Integrand size = 18, antiderivative size = 37 \[ \int \frac {1+x^3}{x \sqrt {-1+x^6}} \, dx=\frac {2}{3} \arctan \left (x^3+\sqrt {-1+x^6}\right )+\frac {1}{3} \log \left (x^3+\sqrt {-1+x^6}\right ) \] Output:
2/3*arctan(x^3+(x^6-1)^(1/2))+1/3*ln(x^3+(x^6-1)^(1/2))
Time = 0.09 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.11 \[ \int \frac {1+x^3}{x \sqrt {-1+x^6}} \, dx=-\frac {2}{3} \arctan \left (x^3-\sqrt {-1+x^6}\right )-\frac {1}{3} \log \left (-x^3+\sqrt {-1+x^6}\right ) \] Input:
Integrate[(1 + x^3)/(x*Sqrt[-1 + x^6]),x]
Output:
(-2*ArcTan[x^3 - Sqrt[-1 + x^6]])/3 - Log[-x^3 + Sqrt[-1 + x^6]]/3
Time = 0.19 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.51, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1396, 948, 140, 43, 103, 216}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x^3+1}{x \sqrt {x^6-1}} \, dx\) |
\(\Big \downarrow \) 1396 |
\(\displaystyle \frac {\sqrt {x^3-1} \sqrt {x^3+1} \int \frac {\sqrt {x^3+1}}{x \sqrt {x^3-1}}dx}{\sqrt {x^6-1}}\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {\sqrt {x^3-1} \sqrt {x^3+1} \int \frac {\sqrt {x^3+1}}{x^3 \sqrt {x^3-1}}dx^3}{3 \sqrt {x^6-1}}\) |
\(\Big \downarrow \) 140 |
\(\displaystyle \frac {\sqrt {x^3-1} \sqrt {x^3+1} \left (\int \frac {1}{\sqrt {x^3-1} \sqrt {x^3+1}}dx^3+\int \frac {1}{x^3 \sqrt {x^3-1} \sqrt {x^3+1}}dx^3\right )}{3 \sqrt {x^6-1}}\) |
\(\Big \downarrow \) 43 |
\(\displaystyle \frac {\sqrt {x^3-1} \sqrt {x^3+1} \left (\int \frac {1}{x^3 \sqrt {x^3-1} \sqrt {x^3+1}}dx^3+\text {arccosh}\left (x^3\right )\right )}{3 \sqrt {x^6-1}}\) |
\(\Big \downarrow \) 103 |
\(\displaystyle \frac {\sqrt {x^3-1} \sqrt {x^3+1} \left (\int \frac {1}{x^6+1}d\left (\sqrt {x^3-1} \sqrt {x^3+1}\right )+\text {arccosh}\left (x^3\right )\right )}{3 \sqrt {x^6-1}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\sqrt {x^3-1} \sqrt {x^3+1} \left (\text {arccosh}\left (x^3\right )+\arctan \left (\sqrt {x^3-1} \sqrt {x^3+1}\right )\right )}{3 \sqrt {x^6-1}}\) |
Input:
Int[(1 + x^3)/(x*Sqrt[-1 + x^6]),x]
Output:
(Sqrt[-1 + x^3]*Sqrt[1 + x^3]*(ArcCosh[x^3] + ArcTan[Sqrt[-1 + x^3]*Sqrt[1 + x^3]]))/(3*Sqrt[-1 + x^6])
Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ ArcCosh[b*(x/a)]/(b*Sqrt[d/b]), x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c + a *d, 0] && GtQ[a, 0] && GtQ[d/b, 0]
Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_ ))), x_] :> Simp[b*f Subst[Int[1/(d*(b*e - a*f)^2 + b*f^2*x^2), x], x, Sq rt[a + b*x]*Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[2*b*d *e - f*(b*c + a*d), 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[b*d^(m + n)*f^p Int[(a + b*x)^(m - 1)/(c + d*x)^m, x] , x] + Int[(a + b*x)^(m - 1)*((e + f*x)^p/(c + d*x)^m)*ExpandToSum[(a + b*x )*(c + d*x)^(-p - 1) - (b*d^(-p - 1)*f^p)/(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && EqQ[m + n + p + 1, 0] && ILtQ[p, 0] && (GtQ[m, 0] || SumSimplerQ[m, -1] || !(GtQ[n, 0] || SumSimplerQ[n, -1]))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[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]]
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_)*((d_) + (e_.)*(x_)^(n_))^(q_.), x _Symbol] :> Simp[(a + c*x^(2*n))^FracPart[p]/((d + e*x^n)^FracPart[p]*(a/d + c*(x^n/e))^FracPart[p]) Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^p, x], x] /; FreeQ[{a, c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 + a* e^2, 0] && !IntegerQ[p] && !(EqQ[q, 1] && EqQ[n, 2])
Time = 0.33 (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 | \(\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )+\sqrt {x^{6}-1}}{x^{3}}\right )}{3}-\frac {\ln \left (-x^{3}+\sqrt {x^{6}-1}\right )}{3}\) | \(45\) |
meijerg | \(\frac {\sqrt {-\operatorname {signum}\left (x^{6}-1\right )}\, \arcsin \left (x^{3}\right )}{3 \sqrt {\operatorname {signum}\left (x^{6}-1\right )}}+\frac {\sqrt {-\operatorname {signum}\left (x^{6}-1\right )}\, \left (\left (-2 \ln \left (2\right )+6 \ln \left (x \right )+i \pi \right ) \sqrt {\pi }-2 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-x^{6}+1}}{2}\right )\right )}{6 \sqrt {\pi }\, \sqrt {\operatorname {signum}\left (x^{6}-1\right )}}\) | \(86\) |
Input:
int((x^3+1)/x/(x^6-1)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/3*ln(x^3+(x^6-1)^(1/2))-1/3*arctan(1/(x^6-1)^(1/2))
Time = 0.10 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.89 \[ \int \frac {1+x^3}{x \sqrt {-1+x^6}} \, dx=\frac {2}{3} \, \arctan \left (-x^{3} + \sqrt {x^{6} - 1}\right ) - \frac {1}{3} \, \log \left (-x^{3} + \sqrt {x^{6} - 1}\right ) \] Input:
integrate((x^3+1)/x/(x^6-1)^(1/2),x, algorithm="fricas")
Output:
2/3*arctan(-x^3 + sqrt(x^6 - 1)) - 1/3*log(-x^3 + sqrt(x^6 - 1))
Time = 3.63 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.92 \[ \int \frac {1+x^3}{x \sqrt {-1+x^6}} \, dx=\frac {\begin {cases} \operatorname {acos}{\left (\frac {1}{x^{3}} \right )} & \text {for}\: x^{3} > -1 \wedge x^{3} < 1 \end {cases}}{3} + \frac {\log {\left (2 x^{3} + 2 \sqrt {x^{6} - 1} \right )}}{3} \] Input:
integrate((x**3+1)/x/(x**6-1)**(1/2),x)
Output:
Piecewise((acos(x**(-3)), (x**3 > -1) & (x**3 < 1)))/3 + log(2*x**3 + 2*sq rt(x**6 - 1))/3
Time = 0.12 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.16 \[ \int \frac {1+x^3}{x \sqrt {-1+x^6}} \, dx=\frac {1}{3} \, \arctan \left (\sqrt {x^{6} - 1}\right ) + \frac {1}{6} \, \log \left (\frac {\sqrt {x^{6} - 1}}{x^{3}} + 1\right ) - \frac {1}{6} \, \log \left (\frac {\sqrt {x^{6} - 1}}{x^{3}} - 1\right ) \] Input:
integrate((x^3+1)/x/(x^6-1)^(1/2),x, algorithm="maxima")
Output:
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)
\[ \int \frac {1+x^3}{x \sqrt {-1+x^6}} \, dx=\int { \frac {x^{3} + 1}{\sqrt {x^{6} - 1} x} \,d x } \] Input:
integrate((x^3+1)/x/(x^6-1)^(1/2),x, algorithm="giac")
Output:
integrate((x^3 + 1)/(sqrt(x^6 - 1)*x), x)
Time = 7.41 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.68 \[ \int \frac {1+x^3}{x \sqrt {-1+x^6}} \, dx=\frac {\ln \left (\sqrt {x^6-1}+x^3\right )}{3}+\frac {\mathrm {atan}\left (\sqrt {x^6-1}\right )}{3} \] Input:
int((x^3 + 1)/(x*(x^6 - 1)^(1/2)),x)
Output:
log((x^6 - 1)^(1/2) + x^3)/3 + atan((x^6 - 1)^(1/2))/3
Time = 0.16 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.65 \[ \int \frac {1+x^3}{x \sqrt {-1+x^6}} \, dx=\frac {\mathit {atan} \left (\frac {\sqrt {x^{6}-1}\, x^{6}-2 \sqrt {x^{6}-1}}{2 x^{6}-2}\right )}{6}+\frac {\mathrm {log}\left (\sqrt {x^{6}-1}+x^{3}\right )}{6}-\frac {\mathrm {log}\left (\sqrt {x^{6}-1}-x^{3}\right )}{6} \] Input:
int((x^3+1)/x/(x^6-1)^(1/2),x)
Output:
(atan((sqrt(x**6 - 1)*x**6 - 2*sqrt(x**6 - 1))/(2*x**6 - 2)) + log(sqrt(x* *6 - 1) + x**3) - log(sqrt(x**6 - 1) - x**3))/6