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

Optimal. Leaf size=50 \[ \frac {1}{8} \text {RootSum}\left [-2 \text {$\#$1}^8+2 \text {$\#$1}^4 a+b\& ,\frac {\log \left (\sqrt [4]{a x^4+b}-\text {$\#$1} x\right )-\log (x)}{\text {$\#$1}}\& \right ] \]

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Rubi [B]  time = 0.52, antiderivative size = 397, normalized size of antiderivative = 7.94, 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 {\sqrt [4]{a-\sqrt {a^2+2 b}} \tan ^{-1}\left (\frac {x \sqrt [4]{-a \sqrt {a^2+2 b}+a^2+b}}{\sqrt [4]{a-\sqrt {a^2+2 b}} \sqrt [4]{a x^4+b}}\right )}{4 \sqrt [4]{-a \sqrt {a^2+2 b}+a^2+b}}-\frac {\sqrt [4]{\sqrt {a^2+2 b}+a} \tan ^{-1}\left (\frac {x \sqrt [4]{a \sqrt {a^2+2 b}+a^2+b}}{\sqrt [4]{\sqrt {a^2+2 b}+a} \sqrt [4]{a x^4+b}}\right )}{4 \sqrt [4]{a \sqrt {a^2+2 b}+a^2+b}}-\frac {\sqrt [4]{a-\sqrt {a^2+2 b}} \tanh ^{-1}\left (\frac {x \sqrt [4]{-a \sqrt {a^2+2 b}+a^2+b}}{\sqrt [4]{a-\sqrt {a^2+2 b}} \sqrt [4]{a x^4+b}}\right )}{4 \sqrt [4]{-a \sqrt {a^2+2 b}+a^2+b}}-\frac {\sqrt [4]{\sqrt {a^2+2 b}+a} \tanh ^{-1}\left (\frac {x \sqrt [4]{a \sqrt {a^2+2 b}+a^2+b}}{\sqrt [4]{\sqrt {a^2+2 b}+a} \sqrt [4]{a x^4+b}}\right )}{4 \sqrt [4]{a \sqrt {a^2+2 b}+a^2+b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [B]  time = 0.11, size = 393, normalized size = 7.86 \begin {gather*} \frac {1}{4} \left (-\frac {\sqrt [4]{a-\sqrt {a^2+2 b}} \tan ^{-1}\left (\frac {x \sqrt [4]{-a \sqrt {a^2+2 b}+a^2+b}}{\sqrt [4]{a-\sqrt {a^2+2 b}} \sqrt [4]{a x^4+b}}\right )}{\sqrt [4]{-a \sqrt {a^2+2 b}+a^2+b}}-\frac {\sqrt [4]{\sqrt {a^2+2 b}+a} \tan ^{-1}\left (\frac {x \sqrt [4]{a \sqrt {a^2+2 b}+a^2+b}}{\sqrt [4]{\sqrt {a^2+2 b}+a} \sqrt [4]{a x^4+b}}\right )}{\sqrt [4]{a \sqrt {a^2+2 b}+a^2+b}}-\frac {\sqrt [4]{a-\sqrt {a^2+2 b}} \tanh ^{-1}\left (\frac {x \sqrt [4]{-a \sqrt {a^2+2 b}+a^2+b}}{\sqrt [4]{a-\sqrt {a^2+2 b}} \sqrt [4]{a x^4+b}}\right )}{\sqrt [4]{-a \sqrt {a^2+2 b}+a^2+b}}-\frac {\sqrt [4]{\sqrt {a^2+2 b}+a} \tanh ^{-1}\left (\frac {x \sqrt [4]{a \sqrt {a^2+2 b}+a^2+b}}{\sqrt [4]{\sqrt {a^2+2 b}+a} \sqrt [4]{a x^4+b}}\right )}{\sqrt [4]{a \sqrt {a^2+2 b}+a^2+b}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.00, size = 50, normalized size = 1.00 \begin {gather*} \frac {1}{8} \text {RootSum}\left [b+2 a \text {$\#$1}^4-2 \text {$\#$1}^8\&,\frac {-\log (x)+\log \left (\sqrt [4]{b+a x^4}-x \text {$\#$1}\right )}{\text {$\#$1}}\&\right ] \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int -\frac {a\,x^4+2\,b}{{\left (a\,x^4+b\right )}^{1/4}\,\left (-x^8+2\,a\,x^4+2\,b\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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