3.5.66 \(\int \frac {1+2 x^3}{x \sqrt {-1+x^6}} \, dx\)

Optimal. Leaf size=37 \[ \frac {2}{3} \log \left (\sqrt {x^6-1}+x^3\right )+\frac {2}{3} \tan ^{-1}\left (\sqrt {x^6-1}+x^3\right ) \]

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Rubi [A]  time = 0.03, antiderivative size = 33, normalized size of antiderivative = 0.89, number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {1475, 844, 217, 206, 266, 63, 203} \begin {gather*} \frac {1}{3} \tan ^{-1}\left (\sqrt {x^6-1}\right )+\frac {2}{3} \tanh ^{-1}\left (\frac {x^3}{\sqrt {x^6-1}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + 2*x^3)/(x*Sqrt[-1 + x^6]),x]

[Out]

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

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 203

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

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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rule 1475

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(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] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 31, normalized size = 0.84 \begin {gather*} \frac {1}{3} \left (\tan ^{-1}\left (\sqrt {x^6-1}\right )+2 \tanh ^{-1}\left (\frac {x^3}{\sqrt {x^6-1}}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + 2*x^3)/(x*Sqrt[-1 + x^6]),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.13, size = 41, normalized size = 1.11 \begin {gather*} -\frac {2}{3} \tan ^{-1}\left (x^3-\sqrt {-1+x^6}\right )-\frac {2}{3} \log \left (-x^3+\sqrt {-1+x^6}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(1 + 2*x^3)/(x*Sqrt[-1 + x^6]),x]

[Out]

(-2*ArcTan[x^3 - Sqrt[-1 + x^6]])/3 - (2*Log[-x^3 + Sqrt[-1 + x^6]])/3

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fricas [A]  time = 0.48, size = 33, normalized size = 0.89 \begin {gather*} \frac {2}{3} \, \arctan \left (-x^{3} + \sqrt {x^{6} - 1}\right ) - \frac {2}{3} \, \log \left (-x^{3} + \sqrt {x^{6} - 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, x^{3} + 1}{\sqrt {x^{6} - 1} x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((2*x^3 + 1)/(sqrt(x^6 - 1)*x), x)

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maple [C]  time = 0.44, size = 43, normalized size = 1.16

method	result	size
trager	\(\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {\RootOf \left (\textit {\_Z}^{2}+1\right )+\sqrt {x^{6}-1}}{x^{3}}\right )}{3}+\frac {2 \ln \left (x^{3}+\sqrt {x^{6}-1}\right )}{3}\)	\(43\)
meijerg	\(\frac {2 \sqrt {-\mathrm {signum}\left (x^{6}-1\right )}\, \arcsin \left (x^{3}\right )}{3 \sqrt {\mathrm {signum}\left (x^{6}-1\right )}}+\frac {\sqrt {-\mathrm {signum}\left (x^{6}-1\right )}\, \left (\left (-2 \ln \relax (2)+6 \ln \relax (x )+i \pi \right ) \sqrt {\pi }-2 \ln \left (\frac {1}{2}+\frac {\sqrt {-x^{6}+1}}{2}\right ) \sqrt {\pi }\right )}{6 \sqrt {\mathrm {signum}\left (x^{6}-1\right )}\, \sqrt {\pi }}\)	\(86\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*RootOf(_Z^2+1)*ln((RootOf(_Z^2+1)+(x^6-1)^(1/2))/x^3)+2/3*ln(x^3+(x^6-1)^(1/2))

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maxima [A]  time = 0.92, size = 43, normalized size = 1.16 \begin {gather*} \frac {1}{3} \, \arctan \left (\sqrt {x^{6} - 1}\right ) + \frac {1}{3} \, \log \left (\frac {\sqrt {x^{6} - 1}}{x^{3}} + 1\right ) - \frac {1}{3} \, \log \left (\frac {\sqrt {x^{6} - 1}}{x^{3}} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.29, size = 25, normalized size = 0.68 \begin {gather*} \frac {2\,\ln \left (\sqrt {x^6-1}+x^3\right )}{3}+\frac {\mathrm {atan}\left (\sqrt {x^6-1}\right )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^3 + 1)/(x*(x^6 - 1)^(1/2)),x)

[Out]

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

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sympy [A]  time = 7.05, size = 20, normalized size = 0.54 \begin {gather*} \frac {\begin {cases} \operatorname {acos}{\left (\frac {1}{x^{3}} \right )} & \text {for}\: x > -1 \wedge x < 1 \end {cases}}{3} + \frac {2 \operatorname {acosh}{\left (x^{3} \right )}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((acos(x**(-3)), (x > -1) & (x < 1)))/3 + 2*acosh(x**3)/3

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