3.2102 \(\int \frac{x^2}{(a+\frac{b}{x^4})^{5/2}} \, dx\)

Optimal. Leaf size=152 \[ \frac{5 b^{3/4} \sqrt{\frac{a+\frac{b}{x^4}}{\left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right )^2}} \left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right ) \text{EllipticF}\left (2 \cot ^{-1}\left (\frac{\sqrt [4]{a} x}{\sqrt [4]{b}}\right ),\frac{1}{2}\right )}{8 a^{13/4} \sqrt{a+\frac{b}{x^4}}}+\frac{5 x^3 \sqrt{a+\frac{b}{x^4}}}{4 a^3}-\frac{3 x^3}{4 a^2 \sqrt{a+\frac{b}{x^4}}}-\frac{x^3}{6 a \left (a+\frac{b}{x^4}\right )^{3/2}} \]

[Out]

-x^3/(6*a*(a + b/x^4)^(3/2)) - (3*x^3)/(4*a^2*Sqrt[a + b/x^4]) + (5*Sqrt[a + b/x^4]*x^3)/(4*a^3) + (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])/(8*a^(13/4)*Sqrt[a + b/x^4])

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Rubi [A]  time = 0.0772644, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {335, 290, 325, 220} \[ \frac{5 b^{3/4} \sqrt{\frac{a+\frac{b}{x^4}}{\left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right )^2}} \left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right ) F\left (2 \cot ^{-1}\left (\frac{\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{8 a^{13/4} \sqrt{a+\frac{b}{x^4}}}+\frac{5 x^3 \sqrt{a+\frac{b}{x^4}}}{4 a^3}-\frac{3 x^3}{4 a^2 \sqrt{a+\frac{b}{x^4}}}-\frac{x^3}{6 a \left (a+\frac{b}{x^4}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(a + b/x^4)^(5/2),x]

[Out]

-x^3/(6*a*(a + b/x^4)^(3/2)) - (3*x^3)/(4*a^2*Sqrt[a + b/x^4]) + (5*Sqrt[a + b/x^4]*x^3)/(4*a^3) + (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])/(8*a^(13/4)*Sqrt[a + b/x^4])

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]

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 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]

Rubi steps

\begin{align*} \int \frac{x^2}{\left (a+\frac{b}{x^4}\right )^{5/2}} \, dx &=-\operatorname{Subst}\left (\int \frac{1}{x^4 \left (a+b x^4\right )^{5/2}} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{x^3}{6 a \left (a+\frac{b}{x^4}\right )^{3/2}}-\frac{3 \operatorname{Subst}\left (\int \frac{1}{x^4 \left (a+b x^4\right )^{3/2}} \, dx,x,\frac{1}{x}\right )}{2 a}\\ &=-\frac{x^3}{6 a \left (a+\frac{b}{x^4}\right )^{3/2}}-\frac{3 x^3}{4 a^2 \sqrt{a+\frac{b}{x^4}}}-\frac{15 \operatorname{Subst}\left (\int \frac{1}{x^4 \sqrt{a+b x^4}} \, dx,x,\frac{1}{x}\right )}{4 a^2}\\ &=-\frac{x^3}{6 a \left (a+\frac{b}{x^4}\right )^{3/2}}-\frac{3 x^3}{4 a^2 \sqrt{a+\frac{b}{x^4}}}+\frac{5 \sqrt{a+\frac{b}{x^4}} x^3}{4 a^3}+\frac{(5 b) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^4}} \, dx,x,\frac{1}{x}\right )}{4 a^3}\\ &=-\frac{x^3}{6 a \left (a+\frac{b}{x^4}\right )^{3/2}}-\frac{3 x^3}{4 a^2 \sqrt{a+\frac{b}{x^4}}}+\frac{5 \sqrt{a+\frac{b}{x^4}} x^3}{4 a^3}+\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 )}{8 a^{13/4} \sqrt{a+\frac{b}{x^4}}}\\ \end{align*}

Mathematica [C]  time = 0.0403472, size = 94, normalized size = 0.62 \[ \frac{4 a^2 x^8-15 b \left (a x^4+b\right ) \sqrt{\frac{a x^4}{b}+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{a x^4}{b}\right )+21 a b x^4+15 b^2}{12 a^3 x \sqrt{a+\frac{b}{x^4}} \left (a x^4+b\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a + b/x^4)^(5/2),x]

[Out]

(15*b^2 + 21*a*b*x^4 + 4*a^2*x^8 - 15*b*(b + a*x^4)*Sqrt[1 + (a*x^4)/b]*Hypergeometric2F1[1/4, 1/2, 5/4, -((a*
x^4)/b)])/(12*a^3*Sqrt[a + b/x^4]*x*(b + a*x^4))

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Maple [C]  time = 0.018, size = 304, normalized size = 2. \begin{align*}{\frac{1}{12\,{a}^{3}{x}^{10}} \left ( 4\,\sqrt{{\frac{i\sqrt{a}}{\sqrt{b}}}}{x}^{13}{a}^{3}-15\,\sqrt{-{\frac{i\sqrt{a}{x}^{2}-\sqrt{b}}{\sqrt{b}}}}\sqrt{{\frac{i\sqrt{a}{x}^{2}+\sqrt{b}}{\sqrt{b}}}}{\it EllipticF} \left ( x\sqrt{{\frac{i\sqrt{a}}{\sqrt{b}}}},i \right ){x}^{8}{a}^{2}b+25\,\sqrt{{\frac{i\sqrt{a}}{\sqrt{b}}}}{x}^{9}{a}^{2}b-30\,\sqrt{-{\frac{i\sqrt{a}{x}^{2}-\sqrt{b}}{\sqrt{b}}}}\sqrt{{\frac{i\sqrt{a}{x}^{2}+\sqrt{b}}{\sqrt{b}}}}{\it EllipticF} \left ( x\sqrt{{\frac{i\sqrt{a}}{\sqrt{b}}}},i \right ){x}^{4}a{b}^{2}+36\,\sqrt{{\frac{i\sqrt{a}}{\sqrt{b}}}}{x}^{5}a{b}^{2}-15\,\sqrt{-{\frac{i\sqrt{a}{x}^{2}-\sqrt{b}}{\sqrt{b}}}}\sqrt{{\frac{i\sqrt{a}{x}^{2}+\sqrt{b}}{\sqrt{b}}}}{\it EllipticF} \left ( x\sqrt{{\frac{i\sqrt{a}}{\sqrt{b}}}},i \right ){b}^{3}+15\,\sqrt{{\frac{i\sqrt{a}}{\sqrt{b}}}}x{b}^{3} \right ) \left ({\frac{a{x}^{4}+b}{{x}^{4}}} \right ) ^{-{\frac{5}{2}}}{\frac{1}{\sqrt{{i\sqrt{a}{\frac{1}{\sqrt{b}}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a+b/x^4)^(5/2),x)

[Out]

1/12*(4*(I*a^(1/2)/b^(1/2))^(1/2)*x^13*a^3-15*(-(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(x*(I*a^(1/2)/b^(1/2))^(1/2),I)*x^8*a^2*b+25*(I*a^(1/2)/b^(1/2))^(1/2)*x^9*a^2*b-30*
(-(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(x*(I*a^(1/2)/b^(1/2
))^(1/2),I)*x^4*a*b^2+36*(I*a^(1/2)/b^(1/2))^(1/2)*x^5*a*b^2-15*(-(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(x*(I*a^(1/2)/b^(1/2))^(1/2),I)*b^3+15*(I*a^(1/2)/b^(1/2))^(1/2)*x
*b^3)/a^3/((a*x^4+b)/x^4)^(5/2)/x^10/(I*a^(1/2)/b^(1/2))^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{{\left (a + \frac{b}{x^{4}}\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a+b/x^4)^(5/2),x, algorithm="maxima")

[Out]

integrate(x^2/(a + b/x^4)^(5/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x^{14} \sqrt{\frac{a x^{4} + b}{x^{4}}}}{a^{3} x^{12} + 3 \, a^{2} b x^{8} + 3 \, a b^{2} x^{4} + b^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a+b/x^4)^(5/2),x, algorithm="fricas")

[Out]

integral(x^14*sqrt((a*x^4 + b)/x^4)/(a^3*x^12 + 3*a^2*b*x^8 + 3*a*b^2*x^4 + b^3), x)

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Sympy [C]  time = 1.68666, size = 42, normalized size = 0.28 \begin{align*} - \frac{x^{3} \Gamma \left (- \frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{4}, \frac{5}{2} \\ \frac{1}{4} \end{matrix}\middle |{\frac{b e^{i \pi }}{a x^{4}}} \right )}}{4 a^{\frac{5}{2}} \Gamma \left (\frac{1}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(a+b/x**4)**(5/2),x)

[Out]

-x**3*gamma(-3/4)*hyper((-3/4, 5/2), (1/4,), b*exp_polar(I*pi)/(a*x**4))/(4*a**(5/2)*gamma(1/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{{\left (a + \frac{b}{x^{4}}\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a+b/x^4)^(5/2),x, algorithm="giac")

[Out]

integrate(x^2/(a + b/x^4)^(5/2), x)