3.9.57 \(\int \frac {1}{(-2 b+a x^4) \sqrt [4]{b x^2+a x^4}} \, dx\)

Optimal. Leaf size=65 \[ \frac {\text {RootSum}\left [2 \text {$\#$1}^8-4 \text {$\#$1}^4 a+2 a^2-a b\& ,\frac {\log \left (\sqrt [4]{a x^4+b x^2}-\text {$\#$1} x\right )-\log (x)}{\text {$\#$1}}\& \right ]}{8 b} \]

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Rubi [B]  time = 0.41, antiderivative size = 481, normalized size of antiderivative = 7.40, number of steps used = 11, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2056, 1270, 1429, 377, 212, 206, 203} \begin {gather*} -\frac {\sqrt {x} \sqrt [4]{a x^2+b} \tan ^{-1}\left (\frac {\sqrt [8]{a} \sqrt {x} \sqrt [4]{\sqrt {2} \sqrt {a}-\sqrt {b}}}{\sqrt [8]{2} \sqrt [4]{a x^2+b}}\right )}{2\ 2^{7/8} \sqrt [8]{a} b \sqrt [4]{\sqrt {2} \sqrt {a}-\sqrt {b}} \sqrt [4]{a x^4+b x^2}}-\frac {\sqrt {x} \sqrt [4]{a x^2+b} \tan ^{-1}\left (\frac {\sqrt [8]{a} \sqrt {x} \sqrt [4]{\sqrt {2} \sqrt {a}+\sqrt {b}}}{\sqrt [8]{2} \sqrt [4]{a x^2+b}}\right )}{2\ 2^{7/8} \sqrt [8]{a} b \sqrt [4]{\sqrt {2} \sqrt {a}+\sqrt {b}} \sqrt [4]{a x^4+b x^2}}-\frac {\sqrt {x} \sqrt [4]{a x^2+b} \tanh ^{-1}\left (\frac {\sqrt [8]{a} \sqrt {x} \sqrt [4]{\sqrt {2} \sqrt {a}-\sqrt {b}}}{\sqrt [8]{2} \sqrt [4]{a x^2+b}}\right )}{2\ 2^{7/8} \sqrt [8]{a} b \sqrt [4]{\sqrt {2} \sqrt {a}-\sqrt {b}} \sqrt [4]{a x^4+b x^2}}-\frac {\sqrt {x} \sqrt [4]{a x^2+b} \tanh ^{-1}\left (\frac {\sqrt [8]{a} \sqrt {x} \sqrt [4]{\sqrt {2} \sqrt {a}+\sqrt {b}}}{\sqrt [8]{2} \sqrt [4]{a x^2+b}}\right )}{2\ 2^{7/8} \sqrt [8]{a} b \sqrt [4]{\sqrt {2} \sqrt {a}+\sqrt {b}} \sqrt [4]{a x^4+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(Sqrt[x]*(b + a*x^2)^(1/4)*ArcTan[(a^(1/8)*(Sqrt[2]*Sqrt[a] - Sqrt[b])^(1/4)*Sqrt[x])/(2^(1/8)*(b + a*x^2
)^(1/4))])/(2^(7/8)*a^(1/8)*(Sqrt[2]*Sqrt[a] - Sqrt[b])^(1/4)*b*(b*x^2 + a*x^4)^(1/4)) - (Sqrt[x]*(b + a*x^2)^
(1/4)*ArcTan[(a^(1/8)*(Sqrt[2]*Sqrt[a] + Sqrt[b])^(1/4)*Sqrt[x])/(2^(1/8)*(b + a*x^2)^(1/4))])/(2*2^(7/8)*a^(1
/8)*(Sqrt[2]*Sqrt[a] + Sqrt[b])^(1/4)*b*(b*x^2 + a*x^4)^(1/4)) - (Sqrt[x]*(b + a*x^2)^(1/4)*ArcTanh[(a^(1/8)*(
Sqrt[2]*Sqrt[a] - Sqrt[b])^(1/4)*Sqrt[x])/(2^(1/8)*(b + a*x^2)^(1/4))])/(2*2^(7/8)*a^(1/8)*(Sqrt[2]*Sqrt[a] -
Sqrt[b])^(1/4)*b*(b*x^2 + a*x^4)^(1/4)) - (Sqrt[x]*(b + a*x^2)^(1/4)*ArcTanh[(a^(1/8)*(Sqrt[2]*Sqrt[a] + Sqrt[
b])^(1/4)*Sqrt[x])/(2^(1/8)*(b + a*x^2)^(1/4))])/(2*2^(7/8)*a^(1/8)*(Sqrt[2]*Sqrt[a] + Sqrt[b])^(1/4)*b*(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 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 1270

Int[((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{k = Denominat
or[m]}, Dist[k/f, Subst[Int[x^(k*(m + 1) - 1)*(d + (e*x^(2*k))/f)^q*(a + (c*x^(4*k))/f)^p, x], x, (f*x)^(1/k)]
, x]] /; FreeQ[{a, c, d, e, f, p, q}, x] && FractionQ[m] && IntegerQ[p]

Rule 1429

Int[((d_) + (e_.)*(x_)^(n_))^(q_)/((a_) + (c_.)*(x_)^(n2_)), x_Symbol] :> With[{r = Rt[-(a*c), 2]}, -Dist[c/(2
*r), Int[(d + e*x^n)^q/(r - c*x^n), x], x] - Dist[c/(2*r), Int[(d + e*x^n)^q/(r + c*x^n), x], x]] /; FreeQ[{a,
c, d, e, n, q}, x] && EqQ[n2, 2*n] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[q]

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 {1}{\left (-2 b+a x^4\right ) \sqrt [4]{b x^2+a x^4}} \, dx &=\frac {\left (\sqrt {x} \sqrt [4]{b+a x^2}\right ) \int \frac {1}{\sqrt {x} \sqrt [4]{b+a x^2} \left (-2 b+a x^4\right )} \, dx}{\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} \left (-2 b+a x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^4}}\\ &=-\frac {\left (\sqrt {a} \sqrt {x} \sqrt [4]{b+a x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\sqrt {2} \sqrt {a} \sqrt {b}-a x^4\right ) \sqrt [4]{b+a x^4}} \, dx,x,\sqrt {x}\right )}{\sqrt {2} \sqrt {b} \sqrt [4]{b x^2+a x^4}}-\frac {\left (\sqrt {a} \sqrt {x} \sqrt [4]{b+a x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\sqrt {2} \sqrt {a} \sqrt {b}+a x^4\right ) \sqrt [4]{b+a x^4}} \, dx,x,\sqrt {x}\right )}{\sqrt {2} \sqrt {b} \sqrt [4]{b x^2+a x^4}}\\ &=-\frac {\left (\sqrt {a} \sqrt {x} \sqrt [4]{b+a x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {2} \sqrt {a} \sqrt {b}-\left (\sqrt {2} a^{3/2} \sqrt {b}-a b\right ) x^4} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{\sqrt {2} \sqrt {b} \sqrt [4]{b x^2+a x^4}}-\frac {\left (\sqrt {a} \sqrt {x} \sqrt [4]{b+a x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {2} \sqrt {a} \sqrt {b}-\left (\sqrt {2} a^{3/2} \sqrt {b}+a b\right ) x^4} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{\sqrt {2} \sqrt {b} \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}{\sqrt [4]{2}-\sqrt [4]{a} \sqrt {\sqrt {2} \sqrt {a}-\sqrt {b}} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{2\ 2^{3/4} b \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}{\sqrt [4]{2}+\sqrt [4]{a} \sqrt {\sqrt {2} \sqrt {a}-\sqrt {b}} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{2\ 2^{3/4} b \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}{\sqrt [4]{2}-\sqrt [4]{a} \sqrt {\sqrt {2} \sqrt {a}+\sqrt {b}} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{2\ 2^{3/4} b \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}{\sqrt [4]{2}+\sqrt [4]{a} \sqrt {\sqrt {2} \sqrt {a}+\sqrt {b}} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{2\ 2^{3/4} b \sqrt [4]{b x^2+a x^4}}\\ &=-\frac {\sqrt {x} \sqrt [4]{b+a x^2} \tan ^{-1}\left (\frac {\sqrt [8]{a} \sqrt [4]{\sqrt {2} \sqrt {a}-\sqrt {b}} \sqrt {x}}{\sqrt [8]{2} \sqrt [4]{b+a x^2}}\right )}{2\ 2^{7/8} \sqrt [8]{a} \sqrt [4]{\sqrt {2} \sqrt {a}-\sqrt {b}} b \sqrt [4]{b x^2+a x^4}}-\frac {\sqrt {x} \sqrt [4]{b+a x^2} \tan ^{-1}\left (\frac {\sqrt [8]{a} \sqrt [4]{\sqrt {2} \sqrt {a}+\sqrt {b}} \sqrt {x}}{\sqrt [8]{2} \sqrt [4]{b+a x^2}}\right )}{2\ 2^{7/8} \sqrt [8]{a} \sqrt [4]{\sqrt {2} \sqrt {a}+\sqrt {b}} b \sqrt [4]{b x^2+a x^4}}-\frac {\sqrt {x} \sqrt [4]{b+a x^2} \tanh ^{-1}\left (\frac {\sqrt [8]{a} \sqrt [4]{\sqrt {2} \sqrt {a}-\sqrt {b}} \sqrt {x}}{\sqrt [8]{2} \sqrt [4]{b+a x^2}}\right )}{2\ 2^{7/8} \sqrt [8]{a} \sqrt [4]{\sqrt {2} \sqrt {a}-\sqrt {b}} b \sqrt [4]{b x^2+a x^4}}-\frac {\sqrt {x} \sqrt [4]{b+a x^2} \tanh ^{-1}\left (\frac {\sqrt [8]{a} \sqrt [4]{\sqrt {2} \sqrt {a}+\sqrt {b}} \sqrt {x}}{\sqrt [8]{2} \sqrt [4]{b+a x^2}}\right )}{2\ 2^{7/8} \sqrt [8]{a} \sqrt [4]{\sqrt {2} \sqrt {a}+\sqrt {b}} b \sqrt [4]{b x^2+a x^4}}\\ \end {align*}

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Mathematica [B]  time = 0.40, size = 350, normalized size = 5.38 \begin {gather*} -\frac {x \sqrt [4]{a+\frac {b}{x^2}} \left (-\sqrt [4]{2 \sqrt {a}+\sqrt {2} \sqrt {b}} \tan ^{-1}\left (\frac {\sqrt [4]{a+\frac {b}{x^2}}}{\sqrt [8]{a} \sqrt [4]{\sqrt {a}-\frac {\sqrt {b}}{\sqrt {2}}}}\right )-\sqrt [4]{2 \sqrt {a}-\sqrt {2} \sqrt {b}} \tan ^{-1}\left (\frac {\sqrt [4]{a+\frac {b}{x^2}}}{\sqrt [8]{a} \sqrt [4]{\sqrt {a}+\frac {\sqrt {b}}{\sqrt {2}}}}\right )+\sqrt [4]{2 \sqrt {a}+\sqrt {2} \sqrt {b}} \tanh ^{-1}\left (\frac {\sqrt [4]{a+\frac {b}{x^2}}}{\sqrt [8]{a} \sqrt [4]{\sqrt {a}-\frac {\sqrt {b}}{\sqrt {2}}}}\right )+\sqrt [4]{2 \sqrt {a}-\sqrt {2} \sqrt {b}} \tanh ^{-1}\left (\frac {\sqrt [4]{a+\frac {b}{x^2}}}{\sqrt [8]{a} \sqrt [4]{\sqrt {a}+\frac {\sqrt {b}}{\sqrt {2}}}}\right )\right )}{2\ 2^{3/4} \sqrt [8]{a} b \sqrt [4]{2 \sqrt {a}-\sqrt {2} \sqrt {b}} \sqrt [4]{2 \sqrt {a}+\sqrt {2} \sqrt {b}} \sqrt [4]{x^2 \left (a x^2+b\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((a + b/x^2)^(1/4)*x*(-((2*Sqrt[a] + Sqrt[2]*Sqrt[b])^(1/4)*ArcTan[(a + b/x^2)^(1/4)/(a^(1/8)*(Sqrt[a] -
Sqrt[b]/Sqrt[2])^(1/4))]) - (2*Sqrt[a] - Sqrt[2]*Sqrt[b])^(1/4)*ArcTan[(a + b/x^2)^(1/4)/(a^(1/8)*(Sqrt[a] + S
qrt[b]/Sqrt[2])^(1/4))] + (2*Sqrt[a] + Sqrt[2]*Sqrt[b])^(1/4)*ArcTanh[(a + b/x^2)^(1/4)/(a^(1/8)*(Sqrt[a] - Sq
rt[b]/Sqrt[2])^(1/4))] + (2*Sqrt[a] - Sqrt[2]*Sqrt[b])^(1/4)*ArcTanh[(a + b/x^2)^(1/4)/(a^(1/8)*(Sqrt[a] + Sqr
t[b]/Sqrt[2])^(1/4))]))/(2^(3/4)*a^(1/8)*(2*Sqrt[a] - Sqrt[2]*Sqrt[b])^(1/4)*(2*Sqrt[a] + Sqrt[2]*Sqrt[b])^(1/
4)*b*(x^2*(b + a*x^2))^(1/4))

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IntegrateAlgebraic [A]  time = 0.00, size = 65, normalized size = 1.00 \begin {gather*} \frac {\text {RootSum}\left [2 a^2-a b-4 a \text {$\#$1}^4+2 \text {$\#$1}^8\&,\frac {-\log (x)+\log \left (\sqrt [4]{b x^2+a x^4}-x \text {$\#$1}\right )}{\text {$\#$1}}\&\right ]}{8 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

RootSum[2*a^2 - a*b - 4*a*#1^4 + 2*#1^8 & , (-Log[x] + Log[(b*x^2 + a*x^4)^(1/4) - x*#1])/#1 & ]/(8*b)

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/(a*x^4-2*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 {1}{{\left (a x^{4} + b x^{2}\right )}^{\frac {1}{4}} {\left (a x^{4} - 2 \, b\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} -\int \frac {1}{{\left (a\,x^4+b\,x^2\right )}^{1/4}\,\left (2\,b-a\,x^4\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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