\(\int \frac {1+x^3}{x \sqrt {-1+x^6}} \, dx\) [464]

Optimal result

Integrand size = 18, antiderivative size = 37 \[ \int \frac {1+x^3}{x \sqrt {-1+x^6}} \, dx=\frac {2}{3} \arctan \left (x^3+\sqrt {-1+x^6}\right )+\frac {1}{3} \log \left (x^3+\sqrt {-1+x^6}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.89, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {1489, 858, 223, 212, 272, 65, 209} \[ \int \frac {1+x^3}{x \sqrt {-1+x^6}} \, dx=\frac {1}{3} \arctan \left (\sqrt {x^6-1}\right )+\frac {1}{3} \text {arctanh}\left (\frac {x^3}{\sqrt {x^6-1}}\right ) \]

[In]

Int[(1 + x^3)/(x*Sqrt[-1 + x^6]),x]

[Out]

ArcTan[Sqrt[-1 + x^6]]/3 + ArcTanh[x^3/Sqrt[-1 + x^6]]/3

Rule 65

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 209

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

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 223

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 272

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 858

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 1489

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*} \text {integral}& = \frac {1}{3} \text {Subst}\left (\int \frac {1+x}{x \sqrt {-1+x^2}} \, dx,x,x^3\right ) \\ & = \frac {1}{3} \text {Subst}\left (\int \frac {1}{\sqrt {-1+x^2}} \, dx,x,x^3\right )+\frac {1}{3} \text {Subst}\left (\int \frac {1}{x \sqrt {-1+x^2}} \, dx,x,x^3\right ) \\ & = \frac {1}{6} \text {Subst}\left (\int \frac {1}{\sqrt {-1+x} x} \, dx,x,x^6\right )+\frac {1}{3} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x^3}{\sqrt {-1+x^6}}\right ) \\ & = \frac {1}{3} \text {arctanh}\left (\frac {x^3}{\sqrt {-1+x^6}}\right )+\frac {1}{3} \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt {-1+x^6}\right ) \\ & = \frac {1}{3} \arctan \left (\sqrt {-1+x^6}\right )+\frac {1}{3} \text {arctanh}\left (\frac {x^3}{\sqrt {-1+x^6}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.11 \[ \int \frac {1+x^3}{x \sqrt {-1+x^6}} \, dx=-\frac {2}{3} \arctan \left (x^3-\sqrt {-1+x^6}\right )-\frac {1}{3} \log \left (-x^3+\sqrt {-1+x^6}\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.95 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.70

method	result	size
pseudoelliptic	\(\frac {\ln \left (x^{3}+\sqrt {x^{6}-1}\right )}{3}-\frac {\arctan \left (\frac {1}{\sqrt {x^{6}-1}}\right )}{3}\)	\(26\)
trager	\(\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )+\sqrt {x^{6}-1}}{x^{3}}\right )}{3}+\frac {\ln \left (x^{3}+\sqrt {x^{6}-1}\right )}{3}\)	\(43\)
meijerg	\(\frac {\sqrt {-\operatorname {signum}\left (x^{6}-1\right )}\, \arcsin \left (x^{3}\right )}{3 \sqrt {\operatorname {signum}\left (x^{6}-1\right )}}+\frac {\sqrt {-\operatorname {signum}\left (x^{6}-1\right )}\, \left (-2 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-x^{6}+1}}{2}\right )+\left (-2 \ln \left (2\right )+6 \ln \left (x \right )+i \pi \right ) \sqrt {\pi }\right )}{6 \sqrt {\pi }\, \sqrt {\operatorname {signum}\left (x^{6}-1\right )}}\)	\(86\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.89 \[ \int \frac {1+x^3}{x \sqrt {-1+x^6}} \, dx=\frac {2}{3} \, \arctan \left (-x^{3} + \sqrt {x^{6} - 1}\right ) - \frac {1}{3} \, \log \left (-x^{3} + \sqrt {x^{6} - 1}\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 3.63 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.92 \[ \int \frac {1+x^3}{x \sqrt {-1+x^6}} \, dx=\frac {\begin {cases} \operatorname {acos}{\left (\frac {1}{x^{3}} \right )} & \text {for}\: x^{3} > -1 \wedge x^{3} < 1 \end {cases}}{3} + \frac {\log {\left (2 x^{3} + 2 \sqrt {x^{6} - 1} \right )}}{3} \]

[In]

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

[Out]

Piecewise((acos(x**(-3)), (x**3 > -1) & (x**3 < 1)))/3 + log(2*x**3 + 2*sqrt(x**6 - 1))/3

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.16 \[ \int \frac {1+x^3}{x \sqrt {-1+x^6}} \, dx=\frac {1}{3} \, \arctan \left (\sqrt {x^{6} - 1}\right ) + \frac {1}{6} \, \log \left (\frac {\sqrt {x^{6} - 1}}{x^{3}} + 1\right ) - \frac {1}{6} \, \log \left (\frac {\sqrt {x^{6} - 1}}{x^{3}} - 1\right ) \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {1+x^3}{x \sqrt {-1+x^6}} \, dx=\int { \frac {x^{3} + 1}{\sqrt {x^{6} - 1} x} \,d x } \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.99 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.68 \[ \int \frac {1+x^3}{x \sqrt {-1+x^6}} \, dx=\frac {\ln \left (\sqrt {x^6-1}+x^3\right )}{3}+\frac {\mathrm {atan}\left (\sqrt {x^6-1}\right )}{3} \]

[In]

int((x^3 + 1)/(x*(x^6 - 1)^(1/2)),x)

[Out]

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