3.14.31 \(\int \frac {x^2}{(-b+a x^2) \sqrt [4]{-b x^2+a x^4}} \, dx\)

Optimal. Leaf size=96 \[ \frac {\tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b x^2}}\right )}{a^{5/4}}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b x^2}}\right )}{a^{5/4}}-\frac {2 \left (a x^4-b x^2\right )^{3/4}}{a x \left (a x^2-b\right )} \]

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Rubi [A]  time = 0.33, antiderivative size = 153, normalized size of antiderivative = 1.59, number of steps used = 7, number of rules used = 7, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {2056, 288, 329, 240, 212, 206, 203} \begin {gather*} \frac {\sqrt {x} \sqrt [4]{a x^2-b} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2-b}}\right )}{a^{5/4} \sqrt [4]{a x^4-b x^2}}+\frac {\sqrt {x} \sqrt [4]{a x^2-b} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2-b}}\right )}{a^{5/4} \sqrt [4]{a x^4-b x^2}}-\frac {2 x}{a \sqrt [4]{a x^4-b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/((-b + a*x^2)*(-(b*x^2) + a*x^4)^(1/4)),x]

[Out]

(-2*x)/(a*(-(b*x^2) + a*x^4)^(1/4)) + (Sqrt[x]*(-b + a*x^2)^(1/4)*ArcTan[(a^(1/4)*Sqrt[x])/(-b + a*x^2)^(1/4)]
)/(a^(5/4)*(-(b*x^2) + a*x^4)^(1/4)) + (Sqrt[x]*(-b + a*x^2)^(1/4)*ArcTanh[(a^(1/4)*Sqrt[x])/(-b + a*x^2)^(1/4
)])/(a^(5/4)*(-(b*x^2) + a*x^4)^(1/4))

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 240

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 1/n + 1), x], x
, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p
+ 1/n]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2056

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rubi steps

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

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Mathematica [A]  time = 0.19, size = 136, normalized size = 1.42 \begin {gather*} \frac {\left (a x^4-b x^2\right )^{3/4} \left (-2 \sqrt [4]{a} \sqrt {x} \left (a x^2-b\right )^{3/4}+\left (a x^2-b\right ) \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2-b}}\right )+\left (a x^2-b\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2-b}}\right )\right )}{a^{5/4} x^{3/2} \left (a x^2-b\right )^{7/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/((-b + a*x^2)*(-(b*x^2) + a*x^4)^(1/4)),x]

[Out]

((-(b*x^2) + a*x^4)^(3/4)*(-2*a^(1/4)*Sqrt[x]*(-b + a*x^2)^(3/4) + (-b + a*x^2)*ArcTan[(a^(1/4)*Sqrt[x])/(-b +
a*x^2)^(1/4)] + (-b + a*x^2)*ArcTanh[(a^(1/4)*Sqrt[x])/(-b + a*x^2)^(1/4)]))/(a^(5/4)*x^(3/2)*(-b + a*x^2)^(7
/4))

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IntegrateAlgebraic [A]  time = 0.29, size = 96, normalized size = 1.00 \begin {gather*} -\frac {2 \left (-b x^2+a x^4\right )^{3/4}}{a x \left (-b+a x^2\right )}+\frac {\tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b x^2+a x^4}}\right )}{a^{5/4}}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b x^2+a x^4}}\right )}{a^{5/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^2/((-b + a*x^2)*(-(b*x^2) + a*x^4)^(1/4)),x]

[Out]

(-2*(-(b*x^2) + a*x^4)^(3/4))/(a*x*(-b + a*x^2)) + ArcTan[(a^(1/4)*x)/(-(b*x^2) + a*x^4)^(1/4)]/a^(5/4) + ArcT
anh[(a^(1/4)*x)/(-(b*x^2) + a*x^4)^(1/4)]/a^(5/4)

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [B]  time = 0.22, size = 207, normalized size = 2.16 \begin {gather*} \frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{2 \, a^{2}} + \frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{2 \, a^{2}} - \frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} \log \left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a - \frac {b}{x^{2}}}\right )}{4 \, a^{2}} + \frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a - \frac {b}{x^{2}}}\right )}{4 \, a^{2}} - \frac {2}{{\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}} a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(2)*(-a)^(3/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(a - b/x^2)^(1/4))/(-a)^(1/4))/a^2 + 1/2*sqr
t(2)*(-a)^(3/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(a - b/x^2)^(1/4))/(-a)^(1/4))/a^2 - 1/4*sqrt(2)*(
-a)^(3/4)*log(sqrt(2)*(-a)^(1/4)*(a - b/x^2)^(1/4) + sqrt(-a) + sqrt(a - b/x^2))/a^2 + 1/4*sqrt(2)*(-a)^(3/4)*
log(-sqrt(2)*(-a)^(1/4)*(a - b/x^2)^(1/4) + sqrt(-a) + sqrt(a - b/x^2))/a^2 - 2/((a - b/x^2)^(1/4)*a)

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {x^{2}}{\left (a \,x^{2}-b \right ) \left (a \,x^{4}-b \,x^{2}\right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2}}{{\left (a x^{4} - b x^{2}\right )}^{\frac {1}{4}} {\left (a x^{2} - b\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/((a*x^4 - b*x^2)^(1/4)*(a*x^2 - b)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} -\int \frac {x^2}{\left (b-a\,x^2\right )\,{\left (a\,x^4-b\,x^2\right )}^{1/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(a*x**2-b)/(a*x**4-b*x**2)**(1/4),x)

[Out]

Integral(x**2/((x**2*(a*x**2 - b))**(1/4)*(a*x**2 - b)), x)

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