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

Optimal. Leaf size=69 \[ -\frac {1}{2} \text {RootSum}\left [\text {$\#$1}^8-\text {$\#$1}^4 a+b\& ,\frac {\text {$\#$1}^3 \log \left (\sqrt [4]{a x^4+b x^2}-\text {$\#$1} x\right )-\text {$\#$1}^3 \log (x)}{2 \text {$\#$1}^4-a}\& \right ] \]

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Rubi [B]  time = 0.91, antiderivative size = 587, normalized size of antiderivative = 8.51, number of steps used = 21, number of rules used = 11, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.314, Rules used = {2056, 1269, 1428, 408, 240, 212, 206, 203, 377, 208, 205} \begin {gather*} -\frac {\sqrt {x} \left (-a \sqrt {a^2-4 b}+a^2-2 b\right )^{3/4} \sqrt [4]{a x^2+b} \tan ^{-1}\left (\frac {\sqrt {x} \sqrt [4]{-a \sqrt {a^2-4 b}+a^2-2 b}}{\sqrt [4]{a-\sqrt {a^2-4 b}} \sqrt [4]{a x^2+b}}\right )}{\left (a-\sqrt {a^2-4 b}\right )^{3/4} \sqrt {a^2-4 b} \sqrt [4]{a x^4+b x^2}}+\frac {\sqrt {x} \left (a \sqrt {a^2-4 b}+a^2-2 b\right )^{3/4} \sqrt [4]{a x^2+b} \tan ^{-1}\left (\frac {\sqrt {x} \sqrt [4]{a \sqrt {a^2-4 b}+a^2-2 b}}{\sqrt [4]{\sqrt {a^2-4 b}+a} \sqrt [4]{a x^2+b}}\right )}{\left (\sqrt {a^2-4 b}+a\right )^{3/4} \sqrt {a^2-4 b} \sqrt [4]{a x^4+b x^2}}-\frac {\sqrt {x} \left (-a \sqrt {a^2-4 b}+a^2-2 b\right )^{3/4} \sqrt [4]{a x^2+b} \tanh ^{-1}\left (\frac {\sqrt {x} \sqrt [4]{-a \sqrt {a^2-4 b}+a^2-2 b}}{\sqrt [4]{a-\sqrt {a^2-4 b}} \sqrt [4]{a x^2+b}}\right )}{\left (a-\sqrt {a^2-4 b}\right )^{3/4} \sqrt {a^2-4 b} \sqrt [4]{a x^4+b x^2}}+\frac {\sqrt {x} \left (a \sqrt {a^2-4 b}+a^2-2 b\right )^{3/4} \sqrt [4]{a x^2+b} \tanh ^{-1}\left (\frac {\sqrt {x} \sqrt [4]{a \sqrt {a^2-4 b}+a^2-2 b}}{\sqrt [4]{\sqrt {a^2-4 b}+a} \sqrt [4]{a x^2+b}}\right )}{\left (\sqrt {a^2-4 b}+a\right )^{3/4} \sqrt {a^2-4 b} \sqrt [4]{a x^4+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a^2 - a*Sqrt[a^2 - 4*b] - 2*b)^(3/4)*Sqrt[x]*(b + a*x^2)^(1/4)*ArcTan[((a^2 - a*Sqrt[a^2 - 4*b] - 2*b)^(1/
4)*Sqrt[x])/((a - Sqrt[a^2 - 4*b])^(1/4)*(b + a*x^2)^(1/4))])/((a - Sqrt[a^2 - 4*b])^(3/4)*Sqrt[a^2 - 4*b]*(b*
x^2 + a*x^4)^(1/4))) + ((a^2 + a*Sqrt[a^2 - 4*b] - 2*b)^(3/4)*Sqrt[x]*(b + a*x^2)^(1/4)*ArcTan[((a^2 + a*Sqrt[
a^2 - 4*b] - 2*b)^(1/4)*Sqrt[x])/((a + Sqrt[a^2 - 4*b])^(1/4)*(b + a*x^2)^(1/4))])/((a + Sqrt[a^2 - 4*b])^(3/4
)*Sqrt[a^2 - 4*b]*(b*x^2 + a*x^4)^(1/4)) - ((a^2 - a*Sqrt[a^2 - 4*b] - 2*b)^(3/4)*Sqrt[x]*(b + a*x^2)^(1/4)*Ar
cTanh[((a^2 - a*Sqrt[a^2 - 4*b] - 2*b)^(1/4)*Sqrt[x])/((a - Sqrt[a^2 - 4*b])^(1/4)*(b + a*x^2)^(1/4))])/((a -
Sqrt[a^2 - 4*b])^(3/4)*Sqrt[a^2 - 4*b]*(b*x^2 + a*x^4)^(1/4)) + ((a^2 + a*Sqrt[a^2 - 4*b] - 2*b)^(3/4)*Sqrt[x]
*(b + a*x^2)^(1/4)*ArcTanh[((a^2 + a*Sqrt[a^2 - 4*b] - 2*b)^(1/4)*Sqrt[x])/((a + Sqrt[a^2 - 4*b])^(1/4)*(b + a
*x^2)^(1/4))])/((a + Sqrt[a^2 - 4*b])^(3/4)*Sqrt[a^2 - 4*b]*(b*x^2 + a*x^4)^(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 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 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 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 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 408

Int[((a_) + (b_.)*(x_)^4)^(p_)/((c_) + (d_.)*(x_)^4), x_Symbol] :> Dist[b/d, Int[(a + b*x^4)^(p - 1), x], x] -
Dist[(b*c - a*d)/d, Int[(a + b*x^4)^(p - 1)/(c + d*x^4), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0
] && (EqQ[p, 3/4] || EqQ[p, 5/4])

Rule 1269

Int[((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With
[{k = Denominator[m]}, Dist[k/f, Subst[Int[x^(k*(m + 1) - 1)*(d + (e*x^(2*k))/f^2)^q*(a + (b*x^(2*k))/f^k + (c
*x^(4*k))/f^4)^p, x], x, (f*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, f, p, q}, x] && NeQ[b^2 - 4*a*c, 0] && Fra
ctionQ[m] && IntegerQ[p]

Rule 1428

Int[((d_) + (e_.)*(x_)^(n_))^(q_)/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x_Symbol] :> With[{r = Rt[b^2 -
4*a*c, 2]}, Dist[(2*c)/r, Int[(d + e*x^n)^q/(b - r + 2*c*x^n), x], x] - Dist[(2*c)/r, Int[(d + e*x^n)^q/(b +
r + 2*c*x^n), x], x]] /; FreeQ[{a, b, c, d, e, n, q}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] &&  !IntegerQ[q]

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]

Rubi steps

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

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Mathematica [B]  time = 0.07, size = 396, normalized size = 5.74 \begin {gather*} \frac {\left (x^2 \left (a x^2+b\right )\right )^{3/4} \left (\left (a-\sqrt {a^2-4 b}\right )^{3/4} \log \left (2^{3/4} \sqrt [4]{a-\sqrt {a^2-4 b}}-2 \sqrt [4]{a+\frac {b}{x^2}}\right )-\left (\sqrt {a^2-4 b}+a\right )^{3/4} \log \left (2^{3/4} \sqrt [4]{\sqrt {a^2-4 b}+a}-2 \sqrt [4]{a+\frac {b}{x^2}}\right )-\left (a-\sqrt {a^2-4 b}\right )^{3/4} \log \left (2^{3/4} \sqrt [4]{a-\sqrt {a^2-4 b}}+2 \sqrt [4]{a+\frac {b}{x^2}}\right )+\left (\sqrt {a^2-4 b}+a\right )^{3/4} \log \left (2^{3/4} \sqrt [4]{\sqrt {a^2-4 b}+a}+2 \sqrt [4]{a+\frac {b}{x^2}}\right )+2 \left (a-\sqrt {a^2-4 b}\right )^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a+\frac {b}{x^2}}}{\sqrt [4]{a-\sqrt {a^2-4 b}}}\right )-2 \left (\sqrt {a^2-4 b}+a\right )^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a+\frac {b}{x^2}}}{\sqrt [4]{\sqrt {a^2-4 b}+a}}\right )\right )}{2\ 2^{3/4} x^3 \sqrt {a^2-4 b} \left (a+\frac {b}{x^2}\right )^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((x^2*(b + a*x^2))^(3/4)*(2*(a - Sqrt[a^2 - 4*b])^(3/4)*ArcTan[(2^(1/4)*(a + b/x^2)^(1/4))/(a - Sqrt[a^2 - 4*b
])^(1/4)] - 2*(a + Sqrt[a^2 - 4*b])^(3/4)*ArcTan[(2^(1/4)*(a + b/x^2)^(1/4))/(a + Sqrt[a^2 - 4*b])^(1/4)] + (a
- Sqrt[a^2 - 4*b])^(3/4)*Log[2^(3/4)*(a - Sqrt[a^2 - 4*b])^(1/4) - 2*(a + b/x^2)^(1/4)] - (a + Sqrt[a^2 - 4*b
])^(3/4)*Log[2^(3/4)*(a + Sqrt[a^2 - 4*b])^(1/4) - 2*(a + b/x^2)^(1/4)] - (a - Sqrt[a^2 - 4*b])^(3/4)*Log[2^(3
/4)*(a - Sqrt[a^2 - 4*b])^(1/4) + 2*(a + b/x^2)^(1/4)] + (a + Sqrt[a^2 - 4*b])^(3/4)*Log[2^(3/4)*(a + Sqrt[a^2
- 4*b])^(1/4) + 2*(a + b/x^2)^(1/4)]))/(2*2^(3/4)*Sqrt[a^2 - 4*b]*(a + b/x^2)^(3/4)*x^3)

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IntegrateAlgebraic [A]  time = 0.00, size = 69, normalized size = 1.00 \begin {gather*} -\frac {1}{2} \text {RootSum}\left [b-a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-\log (x) \text {$\#$1}^3+\log \left (\sqrt [4]{b x^2+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^3}{-a+2 \text {$\#$1}^4}\&\right ] \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*RootSum[b - a*#1^4 + #1^8 & , (-(Log[x]*#1^3) + Log[(b*x^2 + a*x^4)^(1/4) - x*#1]*#1^3)/(-a + 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^2+b)/(x^4+a*x^2+b)/(a*x^4+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^{2} + b}{{\left (a x^{4} + b x^{2}\right )}^{\frac {1}{4}} {\left (x^{4} + a x^{2} + b\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((b + a*x^2)/((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 {a x^{2} + b}{\sqrt [4]{x^{2} \left (a x^{2} + b\right )} \left (a x^{2} + b + x^{4}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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