3.833 \(\int \frac {1+2 x^8}{x (1+x^8)^{3/2}} \, dx\)

Optimal. Leaf size=28 \[ -\frac {1}{4 \sqrt {x^8+1}}-\frac {1}{4} \tanh ^{-1}\left (\sqrt {x^8+1}\right ) \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {446, 78, 63, 207} \[ -\frac {1}{4 \sqrt {x^8+1}}-\frac {1}{4} \tanh ^{-1}\left (\sqrt {x^8+1}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(1 + 2*x^8)/(x*(1 + x^8)^(3/2)),x]

[Out]

-1/(4*Sqrt[1 + x^8]) - ArcTanh[Sqrt[1 + x^8]]/4

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 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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])

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 28, normalized size = 1.00 \[ -\frac {1}{4 \sqrt {x^8+1}}-\frac {1}{4} \tanh ^{-1}\left (\sqrt {x^8+1}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + 2*x^8)/(x*(1 + x^8)^(3/2)),x]

[Out]

-1/4*1/Sqrt[1 + x^8] - ArcTanh[Sqrt[1 + x^8]]/4

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fricas [B]  time = 0.41, size = 52, normalized size = 1.86 \[ -\frac {{\left (x^{8} + 1\right )} \log \left (\sqrt {x^{8} + 1} + 1\right ) - {\left (x^{8} + 1\right )} \log \left (\sqrt {x^{8} + 1} - 1\right ) + 2 \, \sqrt {x^{8} + 1}}{8 \, {\left (x^{8} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^8+1)/x/(x^8+1)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.38, size = 34, normalized size = 1.21 \[ -\frac {1}{4 \, \sqrt {x^{8} + 1}} - \frac {1}{8} \, \log \left (\sqrt {x^{8} + 1} + 1\right ) + \frac {1}{8} \, \log \left (\sqrt {x^{8} + 1} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^8+1)/x/(x^8+1)^(3/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.03, size = 29, normalized size = 1.04 \[ \frac {\ln \left (\frac {\sqrt {x^{8}+1}-1}{\sqrt {x^{8}}}\right )}{4}-\frac {1}{4 \sqrt {x^{8}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^8+1)/x/(x^8+1)^(3/2),x)

[Out]

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

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maxima [A]  time = 0.96, size = 34, normalized size = 1.21 \[ -\frac {1}{4 \, \sqrt {x^{8} + 1}} - \frac {1}{8} \, \log \left (\sqrt {x^{8} + 1} + 1\right ) + \frac {1}{8} \, \log \left (\sqrt {x^{8} + 1} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^8+1)/x/(x^8+1)^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.85, size = 20, normalized size = 0.71 \[ -\frac {\mathrm {atanh}\left (\sqrt {x^8+1}\right )}{4}-\frac {1}{4\,\sqrt {x^8+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^8 + 1)/(x*(x^8 + 1)^(3/2)),x)

[Out]

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

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sympy [A]  time = 22.80, size = 37, normalized size = 1.32 \[ \frac {\log {\left (\sqrt {x^{8} + 1} - 1 \right )}}{8} - \frac {\log {\left (\sqrt {x^{8} + 1} + 1 \right )}}{8} - \frac {1}{4 \sqrt {x^{8} + 1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**8+1)/x/(x**8+1)**(3/2),x)

[Out]

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

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