3.444 \(\int \frac{1}{x^7 \sqrt{1+x^3}} \, dx\)

Optimal. Leaf size=47 \[ \frac{\sqrt{x^3+1}}{4 x^3}-\frac{\sqrt{x^3+1}}{6 x^6}-\frac{1}{4} \tanh ^{-1}\left (\sqrt{x^3+1}\right ) \]

[Out]

-Sqrt[1 + x^3]/(6*x^6) + Sqrt[1 + x^3]/(4*x^3) - ArcTanh[Sqrt[1 + x^3]]/4

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Rubi [A]  time = 0.016173, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {266, 51, 63, 207} \[ \frac{\sqrt{x^3+1}}{4 x^3}-\frac{\sqrt{x^3+1}}{6 x^6}-\frac{1}{4} \tanh ^{-1}\left (\sqrt{x^3+1}\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(x^7*Sqrt[1 + x^3]),x]

[Out]

-Sqrt[1 + x^3]/(6*x^6) + Sqrt[1 + x^3]/(4*x^3) - ArcTanh[Sqrt[1 + x^3]]/4

Rule 266

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]]

Rule 51

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 63

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 207

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

Rubi steps

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

Mathematica [C]  time = 0.004606, size = 26, normalized size = 0.55 \[ -\frac{2}{3} \sqrt{x^3+1} \, _2F_1\left (\frac{1}{2},3;\frac{3}{2};x^3+1\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^7*Sqrt[1 + x^3]),x]

[Out]

(-2*Sqrt[1 + x^3]*Hypergeometric2F1[1/2, 3, 3/2, 1 + x^3])/3

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Maple [A]  time = 0.018, size = 36, normalized size = 0.8 \begin{align*} -{\frac{1}{4}{\it Artanh} \left ( \sqrt{{x}^{3}+1} \right ) }-{\frac{1}{6\,{x}^{6}}\sqrt{{x}^{3}+1}}+{\frac{1}{4\,{x}^{3}}\sqrt{{x}^{3}+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^7/(x^3+1)^(1/2),x)

[Out]

-1/4*arctanh((x^3+1)^(1/2))-1/6*(x^3+1)^(1/2)/x^6+1/4*(x^3+1)^(1/2)/x^3

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Maxima [A]  time = 0.963444, size = 86, normalized size = 1.83 \begin{align*} -\frac{3 \,{\left (x^{3} + 1\right )}^{\frac{3}{2}} - 5 \, \sqrt{x^{3} + 1}}{12 \,{\left (2 \, x^{3} -{\left (x^{3} + 1\right )}^{2} + 1\right )}} - \frac{1}{8} \, \log \left (\sqrt{x^{3} + 1} + 1\right ) + \frac{1}{8} \, \log \left (\sqrt{x^{3} + 1} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*(x^3 + 1)^(3/2) - 5*sqrt(x^3 + 1))/(2*x^3 - (x^3 + 1)^2 + 1) - 1/8*log(sqrt(x^3 + 1) + 1) + 1/8*log(s
qrt(x^3 + 1) - 1)

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Fricas [A]  time = 1.48204, size = 139, normalized size = 2.96 \begin{align*} -\frac{3 \, x^{6} \log \left (\sqrt{x^{3} + 1} + 1\right ) - 3 \, x^{6} \log \left (\sqrt{x^{3} + 1} - 1\right ) - 2 \,{\left (3 \, x^{3} - 2\right )} \sqrt{x^{3} + 1}}{24 \, x^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 3.76679, size = 65, normalized size = 1.38 \begin{align*} - \frac{\operatorname{asinh}{\left (\frac{1}{x^{\frac{3}{2}}} \right )}}{4} + \frac{1}{4 x^{\frac{3}{2}} \sqrt{1 + \frac{1}{x^{3}}}} + \frac{1}{12 x^{\frac{9}{2}} \sqrt{1 + \frac{1}{x^{3}}}} - \frac{1}{6 x^{\frac{15}{2}} \sqrt{1 + \frac{1}{x^{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-asinh(x**(-3/2))/4 + 1/(4*x**(3/2)*sqrt(1 + x**(-3))) + 1/(12*x**(9/2)*sqrt(1 + x**(-3))) - 1/(6*x**(15/2)*sq
rt(1 + x**(-3)))

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Giac [A]  time = 1.12784, size = 68, normalized size = 1.45 \begin{align*} \frac{3 \,{\left (x^{3} + 1\right )}^{\frac{3}{2}} - 5 \, \sqrt{x^{3} + 1}}{12 \, x^{6}} - \frac{1}{8} \, \log \left (\sqrt{x^{3} + 1} + 1\right ) + \frac{1}{8} \, \log \left ({\left | \sqrt{x^{3} + 1} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*(x^3 + 1)^(3/2) - 5*sqrt(x^3 + 1))/x^6 - 1/8*log(sqrt(x^3 + 1) + 1) + 1/8*log(abs(sqrt(x^3 + 1) - 1))