3.964 \(\int \frac {x^4 \sqrt {-1+3 x^2}}{\sqrt {2-3 x^2}} \, dx\)

Optimal. Leaf size=99 \[ -\frac {7}{135} \sqrt {2-3 x^2} \sqrt {3 x^2-1} x-\frac {1}{15} \sqrt {2-3 x^2} \sqrt {3 x^2-1} x^3-\frac {2 F\left (\left .\cos ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{27 \sqrt {3}}-\frac {8 E\left (\left .\cos ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{45 \sqrt {3}} \]

[Out]

-8/135*(x^2)^(1/2)/x*EllipticE(1/2*(-6*x^2+4)^(1/2),2^(1/2))*3^(1/2)-2/81*(x^2)^(1/2)/x*EllipticF(1/2*(-6*x^2+
4)^(1/2),2^(1/2))*3^(1/2)-7/135*x*(-3*x^2+2)^(1/2)*(3*x^2-1)^(1/2)-1/15*x^3*(-3*x^2+2)^(1/2)*(3*x^2-1)^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {478, 582, 524, 425, 420} \[ -\frac {1}{15} \sqrt {2-3 x^2} \sqrt {3 x^2-1} x^3-\frac {7}{135} \sqrt {2-3 x^2} \sqrt {3 x^2-1} x-\frac {2 F\left (\left .\cos ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{27 \sqrt {3}}-\frac {8 E\left (\left .\cos ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{45 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*Sqrt[-1 + 3*x^2])/Sqrt[2 - 3*x^2],x]

[Out]

(-7*x*Sqrt[2 - 3*x^2]*Sqrt[-1 + 3*x^2])/135 - (x^3*Sqrt[2 - 3*x^2]*Sqrt[-1 + 3*x^2])/15 - (8*EllipticE[ArcCos[
Sqrt[3/2]*x], 2])/(45*Sqrt[3]) - (2*EllipticF[ArcCos[Sqrt[3/2]*x], 2])/(27*Sqrt[3])

Rule 420

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> -Simp[EllipticF[ArcCos[Rt[-(d/c), 2]
*x], (b*c)/(b*c - a*d)]/(Sqrt[c]*Rt[-(d/c), 2]*Sqrt[a - (b*c)/d]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] &
& GtQ[c, 0] && GtQ[a - (b*c)/d, 0]

Rule 425

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> -Simp[(Sqrt[a - (b*c)/d]*EllipticE[ArcCo
s[Rt[-(d/c), 2]*x], (b*c)/(b*c - a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] &&
GtQ[c, 0] && GtQ[a - (b*c)/d, 0]

Rule 478

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(n -
1)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(b*(m + n*(p + q) + 1)), x] - Dist[e^n/(b*(m + n*(p +
q) + 1)), Int[(e*x)^(m - n)*(a + b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[a*c*(m - n + 1) + (a*d*(m - n + 1) - n*q*(b
*c - a*d))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && GtQ[q, 0] &&
GtQ[m - n + 1, 0] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 524

Int[((e_) + (f_.)*(x_)^(n_))/(Sqrt[(a_) + (b_.)*(x_)^(n_)]*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/
b, Int[Sqrt[a + b*x^n]/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/(Sqrt[a + b*x^n]*Sqrt[c + d*x^n]),
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&  !(EqQ[n, 2] && ((PosQ[b/a] && PosQ[d/c]) || (NegQ[b/a] && (PosQ[
d/c] || (GtQ[a, 0] && ( !GtQ[c, 0] || SimplerSqrtQ[-(b/a), -(d/c)]))))))

Rule 582

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(f*g^(n - 1)*(g*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*d*(m + n*(p + q
+ 1) + 1)), x] - Dist[g^n/(b*d*(m + n*(p + q + 1) + 1)), Int[(g*x)^(m - n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*
f*c*(m - n + 1) + (a*f*d*(m + n*q + 1) + b*(f*c*(m + n*p + 1) - e*d*(m + n*(p + q + 1) + 1)))*x^n, x], x], x]
/; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && IGtQ[n, 0] && GtQ[m, n - 1]

Rubi steps

\begin {align*} \int \frac {x^4 \sqrt {-1+3 x^2}}{\sqrt {2-3 x^2}} \, dx &=-\frac {1}{15} x^3 \sqrt {2-3 x^2} \sqrt {-1+3 x^2}+\frac {1}{15} \int \frac {x^2 \left (-6+21 x^2\right )}{\sqrt {2-3 x^2} \sqrt {-1+3 x^2}} \, dx\\ &=-\frac {7}{135} x \sqrt {2-3 x^2} \sqrt {-1+3 x^2}-\frac {1}{15} x^3 \sqrt {2-3 x^2} \sqrt {-1+3 x^2}+\frac {1}{405} \int \frac {-42+216 x^2}{\sqrt {2-3 x^2} \sqrt {-1+3 x^2}} \, dx\\ &=-\frac {7}{135} x \sqrt {2-3 x^2} \sqrt {-1+3 x^2}-\frac {1}{15} x^3 \sqrt {2-3 x^2} \sqrt {-1+3 x^2}+\frac {2}{27} \int \frac {1}{\sqrt {2-3 x^2} \sqrt {-1+3 x^2}} \, dx+\frac {8}{45} \int \frac {\sqrt {-1+3 x^2}}{\sqrt {2-3 x^2}} \, dx\\ &=-\frac {7}{135} x \sqrt {2-3 x^2} \sqrt {-1+3 x^2}-\frac {1}{15} x^3 \sqrt {2-3 x^2} \sqrt {-1+3 x^2}-\frac {8 E\left (\left .\cos ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{45 \sqrt {3}}-\frac {2 F\left (\left .\cos ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{27 \sqrt {3}}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 92, normalized size = 0.93 \[ \frac {10 \sqrt {3-9 x^2} F\left (\left .\sin ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )-24 \sqrt {3-9 x^2} E\left (\left .\sin ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )-3 x \sqrt {2-3 x^2} \left (27 x^4+12 x^2-7\right )}{405 \sqrt {3 x^2-1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^4*Sqrt[-1 + 3*x^2])/Sqrt[2 - 3*x^2],x]

[Out]

(-3*x*Sqrt[2 - 3*x^2]*(-7 + 12*x^2 + 27*x^4) - 24*Sqrt[3 - 9*x^2]*EllipticE[ArcSin[Sqrt[3/2]*x], 2] + 10*Sqrt[
3 - 9*x^2]*EllipticF[ArcSin[Sqrt[3/2]*x], 2])/(405*Sqrt[-1 + 3*x^2])

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fricas [F]  time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {3 \, x^{2} - 1} \sqrt {-3 \, x^{2} + 2} x^{4}}{3 \, x^{2} - 2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(3*x^2-1)^(1/2)/(-3*x^2+2)^(1/2),x, algorithm="fricas")

[Out]

integral(-sqrt(3*x^2 - 1)*sqrt(-3*x^2 + 2)*x^4/(3*x^2 - 2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {3 \, x^{2} - 1} x^{4}}{\sqrt {-3 \, x^{2} + 2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(3*x^2-1)^(1/2)/(-3*x^2+2)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(3*x^2 - 1)*x^4/sqrt(-3*x^2 + 2), x)

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maple [A]  time = 0.20, size = 135, normalized size = 1.36 \[ -\frac {\sqrt {3 x^{2}-1}\, \sqrt {2}\, \sqrt {-6 x^{2}+4}\, \left (243 x^{7}-54 x^{5}-135 x^{3}+42 x -12 \sqrt {3}\, \sqrt {2}\, \sqrt {-6 x^{2}+4}\, \sqrt {-3 x^{2}+1}\, \EllipticE \left (\frac {\sqrt {3}\, \sqrt {2}\, x}{2}, \sqrt {2}\right )+5 \sqrt {3}\, \sqrt {2}\, \sqrt {-6 x^{2}+4}\, \sqrt {-3 x^{2}+1}\, \EllipticF \left (\frac {\sqrt {3}\, \sqrt {2}\, x}{2}, \sqrt {2}\right )\right )}{810 \left (9 x^{4}-9 x^{2}+2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(3*x^2-1)^(1/2)/(-3*x^2+2)^(1/2),x)

[Out]

-1/810*(3*x^2-1)^(1/2)*2^(1/2)*(-6*x^2+4)^(1/2)*(243*x^7-54*x^5+5*3^(1/2)*2^(1/2)*(-6*x^2+4)^(1/2)*(-3*x^2+1)^
(1/2)*EllipticF(1/2*3^(1/2)*2^(1/2)*x,2^(1/2))-12*3^(1/2)*2^(1/2)*(-6*x^2+4)^(1/2)*(-3*x^2+1)^(1/2)*EllipticE(
1/2*3^(1/2)*2^(1/2)*x,2^(1/2))-135*x^3+42*x)/(9*x^4-9*x^2+2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {3 \, x^{2} - 1} x^{4}}{\sqrt {-3 \, x^{2} + 2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(3*x^2-1)^(1/2)/(-3*x^2+2)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(3*x^2 - 1)*x^4/sqrt(-3*x^2 + 2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^4\,\sqrt {3\,x^2-1}}{\sqrt {2-3\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(3*x^2 - 1)^(1/2))/(2 - 3*x^2)^(1/2),x)

[Out]

int((x^4*(3*x^2 - 1)^(1/2))/(2 - 3*x^2)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4} \sqrt {3 x^{2} - 1}}{\sqrt {2 - 3 x^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(3*x**2-1)**(1/2)/(-3*x**2+2)**(1/2),x)

[Out]

Integral(x**4*sqrt(3*x**2 - 1)/sqrt(2 - 3*x**2), x)

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