\(\int \frac {1}{x^2 \sqrt [3]{a^2+2 a b x^2+b^2 x^4}} \, dx\) [661]

Optimal result

Integrand size = 26, antiderivative size = 289 \[ \int \frac {1}{x^2 \sqrt [3]{a^2+2 a b x^2+b^2 x^4}} \, dx=-\frac {a+b x^2}{a x \sqrt [3]{a^2+2 a b x^2+b^2 x^4}}+\frac {\sqrt {2-\sqrt {3}} \left (1+\frac {b x^2}{a}\right )^{2/3} \left (1-\sqrt [3]{1+\frac {b x^2}{a}}\right ) \sqrt {\frac {1+\sqrt [3]{1+\frac {b x^2}{a}}+\left (1+\frac {b x^2}{a}\right )^{2/3}}{\left (1-\sqrt {3}-\sqrt [3]{1+\frac {b x^2}{a}}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1+\sqrt {3}-\sqrt [3]{1+\frac {b x^2}{a}}}{1-\sqrt {3}-\sqrt [3]{1+\frac {b x^2}{a}}}\right ),-7+4 \sqrt {3}\right )}{\sqrt [4]{3} x \sqrt [3]{a^2+2 a b x^2+b^2 x^4} \sqrt {-\frac {1-\sqrt [3]{1+\frac {b x^2}{a}}}{\left (1-\sqrt {3}-\sqrt [3]{1+\frac {b x^2}{a}}\right )^2}}} \]

[Out]

(-b*x^2-a)/a/x/(b^2*x^4+2*a*b*x^2+a^2)^(1/3)+1/3*(1+b*x^2/a)^(2/3)*(1-(1+b*x^2/a)^(1/3))*EllipticF((1-(1+b*x^2
/a)^(1/3)+3^(1/2))/(1-(1+b*x^2/a)^(1/3)-3^(1/2)),2*I-I*3^(1/2))*(1/2*6^(1/2)-1/2*2^(1/2))*((1+(1+b*x^2/a)^(1/3
)+(1+b*x^2/a)^(2/3))/(1-(1+b*x^2/a)^(1/3)-3^(1/2))^2)^(1/2)*3^(3/4)/x/(b^2*x^4+2*a*b*x^2+a^2)^(1/3)/((-1+(1+b*
x^2/a)^(1/3))/(1-(1+b*x^2/a)^(1/3)-3^(1/2))^2)^(1/2)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1127, 331, 242, 225} \[ \int \frac {1}{x^2 \sqrt [3]{a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {\sqrt {2-\sqrt {3}} \left (\frac {b x^2}{a}+1\right )^{2/3} \left (1-\sqrt [3]{\frac {b x^2}{a}+1}\right ) \sqrt {\frac {\left (\frac {b x^2}{a}+1\right )^{2/3}+\sqrt [3]{\frac {b x^2}{a}+1}+1}{\left (-\sqrt [3]{\frac {b x^2}{a}+1}-\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {-\sqrt [3]{\frac {b x^2}{a}+1}+\sqrt {3}+1}{-\sqrt [3]{\frac {b x^2}{a}+1}-\sqrt {3}+1}\right ),-7+4 \sqrt {3}\right )}{\sqrt [4]{3} x \sqrt [3]{a^2+2 a b x^2+b^2 x^4} \sqrt {-\frac {1-\sqrt [3]{\frac {b x^2}{a}+1}}{\left (-\sqrt [3]{\frac {b x^2}{a}+1}-\sqrt {3}+1\right )^2}}}-\frac {a+b x^2}{a x \sqrt [3]{a^2+2 a b x^2+b^2 x^4}} \]

[In]

Int[1/(x^2*(a^2 + 2*a*b*x^2 + b^2*x^4)^(1/3)),x]

[Out]

-((a + b*x^2)/(a*x*(a^2 + 2*a*b*x^2 + b^2*x^4)^(1/3))) + (Sqrt[2 - Sqrt[3]]*(1 + (b*x^2)/a)^(2/3)*(1 - (1 + (b
*x^2)/a)^(1/3))*Sqrt[(1 + (1 + (b*x^2)/a)^(1/3) + (1 + (b*x^2)/a)^(2/3))/(1 - Sqrt[3] - (1 + (b*x^2)/a)^(1/3))
^2]*EllipticF[ArcSin[(1 + Sqrt[3] - (1 + (b*x^2)/a)^(1/3))/(1 - Sqrt[3] - (1 + (b*x^2)/a)^(1/3))], -7 + 4*Sqrt
[3]])/(3^(1/4)*x*(a^2 + 2*a*b*x^2 + b^2*x^4)^(1/3)*Sqrt[-((1 - (1 + (b*x^2)/a)^(1/3))/(1 - Sqrt[3] - (1 + (b*x
^2)/a)^(1/3))^2)])

Rule 225

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt
[2 - Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 - Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sq
rt[(-s)*((s + r*x)/((1 - Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 + Sqrt[3])*s + r*x)/((1 - Sqrt[3])*s + r
*x)], -7 + 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x] && NegQ[a]

Rule 242

Int[((a_) + (b_.)*(x_)^2)^(-2/3), x_Symbol] :> Dist[3*(Sqrt[b*x^2]/(2*b*x)), Subst[Int[1/Sqrt[-a + x^3], x], x
, (a + b*x^2)^(1/3)], x] /; FreeQ[{a, b}, 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]

Rule 1127

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^2 +
c*x^4)^FracPart[p]/(1 + 2*c*(x^2/b))^(2*FracPart[p])), Int[(d*x)^m*(1 + 2*c*(x^2/b))^(2*p), x], x] /; FreeQ[{
a, b, c, d, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[2*p]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (1+\frac {b x^2}{a}\right )^{2/3} \int \frac {1}{x^2 \left (1+\frac {b x^2}{a}\right )^{2/3}} \, dx}{\sqrt [3]{a^2+2 a b x^2+b^2 x^4}} \\ & = -\frac {a+b x^2}{a x \sqrt [3]{a^2+2 a b x^2+b^2 x^4}}-\frac {\left (b \left (1+\frac {b x^2}{a}\right )^{2/3}\right ) \int \frac {1}{\left (1+\frac {b x^2}{a}\right )^{2/3}} \, dx}{3 a \sqrt [3]{a^2+2 a b x^2+b^2 x^4}} \\ & = -\frac {a+b x^2}{a x \sqrt [3]{a^2+2 a b x^2+b^2 x^4}}-\frac {\left (\sqrt {\frac {b x^2}{a}} \left (1+\frac {b x^2}{a}\right )^{2/3}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-1+x^3}} \, dx,x,\sqrt [3]{1+\frac {b x^2}{a}}\right )}{2 x \sqrt [3]{a^2+2 a b x^2+b^2 x^4}} \\ & = -\frac {a+b x^2}{a x \sqrt [3]{a^2+2 a b x^2+b^2 x^4}}+\frac {\sqrt {2-\sqrt {3}} \left (1+\frac {b x^2}{a}\right )^{2/3} \left (1-\sqrt [3]{1+\frac {b x^2}{a}}\right ) \sqrt {\frac {1+\sqrt [3]{1+\frac {b x^2}{a}}+\left (1+\frac {b x^2}{a}\right )^{2/3}}{\left (1-\sqrt {3}-\sqrt [3]{1+\frac {b x^2}{a}}\right )^2}} F\left (\sin ^{-1}\left (\frac {1+\sqrt {3}-\sqrt [3]{1+\frac {b x^2}{a}}}{1-\sqrt {3}-\sqrt [3]{1+\frac {b x^2}{a}}}\right )|-7+4 \sqrt {3}\right )}{\sqrt [4]{3} x \sqrt [3]{a^2+2 a b x^2+b^2 x^4} \sqrt {-\frac {1-\sqrt [3]{1+\frac {b x^2}{a}}}{\left (1-\sqrt {3}-\sqrt [3]{1+\frac {b x^2}{a}}\right )^2}}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.01 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.18 \[ \int \frac {1}{x^2 \sqrt [3]{a^2+2 a b x^2+b^2 x^4}} \, dx=-\frac {\left (1+\frac {b x^2}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {2}{3},\frac {1}{2},-\frac {b x^2}{a}\right )}{x \sqrt [3]{\left (a+b x^2\right )^2}} \]

[In]

Integrate[1/(x^2*(a^2 + 2*a*b*x^2 + b^2*x^4)^(1/3)),x]

[Out]

-(((1 + (b*x^2)/a)^(2/3)*Hypergeometric2F1[-1/2, 2/3, 1/2, -((b*x^2)/a)])/(x*((a + b*x^2)^2)^(1/3)))

Maple [F]

\[\int \frac {1}{x^{2} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {1}{3}}}d x\]

[In]

int(1/x^2/(b^2*x^4+2*a*b*x^2+a^2)^(1/3),x)

[Out]

int(1/x^2/(b^2*x^4+2*a*b*x^2+a^2)^(1/3),x)

Fricas [F]

\[ \int \frac {1}{x^2 \sqrt [3]{a^2+2 a b x^2+b^2 x^4}} \, dx=\int { \frac {1}{{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{3}} x^{2}} \,d x } \]

[In]

integrate(1/x^2/(b^2*x^4+2*a*b*x^2+a^2)^(1/3),x, algorithm="fricas")

[Out]

integral((b^2*x^4 + 2*a*b*x^2 + a^2)^(2/3)/(b^2*x^6 + 2*a*b*x^4 + a^2*x^2), x)

Sympy [F]

\[ \int \frac {1}{x^2 \sqrt [3]{a^2+2 a b x^2+b^2 x^4}} \, dx=\int \frac {1}{x^{2} \sqrt [3]{\left (a + b x^{2}\right )^{2}}}\, dx \]

[In]

integrate(1/x**2/(b**2*x**4+2*a*b*x**2+a**2)**(1/3),x)

[Out]

Integral(1/(x**2*((a + b*x**2)**2)**(1/3)), x)

Maxima [F]

\[ \int \frac {1}{x^2 \sqrt [3]{a^2+2 a b x^2+b^2 x^4}} \, dx=\int { \frac {1}{{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{3}} x^{2}} \,d x } \]

[In]

integrate(1/x^2/(b^2*x^4+2*a*b*x^2+a^2)^(1/3),x, algorithm="maxima")

[Out]

integrate(1/((b^2*x^4 + 2*a*b*x^2 + a^2)^(1/3)*x^2), x)

Giac [F]

\[ \int \frac {1}{x^2 \sqrt [3]{a^2+2 a b x^2+b^2 x^4}} \, dx=\int { \frac {1}{{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{3}} x^{2}} \,d x } \]

[In]

integrate(1/x^2/(b^2*x^4+2*a*b*x^2+a^2)^(1/3),x, algorithm="giac")

[Out]

integrate(1/((b^2*x^4 + 2*a*b*x^2 + a^2)^(1/3)*x^2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^2 \sqrt [3]{a^2+2 a b x^2+b^2 x^4}} \, dx=\int \frac {1}{x^2\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{1/3}} \,d x \]

[In]

int(1/(x^2*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/3)),x)

[Out]

int(1/(x^2*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/3)), x)