3.14.65 \(\int \frac {-b+a x^4}{\sqrt [4]{b+a x^4} (b-a x^4+x^8)} \, dx\) [1365]

Optimal. Leaf size=98 \[ -\frac {1}{4} \text {RootSum}\left [2 a^2+b-3 a \text {$\#$1}^4+\text {$\#$1}^8\& ,\frac {2 a \log (x)-2 a \log \left (\sqrt [4]{b+a x^4}-x \text {$\#$1}\right )-\log (x) \text {$\#$1}^4+\log \left (\sqrt [4]{b+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{3 a \text {$\#$1}-2 \text {$\#$1}^5}\& \right ] \]

[Out]

Unintegrable

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Rubi [B] Leaf count is larger than twice the leaf count of optimal. \(499\) vs. \(2(98)=196\).
time = 0.44, antiderivative size = 499, normalized size of antiderivative = 5.09, number of steps used = 10, number of rules used = 5, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {6860, 385, 218, 214, 211} \begin {gather*} -\frac {\left (a-\frac {a^2-2 b}{\sqrt {a^2-4 b}}\right ) \text {ArcTan}\left (\frac {x \sqrt [4]{-a \sqrt {a^2-4 b}+a^2+2 b}}{\sqrt [4]{a-\sqrt {a^2-4 b}} \sqrt [4]{a x^4+b}}\right )}{2 \left (a-\sqrt {a^2-4 b}\right )^{3/4} \sqrt [4]{-a \sqrt {a^2-4 b}+a^2+2 b}}-\frac {\left (\frac {a^2-2 b}{\sqrt {a^2-4 b}}+a\right ) \text {ArcTan}\left (\frac {x \sqrt [4]{a \sqrt {a^2-4 b}+a^2+2 b}}{\sqrt [4]{\sqrt {a^2-4 b}+a} \sqrt [4]{a x^4+b}}\right )}{2 \left (\sqrt {a^2-4 b}+a\right )^{3/4} \sqrt [4]{a \sqrt {a^2-4 b}+a^2+2 b}}-\frac {\left (a-\frac {a^2-2 b}{\sqrt {a^2-4 b}}\right ) \tanh ^{-1}\left (\frac {x \sqrt [4]{-a \sqrt {a^2-4 b}+a^2+2 b}}{\sqrt [4]{a-\sqrt {a^2-4 b}} \sqrt [4]{a x^4+b}}\right )}{2 \left (a-\sqrt {a^2-4 b}\right )^{3/4} \sqrt [4]{-a \sqrt {a^2-4 b}+a^2+2 b}}-\frac {\left (\frac {a^2-2 b}{\sqrt {a^2-4 b}}+a\right ) \tanh ^{-1}\left (\frac {x \sqrt [4]{a \sqrt {a^2-4 b}+a^2+2 b}}{\sqrt [4]{\sqrt {a^2-4 b}+a} \sqrt [4]{a x^4+b}}\right )}{2 \left (\sqrt {a^2-4 b}+a\right )^{3/4} \sqrt [4]{a \sqrt {a^2-4 b}+a^2+2 b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-b + a*x^4)/((b + a*x^4)^(1/4)*(b - a*x^4 + x^8)),x]

[Out]

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

Rule 211

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

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/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 385

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 6860

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps

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

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Mathematica [A]
time = 0.00, size = 98, normalized size = 1.00 \begin {gather*} -\frac {1}{4} \text {RootSum}\left [2 a^2+b-3 a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-2 a \log (x)+2 a \log \left (\sqrt [4]{b+a x^4}-x \text {$\#$1}\right )+\log (x) \text {$\#$1}^4-\log \left (\sqrt [4]{b+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{-3 a \text {$\#$1}+2 \text {$\#$1}^5}\&\right ] \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-b + a*x^4)/((b + a*x^4)^(1/4)*(b - a*x^4 + x^8)),x]

[Out]

-1/4*RootSum[2*a^2 + b - 3*a*#1^4 + #1^8 & , (-2*a*Log[x] + 2*a*Log[(b + a*x^4)^(1/4) - x*#1] + Log[x]*#1^4 -
Log[(b + a*x^4)^(1/4) - x*#1]*#1^4)/(-3*a*#1 + 2*#1^5) & ]

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Maple [F]
time = 0.00, size = 0, normalized size = 0.00 \[\int \frac {a \,x^{4}-b}{\left (a \,x^{4}+b \right )^{\frac {1}{4}} \left (x^{8}-a \,x^{4}+b \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^4-b)/(a*x^4+b)^(1/4)/(x^8-a*x^4+b),x)

[Out]

int((a*x^4-b)/(a*x^4+b)^(1/4)/(x^8-a*x^4+b),x)

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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((a*x^4-b)/(a*x^4+b)^(1/4)/(x^8-a*x^4+b),x, algorithm="maxima")

[Out]

integrate((a*x^4 - b)/((x^8 - a*x^4 + b)*(a*x^4 + b)^(1/4)), x)

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Fricas [F(-1)] Timed out
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((a*x^4-b)/(a*x^4+b)^(1/4)/(x^8-a*x^4+b),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a x^{4} - b}{\sqrt [4]{a x^{4} + b} \left (- a x^{4} + b + x^{8}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x**4-b)/(a*x**4+b)**(1/4)/(x**8-a*x**4+b),x)

[Out]

Integral((a*x**4 - b)/((a*x**4 + b)**(1/4)*(-a*x**4 + b + x**8)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^4-b)/(a*x^4+b)^(1/4)/(x^8-a*x^4+b),x, algorithm="giac")

[Out]

integrate((a*x^4 - b)/((x^8 - a*x^4 + b)*(a*x^4 + b)^(1/4)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int -\frac {b-a\,x^4}{{\left (a\,x^4+b\right )}^{1/4}\,\left (x^8-a\,x^4+b\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(b - a*x^4)/((b + a*x^4)^(1/4)*(b - a*x^4 + x^8)),x)

[Out]

int(-(b - a*x^4)/((b + a*x^4)^(1/4)*(b - a*x^4 + x^8)), x)

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