3.1085 \(\int \frac{x^{5/2}}{(a+b x^2+c x^4)^3} \, dx\)

Optimal. Leaf size=594 \[ \frac{3 x^{3/2} \left (c x^2 \left (12 a c+b^2\right )+b \left (4 a c+b^2\right )\right )}{16 a \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}-\frac{x^{3/2} \left (b+2 c x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}+\frac{3 \sqrt [4]{c} \left (-\frac{b^3}{\sqrt{b^2-4 a c}}+\frac{68 a b c}{\sqrt{b^2-4 a c}}+12 a c+b^2\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{32\ 2^{3/4} a \left (b^2-4 a c\right )^2 \sqrt [4]{-\sqrt{b^2-4 a c}-b}}+\frac{3 \sqrt [4]{c} \left (\sqrt{b^2-4 a c} \left (12 a c+b^2\right )-68 a b c+b^3\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{32\ 2^{3/4} a \left (b^2-4 a c\right )^{5/2} \sqrt [4]{\sqrt{b^2-4 a c}-b}}-\frac{3 \sqrt [4]{c} \left (-\frac{b^3}{\sqrt{b^2-4 a c}}+\frac{68 a b c}{\sqrt{b^2-4 a c}}+12 a c+b^2\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{32\ 2^{3/4} a \left (b^2-4 a c\right )^2 \sqrt [4]{-\sqrt{b^2-4 a c}-b}}-\frac{3 \sqrt [4]{c} \left (\sqrt{b^2-4 a c} \left (12 a c+b^2\right )-68 a b c+b^3\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{32\ 2^{3/4} a \left (b^2-4 a c\right )^{5/2} \sqrt [4]{\sqrt{b^2-4 a c}-b}} \]

[Out]

-(x^(3/2)*(b + 2*c*x^2))/(4*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)^2) + (3*x^(3/2)*(b*(b^2 + 4*a*c) + c*(b^2 + 12*a
*c)*x^2))/(16*a*(b^2 - 4*a*c)^2*(a + b*x^2 + c*x^4)) + (3*c^(1/4)*(b^2 + 12*a*c - b^3/Sqrt[b^2 - 4*a*c] + (68*
a*b*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*(b^2
- 4*a*c)^2*(-b - Sqrt[b^2 - 4*a*c])^(1/4)) + (3*c^(1/4)*(b^3 - 68*a*b*c + Sqrt[b^2 - 4*a*c]*(b^2 + 12*a*c))*A
rcTan[(2^(1/4)*c^(1/4)*Sqrt[x])/(-b + Sqrt[b^2 - 4*a*c])^(1/4)])/(32*2^(3/4)*a*(b^2 - 4*a*c)^(5/2)*(-b + Sqrt[
b^2 - 4*a*c])^(1/4)) - (3*c^(1/4)*(b^2 + 12*a*c - b^3/Sqrt[b^2 - 4*a*c] + (68*a*b*c)/Sqrt[b^2 - 4*a*c])*ArcTan
h[(2^(1/4)*c^(1/4)*Sqrt[x])/(-b - Sqrt[b^2 - 4*a*c])^(1/4)])/(32*2^(3/4)*a*(b^2 - 4*a*c)^2*(-b - Sqrt[b^2 - 4*
a*c])^(1/4)) - (3*c^(1/4)*(b^3 - 68*a*b*c + Sqrt[b^2 - 4*a*c]*(b^2 + 12*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*(b^2 - 4*a*c)^(5/2)*(-b + Sqrt[b^2 - 4*a*c])^(1/4))

________________________________________________________________________________________

Rubi [A]  time = 2.3102, antiderivative size = 594, 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, 1364, 1500, 1510, 298, 205, 208} \[ \frac{3 x^{3/2} \left (c x^2 \left (12 a c+b^2\right )+b \left (4 a c+b^2\right )\right )}{16 a \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}-\frac{x^{3/2} \left (b+2 c x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}+\frac{3 \sqrt [4]{c} \left (-\frac{b^3}{\sqrt{b^2-4 a c}}+\frac{68 a b c}{\sqrt{b^2-4 a c}}+12 a c+b^2\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{32\ 2^{3/4} a \left (b^2-4 a c\right )^2 \sqrt [4]{-\sqrt{b^2-4 a c}-b}}+\frac{3 \sqrt [4]{c} \left (\sqrt{b^2-4 a c} \left (12 a c+b^2\right )-68 a b c+b^3\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{32\ 2^{3/4} a \left (b^2-4 a c\right )^{5/2} \sqrt [4]{\sqrt{b^2-4 a c}-b}}-\frac{3 \sqrt [4]{c} \left (-\frac{b^3}{\sqrt{b^2-4 a c}}+\frac{68 a b c}{\sqrt{b^2-4 a c}}+12 a c+b^2\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{32\ 2^{3/4} a \left (b^2-4 a c\right )^2 \sqrt [4]{-\sqrt{b^2-4 a c}-b}}-\frac{3 \sqrt [4]{c} \left (\sqrt{b^2-4 a c} \left (12 a c+b^2\right )-68 a b c+b^3\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{32\ 2^{3/4} a \left (b^2-4 a c\right )^{5/2} \sqrt [4]{\sqrt{b^2-4 a c}-b}} \]

Antiderivative was successfully verified.

[In]

Int[x^(5/2)/(a + b*x^2 + c*x^4)^3,x]

[Out]

-(x^(3/2)*(b + 2*c*x^2))/(4*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)^2) + (3*x^(3/2)*(b*(b^2 + 4*a*c) + c*(b^2 + 12*a
*c)*x^2))/(16*a*(b^2 - 4*a*c)^2*(a + b*x^2 + c*x^4)) + (3*c^(1/4)*(b^2 + 12*a*c - b^3/Sqrt[b^2 - 4*a*c] + (68*
a*b*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*(b^2
- 4*a*c)^2*(-b - Sqrt[b^2 - 4*a*c])^(1/4)) + (3*c^(1/4)*(b^3 - 68*a*b*c + Sqrt[b^2 - 4*a*c]*(b^2 + 12*a*c))*A
rcTan[(2^(1/4)*c^(1/4)*Sqrt[x])/(-b + Sqrt[b^2 - 4*a*c])^(1/4)])/(32*2^(3/4)*a*(b^2 - 4*a*c)^(5/2)*(-b + Sqrt[
b^2 - 4*a*c])^(1/4)) - (3*c^(1/4)*(b^2 + 12*a*c - b^3/Sqrt[b^2 - 4*a*c] + (68*a*b*c)/Sqrt[b^2 - 4*a*c])*ArcTan
h[(2^(1/4)*c^(1/4)*Sqrt[x])/(-b - Sqrt[b^2 - 4*a*c])^(1/4)])/(32*2^(3/4)*a*(b^2 - 4*a*c)^2*(-b - Sqrt[b^2 - 4*
a*c])^(1/4)) - (3*c^(1/4)*(b^3 - 68*a*b*c + Sqrt[b^2 - 4*a*c]*(b^2 + 12*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*(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 1364

Int[((d_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(d^(n - 1)*(d*x)^(
m - n + 1)*(b + 2*c*x^n)*(a + b*x^n + c*x^(2*n))^(p + 1))/(n*(p + 1)*(b^2 - 4*a*c)), x] - Dist[d^n/(n*(p + 1)*
(b^2 - 4*a*c)), Int[(d*x)^(m - n)*(b*(m - n + 1) + 2*c*(m + 2*n*(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] && ILtQ[p, -1] && G
tQ[m, n - 1] && LeQ[m, 2*n - 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{x^{5/2}}{\left (a+b x^2+c x^4\right )^3} \, dx &=2 \operatorname{Subst}\left (\int \frac{x^6}{\left (a+b x^4+c x^8\right )^3} \, dx,x,\sqrt{x}\right )\\ &=-\frac{x^{3/2} \left (b+2 c 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^2 \left (3 b-18 c 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^{3/2} \left (b+2 c x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}+\frac{3 x^{3/2} \left (b \left (b^2+4 a c\right )+c \left (b^2+12 a c\right ) x^2\right )}{16 a \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 (-3 b \left (b^2-28 a c\right )-3 c \left (b^2+12 a c\right ) x^4\right )}{a+b x^4+c x^8} \, dx,x,\sqrt{x}\right )}{16 a \left (b^2-4 a c\right )^2}\\ &=-\frac{x^{3/2} \left (b+2 c x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}+\frac{3 x^{3/2} \left (b \left (b^2+4 a c\right )+c \left (b^2+12 a c\right ) x^2\right )}{16 a \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}+\frac{\left (3 c \left (b^2+12 a c+\frac{b^3}{\sqrt{b^2-4 a c}}-\frac{68 a b c}{\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 \left (b^2-4 a c\right )^2}+\frac{\left (3 c \left (b^2+12 a c-\frac{b^3}{\sqrt{b^2-4 a c}}+\frac{68 a b c}{\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 \left (b^2-4 a c\right )^2}\\ &=-\frac{x^{3/2} \left (b+2 c x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}+\frac{3 x^{3/2} \left (b \left (b^2+4 a c\right )+c \left (b^2+12 a c\right ) x^2\right )}{16 a \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}-\frac{\left (3 \sqrt{c} \left (b^2+12 a c+\frac{b^3}{\sqrt{b^2-4 a c}}-\frac{68 a b c}{\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 \left (b^2-4 a c\right )^2}+\frac{\left (3 \sqrt{c} \left (b^2+12 a c+\frac{b^3}{\sqrt{b^2-4 a c}}-\frac{68 a b c}{\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 \left (b^2-4 a c\right )^2}-\frac{\left (3 \sqrt{c} \left (b^2+12 a c-\frac{b^3}{\sqrt{b^2-4 a c}}+\frac{68 a b c}{\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 \left (b^2-4 a c\right )^2}+\frac{\left (3 \sqrt{c} \left (b^2+12 a c-\frac{b^3}{\sqrt{b^2-4 a c}}+\frac{68 a b c}{\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 \left (b^2-4 a c\right )^2}\\ &=-\frac{x^{3/2} \left (b+2 c x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}+\frac{3 x^{3/2} \left (b \left (b^2+4 a c\right )+c \left (b^2+12 a c\right ) x^2\right )}{16 a \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}+\frac{3 \sqrt [4]{c} \left (b^2+12 a c-\frac{b^3}{\sqrt{b^2-4 a c}}+\frac{68 a b c}{\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 \left (b^2-4 a c\right )^2 \sqrt [4]{-b-\sqrt{b^2-4 a c}}}+\frac{3 \sqrt [4]{c} \left (b^2+12 a c+\frac{b^3}{\sqrt{b^2-4 a c}}-\frac{68 a b c}{\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 \left (b^2-4 a c\right )^2 \sqrt [4]{-b+\sqrt{b^2-4 a c}}}-\frac{3 \sqrt [4]{c} \left (b^2+12 a c-\frac{b^3}{\sqrt{b^2-4 a c}}+\frac{68 a b c}{\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 \left (b^2-4 a c\right )^2 \sqrt [4]{-b-\sqrt{b^2-4 a c}}}-\frac{3 \sqrt [4]{c} \left (b^2+12 a c+\frac{b^3}{\sqrt{b^2-4 a c}}-\frac{68 a b c}{\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 \left (b^2-4 a c\right )^2 \sqrt [4]{-b+\sqrt{b^2-4 a c}}}\\ \end{align*}

Mathematica [C]  time = 0.411774, size = 222, normalized size = 0.37 \[ \frac{3 \left (a+b x^2+c x^4\right )^2 \text{RootSum}\left [\text{$\#$1}^4 b+\text{$\#$1}^8 c+a\& ,\frac{12 \text{$\#$1}^4 a c^2 \log \left (\sqrt{x}-\text{$\#$1}\right )+\text{$\#$1}^4 b^2 c \log \left (\sqrt{x}-\text{$\#$1}\right )-28 a b c \log \left (\sqrt{x}-\text{$\#$1}\right )+b^3 \log \left (\sqrt{x}-\text{$\#$1}\right )}{2 \text{$\#$1}^5 c+\text{$\#$1} b}\& \right ]+12 x^{3/2} \left (4 a b c+12 a c^2 x^2+b^2 c x^2+b^3\right ) \left (a+b x^2+c x^4\right )-16 a x^{3/2} \left (b^2-4 a c\right ) \left (b+2 c x^2\right )}{64 a \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)/(a + b*x^2 + c*x^4)^3,x]

[Out]

(-16*a*(b^2 - 4*a*c)*x^(3/2)*(b + 2*c*x^2) + 12*x^(3/2)*(b^3 + 4*a*b*c + b^2*c*x^2 + 12*a*c^2*x^2)*(a + b*x^2
+ c*x^4) + 3*(a + b*x^2 + c*x^4)^2*RootSum[a + b*#1^4 + c*#1^8 & , (b^3*Log[Sqrt[x] - #1] - 28*a*b*c*Log[Sqrt[
x] - #1] + b^2*c*Log[Sqrt[x] - #1]*#1^4 + 12*a*c^2*Log[Sqrt[x] - #1]*#1^4)/(b*#1 + 2*c*#1^5) & ])/(64*a*(b^2 -
4*a*c)^2*(a + b*x^2 + c*x^4)^2)

________________________________________________________________________________________

Maple [C]  time = 0.272, size = 277, normalized size = 0.5 \begin{align*} 2\,{\frac{1}{ \left ( c{x}^{4}+b{x}^{2}+a \right ) ^{2}} \left ( 1/32\,{\frac{b \left ( 28\,ac-{b}^{2} \right ){x}^{3/2}}{16\,{a}^{2}{c}^{2}-8\,ac{b}^{2}+{b}^{4}}}+1/32\,{\frac{ \left ( 68\,{a}^{2}{c}^{2}+7\,ac{b}^{2}+3\,{b}^{4} \right ){x}^{7/2}}{a \left ( 16\,{a}^{2}{c}^{2}-8\,ac{b}^{2}+{b}^{4} \right ) }}+3/16\,{\frac{bc \left ( 8\,ac+{b}^{2} \right ){x}^{11/2}}{a \left ( 16\,{a}^{2}{c}^{2}-8\,ac{b}^{2}+{b}^{4} \right ) }}+{\frac{3\,{c}^{2} \left ( 12\,ac+{b}^{2} \right ){x}^{15/2}}{32\,a \left ( 16\,{a}^{2}{c}^{2}-8\,ac{b}^{2}+{b}^{4} \right ) }} \right ) }+{\frac{3}{64\,a \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{c \left ( 12\,ac+{b}^{2} \right ){{\it \_R}}^{6}+b \left ( -28\,ac+{b}^{2} \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^(5/2)/(c*x^4+b*x^2+a)^3,x)

[Out]

2*(1/32*b*(28*a*c-b^2)/(16*a^2*c^2-8*a*b^2*c+b^4)*x^(3/2)+1/32*(68*a^2*c^2+7*a*b^2*c+3*b^4)/a/(16*a^2*c^2-8*a*
b^2*c+b^4)*x^(7/2)+3/16/a*c*b*(8*a*c+b^2)/(16*a^2*c^2-8*a*b^2*c+b^4)*x^(11/2)+3/32*c^2*(12*a*c+b^2)/a/(16*a^2*
c^2-8*a*b^2*c+b^4)*x^(15/2))/(c*x^4+b*x^2+a)^2+3/64/a/(16*a^2*c^2-8*a*b^2*c+b^4)*sum((c*(12*a*c+b^2)*_R^6+b*(-
28*a*c+b^2)*_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{3 \,{\left (b^{2} c^{2} + 12 \, a c^{3}\right )} x^{\frac{15}{2}} + 6 \,{\left (b^{3} c + 8 \, a b c^{2}\right )} x^{\frac{11}{2}} +{\left (3 \, b^{4} + 7 \, a b^{2} c + 68 \, a^{2} c^{2}\right )} x^{\frac{7}{2}} -{\left (a b^{3} - 28 \, a^{2} b c\right )} x^{\frac{3}{2}}}{16 \,{\left ({\left (a b^{4} c^{2} - 8 \, a^{2} b^{2} c^{3} + 16 \, a^{3} c^{4}\right )} x^{8} + a^{3} b^{4} - 8 \, a^{4} b^{2} c + 16 \, a^{5} c^{2} + 2 \,{\left (a b^{5} c - 8 \, a^{2} b^{3} c^{2} + 16 \, a^{3} b c^{3}\right )} x^{6} +{\left (a b^{6} - 6 \, a^{2} b^{4} c + 32 \, a^{4} c^{3}\right )} x^{4} + 2 \,{\left (a^{2} b^{5} - 8 \, a^{3} b^{3} c + 16 \, a^{4} b c^{2}\right )} x^{2}\right )}} + \int \frac{3 \,{\left ({\left (b^{2} c + 12 \, a c^{2}\right )} x^{\frac{5}{2}} +{\left (b^{3} - 28 \, a b c\right )} \sqrt{x}\right )}}{32 \,{\left (a^{2} b^{4} - 8 \, a^{3} b^{2} c + 16 \, a^{4} c^{2} +{\left (a b^{4} c - 8 \, a^{2} b^{2} c^{2} + 16 \, a^{3} c^{3}\right )} x^{4} +{\left (a b^{5} - 8 \, a^{2} b^{3} c + 16 \, a^{3} b c^{2}\right )} x^{2}\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)/(c*x^4+b*x^2+a)^3,x, algorithm="maxima")

[Out]

1/16*(3*(b^2*c^2 + 12*a*c^3)*x^(15/2) + 6*(b^3*c + 8*a*b*c^2)*x^(11/2) + (3*b^4 + 7*a*b^2*c + 68*a^2*c^2)*x^(7
/2) - (a*b^3 - 28*a^2*b*c)*x^(3/2))/((a*b^4*c^2 - 8*a^2*b^2*c^3 + 16*a^3*c^4)*x^8 + a^3*b^4 - 8*a^4*b^2*c + 16
*a^5*c^2 + 2*(a*b^5*c - 8*a^2*b^3*c^2 + 16*a^3*b*c^3)*x^6 + (a*b^6 - 6*a^2*b^4*c + 32*a^4*c^3)*x^4 + 2*(a^2*b^
5 - 8*a^3*b^3*c + 16*a^4*b*c^2)*x^2) + integrate(3/32*((b^2*c + 12*a*c^2)*x^(5/2) + (b^3 - 28*a*b*c)*sqrt(x))/
(a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2 + (a*b^4*c - 8*a^2*b^2*c^2 + 16*a^3*c^3)*x^4 + (a*b^5 - 8*a^2*b^3*c + 16*a
^3*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^(5/2)/(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**(5/2)/(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^(5/2)/(c*x^4+b*x^2+a)^3,x, algorithm="giac")

[Out]

Timed out