3.9.57 \(\int \frac {x^5}{(-2+3 x^2) (-1+3 x^2)^{3/4}} \, dx\)

Optimal. Leaf size=63 \[ \frac {2}{135} \left (3 x^2-1\right )^{5/4}+\frac {2}{9} \sqrt [4]{3 x^2-1}-\frac {4}{27} \tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right )-\frac {4}{27} \tanh ^{-1}\left (\sqrt [4]{3 x^2-1}\right ) \]

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Rubi [A]  time = 0.05, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {446, 88, 63, 212, 206, 203} \begin {gather*} \frac {2}{135} \left (3 x^2-1\right )^{5/4}+\frac {2}{9} \sqrt [4]{3 x^2-1}-\frac {4}{27} \tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right )-\frac {4}{27} \tanh ^{-1}\left (\sqrt [4]{3 x^2-1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5/((-2 + 3*x^2)*(-1 + 3*x^2)^(3/4)),x]

[Out]

(2*(-1 + 3*x^2)^(1/4))/9 + (2*(-1 + 3*x^2)^(5/4))/135 - (4*ArcTan[(-1 + 3*x^2)^(1/4)])/27 - (4*ArcTanh[(-1 + 3
*x^2)^(1/4)])/27

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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 203

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

Rule 206

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

Rule 212

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] &&
!GtQ[a/b, 0]

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(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] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {x^5}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^2}{(-2+3 x) (-1+3 x)^{3/4}} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {1}{3 (-1+3 x)^{3/4}}+\frac {4}{9 (-2+3 x) (-1+3 x)^{3/4}}+\frac {1}{9} \sqrt [4]{-1+3 x}\right ) \, dx,x,x^2\right )\\ &=\frac {2}{9} \sqrt [4]{-1+3 x^2}+\frac {2}{135} \left (-1+3 x^2\right )^{5/4}+\frac {2}{9} \operatorname {Subst}\left (\int \frac {1}{(-2+3 x) (-1+3 x)^{3/4}} \, dx,x,x^2\right )\\ &=\frac {2}{9} \sqrt [4]{-1+3 x^2}+\frac {2}{135} \left (-1+3 x^2\right )^{5/4}+\frac {8}{27} \operatorname {Subst}\left (\int \frac {1}{-1+x^4} \, dx,x,\sqrt [4]{-1+3 x^2}\right )\\ &=\frac {2}{9} \sqrt [4]{-1+3 x^2}+\frac {2}{135} \left (-1+3 x^2\right )^{5/4}-\frac {4}{27} \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt [4]{-1+3 x^2}\right )-\frac {4}{27} \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt [4]{-1+3 x^2}\right )\\ &=\frac {2}{9} \sqrt [4]{-1+3 x^2}+\frac {2}{135} \left (-1+3 x^2\right )^{5/4}-\frac {4}{27} \tan ^{-1}\left (\sqrt [4]{-1+3 x^2}\right )-\frac {4}{27} \tanh ^{-1}\left (\sqrt [4]{-1+3 x^2}\right )\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 53, normalized size = 0.84 \begin {gather*} \frac {1}{135} \left (2 \sqrt [4]{3 x^2-1} \left (3 x^2+14\right )-20 \tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right )-20 \tanh ^{-1}\left (\sqrt [4]{3 x^2-1}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^5/((-2 + 3*x^2)*(-1 + 3*x^2)^(3/4)),x]

[Out]

(2*(-1 + 3*x^2)^(1/4)*(14 + 3*x^2) - 20*ArcTan[(-1 + 3*x^2)^(1/4)] - 20*ArcTanh[(-1 + 3*x^2)^(1/4)])/135

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IntegrateAlgebraic [A]  time = 0.04, size = 55, normalized size = 0.87 \begin {gather*} \frac {2}{135} \sqrt [4]{3 x^2-1} \left (3 x^2+14\right )-\frac {4}{27} \tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right )-\frac {4}{27} \tanh ^{-1}\left (\sqrt [4]{3 x^2-1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^5/((-2 + 3*x^2)*(-1 + 3*x^2)^(3/4)),x]

[Out]

(2*(-1 + 3*x^2)^(1/4)*(14 + 3*x^2))/135 - (4*ArcTan[(-1 + 3*x^2)^(1/4)])/27 - (4*ArcTanh[(-1 + 3*x^2)^(1/4)])/
27

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fricas [A]  time = 0.94, size = 59, normalized size = 0.94 \begin {gather*} \frac {2}{135} \, {\left (3 \, x^{2} + 14\right )} {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} - \frac {4}{27} \, \arctan \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}\right ) - \frac {2}{27} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {2}{27} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(3*x^2-2)/(3*x^2-1)^(3/4),x, algorithm="fricas")

[Out]

2/135*(3*x^2 + 14)*(3*x^2 - 1)^(1/4) - 4/27*arctan((3*x^2 - 1)^(1/4)) - 2/27*log((3*x^2 - 1)^(1/4) + 1) + 2/27
*log((3*x^2 - 1)^(1/4) - 1)

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giac [A]  time = 0.44, size = 64, normalized size = 1.02 \begin {gather*} \frac {2}{135} \, {\left (3 \, x^{2} - 1\right )}^{\frac {5}{4}} + \frac {2}{9} \, {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} - \frac {4}{27} \, \arctan \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}\right ) - \frac {2}{27} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {2}{27} \, \log \left ({\left | {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(3*x^2-2)/(3*x^2-1)^(3/4),x, algorithm="giac")

[Out]

2/135*(3*x^2 - 1)^(5/4) + 2/9*(3*x^2 - 1)^(1/4) - 4/27*arctan((3*x^2 - 1)^(1/4)) - 2/27*log((3*x^2 - 1)^(1/4)
+ 1) + 2/27*log(abs((3*x^2 - 1)^(1/4) - 1))

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maple [C]  time = 0.83, size = 420, normalized size = 6.67 \begin {gather*} \frac {2 \left (3 x^{2}+14\right ) \left (3 x^{2}-1\right )^{\frac {1}{4}}}{135}+\frac {\left (\frac {2 \RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {27 x^{6}-18 \left (27 x^{6}-27 x^{4}+9 x^{2}-1\right )^{\frac {1}{4}} x^{4} \RootOf \left (\textit {\_Z}^{2}+1\right )-18 x^{4}+12 \left (27 x^{6}-27 x^{4}+9 x^{2}-1\right )^{\frac {1}{4}} x^{2} \RootOf \left (\textit {\_Z}^{2}+1\right )-6 \sqrt {27 x^{6}-27 x^{4}+9 x^{2}-1}\, x^{2}+3 x^{2}+2 \left (27 x^{6}-27 x^{4}+9 x^{2}-1\right )^{\frac {3}{4}} \RootOf \left (\textit {\_Z}^{2}+1\right )-2 \left (27 x^{6}-27 x^{4}+9 x^{2}-1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{2}+1\right )+2 \sqrt {27 x^{6}-27 x^{4}+9 x^{2}-1}}{\left (3 x^{2}-2\right ) \left (3 x^{2}-1\right )^{2}}\right )}{27}-\frac {2 \ln \left (-\frac {27 x^{6}+18 \left (27 x^{6}-27 x^{4}+9 x^{2}-1\right )^{\frac {1}{4}} x^{4}-18 x^{4}+6 \sqrt {27 x^{6}-27 x^{4}+9 x^{2}-1}\, x^{2}-12 \left (27 x^{6}-27 x^{4}+9 x^{2}-1\right )^{\frac {1}{4}} x^{2}+3 x^{2}+2 \left (27 x^{6}-27 x^{4}+9 x^{2}-1\right )^{\frac {3}{4}}-2 \sqrt {27 x^{6}-27 x^{4}+9 x^{2}-1}+2 \left (27 x^{6}-27 x^{4}+9 x^{2}-1\right )^{\frac {1}{4}}}{\left (3 x^{2}-2\right ) \left (3 x^{2}-1\right )^{2}}\right )}{27}\right ) \left (\left (3 x^{2}-1\right )^{3}\right )^{\frac {1}{4}}}{\left (3 x^{2}-1\right )^{\frac {3}{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/135*(3*x^2+14)*(3*x^2-1)^(1/4)+(-2/27*ln(-(27*x^6+18*(27*x^6-27*x^4+9*x^2-1)^(1/4)*x^4-18*x^4+6*(27*x^6-27*x
^4+9*x^2-1)^(1/2)*x^2-12*(27*x^6-27*x^4+9*x^2-1)^(1/4)*x^2+3*x^2+2*(27*x^6-27*x^4+9*x^2-1)^(3/4)-2*(27*x^6-27*
x^4+9*x^2-1)^(1/2)+2*(27*x^6-27*x^4+9*x^2-1)^(1/4))/(3*x^2-2)/(3*x^2-1)^2)+2/27*RootOf(_Z^2+1)*ln(-(-18*RootOf
(_Z^2+1)*(27*x^6-27*x^4+9*x^2-1)^(1/4)*x^4+27*x^6+2*RootOf(_Z^2+1)*(27*x^6-27*x^4+9*x^2-1)^(3/4)-6*(27*x^6-27*
x^4+9*x^2-1)^(1/2)*x^2+12*RootOf(_Z^2+1)*(27*x^6-27*x^4+9*x^2-1)^(1/4)*x^2-18*x^4+2*(27*x^6-27*x^4+9*x^2-1)^(1
/2)-2*RootOf(_Z^2+1)*(27*x^6-27*x^4+9*x^2-1)^(1/4)+3*x^2)/(3*x^2-2)/(3*x^2-1)^2))/(3*x^2-1)^(3/4)*((3*x^2-1)^3
)^(1/4)

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maxima [A]  time = 1.95, size = 63, normalized size = 1.00 \begin {gather*} \frac {2}{135} \, {\left (3 \, x^{2} - 1\right )}^{\frac {5}{4}} + \frac {2}{9} \, {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} - \frac {4}{27} \, \arctan \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}\right ) - \frac {2}{27} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {2}{27} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(3*x^2-2)/(3*x^2-1)^(3/4),x, algorithm="maxima")

[Out]

2/135*(3*x^2 - 1)^(5/4) + 2/9*(3*x^2 - 1)^(1/4) - 4/27*arctan((3*x^2 - 1)^(1/4)) - 2/27*log((3*x^2 - 1)^(1/4)
+ 1) + 2/27*log((3*x^2 - 1)^(1/4) - 1)

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mupad [B]  time = 0.12, size = 51, normalized size = 0.81 \begin {gather*} \frac {2\,{\left (3\,x^2-1\right )}^{1/4}}{9}-\frac {4\,\mathrm {atan}\left ({\left (3\,x^2-1\right )}^{1/4}\right )}{27}+\frac {2\,{\left (3\,x^2-1\right )}^{5/4}}{135}+\frac {\mathrm {atan}\left ({\left (3\,x^2-1\right )}^{1/4}\,1{}\mathrm {i}\right )\,4{}\mathrm {i}}{27} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atan((3*x^2 - 1)^(1/4)*1i)*4i)/27 - (4*atan((3*x^2 - 1)^(1/4)))/27 + (2*(3*x^2 - 1)^(1/4))/9 + (2*(3*x^2 - 1)
^(5/4))/135

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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