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

Optimal. Leaf size=35 \[ \frac {1}{3} \log \left (\sqrt {x^6-1}+x^3\right )-\frac {\sqrt {x^6-1}}{3 x^3} \]

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Rubi [A]  time = 0.01, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {275, 277, 217, 206} \begin {gather*} \frac {1}{3} \tanh ^{-1}\left (\frac {x^3}{\sqrt {x^6-1}}\right )-\frac {\sqrt {x^6-1}}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[-1 + x^6]/x^4,x]

[Out]

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

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 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 36, normalized size = 1.03 \begin {gather*} \frac {1}{3} \sqrt {x^6-1} \left (-\frac {1}{x^3}-\frac {\sin ^{-1}\left (x^3\right )}{\sqrt {1-x^6}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[-1 + x^6]/x^4,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.12, size = 35, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {-1+x^6}}{3 x^3}+\frac {1}{3} \log \left (x^3+\sqrt {-1+x^6}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[-1 + x^6]/x^4,x]

[Out]

-1/3*Sqrt[-1 + x^6]/x^3 + Log[x^3 + Sqrt[-1 + x^6]]/3

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fricas [A]  time = 0.55, size = 34, normalized size = 0.97 \begin {gather*} -\frac {x^{3} \log \left (-x^{3} + \sqrt {x^{6} - 1}\right ) + x^{3} + \sqrt {x^{6} - 1}}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.29, size = 46, normalized size = 1.31 \begin {gather*} -\frac {2 \, \sqrt {-\frac {1}{x^{6}} + 1} - \log \left (\sqrt {-\frac {1}{x^{6}} + 1} + 1\right ) + \log \left (-\sqrt {-\frac {1}{x^{6}} + 1} + 1\right )}{6 \, \mathrm {sgn}\relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.24, size = 30, normalized size = 0.86

method	result	size
trager	\(-\frac {\sqrt {x^{6}-1}}{3 x^{3}}-\frac {\ln \left (-x^{3}+\sqrt {x^{6}-1}\right )}{3}\)	\(30\)
risch	\(-\frac {\sqrt {x^{6}-1}}{3 x^{3}}+\frac {\sqrt {-\mathrm {signum}\left (x^{6}-1\right )}\, \arcsin \left (x^{3}\right )}{3 \sqrt {\mathrm {signum}\left (x^{6}-1\right )}}\)	\(38\)
meijerg	\(-\frac {i \sqrt {\mathrm {signum}\left (x^{6}-1\right )}\, \left (-\frac {4 i \sqrt {\pi }\, \sqrt {-x^{6}+1}}{x^{3}}-4 i \sqrt {\pi }\, \arcsin \left (x^{3}\right )\right )}{12 \sqrt {-\mathrm {signum}\left (x^{6}-1\right )}\, \sqrt {\pi }}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.40, size = 45, normalized size = 1.29 \begin {gather*} -\frac {\sqrt {x^{6} - 1}}{3 \, x^{3}} + \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {\sqrt {x^6-1}}{x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^6 - 1)^(1/2)/x^4,x)

[Out]

int((x^6 - 1)^(1/2)/x^4, x)

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sympy [A]  time = 1.06, size = 76, normalized size = 2.17 \begin {gather*} \begin {cases} - \frac {x^{3}}{3 \sqrt {x^{6} - 1}} + \frac {\operatorname {acosh}{\left (x^{3} \right )}}{3} + \frac {1}{3 x^{3} \sqrt {x^{6} - 1}} & \text {for}\: \left |{x^{6}}\right | > 1 \\\frac {i x^{3}}{3 \sqrt {1 - x^{6}}} - \frac {i \operatorname {asin}{\left (x^{3} \right )}}{3} - \frac {i}{3 x^{3} \sqrt {1 - x^{6}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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