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

Optimal. Leaf size=120 \[ \frac {1}{4} \text {RootSum}\left [2 \text {$\#$1}^8-4 \text {$\#$1}^4 a-\text {$\#$1}^4 c+2 a^2-2 a b+a c\& ,\frac {\text {$\#$1}^4 \log \left (\sqrt [4]{a x^4-b}-\text {$\#$1} x\right )+\text {$\#$1}^4 (-\log (x))+a \log \left (\sqrt [4]{a x^4-b}-\text {$\#$1} x\right )-a \log (x)}{-4 \text {$\#$1}^5+4 \text {$\#$1} a+\text {$\#$1} c}\& \right ] \]

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Rubi [B]  time = 0.99, antiderivative size = 529, normalized size of antiderivative = 4.41, number of steps used = 10, number of rules used = 5, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.119, Rules used = {6728, 377, 212, 208, 205} \begin {gather*} \frac {a^{3/4} \left (\frac {2 b-c}{\sqrt {16 a b+c^2}}+1\right ) \tan ^{-1}\left (\frac {\sqrt [4]{a} x \sqrt [4]{\sqrt {16 a b+c^2}+4 b-c}}{\sqrt [4]{\sqrt {16 a b+c^2}-c} \sqrt [4]{a x^4-b}}\right )}{\left (\sqrt {16 a b+c^2}-c\right )^{3/4} \sqrt [4]{\sqrt {16 a b+c^2}+4 b-c}}-\frac {a^{3/4} \left (1-\frac {2 b-c}{\sqrt {16 a b+c^2}}\right ) \tan ^{-1}\left (\frac {\sqrt [4]{a} x \sqrt [4]{\sqrt {16 a b+c^2}-4 b+c}}{\sqrt [4]{\sqrt {16 a b+c^2}+c} \sqrt [4]{a x^4-b}}\right )}{\left (\sqrt {16 a b+c^2}+c\right )^{3/4} \sqrt [4]{\sqrt {16 a b+c^2}-4 b+c}}+\frac {a^{3/4} \left (\frac {2 b-c}{\sqrt {16 a b+c^2}}+1\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{a} x \sqrt [4]{\sqrt {16 a b+c^2}+4 b-c}}{\sqrt [4]{\sqrt {16 a b+c^2}-c} \sqrt [4]{a x^4-b}}\right )}{\left (\sqrt {16 a b+c^2}-c\right )^{3/4} \sqrt [4]{\sqrt {16 a b+c^2}+4 b-c}}-\frac {a^{3/4} \left (1-\frac {2 b-c}{\sqrt {16 a b+c^2}}\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{a} x \sqrt [4]{\sqrt {16 a b+c^2}-4 b+c}}{\sqrt [4]{\sqrt {16 a b+c^2}+c} \sqrt [4]{a x^4-b}}\right )}{\left (\sqrt {16 a b+c^2}+c\right )^{3/4} \sqrt [4]{\sqrt {16 a b+c^2}-4 b+c}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^(3/4)*(1 + (2*b - c)/Sqrt[16*a*b + c^2])*ArcTan[(a^(1/4)*(4*b - c + Sqrt[16*a*b + c^2])^(1/4)*x)/((-c + Sqr
t[16*a*b + c^2])^(1/4)*(-b + a*x^4)^(1/4))])/((-c + Sqrt[16*a*b + c^2])^(3/4)*(4*b - c + Sqrt[16*a*b + c^2])^(
1/4)) - (a^(3/4)*(1 - (2*b - c)/Sqrt[16*a*b + c^2])*ArcTan[(a^(1/4)*(-4*b + c + Sqrt[16*a*b + c^2])^(1/4)*x)/(
(c + Sqrt[16*a*b + c^2])^(1/4)*(-b + a*x^4)^(1/4))])/((c + Sqrt[16*a*b + c^2])^(3/4)*(-4*b + c + Sqrt[16*a*b +
c^2])^(1/4)) + (a^(3/4)*(1 + (2*b - c)/Sqrt[16*a*b + c^2])*ArcTanh[(a^(1/4)*(4*b - c + Sqrt[16*a*b + c^2])^(1
/4)*x)/((-c + Sqrt[16*a*b + c^2])^(1/4)*(-b + a*x^4)^(1/4))])/((-c + Sqrt[16*a*b + c^2])^(3/4)*(4*b - c + Sqrt
[16*a*b + c^2])^(1/4)) - (a^(3/4)*(1 - (2*b - c)/Sqrt[16*a*b + c^2])*ArcTanh[(a^(1/4)*(-4*b + c + Sqrt[16*a*b
+ c^2])^(1/4)*x)/((c + Sqrt[16*a*b + c^2])^(1/4)*(-b + a*x^4)^(1/4))])/((c + Sqrt[16*a*b + c^2])^(3/4)*(-4*b +
c + Sqrt[16*a*b + c^2])^(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 (-2 b-c x^4+2 a x^8\right )} \, dx &=\int \left (\frac {2 a-\frac {2 a (2 b-c)}{\sqrt {16 a b+c^2}}}{\sqrt [4]{-b+a x^4} \left (-c-\sqrt {16 a b+c^2}+4 a x^4\right )}+\frac {2 a+\frac {2 a (2 b-c)}{\sqrt {16 a b+c^2}}}{\sqrt [4]{-b+a x^4} \left (-c+\sqrt {16 a b+c^2}+4 a x^4\right )}\right ) \, dx\\ &=\left (2 a \left (1-\frac {2 b-c}{\sqrt {16 a b+c^2}}\right )\right ) \int \frac {1}{\sqrt [4]{-b+a x^4} \left (-c-\sqrt {16 a b+c^2}+4 a x^4\right )} \, dx+\left (2 a \left (1+\frac {2 b-c}{\sqrt {16 a b+c^2}}\right )\right ) \int \frac {1}{\sqrt [4]{-b+a x^4} \left (-c+\sqrt {16 a b+c^2}+4 a x^4\right )} \, dx\\ &=\left (2 a \left (1-\frac {2 b-c}{\sqrt {16 a b+c^2}}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-c-\sqrt {16 a b+c^2}-\left (4 a b+a \left (-c-\sqrt {16 a b+c^2}\right )\right ) x^4} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )+\left (2 a \left (1+\frac {2 b-c}{\sqrt {16 a b+c^2}}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-c+\sqrt {16 a b+c^2}-\left (4 a b+a \left (-c+\sqrt {16 a b+c^2}\right )\right ) x^4} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )\\ &=\frac {\left (a \left (1+\frac {2 b-c}{\sqrt {16 a b+c^2}}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-c+\sqrt {16 a b+c^2}}-\sqrt {a} \sqrt {4 b-c+\sqrt {16 a b+c^2}} x^2} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )}{\sqrt {-c+\sqrt {16 a b+c^2}}}+\frac {\left (a \left (1+\frac {2 b-c}{\sqrt {16 a b+c^2}}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-c+\sqrt {16 a b+c^2}}+\sqrt {a} \sqrt {4 b-c+\sqrt {16 a b+c^2}} x^2} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )}{\sqrt {-c+\sqrt {16 a b+c^2}}}-\frac {\left (a \left (1-\frac {2 b-c}{\sqrt {16 a b+c^2}}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c+\sqrt {16 a b+c^2}}-\sqrt {a} \sqrt {-4 b+c+\sqrt {16 a b+c^2}} x^2} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )}{\sqrt {c+\sqrt {16 a b+c^2}}}-\frac {\left (a \left (1-\frac {2 b-c}{\sqrt {16 a b+c^2}}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c+\sqrt {16 a b+c^2}}+\sqrt {a} \sqrt {-4 b+c+\sqrt {16 a b+c^2}} x^2} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )}{\sqrt {c+\sqrt {16 a b+c^2}}}\\ &=\frac {a^{3/4} \left (1+\frac {2 b-c}{\sqrt {16 a b+c^2}}\right ) \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [4]{4 b-c+\sqrt {16 a b+c^2}} x}{\sqrt [4]{-c+\sqrt {16 a b+c^2}} \sqrt [4]{-b+a x^4}}\right )}{\left (-c+\sqrt {16 a b+c^2}\right )^{3/4} \sqrt [4]{4 b-c+\sqrt {16 a b+c^2}}}-\frac {a^{3/4} \left (1-\frac {2 b-c}{\sqrt {16 a b+c^2}}\right ) \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [4]{-4 b+c+\sqrt {16 a b+c^2}} x}{\sqrt [4]{c+\sqrt {16 a b+c^2}} \sqrt [4]{-b+a x^4}}\right )}{\left (c+\sqrt {16 a b+c^2}\right )^{3/4} \sqrt [4]{-4 b+c+\sqrt {16 a b+c^2}}}+\frac {a^{3/4} \left (1+\frac {2 b-c}{\sqrt {16 a b+c^2}}\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [4]{4 b-c+\sqrt {16 a b+c^2}} x}{\sqrt [4]{-c+\sqrt {16 a b+c^2}} \sqrt [4]{-b+a x^4}}\right )}{\left (-c+\sqrt {16 a b+c^2}\right )^{3/4} \sqrt [4]{4 b-c+\sqrt {16 a b+c^2}}}-\frac {a^{3/4} \left (1-\frac {2 b-c}{\sqrt {16 a b+c^2}}\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [4]{-4 b+c+\sqrt {16 a b+c^2}} x}{\sqrt [4]{c+\sqrt {16 a b+c^2}} \sqrt [4]{-b+a x^4}}\right )}{\left (c+\sqrt {16 a b+c^2}\right )^{3/4} \sqrt [4]{-4 b+c+\sqrt {16 a b+c^2}}}\\ \end {align*}

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Mathematica [B]  time = 0.44, size = 527, normalized size = 4.39 \begin {gather*} a^{3/4} \left (\frac {\left (\sqrt {16 a b+c^2}+2 b-c\right ) \tan ^{-1}\left (\frac {\sqrt [4]{a} x \sqrt [4]{\sqrt {16 a b+c^2}+4 b-c}}{\sqrt [4]{\sqrt {16 a b+c^2}-c} \sqrt [4]{a x^4-b}}\right )}{\sqrt {16 a b+c^2} \left (\sqrt {16 a b+c^2}-c\right )^{3/4} \sqrt [4]{\sqrt {16 a b+c^2}+4 b-c}}-\frac {\left (\frac {c-2 b}{\sqrt {16 a b+c^2}}+1\right ) \tan ^{-1}\left (\frac {\sqrt [4]{a} x \sqrt [4]{\sqrt {16 a b+c^2}-4 b+c}}{\sqrt [4]{\sqrt {16 a b+c^2}+c} \sqrt [4]{a x^4-b}}\right )}{\left (\sqrt {16 a b+c^2}+c\right )^{3/4} \sqrt [4]{\sqrt {16 a b+c^2}-4 b+c}}+\frac {\left (\sqrt {16 a b+c^2}+2 b-c\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{a} x \sqrt [4]{\sqrt {16 a b+c^2}+4 b-c}}{\sqrt [4]{\sqrt {16 a b+c^2}-c} \sqrt [4]{a x^4-b}}\right )}{\sqrt {16 a b+c^2} \left (\sqrt {16 a b+c^2}-c\right )^{3/4} \sqrt [4]{\sqrt {16 a b+c^2}+4 b-c}}-\frac {\left (\frac {c-2 b}{\sqrt {16 a b+c^2}}+1\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{a} x \sqrt [4]{\sqrt {16 a b+c^2}-4 b+c}}{\sqrt [4]{\sqrt {16 a b+c^2}+c} \sqrt [4]{a x^4-b}}\right )}{\left (\sqrt {16 a b+c^2}+c\right )^{3/4} \sqrt [4]{\sqrt {16 a b+c^2}-4 b+c}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

a^(3/4)*(((2*b - c + Sqrt[16*a*b + c^2])*ArcTan[(a^(1/4)*(4*b - c + Sqrt[16*a*b + c^2])^(1/4)*x)/((-c + Sqrt[1
6*a*b + c^2])^(1/4)*(-b + a*x^4)^(1/4))])/(Sqrt[16*a*b + c^2]*(-c + Sqrt[16*a*b + c^2])^(3/4)*(4*b - c + Sqrt[
16*a*b + c^2])^(1/4)) - ((1 + (-2*b + c)/Sqrt[16*a*b + c^2])*ArcTan[(a^(1/4)*(-4*b + c + Sqrt[16*a*b + c^2])^(
1/4)*x)/((c + Sqrt[16*a*b + c^2])^(1/4)*(-b + a*x^4)^(1/4))])/((c + Sqrt[16*a*b + c^2])^(3/4)*(-4*b + c + Sqrt
[16*a*b + c^2])^(1/4)) + ((2*b - c + Sqrt[16*a*b + c^2])*ArcTanh[(a^(1/4)*(4*b - c + Sqrt[16*a*b + c^2])^(1/4)
*x)/((-c + Sqrt[16*a*b + c^2])^(1/4)*(-b + a*x^4)^(1/4))])/(Sqrt[16*a*b + c^2]*(-c + Sqrt[16*a*b + c^2])^(3/4)
*(4*b - c + Sqrt[16*a*b + c^2])^(1/4)) - ((1 + (-2*b + c)/Sqrt[16*a*b + c^2])*ArcTanh[(a^(1/4)*(-4*b + c + Sqr
t[16*a*b + c^2])^(1/4)*x)/((c + Sqrt[16*a*b + c^2])^(1/4)*(-b + a*x^4)^(1/4))])/((c + Sqrt[16*a*b + c^2])^(3/4
)*(-4*b + c + Sqrt[16*a*b + c^2])^(1/4)))

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IntegrateAlgebraic [A]  time = 0.00, size = 121, normalized size = 1.01 \begin {gather*} \frac {1}{4} \text {RootSum}\left [2 a^2-2 a b+a c-4 a \text {$\#$1}^4-c \text {$\#$1}^4+2 \text {$\#$1}^8\&,\frac {a \log (x)-a \log \left (\sqrt [4]{-b+a x^4}-x \text {$\#$1}\right )+\log (x) \text {$\#$1}^4-\log \left (\sqrt [4]{-b+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{-4 a \text {$\#$1}-c \text {$\#$1}+4 \text {$\#$1}^5}\&\right ] \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

RootSum[2*a^2 - 2*a*b + a*c - 4*a*#1^4 - c*#1^4 + 2*#1^8 & , (a*Log[x] - a*Log[(-b + a*x^4)^(1/4) - x*#1] + Lo
g[x]*#1^4 - Log[(-b + a*x^4)^(1/4) - x*#1]*#1^4)/(-4*a*#1 - c*#1 + 4*#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)/(2*a*x^8-c*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 {2 \, a x^{4} - b}{{\left (2 \, a x^{8} - c 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((2*a*x^4-b)/(a*x^4-b)^(1/4)/(2*a*x^8-c*x^4-2*b),x, algorithm="giac")

[Out]

integrate((2*a*x^4 - b)/((2*a*x^8 - c*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 \[\int \frac {2 a \,x^{4}-b}{\left (a \,x^{4}-b \right )^{\frac {1}{4}} \left (2 a \,x^{8}-c \,x^{4}-2 b \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {b-2\,a\,x^4}{{\left (a\,x^4-b\right )}^{1/4}\,\left (-2\,a\,x^8+c\,x^4+2\,b\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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