3.1087 \(\int \frac{\sqrt{x}}{(a+b x^2+c x^4)^3} \, dx\)

Optimal. Leaf size=658 \[ \frac{x^{3/2} \left (52 a^2 c^2+b c x^2 \left (5 b^2-44 a c\right )-45 a b^2 c+5 b^4\right )}{16 a^2 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}-\frac{\sqrt [4]{c} \left (520 a^2 c^2-54 a b^2 c-b \left (5 b^2-44 a c\right ) \sqrt{b^2-4 a c}+5 b^4\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{32\ 2^{3/4} a^2 \left (b^2-4 a c\right )^{5/2} \sqrt [4]{-\sqrt{b^2-4 a c}-b}}+\frac{\sqrt [4]{c} \left (520 a^2 c^2-54 a b^2 c+b \left (5 b^2-44 a c\right ) \sqrt{b^2-4 a c}+5 b^4\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{32\ 2^{3/4} a^2 \left (b^2-4 a c\right )^{5/2} \sqrt [4]{\sqrt{b^2-4 a c}-b}}+\frac{\sqrt [4]{c} \left (520 a^2 c^2-54 a b^2 c-b \left (5 b^2-44 a c\right ) \sqrt{b^2-4 a c}+5 b^4\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{32\ 2^{3/4} a^2 \left (b^2-4 a c\right )^{5/2} \sqrt [4]{-\sqrt{b^2-4 a c}-b}}-\frac{\sqrt [4]{c} \left (520 a^2 c^2-54 a b^2 c+b \left (5 b^2-44 a c\right ) \sqrt{b^2-4 a c}+5 b^4\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{32\ 2^{3/4} a^2 \left (b^2-4 a c\right )^{5/2} \sqrt [4]{\sqrt{b^2-4 a c}-b}}+\frac{x^{3/2} \left (-2 a c+b^2+b c x^2\right )}{4 a \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2} \]

[Out]

(x^(3/2)*(b^2 - 2*a*c + b*c*x^2))/(4*a*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)^2) + (x^(3/2)*(5*b^4 - 45*a*b^2*c + 5
2*a^2*c^2 + b*c*(5*b^2 - 44*a*c)*x^2))/(16*a^2*(b^2 - 4*a*c)^2*(a + b*x^2 + c*x^4)) - (c^(1/4)*(5*b^4 - 54*a*b
^2*c + 520*a^2*c^2 - b*(5*b^2 - 44*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^(3/4)*a^2*(b^2 - 4*a*c)^(5/2)*(-b - Sqrt[b^2 - 4*a*c])^(1/4)) + (c^(1/4)*(5*b^4 - 54*a*b^
2*c + 520*a^2*c^2 + b*(5*b^2 - 44*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^(3/4)*a^2*(b^2 - 4*a*c)^(5/2)*(-b + Sqrt[b^2 - 4*a*c])^(1/4)) + (c^(1/4)*(5*b^4 - 54*a*b^2
*c + 520*a^2*c^2 - b*(5*b^2 - 44*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^(3/4)*a^2*(b^2 - 4*a*c)^(5/2)*(-b - Sqrt[b^2 - 4*a*c])^(1/4)) - (c^(1/4)*(5*b^4 - 54*a*b^2
*c + 520*a^2*c^2 + b*(5*b^2 - 44*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^(3/4)*a^2*(b^2 - 4*a*c)^(5/2)*(-b + Sqrt[b^2 - 4*a*c])^(1/4))

________________________________________________________________________________________

Rubi [A]  time = 5.48456, 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, 1366, 1500, 1510, 298, 205, 208} \[ \frac{x^{3/2} \left (52 a^2 c^2+b c x^2 \left (5 b^2-44 a c\right )-45 a b^2 c+5 b^4\right )}{16 a^2 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}-\frac{\sqrt [4]{c} \left (520 a^2 c^2-54 a b^2 c-b \left (5 b^2-44 a c\right ) \sqrt{b^2-4 a c}+5 b^4\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{32\ 2^{3/4} a^2 \left (b^2-4 a c\right )^{5/2} \sqrt [4]{-\sqrt{b^2-4 a c}-b}}+\frac{\sqrt [4]{c} \left (520 a^2 c^2-54 a b^2 c+b \left (5 b^2-44 a c\right ) \sqrt{b^2-4 a c}+5 b^4\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{32\ 2^{3/4} a^2 \left (b^2-4 a c\right )^{5/2} \sqrt [4]{\sqrt{b^2-4 a c}-b}}+\frac{\sqrt [4]{c} \left (520 a^2 c^2-54 a b^2 c-b \left (5 b^2-44 a c\right ) \sqrt{b^2-4 a c}+5 b^4\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{32\ 2^{3/4} a^2 \left (b^2-4 a c\right )^{5/2} \sqrt [4]{-\sqrt{b^2-4 a c}-b}}-\frac{\sqrt [4]{c} \left (520 a^2 c^2-54 a b^2 c+b \left (5 b^2-44 a c\right ) \sqrt{b^2-4 a c}+5 b^4\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{32\ 2^{3/4} a^2 \left (b^2-4 a c\right )^{5/2} \sqrt [4]{\sqrt{b^2-4 a c}-b}}+\frac{x^{3/2} \left (-2 a c+b^2+b c x^2\right )}{4 a \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]/(a + b*x^2 + c*x^4)^3,x]

[Out]

(x^(3/2)*(b^2 - 2*a*c + b*c*x^2))/(4*a*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)^2) + (x^(3/2)*(5*b^4 - 45*a*b^2*c + 5
2*a^2*c^2 + b*c*(5*b^2 - 44*a*c)*x^2))/(16*a^2*(b^2 - 4*a*c)^2*(a + b*x^2 + c*x^4)) - (c^(1/4)*(5*b^4 - 54*a*b
^2*c + 520*a^2*c^2 - b*(5*b^2 - 44*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^(3/4)*a^2*(b^2 - 4*a*c)^(5/2)*(-b - Sqrt[b^2 - 4*a*c])^(1/4)) + (c^(1/4)*(5*b^4 - 54*a*b^
2*c + 520*a^2*c^2 + b*(5*b^2 - 44*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^(3/4)*a^2*(b^2 - 4*a*c)^(5/2)*(-b + Sqrt[b^2 - 4*a*c])^(1/4)) + (c^(1/4)*(5*b^4 - 54*a*b^2
*c + 520*a^2*c^2 - b*(5*b^2 - 44*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^(3/4)*a^2*(b^2 - 4*a*c)^(5/2)*(-b - Sqrt[b^2 - 4*a*c])^(1/4)) - (c^(1/4)*(5*b^4 - 54*a*b^2
*c + 520*a^2*c^2 + b*(5*b^2 - 44*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^(3/4)*a^2*(b^2 - 4*a*c)^(5/2)*(-b + Sqrt[b^2 - 4*a*c])^(1/4))

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 1366

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

Rule 1500

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

Rule 1510

Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_)))/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x_Symbol] :> Wi
th[{q = Rt[b^2 - 4*a*c, 2]}, Dist[e/2 + (2*c*d - b*e)/(2*q), Int[(f*x)^m/(b/2 - q/2 + c*x^n), x], x] + Dist[e/
2 - (2*c*d - b*e)/(2*q), Int[(f*x)^m/(b/2 + q/2 + c*x^n), x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[n2
, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0]

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 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 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]

Rubi steps

\begin{align*} \int \frac{\sqrt{x}}{\left (a+b x^2+c x^4\right )^3} \, dx &=2 \operatorname{Subst}\left (\int \frac{x^2}{\left (a+b x^4+c x^8\right )^3} \, dx,x,\sqrt{x}\right )\\ &=\frac{x^{3/2} \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{x^2 \left (-5 b^2+26 a c-9 b c x^4\right )}{\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{x^{3/2} \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{x^{3/2} \left (5 b^4-45 a b^2 c+52 a^2 c^2+b c \left (5 b^2-44 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{x^2 \left (5 b^4-49 a b^2 c+260 a^2 c^2+b c \left (5 b^2-44 a c\right ) x^4\right )}{a+b x^4+c x^8} \, dx,x,\sqrt{x}\right )}{16 a^2 \left (b^2-4 a c\right )^2}\\ &=\frac{x^{3/2} \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{x^{3/2} \left (5 b^4-45 a b^2 c+52 a^2 c^2+b c \left (5 b^2-44 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 (c \left (5 b^4-54 a b^2 c+520 a^2 c^2-b \left (5 b^2-44 a c\right ) \sqrt{b^2-4 a c}\right )\right ) \operatorname{Subst}\left (\int \frac{x^2}{\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 (c \left (5 b^4-54 a b^2 c+520 a^2 c^2+b \left (5 b^2-44 a c\right ) \sqrt{b^2-4 a c}\right )\right ) \operatorname{Subst}\left (\int \frac{x^2}{\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{x^{3/2} \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{x^{3/2} \left (5 b^4-45 a b^2 c+52 a^2 c^2+b c \left (5 b^2-44 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 (\sqrt{c} \left (5 b^4-54 a b^2 c+520 a^2 c^2-b \left (5 b^2-44 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 \sqrt{2} a^2 \left (b^2-4 a c\right )^{5/2}}-\frac{\left (\sqrt{c} \left (5 b^4-54 a b^2 c+520 a^2 c^2-b \left (5 b^2-44 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 \sqrt{2} a^2 \left (b^2-4 a c\right )^{5/2}}-\frac{\left (\sqrt{c} \left (5 b^4-54 a b^2 c+520 a^2 c^2+b \left (5 b^2-44 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 \sqrt{2} a^2 \left (b^2-4 a c\right )^{5/2}}+\frac{\left (\sqrt{c} \left (5 b^4-54 a b^2 c+520 a^2 c^2+b \left (5 b^2-44 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 \sqrt{2} a^2 \left (b^2-4 a c\right )^{5/2}}\\ &=\frac{x^{3/2} \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{x^{3/2} \left (5 b^4-45 a b^2 c+52 a^2 c^2+b c \left (5 b^2-44 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{\sqrt [4]{c} \left (5 b^4-54 a b^2 c+520 a^2 c^2-b \left (5 b^2-44 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\ 2^{3/4} a^2 \left (b^2-4 a c\right )^{5/2} \sqrt [4]{-b-\sqrt{b^2-4 a c}}}+\frac{\sqrt [4]{c} \left (5 b^4-54 a b^2 c+520 a^2 c^2+b \left (5 b^2-44 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\ 2^{3/4} a^2 \left (b^2-4 a c\right )^{5/2} \sqrt [4]{-b+\sqrt{b^2-4 a c}}}+\frac{\sqrt [4]{c} \left (5 b^4-54 a b^2 c+520 a^2 c^2-b \left (5 b^2-44 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\ 2^{3/4} a^2 \left (b^2-4 a c\right )^{5/2} \sqrt [4]{-b-\sqrt{b^2-4 a c}}}-\frac{\sqrt [4]{c} \left (5 b^4-54 a b^2 c+520 a^2 c^2+b \left (5 b^2-44 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\ 2^{3/4} a^2 \left (b^2-4 a c\right )^{5/2} \sqrt [4]{-b+\sqrt{b^2-4 a c}}}\\ \end{align*}

Mathematica [C]  time = 0.508161, size = 254, normalized size = 0.39 \[ \frac{\text{RootSum}\left [\text{$\#$1}^4 b+\text{$\#$1}^8 c+a\& ,\frac{-44 \text{$\#$1}^4 a b c^2 \log \left (\sqrt{x}-\text{$\#$1}\right )+5 \text{$\#$1}^4 b^3 c \log \left (\sqrt{x}-\text{$\#$1}\right )+260 a^2 c^2 \log \left (\sqrt{x}-\text{$\#$1}\right )-49 a b^2 c \log \left (\sqrt{x}-\text{$\#$1}\right )+5 b^4 \log \left (\sqrt{x}-\text{$\#$1}\right )}{2 \text{$\#$1}^5 c+\text{$\#$1} b}\& \right ]+\frac{4 x^{3/2} \left (52 a^2 c^2-45 a b^2 c-44 a b c^2 x^2+5 b^3 c x^2+5 b^4\right )}{a+b x^2+c x^4}-\frac{16 a x^{3/2} \left (4 a c-b^2\right ) \left (-2 a c+b^2+b c x^2\right )}{\left (a+b x^2+c x^4\right )^2}}{64 a^2 \left (b^2-4 a c\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-16*a*(-b^2 + 4*a*c)*x^(3/2)*(b^2 - 2*a*c + b*c*x^2))/(a + b*x^2 + c*x^4)^2 + (4*x^(3/2)*(5*b^4 - 45*a*b^2*c
+ 52*a^2*c^2 + 5*b^3*c*x^2 - 44*a*b*c^2*x^2))/(a + b*x^2 + c*x^4) + RootSum[a + b*#1^4 + c*#1^8 & , (5*b^4*Lo
g[Sqrt[x] - #1] - 49*a*b^2*c*Log[Sqrt[x] - #1] + 260*a^2*c^2*Log[Sqrt[x] - #1] + 5*b^3*c*Log[Sqrt[x] - #1]*#1^
4 - 44*a*b*c^2*Log[Sqrt[x] - #1]*#1^4)/(b*#1 + 2*c*#1^5) & ])/(64*a^2*(b^2 - 4*a*c)^2)

________________________________________________________________________________________

Maple [C]  time = 0.273, size = 321, normalized size = 0.5 \begin{align*} 2\,{\frac{1}{ \left ( c{x}^{4}+b{x}^{2}+a \right ) ^{2}} \left ({\frac{ \left ( 84\,{a}^{2}{c}^{2}-69\,ac{b}^{2}+9\,{b}^{4} \right ){x}^{3/2}}{ \left ( 512\,{a}^{2}{c}^{2}-256\,ac{b}^{2}+32\,{b}^{4} \right ) a}}-1/32\,{\frac{b \left ( 8\,{a}^{2}{c}^{2}+36\,ac{b}^{2}-5\,{b}^{4} \right ){x}^{7/2}}{{a}^{2} \left ( 16\,{a}^{2}{c}^{2}-8\,ac{b}^{2}+{b}^{4} \right ) }}+1/32\,{\frac{c \left ( 52\,{a}^{2}{c}^{2}-89\,ac{b}^{2}+10\,{b}^{4} \right ){x}^{11/2}}{{a}^{2} \left ( 16\,{a}^{2}{c}^{2}-8\,ac{b}^{2}+{b}^{4} \right ) }}-1/32\,{\frac{b{c}^{2} \left ( 44\,ac-5\,{b}^{2} \right ){x}^{15/2}}{{a}^{2} \left ( 16\,{a}^{2}{c}^{2}-8\,ac{b}^{2}+{b}^{4} \right ) }} \right ) }-{\frac{1}{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 ( 44\,ac-5\,{b}^{2} \right ){{\it \_R}}^{6}+ \left ( -260\,{a}^{2}{c}^{2}+49\,ac{b}^{2}-5\,{b}^{4} \right ){{\it \_R}}^{2}}{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(x^(1/2)/(c*x^4+b*x^2+a)^3,x)

[Out]

2*(3/32*(28*a^2*c^2-23*a*b^2*c+3*b^4)/(16*a^2*c^2-8*a*b^2*c+b^4)/a*x^(3/2)-1/32*b*(8*a^2*c^2+36*a*b^2*c-5*b^4)
/a^2/(16*a^2*c^2-8*a*b^2*c+b^4)*x^(7/2)+1/32/a^2*c*(52*a^2*c^2-89*a*b^2*c+10*b^4)/(16*a^2*c^2-8*a*b^2*c+b^4)*x
^(11/2)-1/32*b*c^2*(44*a*c-5*b^2)/a^2/(16*a^2*c^2-8*a*b^2*c+b^4)*x^(15/2))/(c*x^4+b*x^2+a)^2-1/64/a^2/(16*a^2*
c^2-8*a*b^2*c+b^4)*sum((b*c*(44*a*c-5*b^2)*_R^6+(-260*a^2*c^2+49*a*b^2*c-5*b^4)*_R^2)/(2*_R^7*c+_R^3*b)*ln(x^(
1/2)-_R),_R=RootOf(_Z^8*c+_Z^4*b+a))

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (5 \, b^{3} c^{2} - 44 \, a b c^{3}\right )} x^{\frac{15}{2}} +{\left (10 \, b^{4} c - 89 \, a b^{2} c^{2} + 52 \, a^{2} c^{3}\right )} x^{\frac{11}{2}} +{\left (5 \, b^{5} - 36 \, a b^{3} c - 8 \, a^{2} b c^{2}\right )} x^{\frac{7}{2}} + 3 \,{\left (3 \, a b^{4} - 23 \, a^{2} b^{2} c + 28 \, a^{3} c^{2}\right )} x^{\frac{3}{2}}}{16 \,{\left ({\left (a^{2} b^{4} c^{2} - 8 \, a^{3} b^{2} c^{3} + 16 \, a^{4} c^{4}\right )} x^{8} + a^{4} b^{4} - 8 \, a^{5} b^{2} c + 16 \, a^{6} c^{2} + 2 \,{\left (a^{2} b^{5} c - 8 \, a^{3} b^{3} c^{2} + 16 \, a^{4} b c^{3}\right )} x^{6} +{\left (a^{2} b^{6} - 6 \, a^{3} b^{4} c + 32 \, a^{5} c^{3}\right )} x^{4} + 2 \,{\left (a^{3} b^{5} - 8 \, a^{4} b^{3} c + 16 \, a^{5} b c^{2}\right )} x^{2}\right )}} - \int -\frac{{\left (5 \, b^{3} c - 44 \, a b c^{2}\right )} x^{\frac{5}{2}} +{\left (5 \, b^{4} - 49 \, a b^{2} c + 260 \, a^{2} c^{2}\right )} \sqrt{x}}{32 \,{\left (a^{3} b^{4} - 8 \, a^{4} b^{2} c + 16 \, a^{5} c^{2} +{\left (a^{2} b^{4} c - 8 \, a^{3} b^{2} c^{2} + 16 \, a^{4} c^{3}\right )} x^{4} +{\left (a^{2} b^{5} - 8 \, a^{3} b^{3} c + 16 \, a^{4} b c^{2}\right )} x^{2}\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*((5*b^3*c^2 - 44*a*b*c^3)*x^(15/2) + (10*b^4*c - 89*a*b^2*c^2 + 52*a^2*c^3)*x^(11/2) + (5*b^5 - 36*a*b^3*
c - 8*a^2*b*c^2)*x^(7/2) + 3*(3*a*b^4 - 23*a^2*b^2*c + 28*a^3*c^2)*x^(3/2))/((a^2*b^4*c^2 - 8*a^3*b^2*c^3 + 16
*a^4*c^4)*x^8 + a^4*b^4 - 8*a^5*b^2*c + 16*a^6*c^2 + 2*(a^2*b^5*c - 8*a^3*b^3*c^2 + 16*a^4*b*c^3)*x^6 + (a^2*b
^6 - 6*a^3*b^4*c + 32*a^5*c^3)*x^4 + 2*(a^3*b^5 - 8*a^4*b^3*c + 16*a^5*b*c^2)*x^2) - integrate(-1/32*((5*b^3*c
- 44*a*b*c^2)*x^(5/2) + (5*b^4 - 49*a*b^2*c + 260*a^2*c^2)*sqrt(x))/(a^3*b^4 - 8*a^4*b^2*c + 16*a^5*c^2 + (a^
2*b^4*c - 8*a^3*b^2*c^2 + 16*a^4*c^3)*x^4 + (a^2*b^5 - 8*a^3*b^3*c + 16*a^4*b*c^2)*x^2), x)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

[Out]

Timed out