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

Optimal. Leaf size=346 \[ \frac {1}{2} \text {RootSum}\left [\text {$\#$1}^8-3 \text {$\#$1}^4 a+2 a^2-b\& ,\frac {-\text {$\#$1}^4 a^3 \log \left (\sqrt [4]{a x^4-b x^2}-\text {$\#$1} x\right )+\text {$\#$1}^4 a^3 \log (x)+\text {$\#$1}^4 a b \log \left (\sqrt [4]{a x^4-b x^2}-\text {$\#$1} x\right )-\text {$\#$1}^4 b \log \left (\sqrt [4]{a x^4-b x^2}-\text {$\#$1} x\right )-\text {$\#$1}^4 a b \log (x)+\text {$\#$1}^4 b \log (x)+2 a^4 \log \left (\sqrt [4]{a x^4-b x^2}-\text {$\#$1} x\right )-a^2 b \log \left (\sqrt [4]{a x^4-b x^2}-\text {$\#$1} x\right )-2 a^4 \log (x)+a^2 b \log (x)}{3 \text {$\#$1}^3 a-2 \text {$\#$1}^7}\& \right ]+\frac {1}{4} \left (\sqrt [4]{a} b-4 a^{9/4}\right ) \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b x^2}}\right )+\frac {1}{4} \left (4 a^{9/4}-\sqrt [4]{a} b\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b x^2}}\right )+\frac {1}{2} a x \sqrt [4]{a x^4-b x^2} \]

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Rubi [A]  time = 1.83, antiderivative size = 405, normalized size of antiderivative = 1.17, number of steps used = 17, number of rules used = 11, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.268, Rules used = {2056, 6728, 279, 329, 331, 298, 203, 206, 466, 511, 510} \begin {gather*} -\frac {2 x \left (a^3-a^2 \sqrt {a^2+4 b}+2 a b-2 b\right ) \sqrt [4]{a x^4-b x^2} F_1\left (\frac {3}{4};1,-\frac {1}{4};\frac {7}{4};\frac {2 x^2}{a-\sqrt {a^2+4 b}},\frac {a x^2}{b}\right )}{3 \left (-a \sqrt {a^2+4 b}+a^2+4 b\right ) \sqrt [4]{1-\frac {a x^2}{b}}}-\frac {2 x \left (a^2+\frac {a^3+2 a b-2 b}{\sqrt {a^2+4 b}}\right ) \sqrt [4]{a x^4-b x^2} F_1\left (\frac {3}{4};1,-\frac {1}{4};\frac {7}{4};\frac {2 x^2}{a+\sqrt {a^2+4 b}},\frac {a x^2}{b}\right )}{3 \left (\sqrt {a^2+4 b}+a\right ) \sqrt [4]{1-\frac {a x^2}{b}}}+\frac {1}{2} a x \sqrt [4]{a x^4-b x^2}+\frac {\sqrt [4]{a} b \sqrt [4]{a x^4-b x^2} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2-b}}\right )}{4 \sqrt {x} \sqrt [4]{a x^2-b}}-\frac {\sqrt [4]{a} b \sqrt [4]{a x^4-b x^2} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2-b}}\right )}{4 \sqrt {x} \sqrt [4]{a x^2-b}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a*x*(-(b*x^2) + a*x^4)^(1/4))/2 - (2*(a^3 - 2*b + 2*a*b - a^2*Sqrt[a^2 + 4*b])*x*(-(b*x^2) + a*x^4)^(1/4)*App
ellF1[3/4, 1, -1/4, 7/4, (2*x^2)/(a - Sqrt[a^2 + 4*b]), (a*x^2)/b])/(3*(a^2 + 4*b - a*Sqrt[a^2 + 4*b])*(1 - (a
*x^2)/b)^(1/4)) - (2*(a^2 + (a^3 - 2*b + 2*a*b)/Sqrt[a^2 + 4*b])*x*(-(b*x^2) + a*x^4)^(1/4)*AppellF1[3/4, 1, -
1/4, 7/4, (2*x^2)/(a + Sqrt[a^2 + 4*b]), (a*x^2)/b])/(3*(a + Sqrt[a^2 + 4*b])*(1 - (a*x^2)/b)^(1/4)) + (a^(1/4
)*b*(-(b*x^2) + a*x^4)^(1/4)*ArcTan[(a^(1/4)*Sqrt[x])/(-b + a*x^2)^(1/4)])/(4*Sqrt[x]*(-b + a*x^2)^(1/4)) - (a
^(1/4)*b*(-(b*x^2) + a*x^4)^(1/4)*ArcTanh[(a^(1/4)*Sqrt[x])/(-b + a*x^2)^(1/4)])/(4*Sqrt[x]*(-b + a*x^2)^(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 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
n*p + 1)), x] + Dist[(a*n*p)/(m + n*p + 1), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 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)] && IntegersQ[m, p + (m + 1)/n]

Rule 466

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = Deno
minator[m]}, Dist[k/e, Subst[Int[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/e^n)^p*(c + (d*x^(k*n))/e^n)^q, x], x, (e*
x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && FractionQ[m] && Intege
rQ[p]

Rule 510

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a^p*c^q
*(e*x)^(m + 1)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 511

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(e*x)^m*(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] &&  !(IntegerQ[
p] || GtQ[a, 0])

Rule 2056

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

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 {\left (-b+a x^4\right ) \sqrt [4]{-b x^2+a x^4}}{-b-a x^2+x^4} \, dx &=\frac {\sqrt [4]{-b x^2+a x^4} \int \frac {\sqrt {x} \sqrt [4]{-b+a x^2} \left (-b+a x^4\right )}{-b-a x^2+x^4} \, dx}{\sqrt {x} \sqrt [4]{-b+a x^2}}\\ &=\frac {\sqrt [4]{-b x^2+a x^4} \int \left (a \sqrt {x} \sqrt [4]{-b+a x^2}-\frac {\sqrt {x} \sqrt [4]{-b+a x^2} \left (b-a b-a^2 x^2\right )}{-b-a x^2+x^4}\right ) \, dx}{\sqrt {x} \sqrt [4]{-b+a x^2}}\\ &=-\frac {\sqrt [4]{-b x^2+a x^4} \int \frac {\sqrt {x} \sqrt [4]{-b+a x^2} \left (b-a b-a^2 x^2\right )}{-b-a x^2+x^4} \, dx}{\sqrt {x} \sqrt [4]{-b+a x^2}}+\frac {\left (a \sqrt [4]{-b x^2+a x^4}\right ) \int \sqrt {x} \sqrt [4]{-b+a x^2} \, dx}{\sqrt {x} \sqrt [4]{-b+a x^2}}\\ &=\frac {1}{2} a x \sqrt [4]{-b x^2+a x^4}-\frac {\sqrt [4]{-b x^2+a x^4} \int \left (\frac {\left (-a^2+\frac {-a^3+2 b-2 a b}{\sqrt {a^2+4 b}}\right ) \sqrt {x} \sqrt [4]{-b+a x^2}}{-a-\sqrt {a^2+4 b}+2 x^2}+\frac {\left (-a^2-\frac {-a^3+2 b-2 a b}{\sqrt {a^2+4 b}}\right ) \sqrt {x} \sqrt [4]{-b+a x^2}}{-a+\sqrt {a^2+4 b}+2 x^2}\right ) \, dx}{\sqrt {x} \sqrt [4]{-b+a x^2}}-\frac {\left (a b \sqrt [4]{-b x^2+a x^4}\right ) \int \frac {\sqrt {x}}{\left (-b+a x^2\right )^{3/4}} \, dx}{4 \sqrt {x} \sqrt [4]{-b+a x^2}}\\ &=\frac {1}{2} a x \sqrt [4]{-b x^2+a x^4}-\frac {\left (a b \sqrt [4]{-b x^2+a x^4}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\left (-b+a x^4\right )^{3/4}} \, dx,x,\sqrt {x}\right )}{2 \sqrt {x} \sqrt [4]{-b+a x^2}}-\frac {\left (\left (-a^2-\frac {a^3-2 b+2 a b}{\sqrt {a^2+4 b}}\right ) \sqrt [4]{-b x^2+a x^4}\right ) \int \frac {\sqrt {x} \sqrt [4]{-b+a x^2}}{-a-\sqrt {a^2+4 b}+2 x^2} \, dx}{\sqrt {x} \sqrt [4]{-b+a x^2}}-\frac {\left (\left (-a^2+\frac {a^3-2 b+2 a b}{\sqrt {a^2+4 b}}\right ) \sqrt [4]{-b x^2+a x^4}\right ) \int \frac {\sqrt {x} \sqrt [4]{-b+a x^2}}{-a+\sqrt {a^2+4 b}+2 x^2} \, dx}{\sqrt {x} \sqrt [4]{-b+a x^2}}\\ &=\frac {1}{2} a x \sqrt [4]{-b x^2+a x^4}-\frac {\left (a b \sqrt [4]{-b x^2+a x^4}\right ) \operatorname {Subst}\left (\int \frac {x^2}{1-a x^4} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )}{2 \sqrt {x} \sqrt [4]{-b+a x^2}}-\frac {\left (2 \left (-a^2-\frac {a^3-2 b+2 a b}{\sqrt {a^2+4 b}}\right ) \sqrt [4]{-b x^2+a x^4}\right ) \operatorname {Subst}\left (\int \frac {x^2 \sqrt [4]{-b+a x^4}}{-a-\sqrt {a^2+4 b}+2 x^4} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt [4]{-b+a x^2}}-\frac {\left (2 \left (-a^2+\frac {a^3-2 b+2 a b}{\sqrt {a^2+4 b}}\right ) \sqrt [4]{-b x^2+a x^4}\right ) \operatorname {Subst}\left (\int \frac {x^2 \sqrt [4]{-b+a x^4}}{-a+\sqrt {a^2+4 b}+2 x^4} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt [4]{-b+a x^2}}\\ &=\frac {1}{2} a x \sqrt [4]{-b x^2+a x^4}-\frac {\left (\sqrt {a} b \sqrt [4]{-b x^2+a x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {a} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )}{4 \sqrt {x} \sqrt [4]{-b+a x^2}}+\frac {\left (\sqrt {a} b \sqrt [4]{-b x^2+a x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )}{4 \sqrt {x} \sqrt [4]{-b+a x^2}}-\frac {\left (2 \left (-a^2-\frac {a^3-2 b+2 a b}{\sqrt {a^2+4 b}}\right ) \sqrt [4]{-b x^2+a x^4}\right ) \operatorname {Subst}\left (\int \frac {x^2 \sqrt [4]{1-\frac {a x^4}{b}}}{-a-\sqrt {a^2+4 b}+2 x^4} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt [4]{1-\frac {a x^2}{b}}}-\frac {\left (2 \left (-a^2+\frac {a^3-2 b+2 a b}{\sqrt {a^2+4 b}}\right ) \sqrt [4]{-b x^2+a x^4}\right ) \operatorname {Subst}\left (\int \frac {x^2 \sqrt [4]{1-\frac {a x^4}{b}}}{-a+\sqrt {a^2+4 b}+2 x^4} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt [4]{1-\frac {a x^2}{b}}}\\ &=\frac {1}{2} a x \sqrt [4]{-b x^2+a x^4}-\frac {2 \left (a^2-\frac {a^3-2 b+2 a b}{\sqrt {a^2+4 b}}\right ) x \sqrt [4]{-b x^2+a x^4} F_1\left (\frac {3}{4};1,-\frac {1}{4};\frac {7}{4};\frac {2 x^2}{a-\sqrt {a^2+4 b}},\frac {a x^2}{b}\right )}{3 \left (a-\sqrt {a^2+4 b}\right ) \sqrt [4]{1-\frac {a x^2}{b}}}-\frac {2 \left (a^2+\frac {a^3-2 b+2 a b}{\sqrt {a^2+4 b}}\right ) x \sqrt [4]{-b x^2+a x^4} F_1\left (\frac {3}{4};1,-\frac {1}{4};\frac {7}{4};\frac {2 x^2}{a+\sqrt {a^2+4 b}},\frac {a x^2}{b}\right )}{3 \left (a+\sqrt {a^2+4 b}\right ) \sqrt [4]{1-\frac {a x^2}{b}}}+\frac {\sqrt [4]{a} b \sqrt [4]{-b x^2+a x^4} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )}{4 \sqrt {x} \sqrt [4]{-b+a x^2}}-\frac {\sqrt [4]{a} b \sqrt [4]{-b x^2+a x^4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )}{4 \sqrt {x} \sqrt [4]{-b+a x^2}}\\ \end {align*}

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Mathematica [A]  time = 2.75, size = 447, normalized size = 1.29 \begin {gather*} \frac {\sqrt [4]{a x^4-b x^2} \left (4 a^3 x^5+(4-3 a) b^2 x+a b x^3 \left (a-x^2-4\right )\right ) \left (\frac {3 \sqrt [4]{a} \left (4 a^2-b\right ) \left (\tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2-b}}\right )-\tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2-b}}\right )\right )}{x^{3/2} \sqrt [4]{a x^2-b}}+\frac {8 \left (a^4+\frac {-a^5-2 a^3 b+a^2 b+2 a b^2-2 b^2}{\sqrt {a^2+4 b}}-a b\right ) \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};-\frac {b \left (\frac {a}{b}+\frac {2}{\sqrt {a^2+4 b}-a}\right ) x^2}{b-a x^2}\right )}{\left (a-\sqrt {a^2+4 b}\right ) \left (a x^2-b\right )}+\frac {8 \left (a^4+\frac {a^5+2 a^3 b-a^2 b-2 a b^2+2 b^2}{\sqrt {a^2+4 b}}-a b\right ) \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};-\frac {b \left (\frac {a}{b}-\frac {2}{a+\sqrt {a^2+4 b}}\right ) x^2}{b-a x^2}\right )}{\left (\sqrt {a^2+4 b}+a\right ) \left (a x^2-b\right )}\right )}{12 \left (-4 a^3 x^4+(3 a-4) b^2+a b x^2 \left (-a+x^2+4\right )\right )}+\frac {1}{2} a x \sqrt [4]{a x^4-b x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*x*(-(b*x^2) + a*x^4)^(1/4))/2 + ((-(b*x^2) + a*x^4)^(1/4)*((4 - 3*a)*b^2*x + 4*a^3*x^5 + a*b*x^3*(-4 + a -
x^2))*((3*a^(1/4)*(4*a^2 - b)*(ArcTan[(a^(1/4)*Sqrt[x])/(-b + a*x^2)^(1/4)] - ArcTanh[(a^(1/4)*Sqrt[x])/(-b +
a*x^2)^(1/4)]))/(x^(3/2)*(-b + a*x^2)^(1/4)) + (8*(a^4 - a*b + (-a^5 + a^2*b - 2*a^3*b - 2*b^2 + 2*a*b^2)/Sqrt
[a^2 + 4*b])*Hypergeometric2F1[3/4, 1, 7/4, -((b*(a/b + 2/(-a + Sqrt[a^2 + 4*b]))*x^2)/(b - a*x^2))])/((a - Sq
rt[a^2 + 4*b])*(-b + a*x^2)) + (8*(a^4 - a*b + (a^5 - a^2*b + 2*a^3*b + 2*b^2 - 2*a*b^2)/Sqrt[a^2 + 4*b])*Hype
rgeometric2F1[3/4, 1, 7/4, -((b*(a/b - 2/(a + Sqrt[a^2 + 4*b]))*x^2)/(b - a*x^2))])/((a + Sqrt[a^2 + 4*b])*(-b
+ a*x^2))))/(12*((-4 + 3*a)*b^2 - 4*a^3*x^4 + a*b*x^2*(4 - a + x^2)))

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IntegrateAlgebraic [A]  time = 3.16, size = 346, normalized size = 1.00 \begin {gather*} \frac {1}{2} a x \sqrt [4]{-b x^2+a x^4}+\frac {1}{4} \left (-4 a^{9/4}+\sqrt [4]{a} b\right ) \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b x^2+a x^4}}\right )+\frac {1}{4} \left (4 a^{9/4}-\sqrt [4]{a} b\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b x^2+a x^4}}\right )+\frac {1}{2} \text {RootSum}\left [2 a^2-b-3 a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {2 a^4 \log (x)-a^2 b \log (x)-2 a^4 \log \left (\sqrt [4]{-b x^2+a x^4}-x \text {$\#$1}\right )+a^2 b \log \left (\sqrt [4]{-b x^2+a x^4}-x \text {$\#$1}\right )-a^3 \log (x) \text {$\#$1}^4-b \log (x) \text {$\#$1}^4+a b \log (x) \text {$\#$1}^4+a^3 \log \left (\sqrt [4]{-b x^2+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4+b \log \left (\sqrt [4]{-b x^2+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4-a b \log \left (\sqrt [4]{-b x^2+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{-3 a \text {$\#$1}^3+2 \text {$\#$1}^7}\&\right ] \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*x*(-(b*x^2) + a*x^4)^(1/4))/2 + ((-4*a^(9/4) + a^(1/4)*b)*ArcTan[(a^(1/4)*x)/(-(b*x^2) + a*x^4)^(1/4)])/4 +
((4*a^(9/4) - a^(1/4)*b)*ArcTanh[(a^(1/4)*x)/(-(b*x^2) + a*x^4)^(1/4)])/4 + RootSum[2*a^2 - b - 3*a*#1^4 + #1
^8 & , (2*a^4*Log[x] - a^2*b*Log[x] - 2*a^4*Log[(-(b*x^2) + a*x^4)^(1/4) - x*#1] + a^2*b*Log[(-(b*x^2) + a*x^4
)^(1/4) - x*#1] - a^3*Log[x]*#1^4 - b*Log[x]*#1^4 + a*b*Log[x]*#1^4 + a^3*Log[(-(b*x^2) + a*x^4)^(1/4) - x*#1]
*#1^4 + b*Log[(-(b*x^2) + a*x^4)^(1/4) - x*#1]*#1^4 - a*b*Log[(-(b*x^2) + a*x^4)^(1/4) - x*#1]*#1^4)/(-3*a*#1^
3 + 2*#1^7) & ]/2

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

[Out]

Timed out

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choice was
done assuming [a,b]=[-96,-29]Warning, need to choose a branch for the root of a polynomial with parameters. Th
is might be wrong.The choice was done assuming [a,b]=[22,61]Evaluation time: 8.9Unable to convert to real 1/4
Error: Bad Argument Value

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (a \,x^{4}-b \right ) \left (a \,x^{4}-b \,x^{2}\right )^{\frac {1}{4}}}{x^{4}-a \,x^{2}-b}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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