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

Optimal. Leaf size=59 \[ \frac {1}{6} \sqrt {x^6-1} \left (x^3-2\right )-\frac {2}{3} \tan ^{-1}\left (\frac {x^3+1}{\sqrt {x^6-1}}\right )-\frac {1}{3} \tanh ^{-1}\left (\frac {x^3+1}{\sqrt {x^6-1}}\right ) \]

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Rubi [A]  time = 0.04, antiderivative size = 53, normalized size of antiderivative = 0.90, number of steps used = 8, number of rules used = 8, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {1475, 815, 844, 217, 206, 266, 63, 203} \begin {gather*} \frac {1}{3} \tan ^{-1}\left (\sqrt {x^6-1}\right )-\frac {1}{6} \sqrt {x^6-1} \left (2-x^3\right )-\frac {1}{6} \tanh ^{-1}\left (\frac {x^3}{\sqrt {x^6-1}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-1 + x^3)*Sqrt[-1 + x^6])/x,x]

[Out]

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

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 815

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

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

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Mathematica [A]  time = 0.06, size = 80, normalized size = 1.36 \begin {gather*} \frac {2 \sqrt {-\left (x^6-1\right )^2} \tan ^{-1}\left (\sqrt {x^6-1}\right )+\left (x^6-1\right ) \sin ^{-1}\left (x^3\right )+\sqrt {1-x^6} \left (x^9-2 x^6-x^3+2\right )}{6 \sqrt {-\left (x^6-1\right )^2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[1 - x^6]*(2 - x^3 - 2*x^6 + x^9) + (-1 + x^6)*ArcSin[x^3] + 2*Sqrt[-(-1 + x^6)^2]*ArcTan[Sqrt[-1 + x^6]]
)/(6*Sqrt[-(-1 + x^6)^2])

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IntegrateAlgebraic [A]  time = 0.16, size = 63, normalized size = 1.07 \begin {gather*} \frac {1}{6} \left (-2+x^3\right ) \sqrt {-1+x^6}-\frac {2}{3} \tan ^{-1}\left (\frac {\sqrt {-1+x^6}}{-1+x^3}\right )-\frac {1}{3} \tanh ^{-1}\left (\frac {\sqrt {-1+x^6}}{-1+x^3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.56, size = 47, normalized size = 0.80 \begin {gather*} \frac {1}{6} \, \sqrt {x^{6} - 1} {\left (x^{3} - 2\right )} + \frac {2}{3} \, \arctan \left (-x^{3} + \sqrt {x^{6} - 1}\right ) + \frac {1}{6} \, \log \left (-x^{3} + \sqrt {x^{6} - 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*sqrt(x^6 - 1)*(x^3 - 2) + 2/3*arctan(-x^3 + sqrt(x^6 - 1)) + 1/6*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 {\sqrt {x^{6} - 1} {\left (x^{3} - 1\right )}}{x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.46, size = 58, normalized size = 0.98

method	result	size
trager	\(\left (\frac {x^{3}}{6}-\frac {1}{3}\right ) \sqrt {x^{6}-1}-\frac {\ln \left (x^{3}+\sqrt {x^{6}-1}\right )}{6}+\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {\RootOf \left (\textit {\_Z}^{2}+1\right )+\sqrt {x^{6}-1}}{x^{3}}\right )}{3}\)	\(58\)
meijerg	\(\frac {\sqrt {\mathrm {signum}\left (x^{6}-1\right )}\, \left (4 \sqrt {\pi }-4 \sqrt {\pi }\, \sqrt {-x^{6}+1}+4 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-x^{6}+1}}{2}\right )-2 \left (2-2 \ln \relax (2)+6 \ln \relax (x )+i \pi \right ) \sqrt {\pi }\right )}{12 \sqrt {\pi }\, \sqrt {-\mathrm {signum}\left (x^{6}-1\right )}}+\frac {i \sqrt {\mathrm {signum}\left (x^{6}-1\right )}\, \left (-2 i \sqrt {\pi }\, x^{3} \sqrt {-x^{6}+1}-2 i \sqrt {\pi }\, \arcsin \left (x^{3}\right )\right )}{12 \sqrt {\pi }\, \sqrt {-\mathrm {signum}\left (x^{6}-1\right )}}\)	\(136\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.43, size = 77, normalized size = 1.31 \begin {gather*} -\frac {1}{3} \, \sqrt {x^{6} - 1} - \frac {\sqrt {x^{6} - 1}}{6 \, x^{3} {\left (\frac {x^{6} - 1}{x^{6}} - 1\right )}} + \frac {1}{3} \, \arctan \left (\sqrt {x^{6} - 1}\right ) - \frac {1}{12} \, \log \left (\frac {\sqrt {x^{6} - 1}}{x^{3}} + 1\right ) + \frac {1}{12} \, \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((x^3-1)*(x^6-1)^(1/2)/x,x, algorithm="maxima")

[Out]

-1/3*sqrt(x^6 - 1) - 1/6*sqrt(x^6 - 1)/(x^3*((x^6 - 1)/x^6 - 1)) + 1/3*arctan(sqrt(x^6 - 1)) - 1/12*log(sqrt(x
^6 - 1)/x^3 + 1) + 1/12*log(sqrt(x^6 - 1)/x^3 - 1)

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mupad [B]  time = 1.04, size = 54, normalized size = 0.92 \begin {gather*} \frac {x^3\,\sqrt {x^6-1}}{6}-\frac {\sqrt {x^6-1}}{3}-\frac {\ln \left (\sqrt {x^6-1}+x^3\right )}{6}+\frac {\ln \left (\frac {\sqrt {x^6-1}+1{}\mathrm {i}}{x^3}\right )\,1{}\mathrm {i}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^3 - 1)*(x^6 - 1)^(1/2))/x,x)

[Out]

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

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sympy [C]  time = 10.29, size = 150, normalized size = 2.54 \begin {gather*} \begin {cases} \frac {x^{9}}{6 \sqrt {x^{6} - 1}} - \frac {x^{3}}{6 \sqrt {x^{6} - 1}} - \frac {\operatorname {acosh}{\left (x^{3} \right )}}{6} & \text {for}\: \left |{x^{6}}\right | > 1 \\\frac {i x^{3} \sqrt {1 - x^{6}}}{6} + \frac {i \operatorname {asin}{\left (x^{3} \right )}}{6} & \text {otherwise} \end {cases} - \begin {cases} - \frac {i x^{3}}{3 \sqrt {-1 + \frac {1}{x^{6}}}} - \frac {i \operatorname {acosh}{\left (\frac {1}{x^{3}} \right )}}{3} + \frac {i}{3 x^{3} \sqrt {-1 + \frac {1}{x^{6}}}} & \text {for}\: \frac {1}{\left |{x^{6}}\right |} > 1 \\\frac {x^{3}}{3 \sqrt {1 - \frac {1}{x^{6}}}} + \frac {\operatorname {asin}{\left (\frac {1}{x^{3}} \right )}}{3} - \frac {1}{3 x^{3} \sqrt {1 - \frac {1}{x^{6}}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x**9/(6*sqrt(x**6 - 1)) - x**3/(6*sqrt(x**6 - 1)) - acosh(x**3)/6, Abs(x**6) > 1), (I*x**3*sqrt(1 -
x**6)/6 + I*asin(x**3)/6, True)) - Piecewise((-I*x**3/(3*sqrt(-1 + x**(-6))) - I*acosh(x**(-3))/3 + I/(3*x**3
*sqrt(-1 + x**(-6))), 1/Abs(x**6) > 1), (x**3/(3*sqrt(1 - 1/x**6)) + asin(x**(-3))/3 - 1/(3*x**3*sqrt(1 - 1/x*
*6)), True))

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