3.1081 \(\int \frac{x^7}{(-2+3 x^2) (-1+3 x^2)^{3/4}} \, dx\)

Optimal. Leaf size=78 \[ \frac{2}{729} \left (3 x^2-1\right )^{9/4}+\frac{8}{405} \left (3 x^2-1\right )^{5/4}+\frac{14}{81} \sqrt [4]{3 x^2-1}-\frac{8}{81} \tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right )-\frac{8}{81} \tanh ^{-1}\left (\sqrt [4]{3 x^2-1}\right ) \]

[Out]

(14*(-1 + 3*x^2)^(1/4))/81 + (8*(-1 + 3*x^2)^(5/4))/405 + (2*(-1 + 3*x^2)^(9/4))/729 - (8*ArcTan[(-1 + 3*x^2)^
(1/4)])/81 - (8*ArcTanh[(-1 + 3*x^2)^(1/4)])/81

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Rubi [A]  time = 0.0509349, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {446, 88, 63, 212, 206, 203} \[ \frac{2}{729} \left (3 x^2-1\right )^{9/4}+\frac{8}{405} \left (3 x^2-1\right )^{5/4}+\frac{14}{81} \sqrt [4]{3 x^2-1}-\frac{8}{81} \tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right )-\frac{8}{81} \tanh ^{-1}\left (\sqrt [4]{3 x^2-1}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^7/((-2 + 3*x^2)*(-1 + 3*x^2)^(3/4)),x]

[Out]

(14*(-1 + 3*x^2)^(1/4))/81 + (8*(-1 + 3*x^2)^(5/4))/405 + (2*(-1 + 3*x^2)^(9/4))/729 - (8*ArcTan[(-1 + 3*x^2)^
(1/4)])/81 - (8*ArcTanh[(-1 + 3*x^2)^(1/4)])/81

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]]

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 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 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 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 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])

Rubi steps

\begin{align*} \int \frac{x^7}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^3}{(-2+3 x) (-1+3 x)^{3/4}} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (\frac{7}{27 (-1+3 x)^{3/4}}+\frac{8}{27 (-2+3 x) (-1+3 x)^{3/4}}+\frac{4}{27} \sqrt [4]{-1+3 x}+\frac{1}{27} (-1+3 x)^{5/4}\right ) \, dx,x,x^2\right )\\ &=\frac{14}{81} \sqrt [4]{-1+3 x^2}+\frac{8}{405} \left (-1+3 x^2\right )^{5/4}+\frac{2}{729} \left (-1+3 x^2\right )^{9/4}+\frac{4}{27} \operatorname{Subst}\left (\int \frac{1}{(-2+3 x) (-1+3 x)^{3/4}} \, dx,x,x^2\right )\\ &=\frac{14}{81} \sqrt [4]{-1+3 x^2}+\frac{8}{405} \left (-1+3 x^2\right )^{5/4}+\frac{2}{729} \left (-1+3 x^2\right )^{9/4}+\frac{16}{81} \operatorname{Subst}\left (\int \frac{1}{-1+x^4} \, dx,x,\sqrt [4]{-1+3 x^2}\right )\\ &=\frac{14}{81} \sqrt [4]{-1+3 x^2}+\frac{8}{405} \left (-1+3 x^2\right )^{5/4}+\frac{2}{729} \left (-1+3 x^2\right )^{9/4}-\frac{8}{81} \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sqrt [4]{-1+3 x^2}\right )-\frac{8}{81} \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sqrt [4]{-1+3 x^2}\right )\\ &=\frac{14}{81} \sqrt [4]{-1+3 x^2}+\frac{8}{405} \left (-1+3 x^2\right )^{5/4}+\frac{2}{729} \left (-1+3 x^2\right )^{9/4}-\frac{8}{81} \tan ^{-1}\left (\sqrt [4]{-1+3 x^2}\right )-\frac{8}{81} \tanh ^{-1}\left (\sqrt [4]{-1+3 x^2}\right )\\ \end{align*}

Mathematica [A]  time = 0.0346687, size = 58, normalized size = 0.74 \[ \frac{2 \sqrt [4]{3 x^2-1} \left (45 x^4+78 x^2+284\right )-360 \tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right )-360 \tanh ^{-1}\left (\sqrt [4]{3 x^2-1}\right )}{3645} \]

Antiderivative was successfully verified.

[In]

Integrate[x^7/((-2 + 3*x^2)*(-1 + 3*x^2)^(3/4)),x]

[Out]

(2*(-1 + 3*x^2)^(1/4)*(284 + 78*x^2 + 45*x^4) - 360*ArcTan[(-1 + 3*x^2)^(1/4)] - 360*ArcTanh[(-1 + 3*x^2)^(1/4
)])/3645

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Maple [F]  time = 0.075, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{7}}{3\,{x}^{2}-2} \left ( 3\,{x}^{2}-1 \right ) ^{-{\frac{3}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7/(3*x^2-2)/(3*x^2-1)^(3/4),x)

[Out]

int(x^7/(3*x^2-2)/(3*x^2-1)^(3/4),x)

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Maxima [A]  time = 1.5393, size = 100, normalized size = 1.28 \begin{align*} \frac{2}{729} \,{\left (3 \, x^{2} - 1\right )}^{\frac{9}{4}} + \frac{8}{405} \,{\left (3 \, x^{2} - 1\right )}^{\frac{5}{4}} + \frac{14}{81} \,{\left (3 \, x^{2} - 1\right )}^{\frac{1}{4}} - \frac{8}{81} \, \arctan \left ({\left (3 \, x^{2} - 1\right )}^{\frac{1}{4}}\right ) - \frac{4}{81} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac{1}{4}} + 1\right ) + \frac{4}{81} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac{1}{4}} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(3*x^2-2)/(3*x^2-1)^(3/4),x, algorithm="maxima")

[Out]

2/729*(3*x^2 - 1)^(9/4) + 8/405*(3*x^2 - 1)^(5/4) + 14/81*(3*x^2 - 1)^(1/4) - 8/81*arctan((3*x^2 - 1)^(1/4)) -
4/81*log((3*x^2 - 1)^(1/4) + 1) + 4/81*log((3*x^2 - 1)^(1/4) - 1)

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Fricas [A]  time = 1.53744, size = 204, normalized size = 2.62 \begin{align*} \frac{2}{3645} \,{\left (45 \, x^{4} + 78 \, x^{2} + 284\right )}{\left (3 \, x^{2} - 1\right )}^{\frac{1}{4}} - \frac{8}{81} \, \arctan \left ({\left (3 \, x^{2} - 1\right )}^{\frac{1}{4}}\right ) - \frac{4}{81} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac{1}{4}} + 1\right ) + \frac{4}{81} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac{1}{4}} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(3*x^2-2)/(3*x^2-1)^(3/4),x, algorithm="fricas")

[Out]

2/3645*(45*x^4 + 78*x^2 + 284)*(3*x^2 - 1)^(1/4) - 8/81*arctan((3*x^2 - 1)^(1/4)) - 4/81*log((3*x^2 - 1)^(1/4)
+ 1) + 4/81*log((3*x^2 - 1)^(1/4) - 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{7}}{\left (3 x^{2} - 2\right ) \left (3 x^{2} - 1\right )^{\frac{3}{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**7/(3*x**2-2)/(3*x**2-1)**(3/4),x)

[Out]

Integral(x**7/((3*x**2 - 2)*(3*x**2 - 1)**(3/4)), x)

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Giac [A]  time = 1.2302, size = 101, normalized size = 1.29 \begin{align*} \frac{2}{729} \,{\left (3 \, x^{2} - 1\right )}^{\frac{9}{4}} + \frac{8}{405} \,{\left (3 \, x^{2} - 1\right )}^{\frac{5}{4}} + \frac{14}{81} \,{\left (3 \, x^{2} - 1\right )}^{\frac{1}{4}} - \frac{8}{81} \, \arctan \left ({\left (3 \, x^{2} - 1\right )}^{\frac{1}{4}}\right ) - \frac{4}{81} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac{1}{4}} + 1\right ) + \frac{4}{81} \, \log \left ({\left |{\left (3 \, x^{2} - 1\right )}^{\frac{1}{4}} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(3*x^2-2)/(3*x^2-1)^(3/4),x, algorithm="giac")

[Out]

2/729*(3*x^2 - 1)^(9/4) + 8/405*(3*x^2 - 1)^(5/4) + 14/81*(3*x^2 - 1)^(1/4) - 8/81*arctan((3*x^2 - 1)^(1/4)) -
4/81*log((3*x^2 - 1)^(1/4) + 1) + 4/81*log(abs((3*x^2 - 1)^(1/4) - 1))