3.11.66 \(\int \frac {1}{x^3 (2-3 x^2)^{3/4} (4-3 x^2)} \, dx\) [1066]

Optimal. Leaf size=215 \[ -\frac {\sqrt [4]{2-3 x^2}}{16 x^2}-\frac {15 \tan ^{-1}\left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{32\ 2^{3/4}}-\frac {3 \tan ^{-1}\left (1+\sqrt [4]{4-6 x^2}\right )}{32 \sqrt [4]{2}}+\frac {3 \tan ^{-1}\left (1-\sqrt [4]{2} \sqrt [4]{2-3 x^2}\right )}{32 \sqrt [4]{2}}-\frac {15 \tanh ^{-1}\left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{32\ 2^{3/4}}+\frac {3 \log \left (\sqrt {2}-2^{3/4} \sqrt [4]{2-3 x^2}+\sqrt {2-3 x^2}\right )}{64 \sqrt [4]{2}}-\frac {3 \log \left (\sqrt {2}+2^{3/4} \sqrt [4]{2-3 x^2}+\sqrt {2-3 x^2}\right )}{64 \sqrt [4]{2}} \]

[Out]

-1/16*(-3*x^2+2)^(1/4)/x^2-15/64*arctan(1/2*2^(3/4)*(-3*x^2+2)^(1/4))*2^(1/4)-3/64*2^(3/4)*arctan(1+(-6*x^2+4)
^(1/4))-3/64*2^(3/4)*arctan(-1+2^(1/4)*(-3*x^2+2)^(1/4))-15/64*arctanh(1/2*2^(3/4)*(-3*x^2+2)^(1/4))*2^(1/4)+3
/128*2^(3/4)*ln(-2^(3/4)*(-3*x^2+2)^(1/4)+2^(1/2)+(-3*x^2+2)^(1/2))-3/128*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 = 0.16, antiderivative size = 215, normalized size of antiderivative = 1.00, number of steps used = 24, number of rules used = 14, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {454, 272, 44, 65, 218, 212, 209, 455, 217, 1179, 642, 1176, 631, 210} \begin {gather*} -\frac {15 \text {ArcTan}\left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{32\ 2^{3/4}}-\frac {3 \text {ArcTan}\left (\sqrt [4]{4-6 x^2}+1\right )}{32 \sqrt [4]{2}}+\frac {3 \text {ArcTan}\left (1-\sqrt [4]{2} \sqrt [4]{2-3 x^2}\right )}{32 \sqrt [4]{2}}-\frac {\sqrt [4]{2-3 x^2}}{16 x^2}+\frac {3 \log \left (\sqrt {2-3 x^2}-2^{3/4} \sqrt [4]{2-3 x^2}+\sqrt {2}\right )}{64 \sqrt [4]{2}}-\frac {3 \log \left (\sqrt {2-3 x^2}+2^{3/4} \sqrt [4]{2-3 x^2}+\sqrt {2}\right )}{64 \sqrt [4]{2}}-\frac {15 \tanh ^{-1}\left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{32\ 2^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/16*(2 - 3*x^2)^(1/4)/x^2 - (15*ArcTan[(2 - 3*x^2)^(1/4)/2^(1/4)])/(32*2^(3/4)) - (3*ArcTan[1 + (4 - 6*x^2)^
(1/4)])/(32*2^(1/4)) + (3*ArcTan[1 - 2^(1/4)*(2 - 3*x^2)^(1/4)])/(32*2^(1/4)) - (15*ArcTanh[(2 - 3*x^2)^(1/4)/
2^(1/4)])/(32*2^(3/4)) + (3*Log[Sqrt[2] - 2^(3/4)*(2 - 3*x^2)^(1/4) + Sqrt[2 - 3*x^2]])/(64*2^(1/4)) - (3*Log[
Sqrt[2] + 2^(3/4)*(2 - 3*x^2)^(1/4) + Sqrt[2 - 3*x^2]])/(64*2^(1/4))

Rule 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

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 209

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

Rule 210

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

Rule 212

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

Rule 217

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 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[a/b, 0]

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 454

Int[(x_)^(m_)/(((a_) + (b_.)*(x_)^2)^(3/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Int[ExpandIntegrand[x^m/((a +
b*x^2)^(3/4)*(c + d*x^2)), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c - 2*a*d, 0] && IntegerQ[m] && (PosQ[a]
|| IntegerQ[m/2])

Rule 455

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 631

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 642

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 1176

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 1179

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 {1}{x^3 \left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx &=\int \left (\frac {1}{4 x^3 \left (2-3 x^2\right )^{3/4}}+\frac {3}{16 x \left (2-3 x^2\right )^{3/4}}-\frac {9 x}{16 \left (2-3 x^2\right )^{3/4} \left (-4+3 x^2\right )}\right ) \, dx\\ &=\frac {3}{16} \int \frac {1}{x \left (2-3 x^2\right )^{3/4}} \, dx+\frac {1}{4} \int \frac {1}{x^3 \left (2-3 x^2\right )^{3/4}} \, dx-\frac {9}{16} \int \frac {x}{\left (2-3 x^2\right )^{3/4} \left (-4+3 x^2\right )} \, dx\\ &=\frac {3}{32} \text {Subst}\left (\int \frac {1}{(2-3 x)^{3/4} x} \, dx,x,x^2\right )+\frac {1}{8} \text {Subst}\left (\int \frac {1}{(2-3 x)^{3/4} x^2} \, dx,x,x^2\right )-\frac {9}{32} \text {Subst}\left (\int \frac {1}{(2-3 x)^{3/4} (-4+3 x)} \, dx,x,x^2\right )\\ &=-\frac {\sqrt [4]{2-3 x^2}}{16 x^2}-\frac {1}{8} \text {Subst}\left (\int \frac {1}{\frac {2}{3}-\frac {x^4}{3}} \, dx,x,\sqrt [4]{2-3 x^2}\right )+\frac {9}{64} \text {Subst}\left (\int \frac {1}{(2-3 x)^{3/4} x} \, dx,x,x^2\right )+\frac {3}{8} \text {Subst}\left (\int \frac {1}{-2-x^4} \, dx,x,\sqrt [4]{2-3 x^2}\right )\\ &=-\frac {\sqrt [4]{2-3 x^2}}{16 x^2}-\frac {3}{16} \text {Subst}\left (\int \frac {1}{\frac {2}{3}-\frac {x^4}{3}} \, dx,x,\sqrt [4]{2-3 x^2}\right )-\frac {3 \text {Subst}\left (\int \frac {1}{\sqrt {2}-x^2} \, dx,x,\sqrt [4]{2-3 x^2}\right )}{16 \sqrt {2}}-\frac {3 \text {Subst}\left (\int \frac {1}{\sqrt {2}+x^2} \, dx,x,\sqrt [4]{2-3 x^2}\right )}{16 \sqrt {2}}+\frac {3 \text {Subst}\left (\int \frac {\sqrt {2}-x^2}{-2-x^4} \, dx,x,\sqrt [4]{2-3 x^2}\right )}{16 \sqrt {2}}+\frac {3 \text {Subst}\left (\int \frac {\sqrt {2}+x^2}{-2-x^4} \, dx,x,\sqrt [4]{2-3 x^2}\right )}{16 \sqrt {2}}\\ &=-\frac {\sqrt [4]{2-3 x^2}}{16 x^2}-\frac {3 \tan ^{-1}\left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{16\ 2^{3/4}}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{16\ 2^{3/4}}-\frac {3 \text {Subst}\left (\int \frac {1}{\sqrt {2}-2^{3/4} x+x^2} \, dx,x,\sqrt [4]{2-3 x^2}\right )}{32 \sqrt {2}}-\frac {3 \text {Subst}\left (\int \frac {1}{\sqrt {2}+2^{3/4} x+x^2} \, dx,x,\sqrt [4]{2-3 x^2}\right )}{32 \sqrt {2}}-\frac {9 \text {Subst}\left (\int \frac {1}{\sqrt {2}-x^2} \, dx,x,\sqrt [4]{2-3 x^2}\right )}{32 \sqrt {2}}-\frac {9 \text {Subst}\left (\int \frac {1}{\sqrt {2}+x^2} \, dx,x,\sqrt [4]{2-3 x^2}\right )}{32 \sqrt {2}}+\frac {3 \text {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 )}{64 \sqrt [4]{2}}+\frac {3 \text {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 )}{64 \sqrt [4]{2}}\\ &=-\frac {\sqrt [4]{2-3 x^2}}{16 x^2}-\frac {15 \tan ^{-1}\left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{32\ 2^{3/4}}-\frac {15 \tanh ^{-1}\left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{32\ 2^{3/4}}+\frac {3 \log \left (\sqrt {2}-2^{3/4} \sqrt [4]{2-3 x^2}+\sqrt {2-3 x^2}\right )}{64 \sqrt [4]{2}}-\frac {3 \log \left (\sqrt {2}+2^{3/4} \sqrt [4]{2-3 x^2}+\sqrt {2-3 x^2}\right )}{64 \sqrt [4]{2}}-\frac {3 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt [4]{4-6 x^2}\right )}{32 \sqrt [4]{2}}+\frac {3 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt [4]{4-6 x^2}\right )}{32 \sqrt [4]{2}}\\ &=-\frac {\sqrt [4]{2-3 x^2}}{16 x^2}-\frac {15 \tan ^{-1}\left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{32\ 2^{3/4}}-\frac {3 \tan ^{-1}\left (1+\sqrt [4]{4-6 x^2}\right )}{32 \sqrt [4]{2}}+\frac {3 \tan ^{-1}\left (1-\sqrt [4]{2} \sqrt [4]{2-3 x^2}\right )}{32 \sqrt [4]{2}}-\frac {15 \tanh ^{-1}\left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{32\ 2^{3/4}}+\frac {3 \log \left (\sqrt {2}-2^{3/4} \sqrt [4]{2-3 x^2}+\sqrt {2-3 x^2}\right )}{64 \sqrt [4]{2}}-\frac {3 \log \left (\sqrt {2}+2^{3/4} \sqrt [4]{2-3 x^2}+\sqrt {2-3 x^2}\right )}{64 \sqrt [4]{2}}\\ \end {align*}

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Mathematica [A]
time = 0.25, size = 155, normalized size = 0.72 \begin {gather*} -\frac {4 \sqrt [4]{2-3 x^2}+15 \sqrt [4]{2} x^2 \tan ^{-1}\left (\sqrt [4]{1-\frac {3 x^2}{2}}\right )-3\ 2^{3/4} x^2 \tan ^{-1}\left (\frac {\sqrt {2}-\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )+15 \sqrt [4]{2} x^2 \tanh ^{-1}\left (\sqrt [4]{1-\frac {3 x^2}{2}}\right )+3\ 2^{3/4} x^2 \tanh ^{-1}\left (\frac {2 \sqrt [4]{4-6 x^2}}{2+\sqrt {4-6 x^2}}\right )}{64 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/64*(4*(2 - 3*x^2)^(1/4) + 15*2^(1/4)*x^2*ArcTan[(1 - (3*x^2)/2)^(1/4)] - 3*2^(3/4)*x^2*ArcTan[(Sqrt[2] - Sq
rt[2 - 3*x^2])/(2^(3/4)*(2 - 3*x^2)^(1/4))] + 15*2^(1/4)*x^2*ArcTanh[(1 - (3*x^2)/2)^(1/4)] + 3*2^(3/4)*x^2*Ar
cTanh[(2*(4 - 6*x^2)^(1/4))/(2 + Sqrt[4 - 6*x^2])])/x^2

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 25.01, size = 577, normalized size = 2.68

method	result	size
trager	\(\text {Expression too large to display}\)	\(577\)
risch	\(\text {Expression too large to display}\)	\(1542\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(-3*x^2+2)^(1/4)/x^2+15/128*RootOf(_Z^4-2)*ln(-(3*RootOf(_Z^4-2)^3*x^2-4*RootOf(_Z^4-2)^3+4*RootOf(_Z^4-
2)^2*(-3*x^2+2)^(1/4)-4*(-3*x^2+2)^(1/2)*RootOf(_Z^4-2)+4*(-3*x^2+2)^(3/4))/x^2)-15/128*RootOf(_Z^2+RootOf(_Z^
4-2)^2)*ln(-(3*RootOf(_Z^4-2)^2*RootOf(_Z^2+RootOf(_Z^4-2)^2)*x^2-4*RootOf(_Z^4-2)^2*RootOf(_Z^2+RootOf(_Z^4-2
)^2)+4*RootOf(_Z^2+RootOf(_Z^4-2)^2)*(-3*x^2+2)^(1/2)-4*RootOf(_Z^4-2)^2*(-3*x^2+2)^(1/4)+4*(-3*x^2+2)^(3/4))/
x^2)+3/128*ln(-(4*(-3*x^2+2)^(1/2)*RootOf(_Z^4-2)^3-4*RootOf(_Z^4-2)^2*(-3*x^2+2)^(1/4)-3*RootOf(_Z^4-2)*x^2-4
*(-3*x^2+2)^(3/4)+4*RootOf(_Z^4-2))/(3*x^2-4))*RootOf(_Z^4-2)^3+3/128*ln(-(4*(-3*x^2+2)^(1/2)*RootOf(_Z^4-2)^3
-4*RootOf(_Z^4-2)^2*(-3*x^2+2)^(1/4)-3*RootOf(_Z^4-2)*x^2-4*(-3*x^2+2)^(3/4)+4*RootOf(_Z^4-2))/(3*x^2-4))*Root
Of(_Z^4-2)^2*RootOf(_Z^2+RootOf(_Z^4-2)^2)-3/64*RootOf(_Z^4-2)^2*RootOf(_Z^2+RootOf(_Z^4-2)^2)*ln((2*RootOf(_Z
^4-2)^2*(-3*x^2+2)^(1/2)*RootOf(_Z^2+RootOf(_Z^4-2)^2)-2*(-3*x^2+2)^(1/2)*RootOf(_Z^4-2)^3-4*RootOf(_Z^2+RootO
f(_Z^4-2)^2)*RootOf(_Z^4-2)*(-3*x^2+2)^(1/4)+3*RootOf(_Z^2+RootOf(_Z^4-2)^2)*x^2+3*RootOf(_Z^4-2)*x^2+4*(-3*x^
2+2)^(3/4))/(3*x^2-4))

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 352 vs. \(2 (158) = 316\).
time = 1.64, size = 352, normalized size = 1.64 \begin {gather*} \frac {12 \cdot 8^{\frac {3}{4}} \sqrt {2} x^{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 ) + 12 \cdot 8^{\frac {3}{4}} \sqrt {2} x^{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 ) - 3 \cdot 8^{\frac {3}{4}} \sqrt {2} x^{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 ) + 3 \cdot 8^{\frac {3}{4}} \sqrt {2} x^{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 ) + 60 \cdot 8^{\frac {3}{4}} x^{2} \arctan \left (\frac {1}{2} \cdot 8^{\frac {1}{4}} \sqrt {\sqrt {2} + \sqrt {-3 \, x^{2} + 2}} - \frac {1}{2} \cdot 8^{\frac {1}{4}} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) - 15 \cdot 8^{\frac {3}{4}} x^{2} \log \left (8^{\frac {3}{4}} + 4 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) + 15 \cdot 8^{\frac {3}{4}} x^{2} \log \left (-8^{\frac {3}{4}} + 4 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) - 32 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}}{512 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/512*(12*8^(3/4)*sqrt(2)*x^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) + 12*8^(3/4)*sqrt(2)*x^2*arctan(1/16*8^(1/4
)*sqrt(2)*sqrt(-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) - 3*8^(3/4)*sqrt(2)*x^2*log(16*8^(3/4)*sqrt(2)*(-3*x^2 + 2)^(1/4) + 64*sqrt(2) + 64*
sqrt(-3*x^2 + 2)) + 3*8^(3/4)*sqrt(2)*x^2*log(-16*8^(3/4)*sqrt(2)*(-3*x^2 + 2)^(1/4) + 64*sqrt(2) + 64*sqrt(-3
*x^2 + 2)) + 60*8^(3/4)*x^2*arctan(1/2*8^(1/4)*sqrt(sqrt(2) + sqrt(-3*x^2 + 2)) - 1/2*8^(1/4)*(-3*x^2 + 2)^(1/
4)) - 15*8^(3/4)*x^2*log(8^(3/4) + 4*(-3*x^2 + 2)^(1/4)) + 15*8^(3/4)*x^2*log(-8^(3/4) + 4*(-3*x^2 + 2)^(1/4))
- 32*(-3*x^2 + 2)^(1/4))/x^2

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {1}{3 x^{5} \left (2 - 3 x^{2}\right )^{\frac {3}{4}} - 4 x^{3} \left (2 - 3 x^{2}\right )^{\frac {3}{4}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(1/(3*x**5*(2 - 3*x**2)**(3/4) - 4*x**3*(2 - 3*x**2)**(3/4)), x)

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Giac [A]
time = 0.84, size = 192, normalized size = 0.89 \begin {gather*} -\frac {3}{64} \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 {3}{64} \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 {3}{128} \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 {3}{128} \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 {15}{64} \cdot 2^{\frac {1}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {3}{4}} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) - \frac {15}{128} \cdot 2^{\frac {1}{4}} \log \left (2^{\frac {1}{4}} + {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) + \frac {15}{128} \cdot 2^{\frac {1}{4}} \log \left (2^{\frac {1}{4}} - {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) - \frac {{\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}}{16 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/64*2^(3/4)*arctan(1/2*2^(1/4)*(2^(3/4) + 2*(-3*x^2 + 2)^(1/4))) - 3/64*2^(3/4)*arctan(-1/2*2^(1/4)*(2^(3/4)
- 2*(-3*x^2 + 2)^(1/4))) - 3/128*2^(3/4)*log(2^(3/4)*(-3*x^2 + 2)^(1/4) + sqrt(2) + sqrt(-3*x^2 + 2)) + 3/128
*2^(3/4)*log(-2^(3/4)*(-3*x^2 + 2)^(1/4) + sqrt(2) + sqrt(-3*x^2 + 2)) - 15/64*2^(1/4)*arctan(1/2*2^(3/4)*(-3*
x^2 + 2)^(1/4)) - 15/128*2^(1/4)*log(2^(1/4) + (-3*x^2 + 2)^(1/4)) + 15/128*2^(1/4)*log(2^(1/4) - (-3*x^2 + 2)
^(1/4)) - 1/16*(-3*x^2 + 2)^(1/4)/x^2

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Mupad [B]
time = 0.56, size = 107, normalized size = 0.50 \begin {gather*} -\frac {{\left (2-3\,x^2\right )}^{1/4}}{16\,x^2}-\frac {15\,2^{1/4}\,\mathrm {atan}\left (\frac {2^{3/4}\,{\left (2-3\,x^2\right )}^{1/4}}{2}\right )}{64}+\frac {2^{1/4}\,\mathrm {atan}\left (\frac {2^{3/4}\,{\left (2-3\,x^2\right )}^{1/4}\,1{}\mathrm {i}}{2}\right )\,15{}\mathrm {i}}{64}+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 {3}{64}-\frac {3}{64}{}\mathrm {i}\right )+\frac {{\left (-1\right )}^{1/4}\,2^{1/4}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,2^{3/4}\,{\left (2-3\,x^2\right )}^{1/4}}{2}\right )\,3{}\mathrm {i}}{32} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2^(1/4)*atan((2^(3/4)*(2 - 3*x^2)^(1/4)*1i)/2)*15i)/64 - (15*2^(1/4)*atan((2^(3/4)*(2 - 3*x^2)^(1/4))/2))/64
- (2 - 3*x^2)^(1/4)/(16*x^2) - 2^(3/4)*atan(2^(1/4)*(2 - 3*x^2)^(1/4)*(1/2 - 1i/2))*(3/64 + 3i/64) + ((-1)^(1/
4)*2^(1/4)*atan(((-1)^(1/4)*2^(3/4)*(2 - 3*x^2)^(1/4))/2)*3i)/32

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