3.1080 \(\int \frac{x^{15/2}}{(a+b x^2+c x^4)^3} \, dx\)

Optimal. Leaf size=621 \[ -\frac{3 \left (\frac{-24 a^2 c^2-30 a b^2 c+b^4}{\sqrt{b^2-4 a c}}-28 a b c+b^3\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{32 \sqrt [4]{2} c^{5/4} \left (b^2-4 a c\right )^2 \left (-\sqrt{b^2-4 a c}-b\right )^{3/4}}-\frac{3 \left (-\frac{-24 a^2 c^2-30 a b^2 c+b^4}{\sqrt{b^2-4 a c}}-28 a b c+b^3\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{32 \sqrt [4]{2} c^{5/4} \left (b^2-4 a c\right )^2 \left (\sqrt{b^2-4 a c}-b\right )^{3/4}}-\frac{3 \left (\frac{-24 a^2 c^2-30 a b^2 c+b^4}{\sqrt{b^2-4 a c}}-28 a b c+b^3\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{32 \sqrt [4]{2} c^{5/4} \left (b^2-4 a c\right )^2 \left (-\sqrt{b^2-4 a c}-b\right )^{3/4}}-\frac{3 \left (-\frac{-24 a^2 c^2-30 a b^2 c+b^4}{\sqrt{b^2-4 a c}}-28 a b c+b^3\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{32 \sqrt [4]{2} c^{5/4} \left (b^2-4 a c\right )^2 \left (\sqrt{b^2-4 a c}-b\right )^{3/4}}+\frac{x^{9/2} \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}+\frac{3 x^{5/2} \left (x^2 \left (12 a c+b^2\right )+8 a b\right )}{16 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}-\frac{3 \sqrt{x} \left (12 a c+b^2\right )}{16 c \left (b^2-4 a c\right )^2} \]

[Out]

(-3*(b^2 + 12*a*c)*Sqrt[x])/(16*c*(b^2 - 4*a*c)^2) + (x^(9/2)*(2*a + b*x^2))/(4*(b^2 - 4*a*c)*(a + b*x^2 + c*x
^4)^2) + (3*x^(5/2)*(8*a*b + (b^2 + 12*a*c)*x^2))/(16*(b^2 - 4*a*c)^2*(a + b*x^2 + c*x^4)) - (3*(b^3 - 28*a*b*
c + (b^4 - 30*a*b^2*c - 24*a^2*c^2)/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)*c^(5/4)*(b^2 - 4*a*c)^2*(-b - Sqrt[b^2 - 4*a*c])^(3/4)) - (3*(b^3 - 28*a*b*c - (b^4 - 3
0*a*b^2*c - 24*a^2*c^2)/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)*c^(5/4)*(b^2 - 4*a*c)^2*(-b + Sqrt[b^2 - 4*a*c])^(3/4)) - (3*(b^3 - 28*a*b*c + (b^4 - 30*a*b^2*c -
24*a^2*c^2)/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)*
c^(5/4)*(b^2 - 4*a*c)^2*(-b - Sqrt[b^2 - 4*a*c])^(3/4)) - (3*(b^3 - 28*a*b*c - (b^4 - 30*a*b^2*c - 24*a^2*c^2)
/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)*c^(5/4)*(b^
2 - 4*a*c)^2*(-b + Sqrt[b^2 - 4*a*c])^(3/4))

________________________________________________________________________________________

Rubi [A]  time = 1.77181, antiderivative size = 621, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {1115, 1365, 1498, 1502, 1422, 212, 208, 205} \[ -\frac{3 \left (\frac{-24 a^2 c^2-30 a b^2 c+b^4}{\sqrt{b^2-4 a c}}-28 a b c+b^3\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{32 \sqrt [4]{2} c^{5/4} \left (b^2-4 a c\right )^2 \left (-\sqrt{b^2-4 a c}-b\right )^{3/4}}-\frac{3 \left (-\frac{-24 a^2 c^2-30 a b^2 c+b^4}{\sqrt{b^2-4 a c}}-28 a b c+b^3\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{32 \sqrt [4]{2} c^{5/4} \left (b^2-4 a c\right )^2 \left (\sqrt{b^2-4 a c}-b\right )^{3/4}}-\frac{3 \left (\frac{-24 a^2 c^2-30 a b^2 c+b^4}{\sqrt{b^2-4 a c}}-28 a b c+b^3\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{32 \sqrt [4]{2} c^{5/4} \left (b^2-4 a c\right )^2 \left (-\sqrt{b^2-4 a c}-b\right )^{3/4}}-\frac{3 \left (-\frac{-24 a^2 c^2-30 a b^2 c+b^4}{\sqrt{b^2-4 a c}}-28 a b c+b^3\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{32 \sqrt [4]{2} c^{5/4} \left (b^2-4 a c\right )^2 \left (\sqrt{b^2-4 a c}-b\right )^{3/4}}+\frac{x^{9/2} \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}+\frac{3 x^{5/2} \left (x^2 \left (12 a c+b^2\right )+8 a b\right )}{16 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}-\frac{3 \sqrt{x} \left (12 a c+b^2\right )}{16 c \left (b^2-4 a c\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[x^(15/2)/(a + b*x^2 + c*x^4)^3,x]

[Out]

(-3*(b^2 + 12*a*c)*Sqrt[x])/(16*c*(b^2 - 4*a*c)^2) + (x^(9/2)*(2*a + b*x^2))/(4*(b^2 - 4*a*c)*(a + b*x^2 + c*x
^4)^2) + (3*x^(5/2)*(8*a*b + (b^2 + 12*a*c)*x^2))/(16*(b^2 - 4*a*c)^2*(a + b*x^2 + c*x^4)) - (3*(b^3 - 28*a*b*
c + (b^4 - 30*a*b^2*c - 24*a^2*c^2)/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)*c^(5/4)*(b^2 - 4*a*c)^2*(-b - Sqrt[b^2 - 4*a*c])^(3/4)) - (3*(b^3 - 28*a*b*c - (b^4 - 3
0*a*b^2*c - 24*a^2*c^2)/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)*c^(5/4)*(b^2 - 4*a*c)^2*(-b + Sqrt[b^2 - 4*a*c])^(3/4)) - (3*(b^3 - 28*a*b*c + (b^4 - 30*a*b^2*c -
24*a^2*c^2)/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)*
c^(5/4)*(b^2 - 4*a*c)^2*(-b - Sqrt[b^2 - 4*a*c])^(3/4)) - (3*(b^3 - 28*a*b*c - (b^4 - 30*a*b^2*c - 24*a^2*c^2)
/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)*c^(5/4)*(b^
2 - 4*a*c)^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 1365

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

Rule 1498

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

Rule 1502

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

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{x^{15/2}}{\left (a+b x^2+c x^4\right )^3} \, dx &=2 \operatorname{Subst}\left (\int \frac{x^{16}}{\left (a+b x^4+c x^8\right )^3} \, dx,x,\sqrt{x}\right )\\ &=\frac{x^{9/2} \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac{\operatorname{Subst}\left (\int \frac{x^8 \left (18 a-3 b x^4\right )}{\left (a+b x^4+c x^8\right )^2} \, dx,x,\sqrt{x}\right )}{4 \left (b^2-4 a c\right )}\\ &=\frac{x^{9/2} \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}+\frac{3 x^{5/2} \left (8 a b+\left (b^2+12 a c\right ) x^2\right )}{16 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}+\frac{\operatorname{Subst}\left (\int \frac{x^4 \left (-120 a b-3 \left (b^2+12 a c\right ) x^4\right )}{a+b x^4+c x^8} \, dx,x,\sqrt{x}\right )}{16 \left (b^2-4 a c\right )^2}\\ &=-\frac{3 \left (b^2+12 a c\right ) \sqrt{x}}{16 c \left (b^2-4 a c\right )^2}+\frac{x^{9/2} \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}+\frac{3 x^{5/2} \left (8 a b+\left (b^2+12 a c\right ) x^2\right )}{16 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}-\frac{\operatorname{Subst}\left (\int \frac{-3 a \left (b^2+12 a c\right )-3 b \left (b^2-28 a c\right ) x^4}{a+b x^4+c x^8} \, dx,x,\sqrt{x}\right )}{16 c \left (b^2-4 a c\right )^2}\\ &=-\frac{3 \left (b^2+12 a c\right ) \sqrt{x}}{16 c \left (b^2-4 a c\right )^2}+\frac{x^{9/2} \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}+\frac{3 x^{5/2} \left (8 a b+\left (b^2+12 a c\right ) x^2\right )}{16 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}+\frac{\left (3 \left (b^3-28 a b c-\frac{b^4-30 a b^2 c-24 a^2 c^2}{\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 c \left (b^2-4 a c\right )^2}+\frac{\left (3 \left (b^3-28 a b c+\frac{b^4-30 a b^2 c-24 a^2 c^2}{\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 c \left (b^2-4 a c\right )^2}\\ &=-\frac{3 \left (b^2+12 a c\right ) \sqrt{x}}{16 c \left (b^2-4 a c\right )^2}+\frac{x^{9/2} \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}+\frac{3 x^{5/2} \left (8 a b+\left (b^2+12 a c\right ) x^2\right )}{16 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}-\frac{\left (3 \left (b^3-28 a b c-\frac{b^4-30 a b^2 c-24 a^2 c^2}{\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 c \left (b^2-4 a c\right )^2 \sqrt{-b+\sqrt{b^2-4 a c}}}-\frac{\left (3 \left (b^3-28 a b c-\frac{b^4-30 a b^2 c-24 a^2 c^2}{\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 c \left (b^2-4 a c\right )^2 \sqrt{-b+\sqrt{b^2-4 a c}}}-\frac{\left (3 \left (b^3-28 a b c+\frac{b^4-30 a b^2 c-24 a^2 c^2}{\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 c \left (b^2-4 a c\right )^2 \sqrt{-b-\sqrt{b^2-4 a c}}}-\frac{\left (3 \left (b^3-28 a b c+\frac{b^4-30 a b^2 c-24 a^2 c^2}{\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 c \left (b^2-4 a c\right )^2 \sqrt{-b-\sqrt{b^2-4 a c}}}\\ &=-\frac{3 \left (b^2+12 a c\right ) \sqrt{x}}{16 c \left (b^2-4 a c\right )^2}+\frac{x^{9/2} \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}+\frac{3 x^{5/2} \left (8 a b+\left (b^2+12 a c\right ) x^2\right )}{16 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}-\frac{3 \left (b^3-28 a b c+\frac{b^4-30 a b^2 c-24 a^2 c^2}{\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} c^{5/4} \left (b^2-4 a c\right )^2 \left (-b-\sqrt{b^2-4 a c}\right )^{3/4}}-\frac{3 \left (b^3-28 a b c-\frac{b^4-30 a b^2 c-24 a^2 c^2}{\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} c^{5/4} \left (b^2-4 a c\right )^2 \left (-b+\sqrt{b^2-4 a c}\right )^{3/4}}-\frac{3 \left (b^3-28 a b c+\frac{b^4-30 a b^2 c-24 a^2 c^2}{\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} c^{5/4} \left (b^2-4 a c\right )^2 \left (-b-\sqrt{b^2-4 a c}\right )^{3/4}}-\frac{3 \left (b^3-28 a b c-\frac{b^4-30 a b^2 c-24 a^2 c^2}{\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} c^{5/4} \left (b^2-4 a c\right )^2 \left (-b+\sqrt{b^2-4 a c}\right )^{3/4}}\\ \end{align*}

Mathematica [C]  time = 0.448081, size = 254, normalized size = 0.41 \[ \frac{3 c \left (a+b x^2+c x^4\right )^2 \text{RootSum}\left [\text{$\#$1}^4 b+\text{$\#$1}^8 c+a\& ,\frac{-28 \text{$\#$1}^4 a b c \log \left (\sqrt{x}-\text{$\#$1}\right )+\text{$\#$1}^4 b^3 \log \left (\sqrt{x}-\text{$\#$1}\right )+12 a^2 c \log \left (\sqrt{x}-\text{$\#$1}\right )+a b^2 \log \left (\sqrt{x}-\text{$\#$1}\right )}{\text{$\#$1}^3 b+2 \text{$\#$1}^7 c}\& \right ]+4 \sqrt{x} \left (-68 a^2 c^2+21 a b^2 c-28 a b c^2 x^2+b^3 c x^2-4 b^4\right ) \left (a+b x^2+c x^4\right )+16 \sqrt{x} \left (b^2-4 a c\right ) \left (-2 a^2 c+a b \left (b-3 c x^2\right )+b^3 x^2\right )}{64 c^2 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(15/2)/(a + b*x^2 + c*x^4)^3,x]

[Out]

(4*Sqrt[x]*(-4*b^4 + 21*a*b^2*c - 68*a^2*c^2 + b^3*c*x^2 - 28*a*b*c^2*x^2)*(a + b*x^2 + c*x^4) + 16*(b^2 - 4*a
*c)*Sqrt[x]*(-2*a^2*c + b^3*x^2 + a*b*(b - 3*c*x^2)) + 3*c*(a + b*x^2 + c*x^4)^2*RootSum[a + b*#1^4 + c*#1^8 &
, (a*b^2*Log[Sqrt[x] - #1] + 12*a^2*c*Log[Sqrt[x] - #1] + b^3*Log[Sqrt[x] - #1]*#1^4 - 28*a*b*c*Log[Sqrt[x] -
#1]*#1^4)/(b*#1^3 + 2*c*#1^7) & ])/(64*c^2*(b^2 - 4*a*c)^2*(a + b*x^2 + c*x^4)^2)

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Maple [C]  time = 0.297, size = 275, normalized size = 0.4 \begin{align*} 2\,{\frac{1}{ \left ( c{x}^{4}+b{x}^{2}+a \right ) ^{2}} \left ( -{\frac{3\,{a}^{2} \left ( 12\,ac+{b}^{2} \right ) \sqrt{x}}{32\,c \left ( 16\,{a}^{2}{c}^{2}-8\,ac{b}^{2}+{b}^{4} \right ) }}-3/16\,{\frac{ab \left ( 8\,ac+{b}^{2} \right ){x}^{5/2}}{c \left ( 16\,{a}^{2}{c}^{2}-8\,ac{b}^{2}+{b}^{4} \right ) }}-1/32\,{\frac{ \left ( 68\,{a}^{2}{c}^{2}+7\,ac{b}^{2}+3\,{b}^{4} \right ){x}^{9/2}}{c \left ( 16\,{a}^{2}{c}^{2}-8\,ac{b}^{2}+{b}^{4} \right ) }}-1/32\,{\frac{b \left ( 28\,ac-{b}^{2} \right ){x}^{13/2}}{16\,{a}^{2}{c}^{2}-8\,ac{b}^{2}+{b}^{4}}} \right ) }+{\frac{3}{64\,c \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{b \left ( -28\,ac+{b}^{2} \right ){{\it \_R}}^{4}+12\,{a}^{2}c+{b}^{2}a}{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^(15/2)/(c*x^4+b*x^2+a)^3,x)

[Out]

2*(-3/32*a^2*(12*a*c+b^2)/c/(16*a^2*c^2-8*a*b^2*c+b^4)*x^(1/2)-3/16*a/c*b*(8*a*c+b^2)/(16*a^2*c^2-8*a*b^2*c+b^
4)*x^(5/2)-1/32*(68*a^2*c^2+7*a*b^2*c+3*b^4)/c/(16*a^2*c^2-8*a*b^2*c+b^4)*x^(9/2)-1/32*b*(28*a*c-b^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/c/(16*a^2*c^2-8*a*b^2*c+b^4)*sum((b*(-28*a*c+b^2)*_R^4+12*
a^2*c+b^2*a)/(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 (b^{2} c + 12 \, a c^{2}\right )} x^{\frac{17}{2}} +{\left (7 \, b^{3} + 44 \, a b c\right )} x^{\frac{13}{2}} + 24 \, a^{2} b x^{\frac{5}{2}} +{\left (35 \, a b^{2} + 4 \, a^{2} c\right )} x^{\frac{9}{2}}}{16 \,{\left ({\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} x^{8} + 2 \,{\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} x^{6} + a^{2} b^{4} - 8 \, a^{3} b^{2} c + 16 \, a^{4} c^{2} +{\left (b^{6} - 6 \, a b^{4} c + 32 \, a^{3} c^{3}\right )} x^{4} + 2 \,{\left (a b^{5} - 8 \, a^{2} b^{3} c + 16 \, a^{3} b c^{2}\right )} x^{2}\right )}} - \int \frac{3 \,{\left ({\left (b^{2} + 12 \, a c\right )} x^{\frac{7}{2}} + 40 \, a b x^{\frac{3}{2}}\right )}}{32 \,{\left (a b^{4} - 8 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} +{\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{4} +{\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x^{2}\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(15/2)/(c*x^4+b*x^2+a)^3,x, algorithm="maxima")

[Out]

1/16*(3*(b^2*c + 12*a*c^2)*x^(17/2) + (7*b^3 + 44*a*b*c)*x^(13/2) + 24*a^2*b*x^(5/2) + (35*a*b^2 + 4*a^2*c)*x^
(9/2))/((b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*x^8 + 2*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x^6 + a^2*b^4 - 8*a^
3*b^2*c + 16*a^4*c^2 + (b^6 - 6*a*b^4*c + 32*a^3*c^3)*x^4 + 2*(a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*x^2) - inte
grate(3/32*((b^2 + 12*a*c)*x^(7/2) + 40*a*b*x^(3/2))/(a*b^4 - 8*a^2*b^2*c + 16*a^3*c^2 + (b^4*c - 8*a*b^2*c^2
+ 16*a^2*c^3)*x^4 + (b^5 - 8*a*b^3*c + 16*a^2*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(x^(15/2)/(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(x**(15/2)/(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(x^(15/2)/(c*x^4+b*x^2+a)^3,x, algorithm="giac")

[Out]

Timed out