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]

-1/12*2^(3/4)*arctan(1+(-6*x^2+4)^(1/4))-1/12*2^(3/4)*arctan(-1+2^(1/4)*(-3*x^2+2)^(1/4))+1/24*2^(3/4)*ln(-2^(
3/4)*(-3*x^2+2)^(1/4)+2^(1/2)+(-3*x^2+2)^(1/2))-1/24*2^(3/4)*ln(2^(3/4)*(-3*x^2+2)^(1/4)+2^(1/2)+(-3*x^2+2)^(1
/2))

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Rubi [A]  time = 0.12, antiderivative size = 143, normalized size of antiderivative = 1.00, 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 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 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])

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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.03, 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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fricas [B]  time = 1.25, size = 230, normalized size = 1.61 \[ \frac {1}{24} \cdot 8^{\frac {3}{4}} \sqrt {2} \arctan \left (\frac {1}{4} \cdot 8^{\frac {1}{4}} \sqrt {2} \sqrt {8^{\frac {3}{4}} \sqrt {2} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + 4 \, \sqrt {2} + 4 \, \sqrt {-3 \, x^{2} + 2}} - \frac {1}{2} \cdot 8^{\frac {1}{4}} \sqrt {2} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} - 1\right ) + \frac {1}{24} \cdot 8^{\frac {3}{4}} \sqrt {2} \arctan \left (\frac {1}{16} \cdot 8^{\frac {1}{4}} \sqrt {2} \sqrt {-16 \cdot 8^{\frac {3}{4}} \sqrt {2} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + 64 \, \sqrt {2} + 64 \, \sqrt {-3 \, x^{2} + 2}} - \frac {1}{2} \cdot 8^{\frac {1}{4}} \sqrt {2} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + 1\right ) - \frac {1}{96} \cdot 8^{\frac {3}{4}} \sqrt {2} \log \left (16 \cdot 8^{\frac {3}{4}} \sqrt {2} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + 64 \, \sqrt {2} + 64 \, \sqrt {-3 \, x^{2} + 2}\right ) + \frac {1}{96} \cdot 8^{\frac {3}{4}} \sqrt {2} \log \left (-16 \cdot 8^{\frac {3}{4}} \sqrt {2} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + 64 \, \sqrt {2} + 64 \, \sqrt {-3 \, x^{2} + 2}\right ) \]

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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giac [A]  time = 0.38, size = 118, normalized size = 0.83 \[ -\frac {1}{12} \cdot 2^{\frac {3}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {1}{4}} {\left (2^{\frac {3}{4}} + 2 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right )}\right ) - \frac {1}{12} \cdot 2^{\frac {3}{4}} \arctan \left (-\frac {1}{2} \cdot 2^{\frac {1}{4}} {\left (2^{\frac {3}{4}} - 2 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right )}\right ) - \frac {1}{24} \cdot 2^{\frac {3}{4}} \log \left (2^{\frac {3}{4}} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + \sqrt {2} + \sqrt {-3 \, x^{2} + 2}\right ) + \frac {1}{24} \cdot 2^{\frac {3}{4}} \log \left (-2^{\frac {3}{4}} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + \sqrt {2} + \sqrt {-3 \, x^{2} + 2}\right ) \]

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))

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maple [C]  time = 1.80, size = 189, normalized size = 1.32 \[ \frac {\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}+2\right )^{2}\right ) \ln \left (\frac {2 \left (-3 x^{2}+2\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}+2\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}+2\right )^{2}-3 x^{2}-2 \sqrt {-3 x^{2}+2}\, \RootOf \left (\textit {\_Z}^{4}+2\right )^{2}-2 \left (-3 x^{2}+2\right )^{\frac {3}{4}} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}+2\right )^{2}\right )}{3 x^{2}-4}\right )}{12}-\frac {\RootOf \left (\textit {\_Z}^{4}+2\right ) \ln \left (\frac {2 \left (-3 x^{2}+2\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}+2\right )^{3}-3 x^{2}+2 \sqrt {-3 x^{2}+2}\, \RootOf \left (\textit {\_Z}^{4}+2\right )^{2}+2 \left (-3 x^{2}+2\right )^{\frac {3}{4}} \RootOf \left (\textit {\_Z}^{4}+2\right )}{3 x^{2}-4}\right )}{12} \]

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]

1/12*RootOf(_Z^2+RootOf(_Z^4+2)^2)*ln((2*RootOf(_Z^2+RootOf(_Z^4+2)^2)*RootOf(_Z^4+2)^2*(-3*x^2+2)^(1/4)-2*Roo
tOf(_Z^2+RootOf(_Z^4+2)^2)*(-3*x^2+2)^(3/4)-2*RootOf(_Z^4+2)^2*(-3*x^2+2)^(1/2)-3*x^2)/(3*x^2-4))-1/12*RootOf(
_Z^4+2)*ln((2*RootOf(_Z^4+2)^3*(-3*x^2+2)^(1/4)+2*RootOf(_Z^4+2)^2*(-3*x^2+2)^(1/2)+2*RootOf(_Z^4+2)*(-3*x^2+2
)^(3/4)-3*x^2)/(3*x^2-4))

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maxima [A]  time = 2.02, size = 118, normalized size = 0.83 \[ -\frac {1}{12} \cdot 2^{\frac {3}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {1}{4}} {\left (2^{\frac {3}{4}} + 2 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right )}\right ) - \frac {1}{12} \cdot 2^{\frac {3}{4}} \arctan \left (-\frac {1}{2} \cdot 2^{\frac {1}{4}} {\left (2^{\frac {3}{4}} - 2 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right )}\right ) - \frac {1}{24} \cdot 2^{\frac {3}{4}} \log \left (2^{\frac {3}{4}} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + \sqrt {2} + \sqrt {-3 \, x^{2} + 2}\right ) + \frac {1}{24} \cdot 2^{\frac {3}{4}} \log \left (-2^{\frac {3}{4}} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + \sqrt {2} + \sqrt {-3 \, x^{2} + 2}\right ) \]

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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mupad [B]  time = 0.96, size = 49, normalized size = 0.34 \[ 2^{3/4}\,\mathrm {atan}\left (2^{1/4}\,{\left (2-3\,x^2\right )}^{1/4}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{12}-\frac {1}{12}{}\mathrm {i}\right )+2^{3/4}\,\mathrm {atan}\left (2^{1/4}\,{\left (2-3\,x^2\right )}^{1/4}\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{12}+\frac {1}{12}{}\mathrm {i}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- 2^(3/4)*atan(2^(1/4)*(2 - 3*x^2)^(1/4)*(1/2 - 1i/2))*(1/12 + 1i/12) - 2^(3/4)*atan(2^(1/4)*(2 - 3*x^2)^(1/4)
*(1/2 + 1i/2))*(1/12 - 1i/12)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {x}{3 x^{2} \left (2 - 3 x^{2}\right )^{\frac {3}{4}} - 4 \left (2 - 3 x^{2}\right )^{\frac {3}{4}}}\, dx \]

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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