3.19.96 \(\int \frac {-2+a x^4}{\sqrt [4]{b+a x^4} (-b+a x^4+2 x^8)} \, dx\)

Optimal. Leaf size=131 \[ -\frac {\text {RootSum}\left [\text {$\#$1}^8-3 \text {$\#$1}^4 a+2 a^2-2 b\& ,\frac {-2 \text {$\#$1}^4 \log \left (\sqrt [4]{a x^4+b}-\text {$\#$1} x\right )+2 \text {$\#$1}^4 \log (x)+2 a \log \left (\sqrt [4]{a x^4+b}-\text {$\#$1} x\right )+a b \log \left (\sqrt [4]{a x^4+b}-\text {$\#$1} x\right )-a b \log (x)-2 a \log (x)}{3 \text {$\#$1} a-2 \text {$\#$1}^5}\& \right ]}{4 b} \]

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Rubi [B]  time = 1.37, antiderivative size = 491, normalized size of antiderivative = 3.75, number of steps used = 10, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {6728, 377, 212, 208, 205} \begin {gather*} \frac {\left (a-\frac {a^2+8}{\sqrt {a^2+8 b}}\right ) \tan ^{-1}\left (\frac {x \sqrt [4]{-a \sqrt {a^2+8 b}+a^2-4 b}}{\sqrt [4]{a-\sqrt {a^2+8 b}} \sqrt [4]{a x^4+b}}\right )}{2 \left (a-\sqrt {a^2+8 b}\right )^{3/4} \sqrt [4]{-a \sqrt {a^2+8 b}+a^2-4 b}}+\frac {\left (\frac {a^2+8}{\sqrt {a^2+8 b}}+a\right ) \tan ^{-1}\left (\frac {x \sqrt [4]{a \sqrt {a^2+8 b}+a^2-4 b}}{\sqrt [4]{\sqrt {a^2+8 b}+a} \sqrt [4]{a x^4+b}}\right )}{2 \left (\sqrt {a^2+8 b}+a\right )^{3/4} \sqrt [4]{a \sqrt {a^2+8 b}+a^2-4 b}}+\frac {\left (a-\frac {a^2+8}{\sqrt {a^2+8 b}}\right ) \tanh ^{-1}\left (\frac {x \sqrt [4]{-a \sqrt {a^2+8 b}+a^2-4 b}}{\sqrt [4]{a-\sqrt {a^2+8 b}} \sqrt [4]{a x^4+b}}\right )}{2 \left (a-\sqrt {a^2+8 b}\right )^{3/4} \sqrt [4]{-a \sqrt {a^2+8 b}+a^2-4 b}}+\frac {\left (\frac {a^2+8}{\sqrt {a^2+8 b}}+a\right ) \tanh ^{-1}\left (\frac {x \sqrt [4]{a \sqrt {a^2+8 b}+a^2-4 b}}{\sqrt [4]{\sqrt {a^2+8 b}+a} \sqrt [4]{a x^4+b}}\right )}{2 \left (\sqrt {a^2+8 b}+a\right )^{3/4} \sqrt [4]{a \sqrt {a^2+8 b}+a^2-4 b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2 + a*x^4)/((b + a*x^4)^(1/4)*(-b + a*x^4 + 2*x^8)),x]

[Out]

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

Rule 205

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

Rule 208

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

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 6728

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

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Mathematica [B]  time = 0.78, size = 483, normalized size = 3.69 \begin {gather*} \frac {1}{2} \left (\frac {\left (a-\frac {a^2+8}{\sqrt {a^2+8 b}}\right ) \tan ^{-1}\left (\frac {x \sqrt [4]{-a \sqrt {a^2+8 b}+a^2-4 b}}{\sqrt [4]{a-\sqrt {a^2+8 b}} \sqrt [4]{a x^4+b}}\right )}{\left (a-\sqrt {a^2+8 b}\right )^{3/4} \sqrt [4]{-a \sqrt {a^2+8 b}+a^2-4 b}}+\frac {\left (\frac {a^2+8}{\sqrt {a^2+8 b}}+a\right ) \tan ^{-1}\left (\frac {x \sqrt [4]{a \sqrt {a^2+8 b}+a^2-4 b}}{\sqrt [4]{\sqrt {a^2+8 b}+a} \sqrt [4]{a x^4+b}}\right )}{\left (\sqrt {a^2+8 b}+a\right )^{3/4} \sqrt [4]{a \sqrt {a^2+8 b}+a^2-4 b}}+\frac {\left (a-\frac {a^2+8}{\sqrt {a^2+8 b}}\right ) \tanh ^{-1}\left (\frac {x \sqrt [4]{-a \sqrt {a^2+8 b}+a^2-4 b}}{\sqrt [4]{a-\sqrt {a^2+8 b}} \sqrt [4]{a x^4+b}}\right )}{\left (a-\sqrt {a^2+8 b}\right )^{3/4} \sqrt [4]{-a \sqrt {a^2+8 b}+a^2-4 b}}+\frac {\left (\frac {a^2+8}{\sqrt {a^2+8 b}}+a\right ) \tanh ^{-1}\left (\frac {x \sqrt [4]{a \sqrt {a^2+8 b}+a^2-4 b}}{\sqrt [4]{\sqrt {a^2+8 b}+a} \sqrt [4]{a x^4+b}}\right )}{\left (\sqrt {a^2+8 b}+a\right )^{3/4} \sqrt [4]{a \sqrt {a^2+8 b}+a^2-4 b}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 + a*x^4)/((b + a*x^4)^(1/4)*(-b + a*x^4 + 2*x^8)),x]

[Out]

(((a - (8 + a^2)/Sqrt[a^2 + 8*b])*ArcTan[((a^2 - 4*b - a*Sqrt[a^2 + 8*b])^(1/4)*x)/((a - Sqrt[a^2 + 8*b])^(1/4
)*(b + a*x^4)^(1/4))])/((a - Sqrt[a^2 + 8*b])^(3/4)*(a^2 - 4*b - a*Sqrt[a^2 + 8*b])^(1/4)) + ((a + (8 + a^2)/S
qrt[a^2 + 8*b])*ArcTan[((a^2 - 4*b + a*Sqrt[a^2 + 8*b])^(1/4)*x)/((a + Sqrt[a^2 + 8*b])^(1/4)*(b + a*x^4)^(1/4
))])/((a + Sqrt[a^2 + 8*b])^(3/4)*(a^2 - 4*b + a*Sqrt[a^2 + 8*b])^(1/4)) + ((a - (8 + a^2)/Sqrt[a^2 + 8*b])*Ar
cTanh[((a^2 - 4*b - a*Sqrt[a^2 + 8*b])^(1/4)*x)/((a - Sqrt[a^2 + 8*b])^(1/4)*(b + a*x^4)^(1/4))])/((a - Sqrt[a
^2 + 8*b])^(3/4)*(a^2 - 4*b - a*Sqrt[a^2 + 8*b])^(1/4)) + ((a + (8 + a^2)/Sqrt[a^2 + 8*b])*ArcTanh[((a^2 - 4*b
+ a*Sqrt[a^2 + 8*b])^(1/4)*x)/((a + Sqrt[a^2 + 8*b])^(1/4)*(b + a*x^4)^(1/4))])/((a + Sqrt[a^2 + 8*b])^(3/4)*
(a^2 - 4*b + a*Sqrt[a^2 + 8*b])^(1/4)))/2

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IntegrateAlgebraic [A]  time = 1.60, size = 131, normalized size = 1.00 \begin {gather*} -\frac {\text {RootSum}\left [2 a^2-2 b-3 a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {2 a \log (x)+a b \log (x)-2 a \log \left (\sqrt [4]{b+a x^4}-x \text {$\#$1}\right )-a b \log \left (\sqrt [4]{b+a x^4}-x \text {$\#$1}\right )-2 \log (x) \text {$\#$1}^4+2 \log \left (\sqrt [4]{b+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{-3 a \text {$\#$1}+2 \text {$\#$1}^5}\&\right ]}{4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(-2 + a*x^4)/((b + a*x^4)^(1/4)*(-b + a*x^4 + 2*x^8)),x]

[Out]

-1/4*RootSum[2*a^2 - 2*b - 3*a*#1^4 + #1^8 & , (2*a*Log[x] + a*b*Log[x] - 2*a*Log[(b + a*x^4)^(1/4) - x*#1] -
a*b*Log[(b + a*x^4)^(1/4) - x*#1] - 2*Log[x]*#1^4 + 2*Log[(b + a*x^4)^(1/4) - x*#1]*#1^4)/(-3*a*#1 + 2*#1^5) &
]/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((a*x^4-2)/(a*x^4+b)^(1/4)/(2*x^8+a*x^4-b),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 {a x^{4} - 2}{{\left (2 \, x^{8} + a x^{4} - b\right )} {\left (a x^{4} + b\right )}^{\frac {1}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^4-2)/(a*x^4+b)^(1/4)/(2*x^8+a*x^4-b),x, algorithm="giac")

[Out]

integrate((a*x^4 - 2)/((2*x^8 + a*x^4 - b)*(a*x^4 + b)^(1/4)), x)

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maple [F]  time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {a \,x^{4}-2}{\left (a \,x^{4}+b \right )^{\frac {1}{4}} \left (2 x^{8}+a \,x^{4}-b \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^4-2)/(a*x^4+b)^(1/4)/(2*x^8+a*x^4-b),x)

[Out]

int((a*x^4-2)/(a*x^4+b)^(1/4)/(2*x^8+a*x^4-b),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a x^{4} - 2}{{\left (2 \, x^{8} + a x^{4} - b\right )} {\left (a x^{4} + b\right )}^{\frac {1}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^4-2)/(a*x^4+b)^(1/4)/(2*x^8+a*x^4-b),x, algorithm="maxima")

[Out]

integrate((a*x^4 - 2)/((2*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 {a\,x^4-2}{{\left (a\,x^4+b\right )}^{1/4}\,\left (2\,x^8+a\,x^4-b\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^4 - 2)/((b + a*x^4)^(1/4)*(a*x^4 - b + 2*x^8)),x)

[Out]

int((a*x^4 - 2)/((b + a*x^4)^(1/4)*(a*x^4 - b + 2*x^8)), x)

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sympy [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((a*x**4-2)/(a*x**4+b)**(1/4)/(2*x**8+a*x**4-b),x)

[Out]

Timed out

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