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

Optimal. Leaf size=100 \[ \frac {1}{4} \text {RootSum}\left [\text {$\#$1}^8-3 \text {$\#$1}^4 a+2 a^2-b\& ,\frac {\text {$\#$1}^4 \log \left (\sqrt [4]{a x^4+b}-\text {$\#$1} x\right )+\text {$\#$1}^4 (-\log (x))-3 a \log \left (\sqrt [4]{a x^4+b}-\text {$\#$1} x\right )+3 a \log (x)}{3 \text {$\#$1} a-2 \text {$\#$1}^5}\& \right ] \]

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Rubi [B]  time = 1.28, antiderivative size = 479, normalized size of antiderivative = 4.79, number of steps used = 10, number of rules used = 5, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {6728, 377, 212, 208, 205} \begin {gather*} \frac {\left (a-\frac {a^2+b}{\sqrt {a^2+4 b}}\right ) \tan ^{-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 )}{\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+b}{\sqrt {a^2+4 b}}+a\right ) \tan ^{-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 )}{\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+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 )}{\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+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 )}{\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 + 2*a*x^4)/((b + a*x^4)^(1/4)*(-b + a*x^4 + x^8)),x]

[Out]

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

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Mathematica [B]  time = 0.90, size = 499, normalized size = 4.99 \begin {gather*} \frac {\left (a-\frac {a^2+b}{\sqrt {a^2+4 b}}\right ) \tan ^{-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 )}{\left (a-\sqrt {a^2+4 b}\right )^{3/4} \sqrt [4]{-a \sqrt {a^2+4 b}+a^2-2 b}}+\frac {\left (a \sqrt {a^2+4 b}+a^2+b\right ) \tan ^{-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 )}{\sqrt {a^2+4 b} \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+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 )}{\left (a-\sqrt {a^2+4 b}\right )^{3/4} \sqrt [4]{-a \sqrt {a^2+4 b}+a^2-2 b}}+\frac {\left (a \sqrt {a^2+4 b}+a^2+b\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 )}{\sqrt {a^2+4 b} \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]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.92, size = 100, normalized size = 1.00 \begin {gather*} \frac {1}{4} \text {RootSum}\left [2 a^2-b-3 a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-3 a \log (x)+3 a \log \left (\sqrt [4]{b+a x^4}-x \text {$\#$1}\right )+\log (x) \text {$\#$1}^4-\log \left (\sqrt [4]{b+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{-3 a \text {$\#$1}+2 \text {$\#$1}^5}\&\right ] \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

RootSum[2*a^2 - b - 3*a*#1^4 + #1^8 & , (-3*a*Log[x] + 3*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) & ]/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((2*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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, a x^{4} - b}{{\left (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((2*a*x^4-b)/(a*x^4+b)^(1/4)/(x^8+a*x^4-b),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {2 a \,x^{4}-b}{\left (a \,x^{4}+b \right )^{\frac {1}{4}} \left (x^{8}+a \,x^{4}-b \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((2*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*} \int \frac {2 \, a x^{4} - b}{{\left (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((2*a*x^4-b)/(a*x^4+b)^(1/4)/(x^8+a*x^4-b),x, algorithm="maxima")

[Out]

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

[Out]

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

[Out]

Timed out

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