3.9.16 \(\int \frac {x^4}{\sqrt [4]{a+b x^2}} \, dx\) [816]

Optimal. Leaf size=122 \[ \frac {8 a^2 x}{15 b^2 \sqrt [4]{a+b x^2}}-\frac {4 a x \left (a+b x^2\right )^{3/4}}{15 b^2}+\frac {2 x^3 \left (a+b x^2\right )^{3/4}}{9 b}-\frac {8 a^{5/2} \sqrt [4]{1+\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{15 b^{5/2} \sqrt [4]{a+b x^2}} \]

[Out]

8/15*a^2*x/b^2/(b*x^2+a)^(1/4)-4/15*a*x*(b*x^2+a)^(3/4)/b^2+2/9*x^3*(b*x^2+a)^(3/4)/b-8/15*a^(5/2)*(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*arcta
n(x*b^(1/2)/a^(1/2))),2^(1/2))/b^(5/2)/(b*x^2+a)^(1/4)

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Rubi [A]
time = 0.03, antiderivative size = 122, normalized size of antiderivative = 1.00, 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 = {327, 235, 233, 202} \begin {gather*} -\frac {8 a^{5/2} \sqrt [4]{\frac {b x^2}{a}+1} E\left (\left .\frac {1}{2} \text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{15 b^{5/2} \sqrt [4]{a+b x^2}}+\frac {8 a^2 x}{15 b^2 \sqrt [4]{a+b x^2}}-\frac {4 a x \left (a+b x^2\right )^{3/4}}{15 b^2}+\frac {2 x^3 \left (a+b x^2\right )^{3/4}}{9 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4/(a + b*x^2)^(1/4),x]

[Out]

(8*a^2*x)/(15*b^2*(a + b*x^2)^(1/4)) - (4*a*x*(a + b*x^2)^(3/4))/(15*b^2) + (2*x^3*(a + b*x^2)^(3/4))/(9*b) -
(8*a^(5/2)*(1 + (b*x^2)/a)^(1/4)*EllipticE[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(15*b^(5/2)*(a + b*x^2)^(1/4))

Rule 202

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

Rule 233

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 235

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 327

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^4}{\sqrt [4]{a+b x^2}} \, dx &=\frac {2 x^3 \left (a+b x^2\right )^{3/4}}{9 b}-\frac {(2 a) \int \frac {x^2}{\sqrt [4]{a+b x^2}} \, dx}{3 b}\\ &=-\frac {4 a x \left (a+b x^2\right )^{3/4}}{15 b^2}+\frac {2 x^3 \left (a+b x^2\right )^{3/4}}{9 b}+\frac {\left (4 a^2\right ) \int \frac {1}{\sqrt [4]{a+b x^2}} \, dx}{15 b^2}\\ &=-\frac {4 a x \left (a+b x^2\right )^{3/4}}{15 b^2}+\frac {2 x^3 \left (a+b x^2\right )^{3/4}}{9 b}+\frac {\left (4 a^2 \sqrt [4]{1+\frac {b x^2}{a}}\right ) \int \frac {1}{\sqrt [4]{1+\frac {b x^2}{a}}} \, dx}{15 b^2 \sqrt [4]{a+b x^2}}\\ &=\frac {8 a^2 x}{15 b^2 \sqrt [4]{a+b x^2}}-\frac {4 a x \left (a+b x^2\right )^{3/4}}{15 b^2}+\frac {2 x^3 \left (a+b x^2\right )^{3/4}}{9 b}-\frac {\left (4 a^2 \sqrt [4]{1+\frac {b x^2}{a}}\right ) \int \frac {1}{\left (1+\frac {b x^2}{a}\right )^{5/4}} \, dx}{15 b^2 \sqrt [4]{a+b x^2}}\\ &=\frac {8 a^2 x}{15 b^2 \sqrt [4]{a+b x^2}}-\frac {4 a x \left (a+b x^2\right )^{3/4}}{15 b^2}+\frac {2 x^3 \left (a+b x^2\right )^{3/4}}{9 b}-\frac {8 a^{5/2} \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 )}{15 b^{5/2} \sqrt [4]{a+b x^2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 6.66, size = 79, normalized size = 0.65 \begin {gather*} \frac {2 \left (-6 a^2 x-a b x^3+5 b^2 x^5+6 a^2 x \sqrt [4]{1+\frac {b x^2}{a}} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {3}{2};-\frac {b x^2}{a}\right )\right )}{45 b^2 \sqrt [4]{a+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4/(a + b*x^2)^(1/4),x]

[Out]

(2*(-6*a^2*x - a*b*x^3 + 5*b^2*x^5 + 6*a^2*x*(1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[1/4, 1/2, 3/2, -((b*x^2)/
a)]))/(45*b^2*(a + b*x^2)^(1/4))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {x^{4}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^4/(b*x^2 + a)^(1/4), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^4/(b*x^2 + a)^(1/4), x)

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Sympy [C] Result contains complex when optimal does not.
time = 0.46, size = 27, normalized size = 0.22 \begin {gather*} \frac {x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5 \sqrt [4]{a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**5*hyper((1/4, 5/2), (7/2,), b*x**2*exp_polar(I*pi)/a)/(5*a**(1/4))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^4/(b*x^2 + a)^(1/4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(a + b*x^2)^(1/4),x)

[Out]

int(x^4/(a + b*x^2)^(1/4), x)

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