3.1064 \(\int \frac{x}{(2-3 x^2)^{3/4} (4-3 x^2)} \, dx\)

Optimal. Leaf size=143 \[ \frac{\log \left (\sqrt{2-3 x^2}-2^{3/4} \sqrt [4]{2-3 x^2}+\sqrt{2}\right )}{12 \sqrt [4]{2}}-\frac{\log \left (\sqrt{2-3 x^2}+2^{3/4} \sqrt [4]{2-3 x^2}+\sqrt{2}\right )}{12 \sqrt [4]{2}}-\frac{\tan ^{-1}\left (\sqrt [4]{4-6 x^2}+1\right )}{6 \sqrt [4]{2}}+\frac{\tan ^{-1}\left (1-\sqrt [4]{2} \sqrt [4]{2-3 x^2}\right )}{6 \sqrt [4]{2}} \]

[Out]

-ArcTan[1 + (4 - 6*x^2)^(1/4)]/(6*2^(1/4)) + ArcTan[1 - 2^(1/4)*(2 - 3*x^2)^(1/4)]/(6*2^(1/4)) + Log[Sqrt[2] -
2^(3/4)*(2 - 3*x^2)^(1/4) + Sqrt[2 - 3*x^2]]/(12*2^(1/4)) - Log[Sqrt[2] + 2^(3/4)*(2 - 3*x^2)^(1/4) + Sqrt[2
- 3*x^2]]/(12*2^(1/4))

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Rubi [A]  time = 0.117628, antiderivative size = 143, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {444, 63, 211, 1165, 628, 1162, 617, 204} \[ \frac{\log \left (\sqrt{2-3 x^2}-2^{3/4} \sqrt [4]{2-3 x^2}+\sqrt{2}\right )}{12 \sqrt [4]{2}}-\frac{\log \left (\sqrt{2-3 x^2}+2^{3/4} \sqrt [4]{2-3 x^2}+\sqrt{2}\right )}{12 \sqrt [4]{2}}-\frac{\tan ^{-1}\left (\sqrt [4]{4-6 x^2}+1\right )}{6 \sqrt [4]{2}}+\frac{\tan ^{-1}\left (1-\sqrt [4]{2} \sqrt [4]{2-3 x^2}\right )}{6 \sqrt [4]{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcTan[1 + (4 - 6*x^2)^(1/4)]/(6*2^(1/4)) + ArcTan[1 - 2^(1/4)*(2 - 3*x^2)^(1/4)]/(6*2^(1/4)) + Log[Sqrt[2] -
2^(3/4)*(2 - 3*x^2)^(1/4) + Sqrt[2 - 3*x^2]]/(12*2^(1/4)) - Log[Sqrt[2] + 2^(3/4)*(2 - 3*x^2)^(1/4) + Sqrt[2
- 3*x^2]]/(12*2^(1/4))

Rule 444

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rubi steps

\begin{align*} \int \frac{x}{\left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{(2-3 x)^{3/4} (4-3 x)} \, dx,x,x^2\right )\\ &=-\left (\frac{2}{3} \operatorname{Subst}\left (\int \frac{1}{2+x^4} \, dx,x,\sqrt [4]{2-3 x^2}\right )\right )\\ &=-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2}-x^2}{2+x^4} \, dx,x,\sqrt [4]{2-3 x^2}\right )}{3 \sqrt{2}}-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2}+x^2}{2+x^4} \, dx,x,\sqrt [4]{2-3 x^2}\right )}{3 \sqrt{2}}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt{2}-2^{3/4} x+x^2} \, dx,x,\sqrt [4]{2-3 x^2}\right )}{6 \sqrt{2}}-\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt{2}+2^{3/4} x+x^2} \, dx,x,\sqrt [4]{2-3 x^2}\right )}{6 \sqrt{2}}+\frac{\operatorname{Subst}\left (\int \frac{2^{3/4}+2 x}{-\sqrt{2}-2^{3/4} x-x^2} \, dx,x,\sqrt [4]{2-3 x^2}\right )}{12 \sqrt [4]{2}}+\frac{\operatorname{Subst}\left (\int \frac{2^{3/4}-2 x}{-\sqrt{2}+2^{3/4} x-x^2} \, dx,x,\sqrt [4]{2-3 x^2}\right )}{12 \sqrt [4]{2}}\\ &=\frac{\log \left (\sqrt{2}-2^{3/4} \sqrt [4]{2-3 x^2}+\sqrt{2-3 x^2}\right )}{12 \sqrt [4]{2}}-\frac{\log \left (\sqrt{2}+2^{3/4} \sqrt [4]{2-3 x^2}+\sqrt{2-3 x^2}\right )}{12 \sqrt [4]{2}}-\frac{\operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt [4]{4-6 x^2}\right )}{6 \sqrt [4]{2}}+\frac{\operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt [4]{4-6 x^2}\right )}{6 \sqrt [4]{2}}\\ &=-\frac{\tan ^{-1}\left (1+\sqrt [4]{4-6 x^2}\right )}{6 \sqrt [4]{2}}+\frac{\tan ^{-1}\left (1-\sqrt [4]{2} \sqrt [4]{2-3 x^2}\right )}{6 \sqrt [4]{2}}+\frac{\log \left (\sqrt{2}-2^{3/4} \sqrt [4]{2-3 x^2}+\sqrt{2-3 x^2}\right )}{12 \sqrt [4]{2}}-\frac{\log \left (\sqrt{2}+2^{3/4} \sqrt [4]{2-3 x^2}+\sqrt{2-3 x^2}\right )}{12 \sqrt [4]{2}}\\ \end{align*}

Mathematica [A]  time = 0.02572, size = 117, normalized size = 0.82 \[ \frac{\log \left (\sqrt{2-3 x^2}-2^{3/4} \sqrt [4]{2-3 x^2}+\sqrt{2}\right )-\log \left (\sqrt{2-3 x^2}+2^{3/4} \sqrt [4]{2-3 x^2}+\sqrt{2}\right )+2 \tan ^{-1}\left (1-\sqrt [4]{4-6 x^2}\right )-2 \tan ^{-1}\left (\sqrt [4]{4-6 x^2}+1\right )}{12 \sqrt [4]{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*ArcTan[1 - (4 - 6*x^2)^(1/4)] - 2*ArcTan[1 + (4 - 6*x^2)^(1/4)] + Log[Sqrt[2] - 2^(3/4)*(2 - 3*x^2)^(1/4) +
Sqrt[2 - 3*x^2]] - Log[Sqrt[2] + 2^(3/4)*(2 - 3*x^2)^(1/4) + Sqrt[2 - 3*x^2]])/(12*2^(1/4))

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Maple [F]  time = 0.042, size = 0, normalized size = 0. \begin{align*} \int{\frac{x}{-3\,{x}^{2}+4} \left ( -3\,{x}^{2}+2 \right ) ^{-{\frac{3}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.55516, size = 159, normalized size = 1.11 \begin{align*} -\frac{1}{12} \cdot 2^{\frac{3}{4}} \arctan \left (\frac{1}{2} \cdot 2^{\frac{1}{4}}{\left (2^{\frac{3}{4}} + 2 \,{\left (-3 \, x^{2} + 2\right )}^{\frac{1}{4}}\right )}\right ) - \frac{1}{12} \cdot 2^{\frac{3}{4}} \arctan \left (-\frac{1}{2} \cdot 2^{\frac{1}{4}}{\left (2^{\frac{3}{4}} - 2 \,{\left (-3 \, x^{2} + 2\right )}^{\frac{1}{4}}\right )}\right ) - \frac{1}{24} \cdot 2^{\frac{3}{4}} \log \left (2^{\frac{3}{4}}{\left (-3 \, x^{2} + 2\right )}^{\frac{1}{4}} + \sqrt{2} + \sqrt{-3 \, x^{2} + 2}\right ) + \frac{1}{24} \cdot 2^{\frac{3}{4}} \log \left (-2^{\frac{3}{4}}{\left (-3 \, x^{2} + 2\right )}^{\frac{1}{4}} + \sqrt{2} + \sqrt{-3 \, x^{2} + 2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*2^(3/4)*arctan(1/2*2^(1/4)*(2^(3/4) + 2*(-3*x^2 + 2)^(1/4))) - 1/12*2^(3/4)*arctan(-1/2*2^(1/4)*(2^(3/4)
- 2*(-3*x^2 + 2)^(1/4))) - 1/24*2^(3/4)*log(2^(3/4)*(-3*x^2 + 2)^(1/4) + sqrt(2) + sqrt(-3*x^2 + 2)) + 1/24*2
^(3/4)*log(-2^(3/4)*(-3*x^2 + 2)^(1/4) + sqrt(2) + sqrt(-3*x^2 + 2))

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Fricas [B]  time = 1.68018, size = 741, normalized size = 5.18 \begin{align*} \frac{1}{24} \cdot 8^{\frac{3}{4}} \sqrt{2} \arctan \left (\frac{1}{4} \cdot 8^{\frac{1}{4}} \sqrt{2} \sqrt{8^{\frac{3}{4}} \sqrt{2}{\left (-3 \, x^{2} + 2\right )}^{\frac{1}{4}} + 4 \, \sqrt{2} + 4 \, \sqrt{-3 \, x^{2} + 2}} - \frac{1}{2} \cdot 8^{\frac{1}{4}} \sqrt{2}{\left (-3 \, x^{2} + 2\right )}^{\frac{1}{4}} - 1\right ) + \frac{1}{24} \cdot 8^{\frac{3}{4}} \sqrt{2} \arctan \left (\frac{1}{16} \cdot 8^{\frac{1}{4}} \sqrt{2} \sqrt{-16 \cdot 8^{\frac{3}{4}} \sqrt{2}{\left (-3 \, x^{2} + 2\right )}^{\frac{1}{4}} + 64 \, \sqrt{2} + 64 \, \sqrt{-3 \, x^{2} + 2}} - \frac{1}{2} \cdot 8^{\frac{1}{4}} \sqrt{2}{\left (-3 \, x^{2} + 2\right )}^{\frac{1}{4}} + 1\right ) - \frac{1}{96} \cdot 8^{\frac{3}{4}} \sqrt{2} \log \left (16 \cdot 8^{\frac{3}{4}} \sqrt{2}{\left (-3 \, x^{2} + 2\right )}^{\frac{1}{4}} + 64 \, \sqrt{2} + 64 \, \sqrt{-3 \, x^{2} + 2}\right ) + \frac{1}{96} \cdot 8^{\frac{3}{4}} \sqrt{2} \log \left (-16 \cdot 8^{\frac{3}{4}} \sqrt{2}{\left (-3 \, x^{2} + 2\right )}^{\frac{1}{4}} + 64 \, \sqrt{2} + 64 \, \sqrt{-3 \, x^{2} + 2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*8^(3/4)*sqrt(2)*arctan(1/4*8^(1/4)*sqrt(2)*sqrt(8^(3/4)*sqrt(2)*(-3*x^2 + 2)^(1/4) + 4*sqrt(2) + 4*sqrt(-
3*x^2 + 2)) - 1/2*8^(1/4)*sqrt(2)*(-3*x^2 + 2)^(1/4) - 1) + 1/24*8^(3/4)*sqrt(2)*arctan(1/16*8^(1/4)*sqrt(2)*s
qrt(-16*8^(3/4)*sqrt(2)*(-3*x^2 + 2)^(1/4) + 64*sqrt(2) + 64*sqrt(-3*x^2 + 2)) - 1/2*8^(1/4)*sqrt(2)*(-3*x^2 +
2)^(1/4) + 1) - 1/96*8^(3/4)*sqrt(2)*log(16*8^(3/4)*sqrt(2)*(-3*x^2 + 2)^(1/4) + 64*sqrt(2) + 64*sqrt(-3*x^2
+ 2)) + 1/96*8^(3/4)*sqrt(2)*log(-16*8^(3/4)*sqrt(2)*(-3*x^2 + 2)^(1/4) + 64*sqrt(2) + 64*sqrt(-3*x^2 + 2))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{x}{3 x^{2} \left (2 - 3 x^{2}\right )^{\frac{3}{4}} - 4 \left (2 - 3 x^{2}\right )^{\frac{3}{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.21157, size = 159, normalized size = 1.11 \begin{align*} -\frac{1}{12} \cdot 2^{\frac{3}{4}} \arctan \left (\frac{1}{2} \cdot 2^{\frac{1}{4}}{\left (2^{\frac{3}{4}} + 2 \,{\left (-3 \, x^{2} + 2\right )}^{\frac{1}{4}}\right )}\right ) - \frac{1}{12} \cdot 2^{\frac{3}{4}} \arctan \left (-\frac{1}{2} \cdot 2^{\frac{1}{4}}{\left (2^{\frac{3}{4}} - 2 \,{\left (-3 \, x^{2} + 2\right )}^{\frac{1}{4}}\right )}\right ) - \frac{1}{24} \cdot 2^{\frac{3}{4}} \log \left (2^{\frac{3}{4}}{\left (-3 \, x^{2} + 2\right )}^{\frac{1}{4}} + \sqrt{2} + \sqrt{-3 \, x^{2} + 2}\right ) + \frac{1}{24} \cdot 2^{\frac{3}{4}} \log \left (-2^{\frac{3}{4}}{\left (-3 \, x^{2} + 2\right )}^{\frac{1}{4}} + \sqrt{2} + \sqrt{-3 \, x^{2} + 2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*2^(3/4)*arctan(1/2*2^(1/4)*(2^(3/4) + 2*(-3*x^2 + 2)^(1/4))) - 1/12*2^(3/4)*arctan(-1/2*2^(1/4)*(2^(3/4)
- 2*(-3*x^2 + 2)^(1/4))) - 1/24*2^(3/4)*log(2^(3/4)*(-3*x^2 + 2)^(1/4) + sqrt(2) + sqrt(-3*x^2 + 2)) + 1/24*2
^(3/4)*log(-2^(3/4)*(-3*x^2 + 2)^(1/4) + sqrt(2) + sqrt(-3*x^2 + 2))