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

Optimal. Leaf size=105 \[ \frac {1}{8} \text {RootSum}\left [\text {$\#$1}^8-2 \text {$\#$1}^4 a+a^2-a b\& ,\frac {\log \left (\sqrt [4]{a x^4+b}-\text {$\#$1} x\right )-\log (x)}{\text {$\#$1}}\& \right ]+\frac {\tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{\sqrt [4]{a}}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{\sqrt [4]{a}} \]

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Rubi [B]  time = 0.62, antiderivative size = 291, normalized size of antiderivative = 2.77, number of steps used = 15, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {6725, 240, 212, 206, 203, 1429, 377} \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{\sqrt [4]{a}}-\frac {\tan ^{-1}\left (\frac {\sqrt [8]{a} x \sqrt [4]{\sqrt {a}-\sqrt {b}}}{\sqrt [4]{a x^4+b}}\right )}{4 \sqrt [8]{a} \sqrt [4]{\sqrt {a}-\sqrt {b}}}-\frac {\tan ^{-1}\left (\frac {\sqrt [8]{a} x \sqrt [4]{\sqrt {a}+\sqrt {b}}}{\sqrt [4]{a x^4+b}}\right )}{4 \sqrt [8]{a} \sqrt [4]{\sqrt {a}+\sqrt {b}}}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{\sqrt [4]{a}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [8]{a} x \sqrt [4]{\sqrt {a}-\sqrt {b}}}{\sqrt [4]{a x^4+b}}\right )}{4 \sqrt [8]{a} \sqrt [4]{\sqrt {a}-\sqrt {b}}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [8]{a} x \sqrt [4]{\sqrt {a}+\sqrt {b}}}{\sqrt [4]{a x^4+b}}\right )}{4 \sqrt [8]{a} \sqrt [4]{\sqrt {a}+\sqrt {b}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[(a^(1/4)*x)/(b + a*x^4)^(1/4)]/a^(1/4) - ArcTan[(a^(1/8)*(Sqrt[a] - Sqrt[b])^(1/4)*x)/(b + a*x^4)^(1/4)
]/(4*a^(1/8)*(Sqrt[a] - Sqrt[b])^(1/4)) - ArcTan[(a^(1/8)*(Sqrt[a] + Sqrt[b])^(1/4)*x)/(b + a*x^4)^(1/4)]/(4*a
^(1/8)*(Sqrt[a] + Sqrt[b])^(1/4)) + ArcTanh[(a^(1/4)*x)/(b + a*x^4)^(1/4)]/a^(1/4) - ArcTanh[(a^(1/8)*(Sqrt[a]
 - Sqrt[b])^(1/4)*x)/(b + a*x^4)^(1/4)]/(4*a^(1/8)*(Sqrt[a] - Sqrt[b])^(1/4)) - ArcTanh[(a^(1/8)*(Sqrt[a] + Sq
rt[b])^(1/4)*x)/(b + a*x^4)^(1/4)]/(4*a^(1/8)*(Sqrt[a] + Sqrt[b])^(1/4))

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
 FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
 /; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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 240

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 1/n + 1), x], x
, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p
 + 1/n]

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 1429

Int[((d_) + (e_.)*(x_)^(n_))^(q_)/((a_) + (c_.)*(x_)^(n2_)), x_Symbol] :> With[{r = Rt[-(a*c), 2]}, -Dist[c/(2
*r), Int[(d + e*x^n)^q/(r - c*x^n), x], x] - Dist[c/(2*r), Int[(d + e*x^n)^q/(r + c*x^n), x], x]] /; FreeQ[{a,
 c, d, e, n, q}, x] && EqQ[n2, 2*n] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[q]

Rule 6725

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

Rubi steps

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

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Mathematica [F]  time = 0.45, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-b+2 a x^8}{\sqrt [4]{b+a x^4} \left (-b+a x^8\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 1.27, size = 105, normalized size = 1.00 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{\sqrt [4]{a}}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{\sqrt [4]{a}}+\frac {1}{8} \text {RootSum}\left [a^2-a b-2 a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-\log (x)+\log \left (\sqrt [4]{b+a x^4}-x \text {$\#$1}\right )}{\text {$\#$1}}\&\right ] \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.58, size = 1077, normalized size = 10.26

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(-((a - b)*sqrt(b/(a^3 - 2*a^2*b + a*b^2)) - 1)/(a - b))^(1/4)*arctan((x*sqrt((((a^2 - a*b)*x^2*sqrt(b/(a
^3 - 2*a^2*b + a*b^2)) + a*x^2)*sqrt(-((a - b)*sqrt(b/(a^3 - 2*a^2*b + a*b^2)) - 1)/(a - b)) + sqrt(a*x^4 + b)
)/x^2) - (a*x^4 + b)^(1/4))*(-((a - b)*sqrt(b/(a^3 - 2*a^2*b + a*b^2)) - 1)/(a - b))^(1/4)/x) - 1/2*(((a - b)*
sqrt(b/(a^3 - 2*a^2*b + a*b^2)) + 1)/(a - b))^(1/4)*arctan((x*(((a - b)*sqrt(b/(a^3 - 2*a^2*b + a*b^2)) + 1)/(
a - b))^(1/4)*sqrt(-(((a^2 - a*b)*x^2*sqrt(b/(a^3 - 2*a^2*b + a*b^2)) - a*x^2)*sqrt(((a - b)*sqrt(b/(a^3 - 2*a
^2*b + a*b^2)) + 1)/(a - b)) - sqrt(a*x^4 + b))/x^2) - (a*x^4 + b)^(1/4)*(((a - b)*sqrt(b/(a^3 - 2*a^2*b + a*b
^2)) + 1)/(a - b))^(1/4))/x) + 1/8*(((a - b)*sqrt(b/(a^3 - 2*a^2*b + a*b^2)) + 1)/(a - b))^(1/4)*log((((a^2 -
a*b)*x*sqrt(b/(a^3 - 2*a^2*b + a*b^2)) - a*x)*(((a - b)*sqrt(b/(a^3 - 2*a^2*b + a*b^2)) + 1)/(a - b))^(3/4) +
(a*x^4 + b)^(1/4))/x) - 1/8*(((a - b)*sqrt(b/(a^3 - 2*a^2*b + a*b^2)) + 1)/(a - b))^(1/4)*log(-(((a^2 - a*b)*x
*sqrt(b/(a^3 - 2*a^2*b + a*b^2)) - a*x)*(((a - b)*sqrt(b/(a^3 - 2*a^2*b + a*b^2)) + 1)/(a - b))^(3/4) - (a*x^4
 + b)^(1/4))/x) - 1/8*(-((a - b)*sqrt(b/(a^3 - 2*a^2*b + a*b^2)) - 1)/(a - b))^(1/4)*log((((a^2 - a*b)*x*sqrt(
b/(a^3 - 2*a^2*b + a*b^2)) + a*x)*(-((a - b)*sqrt(b/(a^3 - 2*a^2*b + a*b^2)) - 1)/(a - b))^(3/4) + (a*x^4 + b)
^(1/4))/x) + 1/8*(-((a - b)*sqrt(b/(a^3 - 2*a^2*b + a*b^2)) - 1)/(a - b))^(1/4)*log(-(((a^2 - a*b)*x*sqrt(b/(a
^3 - 2*a^2*b + a*b^2)) + a*x)*(-((a - b)*sqrt(b/(a^3 - 2*a^2*b + a*b^2)) - 1)/(a - b))^(3/4) - (a*x^4 + b)^(1/
4))/x) + 2*arctan((x*sqrt((sqrt(a)*x^2 + sqrt(a*x^4 + b))/x^2)/a^(1/4) - (a*x^4 + b)^(1/4)/a^(1/4))/x)/a^(1/4)
 + 1/2*log((a^(1/4)*x + (a*x^4 + b)^(1/4))/x)/a^(1/4) - 1/2*log(-(a^(1/4)*x - (a*x^4 + b)^(1/4))/x)/a^(1/4)

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

[Out]

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

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {2 a \,x^{8}-b}{\left (a \,x^{4}+b \right )^{\frac {1}{4}} \left (a \,x^{8}-b \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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