3.30.36 \(\int \frac {(-b+a x^4) \sqrt [4]{-b x^2+a x^4}}{-b-a x^2+x^4} \, dx\) [2936]

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

[Out]

Unintegrable

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Rubi [A]
time = 1.30, 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 = {2081, 6860, 285, 335, 338, 304, 209, 212, 477, 525, 524} \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 {\sqrt [4]{a} b \sqrt [4]{a x^4-b x^2} \text {ArcTan}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2-b}}\right )}{4 \sqrt {x} \sqrt [4]{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} \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*}

Antiderivative was successfully verified.

[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 209

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

Rule 212

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

Rule 285

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 304

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 335

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 338

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 477

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 524

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)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], 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 525

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPar
t[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 2081

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 6860

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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 = 0.83, size = 380, normalized size = 1.10 \begin {gather*} \frac {\sqrt [4]{-b x^2+a x^4} \left (2 a x^{3/2} \sqrt [4]{-b+a x^2}+\sqrt [4]{a} \left (-4 a^2+b\right ) \text {ArcTan}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )+\sqrt [4]{a} \left (4 a^2-b\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )-\text {RootSum}\left [2 a^2-b-3 a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {2 a^4 \log (x)-a^2 b \log (x)-4 a^4 \log \left (\sqrt [4]{-b+a x^2}-\sqrt {x} \text {$\#$1}\right )+2 a^2 b \log \left (\sqrt [4]{-b+a x^2}-\sqrt {x} \text {$\#$1}\right )-a^3 \log (x) \text {$\#$1}^4-b \log (x) \text {$\#$1}^4+a b \log (x) \text {$\#$1}^4+2 a^3 \log \left (\sqrt [4]{-b+a x^2}-\sqrt {x} \text {$\#$1}\right ) \text {$\#$1}^4+2 b \log \left (\sqrt [4]{-b+a x^2}-\sqrt {x} \text {$\#$1}\right ) \text {$\#$1}^4-2 a b \log \left (\sqrt [4]{-b+a x^2}-\sqrt {x} \text {$\#$1}\right ) \text {$\#$1}^4}{3 a \text {$\#$1}^3-2 \text {$\#$1}^7}\&\right ]\right )}{4 \sqrt {x} \sqrt [4]{-b+a 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]

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

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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*} \text {Failed to integrate} \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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Fricas [F(-1)] Timed out
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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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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Giac [C] Result contains higher order function than in optimal. Order 3 vs. order 1.
time = 38.68, size = 252, normalized size = 0.73 \begin {gather*} \frac {1}{2} \, {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}} a x^{2} + \frac {1}{8} \, \sqrt {2} {\left (4 \, \left (-a\right )^{\frac {1}{4}} a^{2} - \left (-a\right )^{\frac {1}{4}} b\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) + \frac {1}{8} \, \sqrt {2} {\left (4 \, \left (-a\right )^{\frac {1}{4}} a^{2} - \left (-a\right )^{\frac {1}{4}} b\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) + \frac {1}{16} \, \sqrt {2} {\left (4 \, \left (-a\right )^{\frac {1}{4}} a^{2} - \left (-a\right )^{\frac {1}{4}} b\right )} \log \left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a - \frac {b}{x^{2}}}\right ) - \frac {1}{16} \, \sqrt {2} {\left (4 \, \left (-a\right )^{\frac {1}{4}} a^{2} - \left (-a\right )^{\frac {1}{4}} b\right )} \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a - \frac {b}{x^{2}}}\right ) \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]

1/2*(a - b/x^2)^(1/4)*a*x^2 + 1/8*sqrt(2)*(4*(-a)^(1/4)*a^2 - (-a)^(1/4)*b)*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(
1/4) + 2*(a - b/x^2)^(1/4))/(-a)^(1/4)) + 1/8*sqrt(2)*(4*(-a)^(1/4)*a^2 - (-a)^(1/4)*b)*arctan(-1/2*sqrt(2)*(s
qrt(2)*(-a)^(1/4) - 2*(a - b/x^2)^(1/4))/(-a)^(1/4)) + 1/16*sqrt(2)*(4*(-a)^(1/4)*a^2 - (-a)^(1/4)*b)*log(sqrt
(2)*(-a)^(1/4)*(a - b/x^2)^(1/4) + sqrt(-a) + sqrt(a - b/x^2)) - 1/16*sqrt(2)*(4*(-a)^(1/4)*a^2 - (-a)^(1/4)*b
)*log(-sqrt(2)*(-a)^(1/4)*(a - b/x^2)^(1/4) + sqrt(-a) + sqrt(a - b/x^2))

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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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