3.1079 \(\int \frac{1}{x^{3/2} (a+b x^2+c x^4)^2} \, dx\)

Optimal. Leaf size=573 \[ -\frac{5 b^2-18 a c}{2 a^2 \sqrt{x} \left (b^2-4 a c\right )}+\frac{\sqrt [4]{c} \left (-\left (5 b^2-18 a c\right ) \sqrt{b^2-4 a c}-28 a b c+5 b^3\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{4\ 2^{3/4} a^2 \left (b^2-4 a c\right )^{3/2} \sqrt [4]{-\sqrt{b^2-4 a c}-b}}-\frac{\sqrt [4]{c} \left (\left (5 b^2-18 a c\right ) \sqrt{b^2-4 a c}-28 a b c+5 b^3\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{4\ 2^{3/4} a^2 \left (b^2-4 a c\right )^{3/2} \sqrt [4]{\sqrt{b^2-4 a c}-b}}-\frac{\sqrt [4]{c} \left (-\left (5 b^2-18 a c\right ) \sqrt{b^2-4 a c}-28 a b c+5 b^3\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{4\ 2^{3/4} a^2 \left (b^2-4 a c\right )^{3/2} \sqrt [4]{-\sqrt{b^2-4 a c}-b}}+\frac{\sqrt [4]{c} \left (\left (5 b^2-18 a c\right ) \sqrt{b^2-4 a c}-28 a b c+5 b^3\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{4\ 2^{3/4} a^2 \left (b^2-4 a c\right )^{3/2} \sqrt [4]{\sqrt{b^2-4 a c}-b}}+\frac{-2 a c+b^2+b c x^2}{2 a \sqrt{x} \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )} \]

[Out]

-(5*b^2 - 18*a*c)/(2*a^2*(b^2 - 4*a*c)*Sqrt[x]) + (b^2 - 2*a*c + b*c*x^2)/(2*a*(b^2 - 4*a*c)*Sqrt[x]*(a + b*x^
2 + c*x^4)) + (c^(1/4)*(5*b^3 - 28*a*b*c - (5*b^2 - 18*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)])/(4*2^(3/4)*a^2*(b^2 - 4*a*c)^(3/2)*(-b - Sqrt[b^2 - 4*a*c])^(1/4)) - (c^(1/
4)*(5*b^3 - 28*a*b*c + (5*b^2 - 18*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)])/(4*2^(3/4)*a^2*(b^2 - 4*a*c)^(3/2)*(-b + Sqrt[b^2 - 4*a*c])^(1/4)) - (c^(1/4)*(5*b^3 - 28*a*b*c
 - (5*b^2 - 18*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)])/(4*2
^(3/4)*a^2*(b^2 - 4*a*c)^(3/2)*(-b - Sqrt[b^2 - 4*a*c])^(1/4)) + (c^(1/4)*(5*b^3 - 28*a*b*c + (5*b^2 - 18*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)])/(4*2^(3/4)*a^2*(b^2 - 4
*a*c)^(3/2)*(-b + Sqrt[b^2 - 4*a*c])^(1/4))

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Rubi [A]  time = 2.44638, antiderivative size = 573, 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, 1504, 1510, 298, 205, 208} \[ -\frac{5 b^2-18 a c}{2 a^2 \sqrt{x} \left (b^2-4 a c\right )}+\frac{\sqrt [4]{c} \left (-\left (5 b^2-18 a c\right ) \sqrt{b^2-4 a c}-28 a b c+5 b^3\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{4\ 2^{3/4} a^2 \left (b^2-4 a c\right )^{3/2} \sqrt [4]{-\sqrt{b^2-4 a c}-b}}-\frac{\sqrt [4]{c} \left (\left (5 b^2-18 a c\right ) \sqrt{b^2-4 a c}-28 a b c+5 b^3\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{4\ 2^{3/4} a^2 \left (b^2-4 a c\right )^{3/2} \sqrt [4]{\sqrt{b^2-4 a c}-b}}-\frac{\sqrt [4]{c} \left (-\left (5 b^2-18 a c\right ) \sqrt{b^2-4 a c}-28 a b c+5 b^3\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{4\ 2^{3/4} a^2 \left (b^2-4 a c\right )^{3/2} \sqrt [4]{-\sqrt{b^2-4 a c}-b}}+\frac{\sqrt [4]{c} \left (\left (5 b^2-18 a c\right ) \sqrt{b^2-4 a c}-28 a b c+5 b^3\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{4\ 2^{3/4} a^2 \left (b^2-4 a c\right )^{3/2} \sqrt [4]{\sqrt{b^2-4 a c}-b}}+\frac{-2 a c+b^2+b c x^2}{2 a \sqrt{x} \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(5*b^2 - 18*a*c)/(2*a^2*(b^2 - 4*a*c)*Sqrt[x]) + (b^2 - 2*a*c + b*c*x^2)/(2*a*(b^2 - 4*a*c)*Sqrt[x]*(a + b*x^
2 + c*x^4)) + (c^(1/4)*(5*b^3 - 28*a*b*c - (5*b^2 - 18*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)])/(4*2^(3/4)*a^2*(b^2 - 4*a*c)^(3/2)*(-b - Sqrt[b^2 - 4*a*c])^(1/4)) - (c^(1/
4)*(5*b^3 - 28*a*b*c + (5*b^2 - 18*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)])/(4*2^(3/4)*a^2*(b^2 - 4*a*c)^(3/2)*(-b + Sqrt[b^2 - 4*a*c])^(1/4)) - (c^(1/4)*(5*b^3 - 28*a*b*c
 - (5*b^2 - 18*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)])/(4*2
^(3/4)*a^2*(b^2 - 4*a*c)^(3/2)*(-b - Sqrt[b^2 - 4*a*c])^(1/4)) + (c^(1/4)*(5*b^3 - 28*a*b*c + (5*b^2 - 18*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)])/(4*2^(3/4)*a^2*(b^2 - 4
*a*c)^(3/2)*(-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 1504

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_))*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] :>
 Simp[(d*(f*x)^(m + 1)*(a + b*x^n + c*x^(2*n))^(p + 1))/(a*f*(m + 1)), x] + Dist[1/(a*f^n*(m + 1)), Int[(f*x)^
(m + n)*(a + b*x^n + c*x^(2*n))^p*Simp[a*e*(m + 1) - b*d*(m + n*(p + 1) + 1) - c*d*(m + 2*n*(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] && LtQ[m, -
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{1}{x^{3/2} \left (a+b x^2+c x^4\right )^2} \, dx &=2 \operatorname{Subst}\left (\int \frac{1}{x^2 \left (a+b x^4+c x^8\right )^2} \, dx,x,\sqrt{x}\right )\\ &=\frac{b^2-2 a c+b c x^2}{2 a \left (b^2-4 a c\right ) \sqrt{x} \left (a+b x^2+c x^4\right )}-\frac{\operatorname{Subst}\left (\int \frac{-5 b^2+18 a c-5 b c x^4}{x^2 \left (a+b x^4+c x^8\right )} \, dx,x,\sqrt{x}\right )}{2 a \left (b^2-4 a c\right )}\\ &=-\frac{5 b^2-18 a c}{2 a^2 \left (b^2-4 a c\right ) \sqrt{x}}+\frac{b^2-2 a c+b c x^2}{2 a \left (b^2-4 a c\right ) \sqrt{x} \left (a+b x^2+c x^4\right )}+\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (-b \left (5 b^2-23 a c\right )-c \left (5 b^2-18 a c\right ) x^4\right )}{a+b x^4+c x^8} \, dx,x,\sqrt{x}\right )}{2 a^2 \left (b^2-4 a c\right )}\\ &=-\frac{5 b^2-18 a c}{2 a^2 \left (b^2-4 a c\right ) \sqrt{x}}+\frac{b^2-2 a c+b c x^2}{2 a \left (b^2-4 a c\right ) \sqrt{x} \left (a+b x^2+c x^4\right )}-\frac{\left (c \left (5 b^2-18 a c+\frac{5 b^3}{\sqrt{b^2-4 a c}}-\frac{28 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 )}{4 a^2 \left (b^2-4 a c\right )}-\frac{\left (c \left (5 b^2-18 a c-\frac{5 b^3}{\sqrt{b^2-4 a c}}+\frac{28 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 )}{4 a^2 \left (b^2-4 a c\right )}\\ &=-\frac{5 b^2-18 a c}{2 a^2 \left (b^2-4 a c\right ) \sqrt{x}}+\frac{b^2-2 a c+b c x^2}{2 a \left (b^2-4 a c\right ) \sqrt{x} \left (a+b x^2+c x^4\right )}+\frac{\left (\sqrt{c} \left (5 b^2-18 a c+\frac{5 b^3}{\sqrt{b^2-4 a c}}-\frac{28 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 )}{4 \sqrt{2} a^2 \left (b^2-4 a c\right )}-\frac{\left (\sqrt{c} \left (5 b^2-18 a c+\frac{5 b^3}{\sqrt{b^2-4 a c}}-\frac{28 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 )}{4 \sqrt{2} a^2 \left (b^2-4 a c\right )}+\frac{\left (\sqrt{c} \left (5 b^2-18 a c-\frac{5 b^3}{\sqrt{b^2-4 a c}}+\frac{28 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 )}{4 \sqrt{2} a^2 \left (b^2-4 a c\right )}-\frac{\left (\sqrt{c} \left (5 b^2-18 a c-\frac{5 b^3}{\sqrt{b^2-4 a c}}+\frac{28 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 )}{4 \sqrt{2} a^2 \left (b^2-4 a c\right )}\\ &=-\frac{5 b^2-18 a c}{2 a^2 \left (b^2-4 a c\right ) \sqrt{x}}+\frac{b^2-2 a c+b c x^2}{2 a \left (b^2-4 a c\right ) \sqrt{x} \left (a+b x^2+c x^4\right )}-\frac{\sqrt [4]{c} \left (5 b^2-18 a c-\frac{5 b^3}{\sqrt{b^2-4 a c}}+\frac{28 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 )}{4\ 2^{3/4} a^2 \left (b^2-4 a c\right ) \sqrt [4]{-b-\sqrt{b^2-4 a c}}}-\frac{\sqrt [4]{c} \left (5 b^2-18 a c+\frac{5 b^3}{\sqrt{b^2-4 a c}}-\frac{28 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 )}{4\ 2^{3/4} a^2 \left (b^2-4 a c\right ) \sqrt [4]{-b+\sqrt{b^2-4 a c}}}+\frac{\sqrt [4]{c} \left (5 b^2-18 a c-\frac{5 b^3}{\sqrt{b^2-4 a c}}+\frac{28 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 )}{4\ 2^{3/4} a^2 \left (b^2-4 a c\right ) \sqrt [4]{-b-\sqrt{b^2-4 a c}}}+\frac{\sqrt [4]{c} \left (5 b^2-18 a c+\frac{5 b^3}{\sqrt{b^2-4 a c}}-\frac{28 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 )}{4\ 2^{3/4} a^2 \left (b^2-4 a c\right ) \sqrt [4]{-b+\sqrt{b^2-4 a c}}}\\ \end{align*}

Mathematica [C]  time = 0.311675, size = 190, normalized size = 0.33 \[ -\frac{\frac{\text{RootSum}\left [\text{$\#$1}^4 b+\text{$\#$1}^8 c+a\& ,\frac{-18 \text{$\#$1}^4 a c^2 \log \left (\sqrt{x}-\text{$\#$1}\right )+5 \text{$\#$1}^4 b^2 c \log \left (\sqrt{x}-\text{$\#$1}\right )-23 a b c \log \left (\sqrt{x}-\text{$\#$1}\right )+5 b^3 \log \left (\sqrt{x}-\text{$\#$1}\right )}{2 \text{$\#$1}^5 c+\text{$\#$1} b}\& \right ]}{b^2-4 a c}+\frac{4 x^{3/2} \left (-3 a b c-2 a c^2 x^2+b^2 c x^2+b^3\right )}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac{16}{\sqrt{x}}}{8 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(16/Sqrt[x] + (4*x^(3/2)*(b^3 - 3*a*b*c + b^2*c*x^2 - 2*a*c^2*x^2))/((b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) + Roo
tSum[a + b*#1^4 + c*#1^8 & , (5*b^3*Log[Sqrt[x] - #1] - 23*a*b*c*Log[Sqrt[x] - #1] + 5*b^2*c*Log[Sqrt[x] - #1]
*#1^4 - 18*a*c^2*Log[Sqrt[x] - #1]*#1^4)/(b*#1 + 2*c*#1^5) & ]/(b^2 - 4*a*c))/(8*a^2)

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Maple [C]  time = 0.305, size = 245, normalized size = 0.4 \begin{align*} -{\frac{{c}^{2}}{a \left ( c{x}^{4}+b{x}^{2}+a \right ) \left ( 4\,ac-{b}^{2} \right ) }{x}^{{\frac{7}{2}}}}+{\frac{{b}^{2}c}{2\,{a}^{2} \left ( c{x}^{4}+b{x}^{2}+a \right ) \left ( 4\,ac-{b}^{2} \right ) }{x}^{{\frac{7}{2}}}}-{\frac{3\,bc}{2\,a \left ( c{x}^{4}+b{x}^{2}+a \right ) \left ( 4\,ac-{b}^{2} \right ) }{x}^{{\frac{3}{2}}}}+{\frac{{b}^{3}}{2\,{a}^{2} \left ( c{x}^{4}+b{x}^{2}+a \right ) \left ( 4\,ac-{b}^{2} \right ) }{x}^{{\frac{3}{2}}}}-{\frac{1}{8\,{a}^{2} \left ( 4\,ac-{b}^{2} \right ) }\sum _{{\it \_R}={\it RootOf} \left ({{\it \_Z}}^{8}c+{{\it \_Z}}^{4}b+a \right ) }{\frac{c \left ( 18\,ac-5\,{b}^{2} \right ){{\it \_R}}^{6}+b \left ( 23\,ac-5\,{b}^{2} \right ){{\it \_R}}^{2}}{2\,{{\it \_R}}^{7}c+{{\it \_R}}^{3}b}\ln \left ( \sqrt{x}-{\it \_R} \right ) }}-2\,{\frac{1}{{a}^{2}\sqrt{x}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/a/(c*x^4+b*x^2+a)*c^2/(4*a*c-b^2)*x^(7/2)+1/2/a^2/(c*x^4+b*x^2+a)*c/(4*a*c-b^2)*x^(7/2)*b^2-3/2/a/(c*x^4+b*
x^2+a)*b/(4*a*c-b^2)*x^(3/2)*c+1/2/a^2/(c*x^4+b*x^2+a)*b^3/(4*a*c-b^2)*x^(3/2)-1/8/a^2/(4*a*c-b^2)*sum((c*(18*
a*c-5*b^2)*_R^6+b*(23*a*c-5*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))-2/a^2/x^(1
/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{{\left (5 \, b^{2} c - 18 \, a c^{2}\right )} x^{\frac{7}{2}} +{\left (5 \, b^{3} - 19 \, a b c\right )} x^{\frac{3}{2}} + \frac{4 \,{\left (a b^{2} - 4 \, a^{2} c\right )}}{\sqrt{x}}}{2 \,{\left (a^{3} b^{2} - 4 \, a^{4} c +{\left (a^{2} b^{2} c - 4 \, a^{3} c^{2}\right )} x^{4} +{\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} x^{2}\right )}} - \int \frac{{\left (5 \, b^{2} c - 18 \, a c^{2}\right )} x^{\frac{5}{2}} +{\left (5 \, b^{3} - 23 \, a b c\right )} \sqrt{x}}{4 \,{\left (a^{3} b^{2} - 4 \, a^{4} c +{\left (a^{2} b^{2} c - 4 \, a^{3} c^{2}\right )} x^{4} +{\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} x^{2}\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((5*b^2*c - 18*a*c^2)*x^(7/2) + (5*b^3 - 19*a*b*c)*x^(3/2) + 4*(a*b^2 - 4*a^2*c)/sqrt(x))/(a^3*b^2 - 4*a^
4*c + (a^2*b^2*c - 4*a^3*c^2)*x^4 + (a^2*b^3 - 4*a^3*b*c)*x^2) - integrate(1/4*((5*b^2*c - 18*a*c^2)*x^(5/2) +
 (5*b^3 - 23*a*b*c)*sqrt(x))/(a^3*b^2 - 4*a^4*c + (a^2*b^2*c - 4*a^3*c^2)*x^4 + (a^2*b^3 - 4*a^3*b*c)*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^(3/2)/(c*x^4+b*x^2+a)^2,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**(3/2)/(c*x**4+b*x**2+a)**2,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^(3/2)/(c*x^4+b*x^2+a)^2,x, algorithm="giac")

[Out]

Timed out