3.2094 \(\int \frac {x^2}{(a+\frac {b}{x^4})^{3/2}} \, dx\)

Optimal. Leaf size=131 \[ \frac {5 b^{3/4} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) F\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{12 a^{9/4} \sqrt {a+\frac {b}{x^4}}}+\frac {5 x^3 \sqrt {a+\frac {b}{x^4}}}{6 a^2}-\frac {x^3}{2 a \sqrt {a+\frac {b}{x^4}}} \]

[Out]

-1/2*x^3/a/(a+b/x^4)^(1/2)+5/6*x^3*(a+b/x^4)^(1/2)/a^2+5/12*b^(3/4)*(cos(2*arccot(a^(1/4)*x/b^(1/4)))^2)^(1/2)
/cos(2*arccot(a^(1/4)*x/b^(1/4)))*EllipticF(sin(2*arccot(a^(1/4)*x/b^(1/4))),1/2*2^(1/2))*(a^(1/2)+b^(1/2)/x^2
)*((a+b/x^4)/(a^(1/2)+b^(1/2)/x^2)^2)^(1/2)/a^(9/4)/(a+b/x^4)^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {335, 290, 325, 220} \[ \frac {5 b^{3/4} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) F\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{12 a^{9/4} \sqrt {a+\frac {b}{x^4}}}+\frac {5 x^3 \sqrt {a+\frac {b}{x^4}}}{6 a^2}-\frac {x^3}{2 a \sqrt {a+\frac {b}{x^4}}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(a + b/x^4)^(3/2),x]

[Out]

-x^3/(2*a*Sqrt[a + b/x^4]) + (5*Sqrt[a + b/x^4]*x^3)/(6*a^2) + (5*b^(3/4)*Sqrt[(a + b/x^4)/(Sqrt[a] + Sqrt[b]/
x^2)^2]*(Sqrt[a] + Sqrt[b]/x^2)*EllipticF[2*ArcCot[(a^(1/4)*x)/b^(1/4)], 1/2])/(12*a^(9/4)*Sqrt[a + b/x^4])

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 290

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] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 325

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 335

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {x^2}{\left (a+\frac {b}{x^4}\right )^{3/2}} \, dx &=-\operatorname {Subst}\left (\int \frac {1}{x^4 \left (a+b x^4\right )^{3/2}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {x^3}{2 a \sqrt {a+\frac {b}{x^4}}}-\frac {5 \operatorname {Subst}\left (\int \frac {1}{x^4 \sqrt {a+b x^4}} \, dx,x,\frac {1}{x}\right )}{2 a}\\ &=-\frac {x^3}{2 a \sqrt {a+\frac {b}{x^4}}}+\frac {5 \sqrt {a+\frac {b}{x^4}} x^3}{6 a^2}+\frac {(5 b) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^4}} \, dx,x,\frac {1}{x}\right )}{6 a^2}\\ &=-\frac {x^3}{2 a \sqrt {a+\frac {b}{x^4}}}+\frac {5 \sqrt {a+\frac {b}{x^4}} x^3}{6 a^2}+\frac {5 b^{3/4} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) F\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{12 a^{9/4} \sqrt {a+\frac {b}{x^4}}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 67, normalized size = 0.51 \[ \frac {-5 b \sqrt {\frac {a x^4}{b}+1} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {a x^4}{b}\right )+2 a x^4+5 b}{6 a^2 x \sqrt {a+\frac {b}{x^4}}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a + b/x^4)^(3/2),x]

[Out]

(5*b + 2*a*x^4 - 5*b*Sqrt[1 + (a*x^4)/b]*Hypergeometric2F1[1/4, 1/2, 5/4, -((a*x^4)/b)])/(6*a^2*Sqrt[a + b/x^4
]*x)

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fricas [F]  time = 1.01, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{10} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{a^{2} x^{8} + 2 \, a b x^{4} + b^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^10*sqrt((a*x^4 + b)/x^4)/(a^2*x^8 + 2*a*b*x^4 + b^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/(a + b/x^4)^(3/2), x)

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maple [C]  time = 0.02, size = 133, normalized size = 1.02 \[ \frac {\left (a \,x^{4}+b \right ) \left (2 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a \,x^{5}+5 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, b x -5 \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, b \EllipticF \left (\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, x , i\right )\right )}{6 \left (\frac {a \,x^{4}+b}{x^{4}}\right )^{\frac {3}{2}} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a^{2} x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(a*x^4+b)*(2*(I*a^(1/2)/b^(1/2))^(1/2)*a*x^5-5*b*(-(I*a^(1/2)*x^2-b^(1/2))/b^(1/2))^(1/2)*((I*a^(1/2)*x^2+
b^(1/2))/b^(1/2))^(1/2)*EllipticF((I*a^(1/2)/b^(1/2))^(1/2)*x,I)+5*(I*a^(1/2)/b^(1/2))^(1/2)*b*x)/((a*x^4+b)/x
^4)^(3/2)/x^6/a^2/(I*a^(1/2)/b^(1/2))^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/(a + b/x^4)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2}{{\left (a+\frac {b}{x^4}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a + b/x^4)^(3/2),x)

[Out]

int(x^2/(a + b/x^4)^(3/2), x)

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sympy [C]  time = 1.29, size = 42, normalized size = 0.32 \[ - \frac {x^{3} \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {3}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{4}}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {1}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x**3*gamma(-3/4)*hyper((-3/4, 3/2), (1/4,), b*exp_polar(I*pi)/(a*x**4))/(4*a**(3/2)*gamma(1/4))

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