\(\int \frac {1}{x^4 (a+b x^6)} \, dx\) [1330]

Optimal result

Integrand size = 13, antiderivative size = 40 \[ \int \frac {1}{x^4 \left (a+b x^6\right )} \, dx=-\frac {1}{3 a x^3}-\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} x^3}{\sqrt {a}}\right )}{3 a^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {281, 331, 211} \[ \int \frac {1}{x^4 \left (a+b x^6\right )} \, dx=-\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} x^3}{\sqrt {a}}\right )}{3 a^{3/2}}-\frac {1}{3 a x^3} \]

[In]

Int[1/(x^4*(a + b*x^6)),x]

[Out]

-1/3*1/(a*x^3) - (Sqrt[b]*ArcTan[(Sqrt[b]*x^3)/Sqrt[a]])/(3*a^(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 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*} \text {integral}& = \frac {1}{3} \text {Subst}\left (\int \frac {1}{x^2 \left (a+b x^2\right )} \, dx,x,x^3\right ) \\ & = -\frac {1}{3 a x^3}-\frac {b \text {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,x^3\right )}{3 a} \\ & = -\frac {1}{3 a x^3}-\frac {\sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} x^3}{\sqrt {a}}\right )}{3 a^{3/2}} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(101\) vs. \(2(40)=80\).

Time = 0.02 (sec) , antiderivative size = 101, normalized size of antiderivative = 2.52 \[ \int \frac {1}{x^4 \left (a+b x^6\right )} \, dx=\frac {-\sqrt {a}+\sqrt {b} x^3 \arctan \left (\frac {\sqrt [6]{b} x}{\sqrt [6]{a}}\right )+\sqrt {b} x^3 \arctan \left (\sqrt {3}-\frac {2 \sqrt [6]{b} x}{\sqrt [6]{a}}\right )-\sqrt {b} x^3 \arctan \left (\sqrt {3}+\frac {2 \sqrt [6]{b} x}{\sqrt [6]{a}}\right )}{3 a^{3/2} x^3} \]

[In]

Integrate[1/(x^4*(a + b*x^6)),x]

[Out]

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

Maple [A] (verified)

Time = 4.40 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.80

method	result	size
default	\(-\frac {b \arctan \left (\frac {b \,x^{3}}{\sqrt {a b}}\right )}{3 a \sqrt {a b}}-\frac {1}{3 a \,x^{3}}\)	\(32\)
risch	\(-\frac {1}{3 a \,x^{3}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{3} \textit {\_Z}^{2}+b \right )}{\sum }\textit {\_R} \ln \left (\left (-7 a^{3} \textit {\_R}^{2}-6 b \right ) x^{3}-a^{2} \textit {\_R} \right )\right )}{6}\)	\(51\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 90, normalized size of antiderivative = 2.25 \[ \int \frac {1}{x^4 \left (a+b x^6\right )} \, dx=\left [\frac {x^{3} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{6} - 2 \, a x^{3} \sqrt {-\frac {b}{a}} - a}{b x^{6} + a}\right ) - 2}{6 \, a x^{3}}, -\frac {x^{3} \sqrt {\frac {b}{a}} \arctan \left (x^{3} \sqrt {\frac {b}{a}}\right ) + 1}{3 \, a x^{3}}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.78 \[ \int \frac {1}{x^4 \left (a+b x^6\right )} \, dx=\frac {\sqrt {- \frac {b}{a^{3}}} \log {\left (- \frac {a^{2} \sqrt {- \frac {b}{a^{3}}}}{b} + x^{3} \right )}}{6} - \frac {\sqrt {- \frac {b}{a^{3}}} \log {\left (\frac {a^{2} \sqrt {- \frac {b}{a^{3}}}}{b} + x^{3} \right )}}{6} - \frac {1}{3 a x^{3}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x^4 \left (a+b x^6\right )} \, dx=-\frac {b \arctan \left (\frac {b x^{3}}{\sqrt {a b}}\right )}{3 \, \sqrt {a b} a} - \frac {1}{3 \, a x^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x^4 \left (a+b x^6\right )} \, dx=-\frac {b \arctan \left (\frac {b x^{3}}{\sqrt {a b}}\right )}{3 \, \sqrt {a b} a} - \frac {1}{3 \, a x^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.39 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.70 \[ \int \frac {1}{x^4 \left (a+b x^6\right )} \, dx=-\frac {1}{3\,a\,x^3}-\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x^3}{\sqrt {a}}\right )}{3\,a^{3/2}} \]

[In]

int(1/(x^4*(a + b*x^6)),x)

[Out]

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