3.6.54 \(\int \frac {x^{14}}{\sqrt {1+x^6}} \, dx\)

Optimal. Leaf size=43 \[ \frac {1}{8} \log \left (\sqrt {x^6+1}+x^3\right )+\frac {1}{24} \sqrt {x^6+1} \left (2 x^9-3 x^3\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 41, normalized size of antiderivative = 0.95, number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {275, 321, 215} \begin {gather*} \frac {1}{8} \sinh ^{-1}\left (x^3\right )+\frac {1}{12} \sqrt {x^6+1} x^9-\frac {1}{8} \sqrt {x^6+1} x^3 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^14/Sqrt[1 + x^6],x]

[Out]

-1/8*(x^3*Sqrt[1 + x^6]) + (x^9*Sqrt[1 + x^6])/12 + ArcSinh[x^3]/8

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

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 321

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

Rubi steps

\begin {align*} \int \frac {x^{14}}{\sqrt {1+x^6}} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {x^4}{\sqrt {1+x^2}} \, dx,x,x^3\right )\\ &=\frac {1}{12} x^9 \sqrt {1+x^6}-\frac {1}{4} \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1+x^2}} \, dx,x,x^3\right )\\ &=-\frac {1}{8} x^3 \sqrt {1+x^6}+\frac {1}{12} x^9 \sqrt {1+x^6}+\frac {1}{8} \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+x^2}} \, dx,x,x^3\right )\\ &=-\frac {1}{8} x^3 \sqrt {1+x^6}+\frac {1}{12} x^9 \sqrt {1+x^6}+\frac {1}{8} \sinh ^{-1}\left (x^3\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 31, normalized size = 0.72 \begin {gather*} \frac {1}{24} \left (3 \sinh ^{-1}\left (x^3\right )+\sqrt {x^6+1} \left (2 x^6-3\right ) x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^14/Sqrt[1 + x^6],x]

[Out]

(x^3*Sqrt[1 + x^6]*(-3 + 2*x^6) + 3*ArcSinh[x^3])/24

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IntegrateAlgebraic [A]  time = 0.16, size = 43, normalized size = 1.00 \begin {gather*} \frac {1}{24} \sqrt {1+x^6} \left (-3 x^3+2 x^9\right )+\frac {1}{8} \log \left (x^3+\sqrt {1+x^6}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^14/Sqrt[1 + x^6],x]

[Out]

(Sqrt[1 + x^6]*(-3*x^3 + 2*x^9))/24 + Log[x^3 + Sqrt[1 + x^6]]/8

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fricas [A]  time = 0.44, size = 37, normalized size = 0.86 \begin {gather*} \frac {1}{24} \, {\left (2 \, x^{9} - 3 \, x^{3}\right )} \sqrt {x^{6} + 1} - \frac {1}{8} \, \log \left (-x^{3} + \sqrt {x^{6} + 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^14/(x^6+1)^(1/2),x, algorithm="fricas")

[Out]

1/24*(2*x^9 - 3*x^3)*sqrt(x^6 + 1) - 1/8*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 {x^{14}}{\sqrt {x^{6} + 1}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^14/(x^6+1)^(1/2),x, algorithm="giac")

[Out]

integrate(x^14/sqrt(x^6 + 1), x)

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maple [A]  time = 0.16, size = 27, normalized size = 0.63

method	result	size
risch	\(\frac {x^{3} \left (2 x^{6}-3\right ) \sqrt {x^{6}+1}}{24}+\frac {\arcsinh \left (x^{3}\right )}{8}\)	\(27\)
trager	\(\frac {x^{3} \left (2 x^{6}-3\right ) \sqrt {x^{6}+1}}{24}-\frac {\ln \left (x^{3}-\sqrt {x^{6}+1}\right )}{8}\)	\(37\)
meijerg	\(\frac {-\frac {\sqrt {\pi }\, x^{3} \left (-10 x^{6}+15\right ) \sqrt {x^{6}+1}}{20}+\frac {3 \sqrt {\pi }\, \arcsinh \left (x^{3}\right )}{4}}{6 \sqrt {\pi }}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^14/(x^6+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/24*x^3*(2*x^6-3)*(x^6+1)^(1/2)+1/8*arcsinh(x^3)

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maxima [B]  time = 0.32, size = 86, normalized size = 2.00 \begin {gather*} -\frac {\frac {5 \, \sqrt {x^{6} + 1}}{x^{3}} - \frac {3 \, {\left (x^{6} + 1\right )}^{\frac {3}{2}}}{x^{9}}}{24 \, {\left (\frac {2 \, {\left (x^{6} + 1\right )}}{x^{6}} - \frac {{\left (x^{6} + 1\right )}^{2}}{x^{12}} - 1\right )}} + \frac {1}{16} \, \log \left (\frac {\sqrt {x^{6} + 1}}{x^{3}} + 1\right ) - \frac {1}{16} \, \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^14/(x^6+1)^(1/2),x, algorithm="maxima")

[Out]

-1/24*(5*sqrt(x^6 + 1)/x^3 - 3*(x^6 + 1)^(3/2)/x^9)/(2*(x^6 + 1)/x^6 - (x^6 + 1)^2/x^12 - 1) + 1/16*log(sqrt(x
^6 + 1)/x^3 + 1) - 1/16*log(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^{14}}{\sqrt {x^6+1}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^14/(x^6 + 1)^(1/2),x)

[Out]

int(x^14/(x^6 + 1)^(1/2), x)

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sympy [A]  time = 2.74, size = 46, normalized size = 1.07 \begin {gather*} \frac {x^{15}}{12 \sqrt {x^{6} + 1}} - \frac {x^{9}}{24 \sqrt {x^{6} + 1}} - \frac {x^{3}}{8 \sqrt {x^{6} + 1}} + \frac {\operatorname {asinh}{\left (x^{3} \right )}}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**14/(x**6+1)**(1/2),x)

[Out]

x**15/(12*sqrt(x**6 + 1)) - x**9/(24*sqrt(x**6 + 1)) - x**3/(8*sqrt(x**6 + 1)) + asinh(x**3)/8

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