3.6.42 \(\int \frac {1+x^{12}}{x^4 \sqrt {-1+x^6}} \, dx\)

Optimal. Leaf size=42 \[ \frac {\sqrt {x^6-1} \left (x^6+2\right )}{6 x^3}+\frac {1}{3} \tanh ^{-1}\left (\frac {x^3+1}{\sqrt {x^6-1}}\right ) \]

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Rubi [A]  time = 0.03, antiderivative size = 51, normalized size of antiderivative = 1.21, number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {1487, 451, 275, 217, 206} \begin {gather*} \frac {1}{6} \sqrt {x^6-1} x^3+\frac {\sqrt {x^6-1}}{3 x^3}+\frac {1}{6} \tanh ^{-1}\left (\frac {x^3}{\sqrt {x^6-1}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + x^12)/(x^4*Sqrt[-1 + x^6]),x]

[Out]

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

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 451

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

Rule 1487

Int[((f_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Simp[(c^p
*(f*x)^(m + 2*n*p - n + 1)*(d + e*x^n)^(q + 1))/(e*f^(2*n*p - n + 1)*(m + 2*n*p + n*q + 1)), x] + Dist[1/(e*(m
+ 2*n*p + n*q + 1)), Int[(f*x)^m*(d + e*x^n)^q*ExpandToSum[e*(m + 2*n*p + n*q + 1)*((a + c*x^(2*n))^p - c^p*x
^(2*n*p)) - d*c^p*(m + 2*n*p - n + 1)*x^(2*n*p - n), x], x], x] /; FreeQ[{a, c, d, e, f, m, q}, x] && EqQ[n2,
2*n] && IGtQ[n, 0] && IGtQ[p, 0] && GtQ[2*n*p, n - 1] &&  !IntegerQ[q] && NeQ[m + 2*n*p + n*q + 1, 0]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 37, normalized size = 0.88 \begin {gather*} \frac {1}{6} \left (\frac {\sqrt {x^6-1} \left (x^6+2\right )}{x^3}+\tanh ^{-1}\left (\frac {x^3}{\sqrt {x^6-1}}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + x^12)/(x^4*Sqrt[-1 + x^6]),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.25, size = 42, normalized size = 1.00 \begin {gather*} \frac {\sqrt {-1+x^6} \left (2+x^6\right )}{6 x^3}+\frac {1}{3} \tanh ^{-1}\left (\frac {1+x^3}{\sqrt {-1+x^6}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(1 + x^12)/(x^4*Sqrt[-1 + x^6]),x]

[Out]

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

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fricas [A]  time = 0.53, size = 43, normalized size = 1.02 \begin {gather*} -\frac {x^{3} \log \left (-x^{3} + \sqrt {x^{6} - 1}\right ) - 2 \, x^{3} - {\left (x^{6} + 2\right )} \sqrt {x^{6} - 1}}{6 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^12+1)/x^4/(x^6-1)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{12} + 1}{\sqrt {x^{6} - 1} x^{4}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^12+1)/x^4/(x^6-1)^(1/2),x, algorithm="giac")

[Out]

integrate((x^12 + 1)/(sqrt(x^6 - 1)*x^4), x)

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maple [A]  time = 0.23, size = 33, normalized size = 0.79

method	result	size
trager	\(\frac {\sqrt {x^{6}-1}\, \left (x^{6}+2\right )}{6 x^{3}}+\frac {\ln \left (x^{3}+\sqrt {x^{6}-1}\right )}{6}\)	\(33\)
risch	\(\frac {x^{12}+x^{6}-2}{6 x^{3} \sqrt {x^{6}-1}}+\frac {\sqrt {-\mathrm {signum}\left (x^{6}-1\right )}\, \arcsin \left (x^{3}\right )}{6 \sqrt {\mathrm {signum}\left (x^{6}-1\right )}}\)	\(46\)
meijerg	\(\frac {i \sqrt {-\mathrm {signum}\left (x^{6}-1\right )}\, \left (i \sqrt {\pi }\, x^{3} \sqrt {-x^{6}+1}-i \sqrt {\pi }\, \arcsin \left (x^{3}\right )\right )}{6 \sqrt {\mathrm {signum}\left (x^{6}-1\right )}\, \sqrt {\pi }}-\frac {\sqrt {-\mathrm {signum}\left (x^{6}-1\right )}\, \sqrt {-x^{6}+1}}{3 \sqrt {\mathrm {signum}\left (x^{6}-1\right )}\, x^{3}}\)	\(87\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^12+1)/x^4/(x^6-1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.54, size = 70, normalized size = 1.67 \begin {gather*} \frac {\sqrt {x^{6} - 1}}{3 \, x^{3}} - \frac {\sqrt {x^{6} - 1}}{6 \, x^{3} {\left (\frac {x^{6} - 1}{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^12+1)/x^4/(x^6-1)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x^{12}+1}{x^4\,\sqrt {x^6-1}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^12 + 1)/(x^4*(x^6 - 1)^(1/2)),x)

[Out]

int((x^12 + 1)/(x^4*(x^6 - 1)^(1/2)), x)

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sympy [C]  time = 3.54, size = 92, normalized size = 2.19 \begin {gather*} \begin {cases} \frac {i \sqrt {-1 + \frac {1}{x^{6}}}}{3} & \text {for}\: \frac {1}{\left |{x^{6}}\right |} > 1 \\\frac {\sqrt {1 - \frac {1}{x^{6}}}}{3} & \text {otherwise} \end {cases} + \begin {cases} \frac {x^{3} \sqrt {x^{6} - 1}}{6} + \frac {\operatorname {acosh}{\left (x^{3} \right )}}{6} & \text {for}\: \left |{x^{6}}\right | > 1 \\- \frac {i x^{9}}{6 \sqrt {1 - x^{6}}} + \frac {i x^{3}}{6 \sqrt {1 - x^{6}}} - \frac {i \operatorname {asin}{\left (x^{3} \right )}}{6} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**12+1)/x**4/(x**6-1)**(1/2),x)

[Out]

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

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