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

Optimal. Leaf size=98 \[ \frac {4 a^{3/2} \sqrt [4]{\frac {b x^2}{a}+1} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 b^{3/2} \sqrt [4]{a+b x^2}}-\frac {4 a x}{5 b \sqrt [4]{a+b x^2}}+\frac {2 x \left (a+b x^2\right )^{3/4}}{5 b} \]

[Out]

-4/5*a*x/b/(b*x^2+a)^(1/4)+2/5*x*(b*x^2+a)^(3/4)/b+4/5*a^(3/2)*(1+b*x^2/a)^(1/4)*(cos(1/2*arctan(x*b^(1/2)/a^(
1/2)))^2)^(1/2)/cos(1/2*arctan(x*b^(1/2)/a^(1/2)))*EllipticE(sin(1/2*arctan(x*b^(1/2)/a^(1/2))),2^(1/2))/b^(3/
2)/(b*x^2+a)^(1/4)

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Rubi [A]  time = 0.02, antiderivative size = 98, 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 = {321, 229, 227, 196} \[ \frac {4 a^{3/2} \sqrt [4]{\frac {b x^2}{a}+1} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 b^{3/2} \sqrt [4]{a+b x^2}}-\frac {4 a x}{5 b \sqrt [4]{a+b x^2}}+\frac {2 x \left (a+b x^2\right )^{3/4}}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(a + b*x^2)^(1/4),x]

[Out]

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

Rule 196

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2*EllipticE[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(5/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 227

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(2*x)/(a + b*x^2)^(1/4), x] - Dist[a, Int[1/(a + b*x^2)^(5
/4), x], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 229

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Dist[(1 + (b*x^2)/a)^(1/4)/(a + b*x^2)^(1/4), Int[1/(1 + (b*x^2
)/a)^(1/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a]

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]

Rubi steps

\begin {align*} \int \frac {x^2}{\sqrt [4]{a+b x^2}} \, dx &=\frac {2 x \left (a+b x^2\right )^{3/4}}{5 b}-\frac {(2 a) \int \frac {1}{\sqrt [4]{a+b x^2}} \, dx}{5 b}\\ &=\frac {2 x \left (a+b x^2\right )^{3/4}}{5 b}-\frac {\left (2 a \sqrt [4]{1+\frac {b x^2}{a}}\right ) \int \frac {1}{\sqrt [4]{1+\frac {b x^2}{a}}} \, dx}{5 b \sqrt [4]{a+b x^2}}\\ &=-\frac {4 a x}{5 b \sqrt [4]{a+b x^2}}+\frac {2 x \left (a+b x^2\right )^{3/4}}{5 b}+\frac {\left (2 a \sqrt [4]{1+\frac {b x^2}{a}}\right ) \int \frac {1}{\left (1+\frac {b x^2}{a}\right )^{5/4}} \, dx}{5 b \sqrt [4]{a+b x^2}}\\ &=-\frac {4 a x}{5 b \sqrt [4]{a+b x^2}}+\frac {2 x \left (a+b x^2\right )^{3/4}}{5 b}+\frac {4 a^{3/2} \sqrt [4]{1+\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 b^{3/2} \sqrt [4]{a+b x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 62, normalized size = 0.63 \[ \frac {2 x \left (-a \sqrt [4]{\frac {b x^2}{a}+1} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {3}{2};-\frac {b x^2}{a}\right )+a+b x^2\right )}{5 b \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a + b*x^2)^(1/4),x]

[Out]

(2*x*(a + b*x^2 - a*(1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[1/4, 1/2, 3/2, -((b*x^2)/a)]))/(5*b*(a + b*x^2)^(1
/4))

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fricas [F]  time = 0.95, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{2}}{{\left (b x^{2} + a\right )}^{\frac {1}{4}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^2/(b*x^2 + a)^(1/4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/(b*x^2 + a)^(1/4), x)

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maple [F]  time = 0.30, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/(b*x^2 + a)^(1/4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a + b*x^2)^(1/4),x)

[Out]

int(x^2/(a + b*x^2)^(1/4), x)

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sympy [C]  time = 0.80, size = 27, normalized size = 0.28 \[ \frac {x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3 \sqrt [4]{a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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