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

Optimal. Leaf size=298 \[ \frac {3 x \left (a+b x^2\right )}{5 b \sqrt [3]{a^2+2 a b x^2+b^2 x^4}}+\frac {3\ 3^{3/4} \sqrt {2-\sqrt {3}} a^2 \left (\frac {b x^2}{a}+1\right )^{2/3} \left (1-\sqrt [3]{\frac {b x^2}{a}+1}\right ) \sqrt {\frac {\left (\frac {b x^2}{a}+1\right )^{2/3}+\sqrt [3]{\frac {b x^2}{a}+1}+1}{\left (-\sqrt [3]{\frac {b x^2}{a}+1}-\sqrt {3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac {-\sqrt [3]{\frac {b x^2}{a}+1}+\sqrt {3}+1}{-\sqrt [3]{\frac {b x^2}{a}+1}-\sqrt {3}+1}\right )|-7+4 \sqrt {3}\right )}{5 b^2 x \sqrt [3]{a^2+2 a b x^2+b^2 x^4} \sqrt {-\frac {1-\sqrt [3]{\frac {b x^2}{a}+1}}{\left (-\sqrt [3]{\frac {b x^2}{a}+1}-\sqrt {3}+1\right )^2}}} \]

[Out]

3/5*x*(b*x^2+a)/b/(b^2*x^4+2*a*b*x^2+a^2)^(1/3)+3/5*3^(3/4)*a^2*(1+b*x^2/a)^(2/3)*(1-(1+b*x^2/a)^(1/3))*Ellipt
icF((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)/b^2/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)

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Rubi [A]  time = 0.23, antiderivative size = 298, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1113, 321, 236, 219} \[ \frac {3 x \left (a+b x^2\right )}{5 b \sqrt [3]{a^2+2 a b x^2+b^2 x^4}}+\frac {3\ 3^{3/4} \sqrt {2-\sqrt {3}} a^2 \left (\frac {b x^2}{a}+1\right )^{2/3} \left (1-\sqrt [3]{\frac {b x^2}{a}+1}\right ) \sqrt {\frac {\left (\frac {b x^2}{a}+1\right )^{2/3}+\sqrt [3]{\frac {b x^2}{a}+1}+1}{\left (-\sqrt [3]{\frac {b x^2}{a}+1}-\sqrt {3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac {-\sqrt [3]{\frac {b x^2}{a}+1}+\sqrt {3}+1}{-\sqrt [3]{\frac {b x^2}{a}+1}-\sqrt {3}+1}\right )|-7+4 \sqrt {3}\right )}{5 b^2 x \sqrt [3]{a^2+2 a b x^2+b^2 x^4} \sqrt {-\frac {1-\sqrt [3]{\frac {b x^2}{a}+1}}{\left (-\sqrt [3]{\frac {b x^2}{a}+1}-\sqrt {3}+1\right )^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*x*(a + b*x^2))/(5*b*(a^2 + 2*a*b*x^2 + b^2*x^4)^(1/3)) + (3*3^(3/4)*Sqrt[2 - Sqrt[3]]*a^2*(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]])/(5*b^2*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 219

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

Rule 236

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 321

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]

Rule 1113

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*} \int \frac {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} \int \frac {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 {3 x \left (a+b x^2\right )}{5 b \sqrt [3]{a^2+2 a b x^2+b^2 x^4}}-\frac {\left (3 a \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}{5 b \sqrt [3]{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac {3 x \left (a+b x^2\right )}{5 b \sqrt [3]{a^2+2 a b x^2+b^2 x^4}}-\frac {\left (9 a^2 \sqrt {\frac {b x^2}{a}} \left (1+\frac {b x^2}{a}\right )^{2/3}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-1+x^3}} \, dx,x,\sqrt [3]{1+\frac {b x^2}{a}}\right )}{10 b^2 x \sqrt [3]{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac {3 x \left (a+b x^2\right )}{5 b \sqrt [3]{a^2+2 a b x^2+b^2 x^4}}+\frac {3\ 3^{3/4} \sqrt {2-\sqrt {3}} a^2 \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 )}{5 b^2 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*}

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Mathematica [C]  time = 0.03, size = 64, normalized size = 0.21 \[ \frac {3 x \left (-a \left (\frac {b x^2}{a}+1\right )^{2/3} \, _2F_1\left (\frac {1}{2},\frac {2}{3};\frac {3}{2};-\frac {b x^2}{a}\right )+a+b x^2\right )}{5 b \sqrt [3]{\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*x*(a + b*x^2 - a*(1 + (b*x^2)/a)^(2/3)*Hypergeometric2F1[1/2, 2/3, 3/2, -((b*x^2)/a)]))/(5*b*((a + b*x^2)^2
)^(1/3))

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fricas [F]  time = 1.08, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{2}}{{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{3}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.10, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {1}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^2}{{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{1/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\sqrt [3]{\left (a + b x^{2}\right )^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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