3.14.43 \(\int \frac {1}{x^4 (a+b x^6)^2} \, dx\) [1343]

Optimal. Leaf size=59 \[ -\frac {1}{2 a^2 x^3}+\frac {1}{6 a x^3 \left (a+b x^6\right )}-\frac {\sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} x^3}{\sqrt {a}}\right )}{2 a^{5/2}} \]

[Out]

-1/2/a^2/x^3+1/6/a/x^3/(b*x^6+a)-1/2*arctan(x^3*b^(1/2)/a^(1/2))*b^(1/2)/a^(5/2)

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Rubi [A]
time = 0.02, antiderivative size = 59, 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, 296, 331, 211} \begin {gather*} -\frac {\sqrt {b} \text {ArcTan}\left (\frac {\sqrt {b} x^3}{\sqrt {a}}\right )}{2 a^{5/2}}-\frac {1}{2 a^2 x^3}+\frac {1}{6 a x^3 \left (a+b x^6\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(a + b*x^6)^2),x]

[Out]

-1/2*1/(a^2*x^3) + 1/(6*a*x^3*(a + b*x^6)) - (Sqrt[b]*ArcTan[(Sqrt[b]*x^3)/Sqrt[a]])/(2*a^(5/2))

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 296

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

Rule 331

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

Rubi steps

\begin {align*} \int \frac {1}{x^4 \left (a+b x^6\right )^2} \, dx &=\frac {1}{3} \text {Subst}\left (\int \frac {1}{x^2 \left (a+b x^2\right )^2} \, dx,x,x^3\right )\\ &=\frac {1}{6 a x^3 \left (a+b x^6\right )}+\frac {\text {Subst}\left (\int \frac {1}{x^2 \left (a+b x^2\right )} \, dx,x,x^3\right )}{2 a}\\ &=-\frac {1}{2 a^2 x^3}+\frac {1}{6 a x^3 \left (a+b x^6\right )}-\frac {b \text {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,x^3\right )}{2 a^2}\\ &=-\frac {1}{2 a^2 x^3}+\frac {1}{6 a x^3 \left (a+b x^6\right )}-\frac {\sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} x^3}{\sqrt {a}}\right )}{2 a^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 114, normalized size = 1.93 \begin {gather*} \frac {-\frac {2 \sqrt {a}}{x^3}-\frac {\sqrt {a} b x^3}{a+b x^6}+3 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt [6]{b} x}{\sqrt [6]{a}}\right )+3 \sqrt {b} \tan ^{-1}\left (\sqrt {3}-\frac {2 \sqrt [6]{b} x}{\sqrt [6]{a}}\right )-3 \sqrt {b} \tan ^{-1}\left (\sqrt {3}+\frac {2 \sqrt [6]{b} x}{\sqrt [6]{a}}\right )}{6 a^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*(a + b*x^6)^2),x]

[Out]

((-2*Sqrt[a])/x^3 - (Sqrt[a]*b*x^3)/(a + b*x^6) + 3*Sqrt[b]*ArcTan[(b^(1/6)*x)/a^(1/6)] + 3*Sqrt[b]*ArcTan[Sqr
t[3] - (2*b^(1/6)*x)/a^(1/6)] - 3*Sqrt[b]*ArcTan[Sqrt[3] + (2*b^(1/6)*x)/a^(1/6)])/(6*a^(5/2))

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Maple [A]
time = 0.18, size = 49, normalized size = 0.83

method	result	size
default	\(-\frac {b \left (\frac {x^{3}}{2 b \,x^{6}+2 a}+\frac {3 \arctan \left (\frac {b \,x^{3}}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{3 a^{2}}-\frac {1}{3 a^{2} x^{3}}\)	\(49\)
risch	\(\frac {-\frac {b \,x^{6}}{2 a^{2}}-\frac {1}{3 a}}{x^{3} \left (b \,x^{6}+a \right )}+\frac {\left (\munderset {\textit {\_R} =\RootOf \left (a^{5} \textit {\_Z}^{2}+b \right )}{\sum }\textit {\_R} \ln \left (\left (-7 a^{5} \textit {\_R}^{2}-6 b \right ) x^{3}-a^{3} \textit {\_R} \right )\right )}{4}\)	\(71\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^4/(b*x^6+a)^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.51, size = 53, normalized size = 0.90 \begin {gather*} -\frac {3 \, b x^{6} + 2 \, a}{6 \, {\left (a^{2} b x^{9} + a^{3} x^{3}\right )}} - \frac {b \arctan \left (\frac {b x^{3}}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^6+a)^2,x, algorithm="maxima")

[Out]

-1/6*(3*b*x^6 + 2*a)/(a^2*b*x^9 + a^3*x^3) - 1/2*b*arctan(b*x^3/sqrt(a*b))/(sqrt(a*b)*a^2)

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Fricas [A]
time = 0.37, size = 148, normalized size = 2.51 \begin {gather*} \left [-\frac {6 \, b x^{6} - 3 \, {\left (b x^{9} + a x^{3}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{6} - 2 \, a x^{3} \sqrt {-\frac {b}{a}} - a}{b x^{6} + a}\right ) + 4 \, a}{12 \, {\left (a^{2} b x^{9} + a^{3} x^{3}\right )}}, -\frac {3 \, b x^{6} + 3 \, {\left (b x^{9} + a x^{3}\right )} \sqrt {\frac {b}{a}} \arctan \left (x^{3} \sqrt {\frac {b}{a}}\right ) + 2 \, a}{6 \, {\left (a^{2} b x^{9} + a^{3} x^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^6+a)^2,x, algorithm="fricas")

[Out]

[-1/12*(6*b*x^6 - 3*(b*x^9 + a*x^3)*sqrt(-b/a)*log((b*x^6 - 2*a*x^3*sqrt(-b/a) - a)/(b*x^6 + a)) + 4*a)/(a^2*b
*x^9 + a^3*x^3), -1/6*(3*b*x^6 + 3*(b*x^9 + a*x^3)*sqrt(b/a)*arctan(x^3*sqrt(b/a)) + 2*a)/(a^2*b*x^9 + a^3*x^3
)]

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Sympy [A]
time = 0.25, size = 94, normalized size = 1.59 \begin {gather*} \frac {\sqrt {- \frac {b}{a^{5}}} \log {\left (- \frac {a^{3} \sqrt {- \frac {b}{a^{5}}}}{b} + x^{3} \right )}}{4} - \frac {\sqrt {- \frac {b}{a^{5}}} \log {\left (\frac {a^{3} \sqrt {- \frac {b}{a^{5}}}}{b} + x^{3} \right )}}{4} + \frac {- 2 a - 3 b x^{6}}{6 a^{3} x^{3} + 6 a^{2} b x^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**4/(b*x**6+a)**2,x)

[Out]

sqrt(-b/a**5)*log(-a**3*sqrt(-b/a**5)/b + x**3)/4 - sqrt(-b/a**5)*log(a**3*sqrt(-b/a**5)/b + x**3)/4 + (-2*a -
3*b*x**6)/(6*a**3*x**3 + 6*a**2*b*x**9)

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Giac [A]
time = 2.81, size = 51, normalized size = 0.86 \begin {gather*} -\frac {b \arctan \left (\frac {b x^{3}}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{2}} - \frac {3 \, b x^{6} + 2 \, a}{6 \, {\left (b x^{9} + a x^{3}\right )} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^6+a)^2,x, algorithm="giac")

[Out]

-1/2*b*arctan(b*x^3/sqrt(a*b))/(sqrt(a*b)*a^2) - 1/6*(3*b*x^6 + 2*a)/((b*x^9 + a*x^3)*a^2)

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Mupad [B]
time = 0.09, size = 50, normalized size = 0.85 \begin {gather*} -\frac {\frac {1}{3\,a}+\frac {b\,x^6}{2\,a^2}}{b\,x^9+a\,x^3}-\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x^3}{\sqrt {a}}\right )}{2\,a^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*(a + b*x^6)^2),x)

[Out]

- (1/(3*a) + (b*x^6)/(2*a^2))/(a*x^3 + b*x^9) - (b^(1/2)*atan((b^(1/2)*x^3)/a^(1/2)))/(2*a^(5/2))

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