3.15.13 \(\int \frac {1}{x^7 (2+x^6)^{3/2}} \, dx\) [1413]

Optimal. Leaf size=55 \[ -\frac {1}{8 \sqrt {2+x^6}}-\frac {1}{12 x^6 \sqrt {2+x^6}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {2+x^6}}{\sqrt {2}}\right )}{8 \sqrt {2}} \]

[Out]

1/16*arctanh(1/2*(x^6+2)^(1/2)*2^(1/2))*2^(1/2)-1/8/(x^6+2)^(1/2)-1/12/x^6/(x^6+2)^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {272, 44, 53, 65, 213} \begin {gather*} -\frac {1}{12 x^6 \sqrt {x^6+2}}-\frac {1}{8 \sqrt {x^6+2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {x^6+2}}{\sqrt {2}}\right )}{8 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^7*(2 + x^6)^(3/2)),x]

[Out]

-1/8*1/Sqrt[2 + x^6] - 1/(12*x^6*Sqrt[2 + x^6]) + ArcTanh[Sqrt[2 + x^6]/Sqrt[2]]/(8*Sqrt[2])

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 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^7 \left (2+x^6\right )^{3/2}} \, dx &=\frac {1}{6} \text {Subst}\left (\int \frac {1}{x^2 (2+x)^{3/2}} \, dx,x,x^6\right )\\ &=\frac {1}{6 x^6 \sqrt {2+x^6}}+\frac {1}{4} \text {Subst}\left (\int \frac {1}{x^2 \sqrt {2+x}} \, dx,x,x^6\right )\\ &=\frac {1}{6 x^6 \sqrt {2+x^6}}-\frac {\sqrt {2+x^6}}{8 x^6}-\frac {1}{16} \text {Subst}\left (\int \frac {1}{x \sqrt {2+x}} \, dx,x,x^6\right )\\ &=\frac {1}{6 x^6 \sqrt {2+x^6}}-\frac {\sqrt {2+x^6}}{8 x^6}-\frac {1}{8} \text {Subst}\left (\int \frac {1}{-2+x^2} \, dx,x,\sqrt {2+x^6}\right )\\ &=\frac {1}{6 x^6 \sqrt {2+x^6}}-\frac {\sqrt {2+x^6}}{8 x^6}+\frac {\tanh ^{-1}\left (\frac {\sqrt {2+x^6}}{\sqrt {2}}\right )}{8 \sqrt {2}}\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 49, normalized size = 0.89 \begin {gather*} \frac {-2-3 x^6}{24 x^6 \sqrt {2+x^6}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {2+x^6}}{\sqrt {2}}\right )}{8 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^7*(2 + x^6)^(3/2)),x]

[Out]

(-2 - 3*x^6)/(24*x^6*Sqrt[2 + x^6]) + ArcTanh[Sqrt[2 + x^6]/Sqrt[2]]/(8*Sqrt[2])

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Maple [A]
time = 0.35, size = 46, normalized size = 0.84

method	result	size
risch	\(-\frac {3 x^{6}+2}{24 x^{6} \sqrt {x^{6}+2}}-\frac {\sqrt {2}\, \ln \left (\frac {\sqrt {x^{6}+2}-\sqrt {2}}{\sqrt {x^{6}}}\right )}{16}\)	\(46\)
trager	\(-\frac {3 x^{6}+2}{24 x^{6} \sqrt {x^{6}+2}}+\frac {\RootOf \left (\textit {\_Z}^{2}-2\right ) \ln \left (\frac {\sqrt {x^{6}+2}+\RootOf \left (\textit {\_Z}^{2}-2\right )}{x^{3}}\right )}{16}\)	\(48\)
meijerg	\(\frac {\sqrt {2}\, \left (\frac {\sqrt {\pi }\, \left (10 x^{6}+8\right )}{8 x^{6}}-\frac {\sqrt {\pi }\, \left (12 x^{6}+8\right )}{8 x^{6} \sqrt {1+\frac {x^{6}}{2}}}+\frac {3 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {1+\frac {x^{6}}{2}}}{2}\right )}{2}-\frac {3 \left (\frac {5}{3}-3 \ln \left (2\right )+6 \ln \left (x \right )\right ) \sqrt {\pi }}{4}-\frac {\sqrt {\pi }}{x^{6}}\right )}{24 \sqrt {\pi }}\)	\(91\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^7/(x^6+2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/24*(3*x^6+2)/x^6/(x^6+2)^(1/2)-1/16*2^(1/2)*ln(((x^6+2)^(1/2)-2^(1/2))/(x^6)^(1/2))

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Maxima [A]
time = 0.50, size = 63, normalized size = 1.15 \begin {gather*} -\frac {1}{32} \, \sqrt {2} \log \left (-\frac {\sqrt {2} - \sqrt {x^{6} + 2}}{\sqrt {2} + \sqrt {x^{6} + 2}}\right ) - \frac {3 \, x^{6} + 2}{24 \, {\left ({\left (x^{6} + 2\right )}^{\frac {3}{2}} - 2 \, \sqrt {x^{6} + 2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(x^6+2)^(3/2),x, algorithm="maxima")

[Out]

-1/32*sqrt(2)*log(-(sqrt(2) - sqrt(x^6 + 2))/(sqrt(2) + sqrt(x^6 + 2))) - 1/24*(3*x^6 + 2)/((x^6 + 2)^(3/2) -
2*sqrt(x^6 + 2))

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Fricas [A]
time = 0.36, size = 66, normalized size = 1.20 \begin {gather*} \frac {3 \, \sqrt {2} {\left (x^{12} + 2 \, x^{6}\right )} \log \left (\frac {x^{6} + 2 \, \sqrt {2} \sqrt {x^{6} + 2} + 4}{x^{6}}\right ) - 4 \, {\left (3 \, x^{6} + 2\right )} \sqrt {x^{6} + 2}}{96 \, {\left (x^{12} + 2 \, x^{6}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(x^6+2)^(3/2),x, algorithm="fricas")

[Out]

1/96*(3*sqrt(2)*(x^12 + 2*x^6)*log((x^6 + 2*sqrt(2)*sqrt(x^6 + 2) + 4)/x^6) - 4*(3*x^6 + 2)*sqrt(x^6 + 2))/(x^
12 + 2*x^6)

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Sympy [A]
time = 1.48, size = 49, normalized size = 0.89 \begin {gather*} \frac {\sqrt {2} \operatorname {asinh}{\left (\frac {\sqrt {2}}{x^{3}} \right )}}{16} - \frac {1}{8 x^{3} \sqrt {1 + \frac {2}{x^{6}}}} - \frac {1}{12 x^{9} \sqrt {1 + \frac {2}{x^{6}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**7/(x**6+2)**(3/2),x)

[Out]

sqrt(2)*asinh(sqrt(2)/x**3)/16 - 1/(8*x**3*sqrt(1 + 2/x**6)) - 1/(12*x**9*sqrt(1 + 2/x**6))

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Giac [A]
time = 1.54, size = 63, normalized size = 1.15 \begin {gather*} -\frac {1}{32} \, \sqrt {2} \log \left (-\frac {\sqrt {2} - \sqrt {x^{6} + 2}}{\sqrt {2} + \sqrt {x^{6} + 2}}\right ) - \frac {3 \, x^{6} + 2}{24 \, {\left ({\left (x^{6} + 2\right )}^{\frac {3}{2}} - 2 \, \sqrt {x^{6} + 2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(x^6+2)^(3/2),x, algorithm="giac")

[Out]

-1/32*sqrt(2)*log(-(sqrt(2) - sqrt(x^6 + 2))/(sqrt(2) + sqrt(x^6 + 2))) - 1/24*(3*x^6 + 2)/((x^6 + 2)^(3/2) -
2*sqrt(x^6 + 2))

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Mupad [B]
time = 1.30, size = 40, normalized size = 0.73 \begin {gather*} \frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {x^6+2}}{2}\right )}{16}-\frac {1}{8\,\sqrt {x^6+2}}-\frac {1}{12\,x^6\,\sqrt {x^6+2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^7*(x^6 + 2)^(3/2)),x)

[Out]

(2^(1/2)*atanh((2^(1/2)*(x^6 + 2)^(1/2))/2))/16 - 1/(8*(x^6 + 2)^(1/2)) - 1/(12*x^6*(x^6 + 2)^(1/2))

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