3.15.15 \(\int \frac {x^{14}}{(2+x^6)^{3/2}} \, dx\) [1415]

Optimal. Leaf size=45 \[ -\frac {x^9}{3 \sqrt {2+x^6}}+\frac {1}{2} x^3 \sqrt {2+x^6}-\sinh ^{-1}\left (\frac {x^3}{\sqrt {2}}\right ) \]

[Out]

-arcsinh(1/2*x^3*2^(1/2))-1/3*x^9/(x^6+2)^(1/2)+1/2*x^3*(x^6+2)^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 45, 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 = {281, 294, 327, 221} \begin {gather*} -\sinh ^{-1}\left (\frac {x^3}{\sqrt {2}}\right )-\frac {x^9}{3 \sqrt {x^6+2}}+\frac {1}{2} \sqrt {x^6+2} x^3 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^14/(2 + x^6)^(3/2),x]

[Out]

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

Rule 221

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 281

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 294

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*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 327

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

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Mathematica [A]
time = 0.14, size = 38, normalized size = 0.84 \begin {gather*} \frac {x^3 \left (6+x^6\right )}{6 \sqrt {2+x^6}}-\tanh ^{-1}\left (\frac {x^3}{\sqrt {2+x^6}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^14/(2 + x^6)^(3/2),x]

[Out]

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

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Maple [A]
time = 0.18, size = 30, normalized size = 0.67

method	result	size
risch	\(\frac {x^{3} \left (x^{6}+6\right )}{6 \sqrt {x^{6}+2}}-\arcsinh \left (\frac {x^{3} \sqrt {2}}{2}\right )\)	\(30\)
trager	\(\frac {x^{3} \left (x^{6}+6\right )}{6 \sqrt {x^{6}+2}}+\ln \left (x^{3}-\sqrt {x^{6}+2}\right )\)	\(33\)
meijerg	\(\frac {\frac {\sqrt {\pi }\, x^{3} \sqrt {2}\, \left (\frac {5 x^{6}}{2}+15\right )}{30 \sqrt {1+\frac {x^{6}}{2}}}-\sqrt {\pi }\, \arcsinh \left (\frac {x^{3} \sqrt {2}}{2}\right )}{\sqrt {\pi }}\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^14/(x^6+2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (36) = 72\).
time = 0.29, size = 73, normalized size = 1.62 \begin {gather*} -\frac {\frac {3 \, {\left (x^{6} + 2\right )}}{x^{6}} - 2}{3 \, {\left (\frac {\sqrt {x^{6} + 2}}{x^{3}} - \frac {{\left (x^{6} + 2\right )}^{\frac {3}{2}}}{x^{9}}\right )}} - \frac {1}{2} \, \log \left (\frac {\sqrt {x^{6} + 2}}{x^{3}} + 1\right ) + \frac {1}{2} \, \log \left (\frac {\sqrt {x^{6} + 2}}{x^{3}} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^14/(x^6+2)^(3/2),x, algorithm="maxima")

[Out]

-1/3*(3*(x^6 + 2)/x^6 - 2)/(sqrt(x^6 + 2)/x^3 - (x^6 + 2)^(3/2)/x^9) - 1/2*log(sqrt(x^6 + 2)/x^3 + 1) + 1/2*lo
g(sqrt(x^6 + 2)/x^3 - 1)

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Fricas [A]
time = 0.36, size = 54, normalized size = 1.20 \begin {gather*} \frac {4 \, x^{6} + 6 \, {\left (x^{6} + 2\right )} \log \left (-x^{3} + \sqrt {x^{6} + 2}\right ) + {\left (x^{9} + 6 \, x^{3}\right )} \sqrt {x^{6} + 2} + 8}{6 \, {\left (x^{6} + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^14/(x^6+2)^(3/2),x, algorithm="fricas")

[Out]

1/6*(4*x^6 + 6*(x^6 + 2)*log(-x^3 + sqrt(x^6 + 2)) + (x^9 + 6*x^3)*sqrt(x^6 + 2) + 8)/(x^6 + 2)

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Sympy [A]
time = 1.67, size = 36, normalized size = 0.80 \begin {gather*} \frac {x^{9}}{6 \sqrt {x^{6} + 2}} + \frac {x^{3}}{\sqrt {x^{6} + 2}} - \operatorname {asinh}{\left (\frac {\sqrt {2} x^{3}}{2} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**14/(x**6+2)**(3/2),x)

[Out]

x**9/(6*sqrt(x**6 + 2)) + x**3/sqrt(x**6 + 2) - asinh(sqrt(2)*x**3/2)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^14/(x^6+2)^(3/2),x, algorithm="giac")

[Out]

integrate(x^14/(x^6 + 2)^(3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x^{14}}{{\left (x^6+2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^14/(x^6 + 2)^(3/2),x)

[Out]

int(x^14/(x^6 + 2)^(3/2), x)

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