3.14.31 \(\int \frac {x^8}{(a+b x^6)^2} \, dx\) [1331]

Optimal. Leaf size=49 \[ -\frac {x^3}{6 b \left (a+b x^6\right )}+\frac {\tan ^{-1}\left (\frac {\sqrt {b} x^3}{\sqrt {a}}\right )}{6 \sqrt {a} b^{3/2}} \]

[Out]

-1/6*x^3/b/(b*x^6+a)+1/6*arctan(x^3*b^(1/2)/a^(1/2))/b^(3/2)/a^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {281, 294, 211} \begin {gather*} \frac {\text {ArcTan}\left (\frac {\sqrt {b} x^3}{\sqrt {a}}\right )}{6 \sqrt {a} b^{3/2}}-\frac {x^3}{6 b \left (a+b x^6\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^8/(a + b*x^6)^2,x]

[Out]

-1/6*x^3/(b*(a + b*x^6)) + ArcTan[(Sqrt[b]*x^3)/Sqrt[a]]/(6*Sqrt[a]*b^(3/2))

Rule 211

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

Rubi steps

\begin {align*} \int \frac {x^8}{\left (a+b x^6\right )^2} \, dx &=\frac {1}{3} \text {Subst}\left (\int \frac {x^2}{\left (a+b x^2\right )^2} \, dx,x,x^3\right )\\ &=-\frac {x^3}{6 b \left (a+b x^6\right )}+\frac {\text {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,x^3\right )}{6 b}\\ &=-\frac {x^3}{6 b \left (a+b x^6\right )}+\frac {\tan ^{-1}\left (\frac {\sqrt {b} x^3}{\sqrt {a}}\right )}{6 \sqrt {a} b^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 49, normalized size = 1.00 \begin {gather*} -\frac {x^3}{6 b \left (a+b x^6\right )}+\frac {\tan ^{-1}\left (\frac {\sqrt {b} x^3}{\sqrt {a}}\right )}{6 \sqrt {a} b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^8/(a + b*x^6)^2,x]

[Out]

-1/6*x^3/(b*(a + b*x^6)) + ArcTan[(Sqrt[b]*x^3)/Sqrt[a]]/(6*Sqrt[a]*b^(3/2))

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Maple [A]
time = 0.19, size = 40, normalized size = 0.82

method	result	size
default	\(-\frac {x^{3}}{6 b \left (b \,x^{6}+a \right )}+\frac {\arctan \left (\frac {b \,x^{3}}{\sqrt {a b}}\right )}{6 b \sqrt {a b}}\)	\(40\)
risch	\(-\frac {x^{3}}{6 b \left (b \,x^{6}+a \right )}-\frac {\ln \left (x^{3} \sqrt {-a b}-a \right )}{12 \sqrt {-a b}\, b}+\frac {\ln \left (x^{3} \sqrt {-a b}+a \right )}{12 \sqrt {-a b}\, b}\)	\(69\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/(b*x^6+a)^2,x,method=_RETURNVERBOSE)

[Out]

-1/6*x^3/b/(b*x^6+a)+1/6/b/(a*b)^(1/2)*arctan(b*x^3/(a*b)^(1/2))

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Maxima [A]
time = 0.49, size = 40, normalized size = 0.82 \begin {gather*} -\frac {x^{3}}{6 \, {\left (b^{2} x^{6} + a b\right )}} + \frac {\arctan \left (\frac {b x^{3}}{\sqrt {a b}}\right )}{6 \, \sqrt {a b} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^6+a)^2,x, algorithm="maxima")

[Out]

-1/6*x^3/(b^2*x^6 + a*b) + 1/6*arctan(b*x^3/sqrt(a*b))/(sqrt(a*b)*b)

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Fricas [A]
time = 0.37, size = 128, normalized size = 2.61 \begin {gather*} \left [-\frac {2 \, a b x^{3} + {\left (b x^{6} + a\right )} \sqrt {-a b} \log \left (\frac {b x^{6} - 2 \, \sqrt {-a b} x^{3} - a}{b x^{6} + a}\right )}{12 \, {\left (a b^{3} x^{6} + a^{2} b^{2}\right )}}, -\frac {a b x^{3} - {\left (b x^{6} + a\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x^{3}}{a}\right )}{6 \, {\left (a b^{3} x^{6} + a^{2} b^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^6+a)^2,x, algorithm="fricas")

[Out]

[-1/12*(2*a*b*x^3 + (b*x^6 + a)*sqrt(-a*b)*log((b*x^6 - 2*sqrt(-a*b)*x^3 - a)/(b*x^6 + a)))/(a*b^3*x^6 + a^2*b
^2), -1/6*(a*b*x^3 - (b*x^6 + a)*sqrt(a*b)*arctan(sqrt(a*b)*x^3/a))/(a*b^3*x^6 + a^2*b^2)]

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 83 vs. \(2 (39) = 78\).
time = 0.19, size = 83, normalized size = 1.69 \begin {gather*} - \frac {x^{3}}{6 a b + 6 b^{2} x^{6}} - \frac {\sqrt {- \frac {1}{a b^{3}}} \log {\left (- a b \sqrt {- \frac {1}{a b^{3}}} + x^{3} \right )}}{12} + \frac {\sqrt {- \frac {1}{a b^{3}}} \log {\left (a b \sqrt {- \frac {1}{a b^{3}}} + x^{3} \right )}}{12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8/(b*x**6+a)**2,x)

[Out]

-x**3/(6*a*b + 6*b**2*x**6) - sqrt(-1/(a*b**3))*log(-a*b*sqrt(-1/(a*b**3)) + x**3)/12 + sqrt(-1/(a*b**3))*log(
a*b*sqrt(-1/(a*b**3)) + x**3)/12

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Giac [A]
time = 2.91, size = 39, normalized size = 0.80 \begin {gather*} -\frac {x^{3}}{6 \, {\left (b x^{6} + a\right )} b} + \frac {\arctan \left (\frac {b x^{3}}{\sqrt {a b}}\right )}{6 \, \sqrt {a b} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^6+a)^2,x, algorithm="giac")

[Out]

-1/6*x^3/((b*x^6 + a)*b) + 1/6*arctan(b*x^3/sqrt(a*b))/(sqrt(a*b)*b)

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Mupad [B]
time = 0.09, size = 37, normalized size = 0.76 \begin {gather*} \frac {\mathrm {atan}\left (\frac {\sqrt {b}\,x^3}{\sqrt {a}}\right )}{6\,\sqrt {a}\,b^{3/2}}-\frac {x^3}{6\,b\,\left (b\,x^6+a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/(a + b*x^6)^2,x)

[Out]

atan((b^(1/2)*x^3)/a^(1/2))/(6*a^(1/2)*b^(3/2)) - x^3/(6*b*(a + b*x^6))

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