3.9.6 \(\int \frac {-b+a x^4}{(b-2 x^2+a x^4) \sqrt [4]{b x^2+a x^6}} \, dx\)

Optimal. Leaf size=61 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{a x^6+b x^2}}\right )}{\sqrt [4]{2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{a x^6+b x^2}}\right )}{\sqrt [4]{2}} \]

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Rubi [C]  time = 2.11, antiderivative size = 444, normalized size of antiderivative = 7.28, number of steps used = 20, number of rules used = 10, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.256, Rules used = {2056, 6715, 6728, 246, 245, 1438, 430, 429, 511, 510} \begin {gather*} -\frac {2 x \sqrt [4]{\frac {a x^4}{b}+1} F_1\left (\frac {1}{8};1,\frac {1}{4};\frac {9}{8};\frac {a^2 x^4}{-a b-2 \sqrt {1-a b}+2},-\frac {a x^4}{b}\right )}{\sqrt [4]{a x^6+b x^2}}-\frac {2 x \sqrt [4]{\frac {a x^4}{b}+1} F_1\left (\frac {1}{8};1,\frac {1}{4};\frac {9}{8};\frac {a^2 x^4}{-a b+2 \sqrt {1-a b}+2},-\frac {a x^4}{b}\right )}{\sqrt [4]{a x^6+b x^2}}-\frac {2 a x^3 \left (1-\sqrt {1-a b}\right ) \sqrt [4]{\frac {a x^4}{b}+1} F_1\left (\frac {5}{8};1,\frac {1}{4};\frac {13}{8};\frac {a^2 x^4}{-a b-2 \sqrt {1-a b}+2},-\frac {a x^4}{b}\right )}{5 \left (-a b-2 \sqrt {1-a b}+2\right ) \sqrt [4]{a x^6+b x^2}}-\frac {2 a x^3 \left (\sqrt {1-a b}+1\right ) \sqrt [4]{\frac {a x^4}{b}+1} F_1\left (\frac {5}{8};1,\frac {1}{4};\frac {13}{8};\frac {a^2 x^4}{-a b+2 \sqrt {1-a b}+2},-\frac {a x^4}{b}\right )}{5 \left (-a b+2 \sqrt {1-a b}+2\right ) \sqrt [4]{a x^6+b x^2}}+\frac {2 x \sqrt [4]{\frac {a x^4}{b}+1} \, _2F_1\left (\frac {1}{8},\frac {1}{4};\frac {9}{8};-\frac {a x^4}{b}\right )}{\sqrt [4]{a x^6+b x^2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Int[(-b + a*x^4)/((b - 2*x^2 + a*x^4)*(b*x^2 + a*x^6)^(1/4)),x]

[Out]

(-2*x*(1 + (a*x^4)/b)^(1/4)*AppellF1[1/8, 1, 1/4, 9/8, (a^2*x^4)/(2 - a*b - 2*Sqrt[1 - a*b]), -((a*x^4)/b)])/(
b*x^2 + a*x^6)^(1/4) - (2*x*(1 + (a*x^4)/b)^(1/4)*AppellF1[1/8, 1, 1/4, 9/8, (a^2*x^4)/(2 - a*b + 2*Sqrt[1 - a
*b]), -((a*x^4)/b)])/(b*x^2 + a*x^6)^(1/4) - (2*a*(1 - Sqrt[1 - a*b])*x^3*(1 + (a*x^4)/b)^(1/4)*AppellF1[5/8,
1, 1/4, 13/8, (a^2*x^4)/(2 - a*b - 2*Sqrt[1 - a*b]), -((a*x^4)/b)])/(5*(2 - a*b - 2*Sqrt[1 - a*b])*(b*x^2 + a*
x^6)^(1/4)) - (2*a*(1 + Sqrt[1 - a*b])*x^3*(1 + (a*x^4)/b)^(1/4)*AppellF1[5/8, 1, 1/4, 13/8, (a^2*x^4)/(2 - a*
b + 2*Sqrt[1 - a*b]), -((a*x^4)/b)])/(5*(2 - a*b + 2*Sqrt[1 - a*b])*(b*x^2 + a*x^6)^(1/4)) + (2*x*(1 + (a*x^4)
/b)^(1/4)*Hypergeometric2F1[1/8, 1/4, 9/8, -((a*x^4)/b)])/(b*x^2 + a*x^6)^(1/4)

Rule 245

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, -((b*x^n)/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^Fr
acPart[p], Int[(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 429

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

Rule 430

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

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 1438

Int[((d_) + (e_.)*(x_)^(n_))^(q_)*((a_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + c*x^(2
*n))^p, (d/(d^2 - e^2*x^(2*n)) - (e*x^n)/(d^2 - e^2*x^(2*n)))^(-q), x], x] /; FreeQ[{a, c, d, e, n, p}, x] &&
EqQ[n2, 2*n] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[q, 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 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

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+a x^4}{\left (b-2 x^2+a x^4\right ) \sqrt [4]{b x^2+a x^6}} \, dx &=\frac {\left (\sqrt {x} \sqrt [4]{b+a x^4}\right ) \int \frac {-b+a x^4}{\sqrt {x} \sqrt [4]{b+a x^4} \left (b-2 x^2+a x^4\right )} \, dx}{\sqrt [4]{b x^2+a x^6}}\\ &=\frac {\left (2 \sqrt {x} \sqrt [4]{b+a x^4}\right ) \operatorname {Subst}\left (\int \frac {-b+a x^8}{\sqrt [4]{b+a x^8} \left (b-2 x^4+a x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^6}}\\ &=\frac {\left (2 \sqrt {x} \sqrt [4]{b+a x^4}\right ) \operatorname {Subst}\left (\int \left (\frac {1}{\sqrt [4]{b+a x^8}}-\frac {2 \left (b-x^4\right )}{\sqrt [4]{b+a x^8} \left (b-2 x^4+a x^8\right )}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^6}}\\ &=\frac {\left (2 \sqrt {x} \sqrt [4]{b+a x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{b+a x^8}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^6}}-\frac {\left (4 \sqrt {x} \sqrt [4]{b+a x^4}\right ) \operatorname {Subst}\left (\int \frac {b-x^4}{\sqrt [4]{b+a x^8} \left (b-2 x^4+a x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^6}}\\ &=-\frac {\left (4 \sqrt {x} \sqrt [4]{b+a x^4}\right ) \operatorname {Subst}\left (\int \left (\frac {-1-\sqrt {1-a b}}{\left (-2-2 \sqrt {1-a b}+2 a x^4\right ) \sqrt [4]{b+a x^8}}+\frac {-1+\sqrt {1-a b}}{\left (-2+2 \sqrt {1-a b}+2 a x^4\right ) \sqrt [4]{b+a x^8}}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^6}}+\frac {\left (2 \sqrt {x} \sqrt [4]{1+\frac {a x^4}{b}}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{1+\frac {a x^8}{b}}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^6}}\\ &=\frac {2 x \sqrt [4]{1+\frac {a x^4}{b}} \, _2F_1\left (\frac {1}{8},\frac {1}{4};\frac {9}{8};-\frac {a x^4}{b}\right )}{\sqrt [4]{b x^2+a x^6}}-\frac {\left (4 \left (-1-\sqrt {1-a b}\right ) \sqrt {x} \sqrt [4]{b+a x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (-2-2 \sqrt {1-a b}+2 a x^4\right ) \sqrt [4]{b+a x^8}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^6}}-\frac {\left (4 \left (-1+\sqrt {1-a b}\right ) \sqrt {x} \sqrt [4]{b+a x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (-2+2 \sqrt {1-a b}+2 a x^4\right ) \sqrt [4]{b+a x^8}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^6}}\\ &=\frac {2 x \sqrt [4]{1+\frac {a x^4}{b}} \, _2F_1\left (\frac {1}{8},\frac {1}{4};\frac {9}{8};-\frac {a x^4}{b}\right )}{\sqrt [4]{b x^2+a x^6}}-\frac {\left (4 \left (-1-\sqrt {1-a b}\right ) \sqrt {x} \sqrt [4]{b+a x^4}\right ) \operatorname {Subst}\left (\int \left (\frac {-1-\sqrt {1-a b}}{2 \sqrt [4]{b+a x^8} \left (2-a b+2 \sqrt {1-a b}-a^2 x^8\right )}-\frac {a x^4}{2 \sqrt [4]{b+a x^8} \left (2-a b+2 \sqrt {1-a b}-a^2 x^8\right )}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^6}}-\frac {\left (4 \left (-1+\sqrt {1-a b}\right ) \sqrt {x} \sqrt [4]{b+a x^4}\right ) \operatorname {Subst}\left (\int \left (\frac {1-\sqrt {1-a b}}{2 \sqrt [4]{b+a x^8} \left (-2+a b+2 \sqrt {1-a b}+a^2 x^8\right )}+\frac {a x^4}{2 \sqrt [4]{b+a x^8} \left (-2+a b+2 \sqrt {1-a b}+a^2 x^8\right )}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^6}}\\ &=\frac {2 x \sqrt [4]{1+\frac {a x^4}{b}} \, _2F_1\left (\frac {1}{8},\frac {1}{4};\frac {9}{8};-\frac {a x^4}{b}\right )}{\sqrt [4]{b x^2+a x^6}}+\frac {\left (2 a \left (-1-\sqrt {1-a b}\right ) \sqrt {x} \sqrt [4]{b+a x^4}\right ) \operatorname {Subst}\left (\int \frac {x^4}{\sqrt [4]{b+a x^8} \left (2-a b+2 \sqrt {1-a b}-a^2 x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^6}}-\frac {\left (2 \left (-1-\sqrt {1-a b}\right )^2 \sqrt {x} \sqrt [4]{b+a x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{b+a x^8} \left (2-a b+2 \sqrt {1-a b}-a^2 x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^6}}-\frac {\left (2 a \left (-1+\sqrt {1-a b}\right ) \sqrt {x} \sqrt [4]{b+a x^4}\right ) \operatorname {Subst}\left (\int \frac {x^4}{\sqrt [4]{b+a x^8} \left (-2+a b+2 \sqrt {1-a b}+a^2 x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^6}}-\frac {\left (2 \left (1-\sqrt {1-a b}\right ) \left (-1+\sqrt {1-a b}\right ) \sqrt {x} \sqrt [4]{b+a x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{b+a x^8} \left (-2+a b+2 \sqrt {1-a b}+a^2 x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^6}}\\ &=\frac {2 x \sqrt [4]{1+\frac {a x^4}{b}} \, _2F_1\left (\frac {1}{8},\frac {1}{4};\frac {9}{8};-\frac {a x^4}{b}\right )}{\sqrt [4]{b x^2+a x^6}}+\frac {\left (2 a \left (-1-\sqrt {1-a b}\right ) \sqrt {x} \sqrt [4]{1+\frac {a x^4}{b}}\right ) \operatorname {Subst}\left (\int \frac {x^4}{\left (2-a b+2 \sqrt {1-a b}-a^2 x^8\right ) \sqrt [4]{1+\frac {a x^8}{b}}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^6}}-\frac {\left (2 \left (-1-\sqrt {1-a b}\right )^2 \sqrt {x} \sqrt [4]{1+\frac {a x^4}{b}}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (2-a b+2 \sqrt {1-a b}-a^2 x^8\right ) \sqrt [4]{1+\frac {a x^8}{b}}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^6}}-\frac {\left (2 a \left (-1+\sqrt {1-a b}\right ) \sqrt {x} \sqrt [4]{1+\frac {a x^4}{b}}\right ) \operatorname {Subst}\left (\int \frac {x^4}{\left (-2+a b+2 \sqrt {1-a b}+a^2 x^8\right ) \sqrt [4]{1+\frac {a x^8}{b}}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^6}}-\frac {\left (2 \left (1-\sqrt {1-a b}\right ) \left (-1+\sqrt {1-a b}\right ) \sqrt {x} \sqrt [4]{1+\frac {a x^4}{b}}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (-2+a b+2 \sqrt {1-a b}+a^2 x^8\right ) \sqrt [4]{1+\frac {a x^8}{b}}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^6}}\\ &=-\frac {2 x \sqrt [4]{1+\frac {a x^4}{b}} F_1\left (\frac {1}{8};1,\frac {1}{4};\frac {9}{8};\frac {a^2 x^4}{2-a b-2 \sqrt {1-a b}},-\frac {a x^4}{b}\right )}{\sqrt [4]{b x^2+a x^6}}-\frac {2 x \sqrt [4]{1+\frac {a x^4}{b}} F_1\left (\frac {1}{8};1,\frac {1}{4};\frac {9}{8};\frac {a^2 x^4}{2-a b+2 \sqrt {1-a b}},-\frac {a x^4}{b}\right )}{\sqrt [4]{b x^2+a x^6}}-\frac {2 a \left (1-\sqrt {1-a b}\right ) x^3 \sqrt [4]{1+\frac {a x^4}{b}} F_1\left (\frac {5}{8};1,\frac {1}{4};\frac {13}{8};\frac {a^2 x^4}{2-a b-2 \sqrt {1-a b}},-\frac {a x^4}{b}\right )}{5 \left (2-a b-2 \sqrt {1-a b}\right ) \sqrt [4]{b x^2+a x^6}}-\frac {2 a \left (1+\sqrt {1-a b}\right ) x^3 \sqrt [4]{1+\frac {a x^4}{b}} F_1\left (\frac {5}{8};1,\frac {1}{4};\frac {13}{8};\frac {a^2 x^4}{2-a b+2 \sqrt {1-a b}},-\frac {a x^4}{b}\right )}{5 \left (2-a b+2 \sqrt {1-a b}\right ) \sqrt [4]{b x^2+a x^6}}+\frac {2 x \sqrt [4]{1+\frac {a x^4}{b}} \, _2F_1\left (\frac {1}{8},\frac {1}{4};\frac {9}{8};-\frac {a x^4}{b}\right )}{\sqrt [4]{b x^2+a x^6}}\\ \end {align*}

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Mathematica [F]  time = 1.84, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-b+a x^4}{\left (b-2 x^2+a x^4\right ) \sqrt [4]{b x^2+a x^6}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(-b + a*x^4)/((b - 2*x^2 + a*x^4)*(b*x^2 + a*x^6)^(1/4)),x]

[Out]

Integrate[(-b + a*x^4)/((b - 2*x^2 + a*x^4)*(b*x^2 + a*x^6)^(1/4)), x]

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IntegrateAlgebraic [A]  time = 1.39, size = 61, normalized size = 1.00 \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{b x^2+a x^6}}\right )}{\sqrt [4]{2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{b x^2+a x^6}}\right )}{\sqrt [4]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(-b + a*x^4)/((b - 2*x^2 + a*x^4)*(b*x^2 + a*x^6)^(1/4)),x]

[Out]

-(ArcTan[(2^(1/4)*x)/(b*x^2 + a*x^6)^(1/4)]/2^(1/4)) - ArcTanh[(2^(1/4)*x)/(b*x^2 + a*x^6)^(1/4)]/2^(1/4)

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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((a*x^4-b)/(a*x^4-2*x^2+b)/(a*x^6+b*x^2)^(1/4),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} - b}{{\left (a x^{6} + b x^{2}\right )}^{\frac {1}{4}} {\left (a x^{4} - 2 \, x^{2} + b\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^4-b)/(a*x^4-2*x^2+b)/(a*x^6+b*x^2)^(1/4),x, algorithm="giac")

[Out]

integrate((a*x^4 - b)/((a*x^6 + b*x^2)^(1/4)*(a*x^4 - 2*x^2 + b)), x)

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {a \,x^{4}-b}{\left (a \,x^{4}-2 x^{2}+b \right ) \left (a \,x^{6}+b \,x^{2}\right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^4-b)/(a*x^4-2*x^2+b)/(a*x^6+b*x^2)^(1/4),x)

[Out]

int((a*x^4-b)/(a*x^4-2*x^2+b)/(a*x^6+b*x^2)^(1/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a x^{4} - b}{{\left (a x^{6} + b x^{2}\right )}^{\frac {1}{4}} {\left (a x^{4} - 2 \, x^{2} + b\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^4-b)/(a*x^4-2*x^2+b)/(a*x^6+b*x^2)^(1/4),x, algorithm="maxima")

[Out]

integrate((a*x^4 - b)/((a*x^6 + b*x^2)^(1/4)*(a*x^4 - 2*x^2 + b)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int -\frac {b-a\,x^4}{{\left (a\,x^6+b\,x^2\right )}^{1/4}\,\left (a\,x^4-2\,x^2+b\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(b - a*x^4)/((a*x^6 + b*x^2)^(1/4)*(b + a*x^4 - 2*x^2)),x)

[Out]

int(-(b - a*x^4)/((a*x^6 + b*x^2)^(1/4)*(b + a*x^4 - 2*x^2)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x**4-b)/(a*x**4-2*x**2+b)/(a*x**6+b*x**2)**(1/4),x)

[Out]

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

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