3.1088 \(\int \frac{1}{\sqrt{x} (a+b x^2+c x^4)^3} \, dx\)

Optimal. Leaf size=658 \[ \frac{\sqrt{x} \left (60 a^2 c^2+b c x^2 \left (7 b^2-52 a c\right )-55 a b^2 c+7 b^4\right )}{16 a^2 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}+\frac{3 c^{3/4} \left (280 a^2 c^2-66 a b^2 c-b \left (7 b^2-52 a c\right ) \sqrt{b^2-4 a c}+7 b^4\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{32 \sqrt [4]{2} a^2 \left (b^2-4 a c\right )^{5/2} \left (-\sqrt{b^2-4 a c}-b\right )^{3/4}}-\frac{3 c^{3/4} \left (280 a^2 c^2-66 a b^2 c+b \left (7 b^2-52 a c\right ) \sqrt{b^2-4 a c}+7 b^4\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{32 \sqrt [4]{2} a^2 \left (b^2-4 a c\right )^{5/2} \left (\sqrt{b^2-4 a c}-b\right )^{3/4}}+\frac{3 c^{3/4} \left (280 a^2 c^2-66 a b^2 c-b \left (7 b^2-52 a c\right ) \sqrt{b^2-4 a c}+7 b^4\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{32 \sqrt [4]{2} a^2 \left (b^2-4 a c\right )^{5/2} \left (-\sqrt{b^2-4 a c}-b\right )^{3/4}}-\frac{3 c^{3/4} \left (280 a^2 c^2-66 a b^2 c+b \left (7 b^2-52 a c\right ) \sqrt{b^2-4 a c}+7 b^4\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{32 \sqrt [4]{2} a^2 \left (b^2-4 a c\right )^{5/2} \left (\sqrt{b^2-4 a c}-b\right )^{3/4}}+\frac{\sqrt{x} \left (-2 a c+b^2+b c x^2\right )}{4 a \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2} \]

[Out]

(Sqrt[x]*(b^2 - 2*a*c + b*c*x^2))/(4*a*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)^2) + (Sqrt[x]*(7*b^4 - 55*a*b^2*c + 6
0*a^2*c^2 + b*c*(7*b^2 - 52*a*c)*x^2))/(16*a^2*(b^2 - 4*a*c)^2*(a + b*x^2 + c*x^4)) + (3*c^(3/4)*(7*b^4 - 66*a
*b^2*c + 280*a^2*c^2 - b*(7*b^2 - 52*a*c)*Sqrt[b^2 - 4*a*c])*ArcTan[(2^(1/4)*c^(1/4)*Sqrt[x])/(-b - Sqrt[b^2 -
4*a*c])^(1/4)])/(32*2^(1/4)*a^2*(b^2 - 4*a*c)^(5/2)*(-b - Sqrt[b^2 - 4*a*c])^(3/4)) - (3*c^(3/4)*(7*b^4 - 66*
a*b^2*c + 280*a^2*c^2 + b*(7*b^2 - 52*a*c)*Sqrt[b^2 - 4*a*c])*ArcTan[(2^(1/4)*c^(1/4)*Sqrt[x])/(-b + Sqrt[b^2
- 4*a*c])^(1/4)])/(32*2^(1/4)*a^2*(b^2 - 4*a*c)^(5/2)*(-b + Sqrt[b^2 - 4*a*c])^(3/4)) + (3*c^(3/4)*(7*b^4 - 66
*a*b^2*c + 280*a^2*c^2 - b*(7*b^2 - 52*a*c)*Sqrt[b^2 - 4*a*c])*ArcTanh[(2^(1/4)*c^(1/4)*Sqrt[x])/(-b - Sqrt[b^
2 - 4*a*c])^(1/4)])/(32*2^(1/4)*a^2*(b^2 - 4*a*c)^(5/2)*(-b - Sqrt[b^2 - 4*a*c])^(3/4)) - (3*c^(3/4)*(7*b^4 -
66*a*b^2*c + 280*a^2*c^2 + b*(7*b^2 - 52*a*c)*Sqrt[b^2 - 4*a*c])*ArcTanh[(2^(1/4)*c^(1/4)*Sqrt[x])/(-b + Sqrt[
b^2 - 4*a*c])^(1/4)])/(32*2^(1/4)*a^2*(b^2 - 4*a*c)^(5/2)*(-b + Sqrt[b^2 - 4*a*c])^(3/4))

________________________________________________________________________________________

Rubi [A]  time = 5.79179, antiderivative size = 658, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {1115, 1345, 1430, 1422, 212, 208, 205} \[ \frac{\sqrt{x} \left (60 a^2 c^2+b c x^2 \left (7 b^2-52 a c\right )-55 a b^2 c+7 b^4\right )}{16 a^2 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}+\frac{3 c^{3/4} \left (280 a^2 c^2-66 a b^2 c-b \left (7 b^2-52 a c\right ) \sqrt{b^2-4 a c}+7 b^4\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{32 \sqrt [4]{2} a^2 \left (b^2-4 a c\right )^{5/2} \left (-\sqrt{b^2-4 a c}-b\right )^{3/4}}-\frac{3 c^{3/4} \left (280 a^2 c^2-66 a b^2 c+b \left (7 b^2-52 a c\right ) \sqrt{b^2-4 a c}+7 b^4\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{32 \sqrt [4]{2} a^2 \left (b^2-4 a c\right )^{5/2} \left (\sqrt{b^2-4 a c}-b\right )^{3/4}}+\frac{3 c^{3/4} \left (280 a^2 c^2-66 a b^2 c-b \left (7 b^2-52 a c\right ) \sqrt{b^2-4 a c}+7 b^4\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{32 \sqrt [4]{2} a^2 \left (b^2-4 a c\right )^{5/2} \left (-\sqrt{b^2-4 a c}-b\right )^{3/4}}-\frac{3 c^{3/4} \left (280 a^2 c^2-66 a b^2 c+b \left (7 b^2-52 a c\right ) \sqrt{b^2-4 a c}+7 b^4\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{32 \sqrt [4]{2} a^2 \left (b^2-4 a c\right )^{5/2} \left (\sqrt{b^2-4 a c}-b\right )^{3/4}}+\frac{\sqrt{x} \left (-2 a c+b^2+b c x^2\right )}{4 a \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[x]*(a + b*x^2 + c*x^4)^3),x]

[Out]

(Sqrt[x]*(b^2 - 2*a*c + b*c*x^2))/(4*a*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)^2) + (Sqrt[x]*(7*b^4 - 55*a*b^2*c + 6
0*a^2*c^2 + b*c*(7*b^2 - 52*a*c)*x^2))/(16*a^2*(b^2 - 4*a*c)^2*(a + b*x^2 + c*x^4)) + (3*c^(3/4)*(7*b^4 - 66*a
*b^2*c + 280*a^2*c^2 - b*(7*b^2 - 52*a*c)*Sqrt[b^2 - 4*a*c])*ArcTan[(2^(1/4)*c^(1/4)*Sqrt[x])/(-b - Sqrt[b^2 -
4*a*c])^(1/4)])/(32*2^(1/4)*a^2*(b^2 - 4*a*c)^(5/2)*(-b - Sqrt[b^2 - 4*a*c])^(3/4)) - (3*c^(3/4)*(7*b^4 - 66*
a*b^2*c + 280*a^2*c^2 + b*(7*b^2 - 52*a*c)*Sqrt[b^2 - 4*a*c])*ArcTan[(2^(1/4)*c^(1/4)*Sqrt[x])/(-b + Sqrt[b^2
- 4*a*c])^(1/4)])/(32*2^(1/4)*a^2*(b^2 - 4*a*c)^(5/2)*(-b + Sqrt[b^2 - 4*a*c])^(3/4)) + (3*c^(3/4)*(7*b^4 - 66
*a*b^2*c + 280*a^2*c^2 - b*(7*b^2 - 52*a*c)*Sqrt[b^2 - 4*a*c])*ArcTanh[(2^(1/4)*c^(1/4)*Sqrt[x])/(-b - Sqrt[b^
2 - 4*a*c])^(1/4)])/(32*2^(1/4)*a^2*(b^2 - 4*a*c)^(5/2)*(-b - Sqrt[b^2 - 4*a*c])^(3/4)) - (3*c^(3/4)*(7*b^4 -
66*a*b^2*c + 280*a^2*c^2 + b*(7*b^2 - 52*a*c)*Sqrt[b^2 - 4*a*c])*ArcTanh[(2^(1/4)*c^(1/4)*Sqrt[x])/(-b + Sqrt[
b^2 - 4*a*c])^(1/4)])/(32*2^(1/4)*a^2*(b^2 - 4*a*c)^(5/2)*(-b + Sqrt[b^2 - 4*a*c])^(3/4))

Rule 1115

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

Rule 1345

Int[((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(b^2 - 2*a*c + b*c*x^n)*(a + b*x^
n + c*x^(2*n))^(p + 1))/(a*n*(p + 1)*(b^2 - 4*a*c)), x] + Dist[1/(a*n*(p + 1)*(b^2 - 4*a*c)), Int[(b^2 - 2*a*c
+ n*(p + 1)*(b^2 - 4*a*c) + b*c*(n*(2*p + 3) + 1)*x^n)*(a + b*x^n + c*x^(2*n))^(p + 1), x], x] /; FreeQ[{a, b
, c, n}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && ILtQ[p, -1]

Rule 1430

Int[((d_) + (e_.)*(x_)^(n_))*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] :> -Simp[(x*(d*b^2 -
a*b*e - 2*a*c*d + (b*d - 2*a*e)*c*x^n)*(a + b*x^n + c*x^(2*n))^(p + 1))/(a*n*(p + 1)*(b^2 - 4*a*c)), x] + Dist
[1/(a*n*(p + 1)*(b^2 - 4*a*c)), Int[Simp[(n*p + n + 1)*d*b^2 - a*b*e - 2*a*c*d*(2*n*p + 2*n + 1) + (2*n*p + 3*
n + 1)*(d*b - 2*a*e)*c*x^n, x]*(a + b*x^n + c*x^(2*n))^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[
n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && ILtQ[p, -1]

Rule 1422

Int[((d_) + (e_.)*(x_)^(n_))/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x_Symbol] :> With[{q = Rt[b^2 - 4*a*
c, 2]}, Dist[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^n), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), In
t[1/(b/2 + q/2 + c*x^n), x], x]] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && NeQ
[c*d^2 - b*d*e + a*e^2, 0] && (PosQ[b^2 - 4*a*c] ||  !IGtQ[n/2, 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 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 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]

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{x} \left (a+b x^2+c x^4\right )^3} \, dx &=2 \operatorname{Subst}\left (\int \frac{1}{\left (a+b x^4+c x^8\right )^3} \, dx,x,\sqrt{x}\right )\\ &=\frac{\sqrt{x} \left (b^2-2 a c+b c x^2\right )}{4 a \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac{\operatorname{Subst}\left (\int \frac{b^2-2 a c-8 \left (b^2-4 a c\right )-11 b c x^4}{\left (a+b x^4+c x^8\right )^2} \, dx,x,\sqrt{x}\right )}{4 a \left (b^2-4 a c\right )}\\ &=\frac{\sqrt{x} \left (b^2-2 a c+b c x^2\right )}{4 a \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}+\frac{\sqrt{x} \left (7 b^4-55 a b^2 c+60 a^2 c^2+b c \left (7 b^2-52 a c\right ) x^2\right )}{16 a^2 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}+\frac{\operatorname{Subst}\left (\int \frac{3 \left (7 b^4-59 a b^2 c+140 a^2 c^2\right )+3 b c \left (7 b^2-52 a c\right ) x^4}{a+b x^4+c x^8} \, dx,x,\sqrt{x}\right )}{16 a^2 \left (b^2-4 a c\right )^2}\\ &=\frac{\sqrt{x} \left (b^2-2 a c+b c x^2\right )}{4 a \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}+\frac{\sqrt{x} \left (7 b^4-55 a b^2 c+60 a^2 c^2+b c \left (7 b^2-52 a c\right ) x^2\right )}{16 a^2 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}-\frac{\left (3 c \left (7 b^4-66 a b^2 c+280 a^2 c^2-b \left (7 b^2-52 a c\right ) \sqrt{b^2-4 a c}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{b}{2}+\frac{1}{2} \sqrt{b^2-4 a c}+c x^4} \, dx,x,\sqrt{x}\right )}{32 a^2 \left (b^2-4 a c\right )^{5/2}}+\frac{\left (3 c \left (7 b^4-66 a b^2 c+280 a^2 c^2+b \left (7 b^2-52 a c\right ) \sqrt{b^2-4 a c}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{b}{2}-\frac{1}{2} \sqrt{b^2-4 a c}+c x^4} \, dx,x,\sqrt{x}\right )}{32 a^2 \left (b^2-4 a c\right )^{5/2}}\\ &=\frac{\sqrt{x} \left (b^2-2 a c+b c x^2\right )}{4 a \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}+\frac{\sqrt{x} \left (7 b^4-55 a b^2 c+60 a^2 c^2+b c \left (7 b^2-52 a c\right ) x^2\right )}{16 a^2 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}+\frac{\left (3 c \left (7 b^4-66 a b^2 c+280 a^2 c^2-b \left (7 b^2-52 a c\right ) \sqrt{b^2-4 a c}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-b-\sqrt{b^2-4 a c}}-\sqrt{2} \sqrt{c} x^2} \, dx,x,\sqrt{x}\right )}{32 a^2 \left (b^2-4 a c\right )^{5/2} \sqrt{-b-\sqrt{b^2-4 a c}}}+\frac{\left (3 c \left (7 b^4-66 a b^2 c+280 a^2 c^2-b \left (7 b^2-52 a c\right ) \sqrt{b^2-4 a c}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-b-\sqrt{b^2-4 a c}}+\sqrt{2} \sqrt{c} x^2} \, dx,x,\sqrt{x}\right )}{32 a^2 \left (b^2-4 a c\right )^{5/2} \sqrt{-b-\sqrt{b^2-4 a c}}}-\frac{\left (3 c \left (7 b^4-66 a b^2 c+280 a^2 c^2+b \left (7 b^2-52 a c\right ) \sqrt{b^2-4 a c}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-b+\sqrt{b^2-4 a c}}-\sqrt{2} \sqrt{c} x^2} \, dx,x,\sqrt{x}\right )}{32 a^2 \left (b^2-4 a c\right )^{5/2} \sqrt{-b+\sqrt{b^2-4 a c}}}-\frac{\left (3 c \left (7 b^4-66 a b^2 c+280 a^2 c^2+b \left (7 b^2-52 a c\right ) \sqrt{b^2-4 a c}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-b+\sqrt{b^2-4 a c}}+\sqrt{2} \sqrt{c} x^2} \, dx,x,\sqrt{x}\right )}{32 a^2 \left (b^2-4 a c\right )^{5/2} \sqrt{-b+\sqrt{b^2-4 a c}}}\\ &=\frac{\sqrt{x} \left (b^2-2 a c+b c x^2\right )}{4 a \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}+\frac{\sqrt{x} \left (7 b^4-55 a b^2 c+60 a^2 c^2+b c \left (7 b^2-52 a c\right ) x^2\right )}{16 a^2 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}+\frac{3 c^{3/4} \left (7 b^4-66 a b^2 c+280 a^2 c^2-b \left (7 b^2-52 a c\right ) \sqrt{b^2-4 a c}\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-b-\sqrt{b^2-4 a c}}}\right )}{32 \sqrt [4]{2} a^2 \left (b^2-4 a c\right )^{5/2} \left (-b-\sqrt{b^2-4 a c}\right )^{3/4}}-\frac{3 c^{3/4} \left (7 b^4-66 a b^2 c+280 a^2 c^2+b \left (7 b^2-52 a c\right ) \sqrt{b^2-4 a c}\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-b+\sqrt{b^2-4 a c}}}\right )}{32 \sqrt [4]{2} a^2 \left (b^2-4 a c\right )^{5/2} \left (-b+\sqrt{b^2-4 a c}\right )^{3/4}}+\frac{3 c^{3/4} \left (7 b^4-66 a b^2 c+280 a^2 c^2-b \left (7 b^2-52 a c\right ) \sqrt{b^2-4 a c}\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-b-\sqrt{b^2-4 a c}}}\right )}{32 \sqrt [4]{2} a^2 \left (b^2-4 a c\right )^{5/2} \left (-b-\sqrt{b^2-4 a c}\right )^{3/4}}-\frac{3 c^{3/4} \left (7 b^4-66 a b^2 c+280 a^2 c^2+b \left (7 b^2-52 a c\right ) \sqrt{b^2-4 a c}\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-b+\sqrt{b^2-4 a c}}}\right )}{32 \sqrt [4]{2} a^2 \left (b^2-4 a c\right )^{5/2} \left (-b+\sqrt{b^2-4 a c}\right )^{3/4}}\\ \end{align*}

Mathematica [C]  time = 0.47207, size = 258, normalized size = 0.39 \[ \frac{3 \text{RootSum}\left [\text{$\#$1}^4 b+\text{$\#$1}^8 c+a\& ,\frac{-52 \text{$\#$1}^4 a b c^2 \log \left (\sqrt{x}-\text{$\#$1}\right )+7 \text{$\#$1}^4 b^3 c \log \left (\sqrt{x}-\text{$\#$1}\right )+140 a^2 c^2 \log \left (\sqrt{x}-\text{$\#$1}\right )-59 a b^2 c \log \left (\sqrt{x}-\text{$\#$1}\right )+7 b^4 \log \left (\sqrt{x}-\text{$\#$1}\right )}{\text{$\#$1}^3 b+2 \text{$\#$1}^7 c}\& \right ]+\frac{4 \sqrt{x} \left (60 a^2 c^2-55 a b^2 c-52 a b c^2 x^2+7 b^3 c x^2+7 b^4\right )}{a+b x^2+c x^4}-\frac{16 a \sqrt{x} \left (4 a c-b^2\right ) \left (-2 a c+b^2+b c x^2\right )}{\left (a+b x^2+c x^4\right )^2}}{64 a^2 \left (b^2-4 a c\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[x]*(a + b*x^2 + c*x^4)^3),x]

[Out]

((-16*a*(-b^2 + 4*a*c)*Sqrt[x]*(b^2 - 2*a*c + b*c*x^2))/(a + b*x^2 + c*x^4)^2 + (4*Sqrt[x]*(7*b^4 - 55*a*b^2*c
+ 60*a^2*c^2 + 7*b^3*c*x^2 - 52*a*b*c^2*x^2))/(a + b*x^2 + c*x^4) + 3*RootSum[a + b*#1^4 + c*#1^8 & , (7*b^4*
Log[Sqrt[x] - #1] - 59*a*b^2*c*Log[Sqrt[x] - #1] + 140*a^2*c^2*Log[Sqrt[x] - #1] + 7*b^3*c*Log[Sqrt[x] - #1]*#
1^4 - 52*a*b*c^2*Log[Sqrt[x] - #1]*#1^4)/(b*#1^3 + 2*c*#1^7) & ])/(64*a^2*(b^2 - 4*a*c)^2)

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Maple [C]  time = 0.266, size = 316, normalized size = 0.5 \begin{align*} 2\,{\frac{1}{ \left ( c{x}^{4}+b{x}^{2}+a \right ) ^{2}} \left ( 1/32\,{\frac{ \left ( 92\,{a}^{2}{c}^{2}-79\,ac{b}^{2}+11\,{b}^{4} \right ) \sqrt{x}}{ \left ( 16\,{a}^{2}{c}^{2}-8\,ac{b}^{2}+{b}^{4} \right ) a}}-1/32\,{\frac{b \left ( 8\,{a}^{2}{c}^{2}+44\,ac{b}^{2}-7\,{b}^{4} \right ){x}^{5/2}}{{a}^{2} \left ( 16\,{a}^{2}{c}^{2}-8\,ac{b}^{2}+{b}^{4} \right ) }}+1/32\,{\frac{c \left ( 60\,{a}^{2}{c}^{2}-107\,ac{b}^{2}+14\,{b}^{4} \right ){x}^{9/2}}{{a}^{2} \left ( 16\,{a}^{2}{c}^{2}-8\,ac{b}^{2}+{b}^{4} \right ) }}-1/32\,{\frac{b{c}^{2} \left ( 52\,ac-7\,{b}^{2} \right ){x}^{13/2}}{{a}^{2} \left ( 16\,{a}^{2}{c}^{2}-8\,ac{b}^{2}+{b}^{4} \right ) }} \right ) }+{\frac{3}{64\,{a}^{2} \left ( 16\,{a}^{2}{c}^{2}-8\,ac{b}^{2}+{b}^{4} \right ) }\sum _{{\it \_R}={\it RootOf} \left ({{\it \_Z}}^{8}c+{{\it \_Z}}^{4}b+a \right ) }{\frac{bc \left ( -52\,ac+7\,{b}^{2} \right ){{\it \_R}}^{4}+140\,{a}^{2}{c}^{2}-59\,ac{b}^{2}+7\,{b}^{4}}{2\,{{\it \_R}}^{7}c+{{\it \_R}}^{3}b}\ln \left ( \sqrt{x}-{\it \_R} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(1/2)/(c*x^4+b*x^2+a)^3,x)

[Out]

2*(1/32*(92*a^2*c^2-79*a*b^2*c+11*b^4)/(16*a^2*c^2-8*a*b^2*c+b^4)/a*x^(1/2)-1/32*b*(8*a^2*c^2+44*a*b^2*c-7*b^4
)/a^2/(16*a^2*c^2-8*a*b^2*c+b^4)*x^(5/2)+1/32/a^2*c*(60*a^2*c^2-107*a*b^2*c+14*b^4)/(16*a^2*c^2-8*a*b^2*c+b^4)
*x^(9/2)-1/32*b*c^2*(52*a*c-7*b^2)/a^2/(16*a^2*c^2-8*a*b^2*c+b^4)*x^(13/2))/(c*x^4+b*x^2+a)^2+3/64/a^2/(16*a^2
*c^2-8*a*b^2*c+b^4)*sum((b*c*(-52*a*c+7*b^2)*_R^4+140*a^2*c^2-59*a*c*b^2+7*b^4)/(2*_R^7*c+_R^3*b)*ln(x^(1/2)-_
R),_R=RootOf(_Z^8*c+_Z^4*b+a))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{3 \,{\left (7 \, b^{4} c^{2} - 59 \, a b^{2} c^{3} + 140 \, a^{2} c^{4}\right )} x^{\frac{17}{2}} +{\left (42 \, b^{5} c - 347 \, a b^{3} c^{2} + 788 \, a^{2} b c^{3}\right )} x^{\frac{13}{2}} +{\left (21 \, b^{6} - 121 \, a b^{4} c - 41 \, a^{2} b^{2} c^{2} + 900 \, a^{3} c^{3}\right )} x^{\frac{9}{2}} +{\left (49 \, a b^{5} - 398 \, a^{2} b^{3} c + 832 \, a^{3} b c^{2}\right )} x^{\frac{5}{2}} + 32 \,{\left (a^{2} b^{4} - 8 \, a^{3} b^{2} c + 16 \, a^{4} c^{2}\right )} \sqrt{x}}{16 \,{\left (a^{5} b^{4} - 8 \, a^{6} b^{2} c + 16 \, a^{7} c^{2} +{\left (a^{3} b^{4} c^{2} - 8 \, a^{4} b^{2} c^{3} + 16 \, a^{5} c^{4}\right )} x^{8} + 2 \,{\left (a^{3} b^{5} c - 8 \, a^{4} b^{3} c^{2} + 16 \, a^{5} b c^{3}\right )} x^{6} +{\left (a^{3} b^{6} - 6 \, a^{4} b^{4} c + 32 \, a^{6} c^{3}\right )} x^{4} + 2 \,{\left (a^{4} b^{5} - 8 \, a^{5} b^{3} c + 16 \, a^{6} b c^{2}\right )} x^{2}\right )}} - \int \frac{3 \,{\left ({\left (7 \, b^{4} c - 59 \, a b^{2} c^{2} + 140 \, a^{2} c^{3}\right )} x^{\frac{7}{2}} +{\left (7 \, b^{5} - 66 \, a b^{3} c + 192 \, a^{2} b c^{2}\right )} x^{\frac{3}{2}}\right )}}{32 \,{\left (a^{4} b^{4} - 8 \, a^{5} b^{2} c + 16 \, a^{6} c^{2} +{\left (a^{3} b^{4} c - 8 \, a^{4} b^{2} c^{2} + 16 \, a^{5} c^{3}\right )} x^{4} +{\left (a^{3} b^{5} - 8 \, a^{4} b^{3} c + 16 \, a^{5} b c^{2}\right )} x^{2}\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(1/2)/(c*x^4+b*x^2+a)^3,x, algorithm="maxima")

[Out]

1/16*(3*(7*b^4*c^2 - 59*a*b^2*c^3 + 140*a^2*c^4)*x^(17/2) + (42*b^5*c - 347*a*b^3*c^2 + 788*a^2*b*c^3)*x^(13/2
) + (21*b^6 - 121*a*b^4*c - 41*a^2*b^2*c^2 + 900*a^3*c^3)*x^(9/2) + (49*a*b^5 - 398*a^2*b^3*c + 832*a^3*b*c^2)
*x^(5/2) + 32*(a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2)*sqrt(x))/(a^5*b^4 - 8*a^6*b^2*c + 16*a^7*c^2 + (a^3*b^4*c^2
- 8*a^4*b^2*c^3 + 16*a^5*c^4)*x^8 + 2*(a^3*b^5*c - 8*a^4*b^3*c^2 + 16*a^5*b*c^3)*x^6 + (a^3*b^6 - 6*a^4*b^4*c
+ 32*a^6*c^3)*x^4 + 2*(a^4*b^5 - 8*a^5*b^3*c + 16*a^6*b*c^2)*x^2) - integrate(3/32*((7*b^4*c - 59*a*b^2*c^2 +
140*a^2*c^3)*x^(7/2) + (7*b^5 - 66*a*b^3*c + 192*a^2*b*c^2)*x^(3/2))/(a^4*b^4 - 8*a^5*b^2*c + 16*a^6*c^2 + (a
^3*b^4*c - 8*a^4*b^2*c^2 + 16*a^5*c^3)*x^4 + (a^3*b^5 - 8*a^4*b^3*c + 16*a^5*b*c^2)*x^2), x)

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(1/2)/(c*x^4+b*x^2+a)^3,x, algorithm="fricas")

[Out]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**(1/2)/(c*x**4+b*x**2+a)**3,x)

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(1/2)/(c*x^4+b*x^2+a)^3,x, algorithm="giac")

[Out]

Timed out