3.12.45 \(\int \frac {\sqrt {-1+x^6} (-1+2 x^6)^2}{x^4 (-1+4 x^6)} \, dx\)

Optimal. Leaf size=85 \[ \frac {\sqrt {x^6-1} \left (x^6+2\right )}{6 x^3}-\frac {5}{12} \log \left (\sqrt {x^6-1}+x^3\right )+\frac {\tan ^{-1}\left (-\frac {4 x^6}{\sqrt {3}}-\frac {4 \sqrt {x^6-1} x^3}{\sqrt {3}}+\frac {1}{\sqrt {3}}\right )}{4 \sqrt {3}} \]

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Rubi [A]  time = 0.23, antiderivative size = 79, normalized size of antiderivative = 0.93, number of steps used = 14, number of rules used = 9, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, Rules used = {575, 586, 580, 523, 217, 206, 377, 204, 528} \begin {gather*} \frac {1}{6} \sqrt {x^6-1} x^3+\frac {\sqrt {x^6-1}}{3 x^3}+\frac {\tan ^{-1}\left (\frac {\sqrt {3} x^3}{\sqrt {x^6-1}}\right )}{4 \sqrt {3}}-\frac {5}{12} \tanh ^{-1}\left (\frac {x^3}{\sqrt {x^6-1}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sqrt[-1 + x^6]*(-1 + 2*x^6)^2)/(x^4*(-1 + 4*x^6)),x]

[Out]

Sqrt[-1 + x^6]/(3*x^3) + (x^3*Sqrt[-1 + x^6])/6 + ArcTan[(Sqrt[3]*x^3)/Sqrt[-1 + x^6]]/(4*Sqrt[3]) - (5*ArcTan
h[x^3/Sqrt[-1 + x^6]])/12

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[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 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 523

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rule 528

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[
(f*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(b*(n*(p + q + 1) + 1)), x] + Dist[1/(b*(n*(p + q + 1) + 1)), Int[(a +
b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*(b*e - a*f + b*e*n*(p + q + 1)) + (d*(b*e - a*f) + f*n*q*(b*c - a*d) + b*
d*e*n*(p + q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && GtQ[q, 0] && NeQ[n*(p + q + 1) + 1
, 0]

Rule 575

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_))^(r_.), x
_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p*(c + d*x^(n/k))^
q*(e + f*x^(n/k))^r, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, c, d, e, f, p, q, r}, x] && IGtQ[n, 0] && Inte
gerQ[m]

Rule 580

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(a*g*(m + 1)), x] - Dist[1/(a*g^n*(m + 1
)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*(b*e - a*f)*(m + 1) + e*n*(b*c*(p + 1) + a*d*q)
+ d*((b*e - a*f)*(m + 1) + b*e*n*(p + q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && IGtQ[n
, 0] && GtQ[q, 0] && LtQ[m, -1] &&  !(EqQ[q, 1] && SimplerQ[e + f*x^n, c + d*x^n])

Rule 586

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_))^
(r_.), x_Symbol] :> Dist[e, Int[(g*x)^m*(a + b*x^n)^p*(c + d*x^n)^q*(e + f*x^n)^(r - 1), x], x] + Dist[f/e^n,
Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*(e + f*x^n)^(r - 1), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p,
q}, x] && IGtQ[n, 0] && IGtQ[r, 0]

Rubi steps

\begin {align*} \int \frac {\sqrt {-1+x^6} \left (-1+2 x^6\right )^2}{x^4 \left (-1+4 x^6\right )} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {\sqrt {-1+x^2} \left (-1+2 x^2\right )^2}{x^2 \left (-1+4 x^2\right )} \, dx,x,x^3\right )\\ &=-\left (\frac {1}{3} \operatorname {Subst}\left (\int \frac {\sqrt {-1+x^2} \left (-1+2 x^2\right )}{x^2 \left (-1+4 x^2\right )} \, dx,x,x^3\right )\right )+\frac {2}{3} \operatorname {Subst}\left (\int \frac {\sqrt {-1+x^2} \left (-1+2 x^2\right )}{-1+4 x^2} \, dx,x,x^3\right )\\ &=\frac {\sqrt {-1+x^6}}{3 x^3}+\frac {1}{6} x^3 \sqrt {-1+x^6}+\frac {1}{12} \operatorname {Subst}\left (\int \frac {6-12 x^2}{\sqrt {-1+x^2} \left (-1+4 x^2\right )} \, dx,x,x^3\right )+\frac {1}{3} \operatorname {Subst}\left (\int \frac {-1-2 x^2}{\sqrt {-1+x^2} \left (-1+4 x^2\right )} \, dx,x,x^3\right )\\ &=\frac {\sqrt {-1+x^6}}{3 x^3}+\frac {1}{6} x^3 \sqrt {-1+x^6}-\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{\sqrt {-1+x^2}} \, dx,x,x^3\right )-\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{\sqrt {-1+x^2}} \, dx,x,x^3\right )+\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{\sqrt {-1+x^2} \left (-1+4 x^2\right )} \, dx,x,x^3\right )-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{\sqrt {-1+x^2} \left (-1+4 x^2\right )} \, dx,x,x^3\right )\\ &=\frac {\sqrt {-1+x^6}}{3 x^3}+\frac {1}{6} x^3 \sqrt {-1+x^6}-\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x^3}{\sqrt {-1+x^6}}\right )+\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{-1-3 x^2} \, dx,x,\frac {x^3}{\sqrt {-1+x^6}}\right )-\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x^3}{\sqrt {-1+x^6}}\right )-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{-1-3 x^2} \, dx,x,\frac {x^3}{\sqrt {-1+x^6}}\right )\\ &=\frac {\sqrt {-1+x^6}}{3 x^3}+\frac {1}{6} x^3 \sqrt {-1+x^6}+\frac {\tan ^{-1}\left (\frac {\sqrt {3} x^3}{\sqrt {-1+x^6}}\right )}{4 \sqrt {3}}-\frac {5}{12} \tanh ^{-1}\left (\frac {x^3}{\sqrt {-1+x^6}}\right )\\ \end {align*}

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Mathematica [B]  time = 0.18, size = 201, normalized size = 2.36 \begin {gather*} \frac {4 \sqrt {1-x^6} x^6-8 \sqrt {1-x^6}+4 \sqrt {1-x^6} x^{12}+12 \left (x^6-1\right ) x^3 \sin ^{-1}\left (x^3\right )+\sqrt {3} \sqrt {-\left (x^6-1\right )^2} x^3 \tan ^{-1}\left (\frac {2-x^3}{\sqrt {3} \sqrt {x^6-1}}\right )-\sqrt {3} \sqrt {-\left (x^6-1\right )^2} x^3 \tan ^{-1}\left (\frac {x^3+2}{\sqrt {3} \sqrt {x^6-1}}\right )+2 \sqrt {-\left (x^6-1\right )^2} x^3 \tanh ^{-1}\left (\frac {x^3}{\sqrt {x^6-1}}\right )}{24 x^3 \sqrt {-\left (x^6-1\right )^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[-1 + x^6]*(-1 + 2*x^6)^2)/(x^4*(-1 + 4*x^6)),x]

[Out]

(-8*Sqrt[1 - x^6] + 4*x^6*Sqrt[1 - x^6] + 4*x^12*Sqrt[1 - x^6] + 12*x^3*(-1 + x^6)*ArcSin[x^3] + Sqrt[3]*x^3*S
qrt[-(-1 + x^6)^2]*ArcTan[(2 - x^3)/(Sqrt[3]*Sqrt[-1 + x^6])] - Sqrt[3]*x^3*Sqrt[-(-1 + x^6)^2]*ArcTan[(2 + x^
3)/(Sqrt[3]*Sqrt[-1 + x^6])] + 2*x^3*Sqrt[-(-1 + x^6)^2]*ArcTanh[x^3/Sqrt[-1 + x^6]])/(24*x^3*Sqrt[-(-1 + x^6)
^2])

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IntegrateAlgebraic [A]  time = 0.19, size = 87, normalized size = 1.02 \begin {gather*} \frac {\sqrt {-1+x^6} \left (2+x^6\right )}{6 x^3}-\frac {\tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {4 x^6}{\sqrt {3}}+\frac {4 x^3 \sqrt {-1+x^6}}{\sqrt {3}}\right )}{4 \sqrt {3}}+\frac {5}{12} \log \left (-x^3+\sqrt {-1+x^6}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(Sqrt[-1 + x^6]*(-1 + 2*x^6)^2)/(x^4*(-1 + 4*x^6)),x]

[Out]

(Sqrt[-1 + x^6]*(2 + x^6))/(6*x^3) - ArcTan[1/Sqrt[3] - (4*x^6)/Sqrt[3] + (4*x^3*Sqrt[-1 + x^6])/Sqrt[3]]/(4*S
qrt[3]) + (5*Log[-x^3 + Sqrt[-1 + x^6]])/12

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fricas [A]  time = 0.49, size = 80, normalized size = 0.94 \begin {gather*} -\frac {\sqrt {3} x^{3} \arctan \left (\frac {4}{3} \, \sqrt {3} \sqrt {x^{6} - 1} x^{3} - \frac {1}{3} \, \sqrt {3} {\left (4 \, x^{6} - 1\right )}\right ) - 5 \, x^{3} \log \left (-x^{3} + \sqrt {x^{6} - 1}\right ) - 4 \, x^{3} - 2 \, {\left (x^{6} + 2\right )} \sqrt {x^{6} - 1}}{12 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^6-1)^(1/2)*(2*x^6-1)^2/x^4/(4*x^6-1),x, algorithm="fricas")

[Out]

-1/12*(sqrt(3)*x^3*arctan(4/3*sqrt(3)*sqrt(x^6 - 1)*x^3 - 1/3*sqrt(3)*(4*x^6 - 1)) - 5*x^3*log(-x^3 + sqrt(x^6
- 1)) - 4*x^3 - 2*(x^6 + 2)*sqrt(x^6 - 1))/x^3

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^6-1)^(1/2)*(2*x^6-1)^2/x^4/(4*x^6-1),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:root
of minimal polynomial must be unitary Error: Bad Argument Valuerootof minimal polynomial must be unitary Error
: Bad Argument Valuerootof minimal polynomial must be unitary Error: Bad Argument Valuerootof minimal polynomi
al must be unitary Error: Bad Argument Valuerootof minimal polynomial must be unitary Error: Bad Argument Valu
erootof minimal polynomial must be unitary Error: Bad Argument Valuerootof minimal polynomial must be unitary
Error: Bad Argument Valuerootof minimal polynomial must be unitary Error: Bad Argument Valuerootof minimal pol
ynomial must be unitary Error: Bad Argument Valuerootof minimal polynomial must be unitary Error: Bad Argument
Valuerootof minimal polynomial must be unitary Error: Bad Argument Valuerootof minimal polynomial must be uni
tary Error: Bad Argument ValueDone

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maple [C]  time = 1.16, size = 96, normalized size = 1.13

method	result	size
trager	\(\frac {\sqrt {x^{6}-1}\, \left (x^{6}+2\right )}{6 x^{3}}+\frac {5 \ln \left (-x^{3}+\sqrt {x^{6}-1}\right )}{12}+\frac {\RootOf \left (\textit {\_Z}^{2}+3\right ) \ln \left (-\frac {-2 \RootOf \left (\textit {\_Z}^{2}+3\right ) x^{6}+6 x^{3} \sqrt {x^{6}-1}-\RootOf \left (\textit {\_Z}^{2}+3\right )}{\left (2 x^{3}-1\right ) \left (2 x^{3}+1\right )}\right )}{24}\)	\(96\)
risch	\(\frac {x^{12}+x^{6}-2}{6 x^{3} \sqrt {x^{6}-1}}+\frac {5 \ln \left (x^{3}-\sqrt {x^{6}-1}\right )}{12}-\frac {\RootOf \left (\textit {\_Z}^{2}+3\right ) \ln \left (-\frac {2 \RootOf \left (\textit {\_Z}^{2}+3\right ) x^{6}+6 x^{3} \sqrt {x^{6}-1}+\RootOf \left (\textit {\_Z}^{2}+3\right )}{\left (2 x^{3}-1\right ) \left (2 x^{3}+1\right )}\right )}{24}\)	\(97\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^6-1)^(1/2)*(2*x^6-1)^2/x^4/(4*x^6-1),x,method=_RETURNVERBOSE)

[Out]

1/6*(x^6-1)^(1/2)*(x^6+2)/x^3+5/12*ln(-x^3+(x^6-1)^(1/2))+1/24*RootOf(_Z^2+3)*ln(-(-2*RootOf(_Z^2+3)*x^6+6*x^3
*(x^6-1)^(1/2)-RootOf(_Z^2+3))/(2*x^3-1)/(2*x^3+1))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (2 \, x^{6} - 1\right )}^{2} \sqrt {x^{6} - 1}}{{\left (4 \, x^{6} - 1\right )} x^{4}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^6-1)^(1/2)*(2*x^6-1)^2/x^4/(4*x^6-1),x, algorithm="maxima")

[Out]

integrate((2*x^6 - 1)^2*sqrt(x^6 - 1)/((4*x^6 - 1)*x^4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {x^6-1}\,{\left (2\,x^6-1\right )}^2}{x^4\,\left (4\,x^6-1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^6 - 1)^(1/2)*(2*x^6 - 1)^2)/(x^4*(4*x^6 - 1)),x)

[Out]

int(((x^6 - 1)^(1/2)*(2*x^6 - 1)^2)/(x^4*(4*x^6 - 1)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} - x + 1\right ) \left (x^{2} + x + 1\right )} \left (2 x^{6} - 1\right )^{2}}{x^{4} \left (2 x^{3} - 1\right ) \left (2 x^{3} + 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**6-1)**(1/2)*(2*x**6-1)**2/x**4/(4*x**6-1),x)

[Out]

Integral(sqrt((x - 1)*(x + 1)*(x**2 - x + 1)*(x**2 + x + 1))*(2*x**6 - 1)**2/(x**4*(2*x**3 - 1)*(2*x**3 + 1)),
x)

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