^8+1)^(1/2),x, algorithm="giac")

[Out]

-1/8*sqrt(x^8 + 1)/x^8 + 1/16*log(sqrt(x^8 + 1) + 1) - 1/16*log(sqrt(x^8 + 1) - 1)

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Mupad [B]
time = 1.41, size = 23, normalized size = 0.74 \begin {gather*} \frac {\mathrm {atanh}\left (\sqrt {x^8+1}\right )}{8}-\frac {\sqrt {x^8+1}}{8\,x^8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^9*(x^8 + 1)^(1/2)),x)

[Out]

atanh((x^8 + 1)^(1/2))/8 - (x^8 + 1)^(1/2)/(8*x^8)

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