3.4.8 \(\int \frac {x^3}{(-1+x^4) \sqrt {1+2 x^8}} \, dx\) [308]

Optimal. Leaf size=34 \[ -\frac {\tanh ^{-1}\left (\frac {1+2 x^4}{\sqrt {3} \sqrt {1+2 x^8}}\right )}{4 \sqrt {3}} \]

[Out]

-1/12*arctanh(1/3*(2*x^4+1)*3^(1/2)/(2*x^8+1)^(1/2))*3^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1483, 739, 212} \begin {gather*} -\frac {\tanh ^{-1}\left (\frac {2 x^4+1}{\sqrt {3} \sqrt {2 x^8+1}}\right )}{4 \sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3/((-1 + x^4)*Sqrt[1 + 2*x^8]),x]

[Out]

-1/4*ArcTanh[(1 + 2*x^4)/(Sqrt[3]*Sqrt[1 + 2*x^8])]/Sqrt[3]

Rule 212

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

Rule 739

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 1483

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[I
nt[(d + e*x)^q*(a + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, c, d, e, m, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[Sim
plify[m - n + 1], 0]

Rubi steps

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

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Mathematica [A]
time = 0.17, size = 41, normalized size = 1.21 \begin {gather*} -\frac {\tanh ^{-1}\left (\frac {1}{3} \left (\sqrt {6}-\sqrt {6} x^4+\sqrt {3+6 x^8}\right )\right )}{2 \sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3/((-1 + x^4)*Sqrt[1 + 2*x^8]),x]

[Out]

-1/2*ArcTanh[(Sqrt[6] - Sqrt[6]*x^4 + Sqrt[3 + 6*x^8])/3]/Sqrt[3]

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.15, size = 60, normalized size = 1.76

method	result	size
trager	\(\frac {\RootOf \left (\textit {\_Z}^{2}-3\right ) \ln \left (-\frac {-2 \RootOf \left (\textit {\_Z}^{2}-3\right ) x^{4}+3 \sqrt {2 x^{8}+1}-\RootOf \left (\textit {\_Z}^{2}-3\right )}{\left (-1+x \right ) \left (1+x \right ) \left (x^{2}+1\right )}\right )}{12}\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(x^4-1)/(2*x^8+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/12*RootOf(_Z^2-3)*ln(-(-2*RootOf(_Z^2-3)*x^4+3*(2*x^8+1)^(1/2)-RootOf(_Z^2-3))/(-1+x)/(1+x)/(x^2+1))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(x^4-1)/(2*x^8+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^3/(sqrt(2*x^8 + 1)*(x^4 - 1)), x)

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Fricas [A]
time = 1.23, size = 49, normalized size = 1.44 \begin {gather*} \frac {1}{12} \, \sqrt {3} \log \left (\frac {2 \, x^{4} - \sqrt {3} {\left (2 \, x^{4} + 1\right )} - \sqrt {2 \, x^{8} + 1} {\left (\sqrt {3} - 3\right )} + 1}{x^{4} - 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(x^4-1)/(2*x^8+1)^(1/2),x, algorithm="fricas")

[Out]

1/12*sqrt(3)*log((2*x^4 - sqrt(3)*(2*x^4 + 1) - sqrt(2*x^8 + 1)*(sqrt(3) - 3) + 1)/(x^4 - 1))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3}}{\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right ) \sqrt {2 x^{8} + 1}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/(x**4-1)/(2*x**8+1)**(1/2),x)

[Out]

Integral(x**3/((x - 1)*(x + 1)*(x**2 + 1)*sqrt(2*x**8 + 1)), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (27) = 54\).
time = 0.87, size = 70, normalized size = 2.06 \begin {gather*} \frac {1}{12} \, \sqrt {3} \log \left (-\frac {{\left | -2 \, \sqrt {2} x^{4} - 2 \, \sqrt {3} + 2 \, \sqrt {2} + 2 \, \sqrt {2 \, x^{8} + 1} \right |}}{2 \, {\left (\sqrt {2} x^{4} - \sqrt {3} - \sqrt {2} - \sqrt {2 \, x^{8} + 1}\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(x^4-1)/(2*x^8+1)^(1/2),x, algorithm="giac")

[Out]

1/12*sqrt(3)*log(-1/2*abs(-2*sqrt(2)*x^4 - 2*sqrt(3) + 2*sqrt(2) + 2*sqrt(2*x^8 + 1))/(sqrt(2)*x^4 - sqrt(3) -
sqrt(2) - sqrt(2*x^8 + 1)))

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Mupad [B]
time = 0.52, size = 35, normalized size = 1.03 \begin {gather*} -\frac {\sqrt {3}\,\left (\ln \left (x^4+\frac {\sqrt {2}\,\sqrt {3}\,\sqrt {x^8+\frac {1}{2}}}{2}+\frac {1}{2}\right )-\ln \left (x^4-1\right )\right )}{12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/((x^4 - 1)*(2*x^8 + 1)^(1/2)),x)

[Out]

-(3^(1/2)*(log(x^4 + (2^(1/2)*3^(1/2)*(x^8 + 1/2)^(1/2))/2 + 1/2) - log(x^4 - 1)))/12

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