3.308 \(\int \frac {x^3}{(-1+x^4) \sqrt {1+2 x^8}} \, dx\)

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

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Rubi [A]  time = 0.03, 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 = {1469, 725, 206} \[ -\frac {\tanh ^{-1}\left (\frac {2 x^4+1}{\sqrt {3} \sqrt {2 x^8+1}}\right )}{4 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 206

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

Rule 725

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 1469

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} \operatorname {Subst}\left (\int \frac {1}{(-1+x) \sqrt {1+2 x^2}} \, dx,x,x^4\right )\\ &=-\left (\frac {1}{4} \operatorname {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 = 0.01, size = 29, normalized size = 0.85 \[ -\frac {\tanh ^{-1}\left (\frac {2 x^4+1}{\sqrt {6 x^8+3}}\right )}{4 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.42, size = 48, normalized size = 1.41 \[ \frac {\tanh ^{-1}\left (-\frac {\sqrt {2 x^8+1}}{\sqrt {3}}-\sqrt {\frac {2}{3}} x^4+\sqrt {\frac {2}{3}}\right )}{2 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.61, size = 49, normalized size = 1.44 \[ \frac {1}{12} \, \sqrt {3} \log \left (\frac {2 \, x^{4} - \sqrt {3} {\left (2 \, x^{4} + 1\right )} - \sqrt {2 \, x^{8} + 1} {\left (\sqrt {3} - 3\right )} + 1}{x^{4} - 1}\right ) \]

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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giac [B]  time = 0.66, size = 70, normalized size = 2.06 \[ \frac {1}{12} \, \sqrt {3} \log \left (-\frac {{\left | -2 \, \sqrt {2} x^{4} - 2 \, \sqrt {3} + 2 \, \sqrt {2} + 2 \, \sqrt {2 \, x^{8} + 1} \right |}}{2 \, {\left (\sqrt {2} x^{4} - \sqrt {3} - \sqrt {2} - \sqrt {2 \, x^{8} + 1}\right )}}\right ) \]

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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maple [C]  time = 0.30, size = 58, normalized size = 1.71

method	result	size
trager	\(-\frac {\RootOf \left (\textit {\_Z}^{2}-3\right ) \ln \left (-\frac {2 \RootOf \left (\textit {\_Z}^{2}-3\right ) x^{4}+\RootOf \left (\textit {\_Z}^{2}-3\right )+3 \sqrt {2 x^{8}+1}}{\left (-1+x \right ) \left (1+x \right ) \left (x^{2}+1\right )}\right )}{12}\)	\(58\)

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\sqrt {2 \, x^{8} + 1} {\left (x^{4} - 1\right )}}\,{d x} \]

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

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

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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