3.9.78 \(\int \frac {x^4}{(2+3 x^2)^{3/4}} \, dx\) [878]

Optimal. Leaf size=65 \[ -\frac {8}{63} x \sqrt [4]{2+3 x^2}+\frac {2}{21} x^3 \sqrt [4]{2+3 x^2}+\frac {16\ 2^{3/4} F\left (\left .\frac {1}{2} \tan ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{63 \sqrt {3}} \]

[Out]

-8/63*x*(3*x^2+2)^(1/4)+2/21*x^3*(3*x^2+2)^(1/4)+16/189*2^(3/4)*(cos(1/2*arctan(1/2*x*6^(1/2)))^2)^(1/2)/cos(1
/2*arctan(1/2*x*6^(1/2)))*EllipticF(sin(1/2*arctan(1/2*x*6^(1/2))),2^(1/2))*3^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.01, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {327, 237} \begin {gather*} \frac {16\ 2^{3/4} F\left (\left .\frac {1}{2} \text {ArcTan}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{63 \sqrt {3}}-\frac {8}{63} \sqrt [4]{3 x^2+2} x+\frac {2}{21} \sqrt [4]{3 x^2+2} x^3 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4/(2 + 3*x^2)^(3/4),x]

[Out]

(-8*x*(2 + 3*x^2)^(1/4))/63 + (2*x^3*(2 + 3*x^2)^(1/4))/21 + (16*2^(3/4)*EllipticF[ArcTan[Sqrt[3/2]*x]/2, 2])/
(63*Sqrt[3])

Rule 237

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]))*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]
*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/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}{\left (2+3 x^2\right )^{3/4}} \, dx &=\frac {2}{21} x^3 \sqrt [4]{2+3 x^2}-\frac {4}{7} \int \frac {x^2}{\left (2+3 x^2\right )^{3/4}} \, dx\\ &=-\frac {8}{63} x \sqrt [4]{2+3 x^2}+\frac {2}{21} x^3 \sqrt [4]{2+3 x^2}+\frac {16}{63} \int \frac {1}{\left (2+3 x^2\right )^{3/4}} \, dx\\ &=-\frac {8}{63} x \sqrt [4]{2+3 x^2}+\frac {2}{21} x^3 \sqrt [4]{2+3 x^2}+\frac {16\ 2^{3/4} F\left (\left .\frac {1}{2} \tan ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{63 \sqrt {3}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 5.92, size = 49, normalized size = 0.75 \begin {gather*} \frac {2}{63} x \left (\left (-4+3 x^2\right ) \sqrt [4]{2+3 x^2}+4 \sqrt [4]{2} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {3}{2};-\frac {3 x^2}{2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4/(2 + 3*x^2)^(3/4),x]

[Out]

(2*x*((-4 + 3*x^2)*(2 + 3*x^2)^(1/4) + 4*2^(1/4)*Hypergeometric2F1[1/2, 3/4, 3/2, (-3*x^2)/2]))/63

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 5 vs. order 4.
time = 0.06, size = 20, normalized size = 0.31

method	result	size
meijerg	\(\frac {2^{\frac {1}{4}} x^{5} \hypergeom \left (\left [\frac {3}{4}, \frac {5}{2}\right ], \left [\frac {7}{2}\right ], -\frac {3 x^{2}}{2}\right )}{10}\)	\(20\)
risch	\(\frac {2 x \left (3 x^{2}-4\right ) \left (3 x^{2}+2\right )^{\frac {1}{4}}}{63}+\frac {8 \,2^{\frac {1}{4}} x \hypergeom \left (\left [\frac {1}{2}, \frac {3}{4}\right ], \left [\frac {3}{2}\right ], -\frac {3 x^{2}}{2}\right )}{63}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(3*x^2+2)^(3/4),x,method=_RETURNVERBOSE)

[Out]

1/10*2^(1/4)*x^5*hypergeom([3/4,5/2],[7/2],-3/2*x^2)

________________________________________________________________________________________

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/(3*x^2+2)^(3/4),x, algorithm="maxima")

[Out]

integrate(x^4/(3*x^2 + 2)^(3/4), x)

________________________________________________________________________________________

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/(3*x^2+2)^(3/4),x, algorithm="fricas")

[Out]

integral(x^4/(3*x^2 + 2)^(3/4), x)

________________________________________________________________________________________

Sympy [C] Result contains complex when optimal does not.
time = 0.41, size = 27, normalized size = 0.42 \begin {gather*} \frac {\sqrt [4]{2} x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {3 x^{2} e^{i \pi }}{2}} \right )}}{10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2**(1/4)*x**5*hyper((3/4, 5/2), (7/2,), 3*x**2*exp_polar(I*pi)/2)/10

________________________________________________________________________________________

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/(3*x^2+2)^(3/4),x, algorithm="giac")

[Out]

integrate(x^4/(3*x^2 + 2)^(3/4), x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x^4}{{\left (3\,x^2+2\right )}^{3/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(3*x^2 + 2)^(3/4),x)

[Out]

int(x^4/(3*x^2 + 2)^(3/4), x)

________________________________________________________________________________________