3.1084 \(\int \frac{x^{7/2}}{(a+b x^2+c x^4)^3} \, dx\)

Optimal. Leaf size=533 \[ \frac{c^{3/4} \left (36 b \sqrt{b^2-4 a c}+28 a c+41 b^2\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{16 \sqrt [4]{2} \left (b^2-4 a c\right )^{5/2} \left (-\sqrt{b^2-4 a c}-b\right )^{3/4}}-\frac{c^{3/4} \left (-36 b \sqrt{b^2-4 a c}+28 a c+41 b^2\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{16 \sqrt [4]{2} \left (b^2-4 a c\right )^{5/2} \left (\sqrt{b^2-4 a c}-b\right )^{3/4}}+\frac{c^{3/4} \left (36 b \sqrt{b^2-4 a c}+28 a c+41 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{16 \sqrt [4]{2} \left (b^2-4 a c\right )^{5/2} \left (-\sqrt{b^2-4 a c}-b\right )^{3/4}}-\frac{c^{3/4} \left (-36 b \sqrt{b^2-4 a c}+28 a c+41 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{16 \sqrt [4]{2} \left (b^2-4 a c\right )^{5/2} \left (\sqrt{b^2-4 a c}-b\right )^{3/4}}+\frac{\sqrt{x} \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac{\sqrt{x} \left (-4 a c+13 b^2+24 b c x^2\right )}{16 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )} \]

[Out]

(Sqrt[x]*(2*a + b*x^2))/(4*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)^2) - (Sqrt[x]*(13*b^2 - 4*a*c + 24*b*c*x^2))/(16*
(b^2 - 4*a*c)^2*(a + b*x^2 + c*x^4)) + (c^(3/4)*(41*b^2 + 28*a*c + 36*b*Sqrt[b^2 - 4*a*c])*ArcTan[(2^(1/4)*c^(
1/4)*Sqrt[x])/(-b - Sqrt[b^2 - 4*a*c])^(1/4)])/(16*2^(1/4)*(b^2 - 4*a*c)^(5/2)*(-b - Sqrt[b^2 - 4*a*c])^(3/4))
- (c^(3/4)*(41*b^2 + 28*a*c - 36*b*Sqrt[b^2 - 4*a*c])*ArcTan[(2^(1/4)*c^(1/4)*Sqrt[x])/(-b + Sqrt[b^2 - 4*a*c
])^(1/4)])/(16*2^(1/4)*(b^2 - 4*a*c)^(5/2)*(-b + Sqrt[b^2 - 4*a*c])^(3/4)) + (c^(3/4)*(41*b^2 + 28*a*c + 36*b*
Sqrt[b^2 - 4*a*c])*ArcTanh[(2^(1/4)*c^(1/4)*Sqrt[x])/(-b - Sqrt[b^2 - 4*a*c])^(1/4)])/(16*2^(1/4)*(b^2 - 4*a*c
)^(5/2)*(-b - Sqrt[b^2 - 4*a*c])^(3/4)) - (c^(3/4)*(41*b^2 + 28*a*c - 36*b*Sqrt[b^2 - 4*a*c])*ArcTanh[(2^(1/4)
*c^(1/4)*Sqrt[x])/(-b + Sqrt[b^2 - 4*a*c])^(1/4)])/(16*2^(1/4)*(b^2 - 4*a*c)^(5/2)*(-b + Sqrt[b^2 - 4*a*c])^(3
/4))

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Rubi [A]  time = 1.36301, antiderivative size = 533, 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, 1365, 1430, 1422, 212, 208, 205} \[ \frac{c^{3/4} \left (36 b \sqrt{b^2-4 a c}+28 a c+41 b^2\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{16 \sqrt [4]{2} \left (b^2-4 a c\right )^{5/2} \left (-\sqrt{b^2-4 a c}-b\right )^{3/4}}-\frac{c^{3/4} \left (-36 b \sqrt{b^2-4 a c}+28 a c+41 b^2\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{16 \sqrt [4]{2} \left (b^2-4 a c\right )^{5/2} \left (\sqrt{b^2-4 a c}-b\right )^{3/4}}+\frac{c^{3/4} \left (36 b \sqrt{b^2-4 a c}+28 a c+41 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{16 \sqrt [4]{2} \left (b^2-4 a c\right )^{5/2} \left (-\sqrt{b^2-4 a c}-b\right )^{3/4}}-\frac{c^{3/4} \left (-36 b \sqrt{b^2-4 a c}+28 a c+41 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{16 \sqrt [4]{2} \left (b^2-4 a c\right )^{5/2} \left (\sqrt{b^2-4 a c}-b\right )^{3/4}}+\frac{\sqrt{x} \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac{\sqrt{x} \left (-4 a c+13 b^2+24 b c x^2\right )}{16 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^(7/2)/(a + b*x^2 + c*x^4)^3,x]

[Out]

(Sqrt[x]*(2*a + b*x^2))/(4*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)^2) - (Sqrt[x]*(13*b^2 - 4*a*c + 24*b*c*x^2))/(16*
(b^2 - 4*a*c)^2*(a + b*x^2 + c*x^4)) + (c^(3/4)*(41*b^2 + 28*a*c + 36*b*Sqrt[b^2 - 4*a*c])*ArcTan[(2^(1/4)*c^(
1/4)*Sqrt[x])/(-b - Sqrt[b^2 - 4*a*c])^(1/4)])/(16*2^(1/4)*(b^2 - 4*a*c)^(5/2)*(-b - Sqrt[b^2 - 4*a*c])^(3/4))
- (c^(3/4)*(41*b^2 + 28*a*c - 36*b*Sqrt[b^2 - 4*a*c])*ArcTan[(2^(1/4)*c^(1/4)*Sqrt[x])/(-b + Sqrt[b^2 - 4*a*c
])^(1/4)])/(16*2^(1/4)*(b^2 - 4*a*c)^(5/2)*(-b + Sqrt[b^2 - 4*a*c])^(3/4)) + (c^(3/4)*(41*b^2 + 28*a*c + 36*b*
Sqrt[b^2 - 4*a*c])*ArcTanh[(2^(1/4)*c^(1/4)*Sqrt[x])/(-b - Sqrt[b^2 - 4*a*c])^(1/4)])/(16*2^(1/4)*(b^2 - 4*a*c
)^(5/2)*(-b - Sqrt[b^2 - 4*a*c])^(3/4)) - (c^(3/4)*(41*b^2 + 28*a*c - 36*b*Sqrt[b^2 - 4*a*c])*ArcTanh[(2^(1/4)
*c^(1/4)*Sqrt[x])/(-b + Sqrt[b^2 - 4*a*c])^(1/4)])/(16*2^(1/4)*(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 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 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{x^{7/2}}{\left (a+b x^2+c x^4\right )^3} \, dx &=2 \operatorname{Subst}\left (\int \frac{x^8}{\left (a+b x^4+c x^8\right )^3} \, dx,x,\sqrt{x}\right )\\ &=\frac{\sqrt{x} \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{2 a-11 b x^4}{\left (a+b x^4+c x^8\right )^2} \, dx,x,\sqrt{x}\right )}{4 \left (b^2-4 a c\right )}\\ &=\frac{\sqrt{x} \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{\sqrt{x} \left (13 b^2-4 a c+24 b c 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{a \left (5 b^2+28 a c\right )-72 a b c x^4}{a+b x^4+c x^8} \, dx,x,\sqrt{x}\right )}{16 a \left (b^2-4 a c\right )^2}\\ &=\frac{\sqrt{x} \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{\sqrt{x} \left (13 b^2-4 a c+24 b c x^2\right )}{16 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}+\frac{\left (c \left (41 b^2+28 a c-36 b \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 )}{16 \left (b^2-4 a c\right )^{5/2}}-\frac{\left (c \left (41 b^2+28 a c+36 b \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 )}{16 \left (b^2-4 a c\right )^{5/2}}\\ &=\frac{\sqrt{x} \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{\sqrt{x} \left (13 b^2-4 a c+24 b c x^2\right )}{16 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}-\frac{\left (c \left (41 b^2+28 a c-36 b \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 )}{16 \left (b^2-4 a c\right )^{5/2} \sqrt{-b+\sqrt{b^2-4 a c}}}-\frac{\left (c \left (41 b^2+28 a c-36 b \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 )}{16 \left (b^2-4 a c\right )^{5/2} \sqrt{-b+\sqrt{b^2-4 a c}}}+\frac{\left (c \left (41 b^2+28 a c+36 b \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 )}{16 \left (b^2-4 a c\right )^{5/2} \sqrt{-b-\sqrt{b^2-4 a c}}}+\frac{\left (c \left (41 b^2+28 a c+36 b \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 )}{16 \left (b^2-4 a c\right )^{5/2} \sqrt{-b-\sqrt{b^2-4 a c}}}\\ &=\frac{\sqrt{x} \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{\sqrt{x} \left (13 b^2-4 a c+24 b c x^2\right )}{16 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}+\frac{c^{3/4} \left (41 b^2+28 a c+36 b \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 )}{16 \sqrt [4]{2} \left (b^2-4 a c\right )^{5/2} \left (-b-\sqrt{b^2-4 a c}\right )^{3/4}}-\frac{c^{3/4} \left (41 b^2+28 a c-36 b \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 )}{16 \sqrt [4]{2} \left (b^2-4 a c\right )^{5/2} \left (-b+\sqrt{b^2-4 a c}\right )^{3/4}}+\frac{c^{3/4} \left (41 b^2+28 a c+36 b \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 )}{16 \sqrt [4]{2} \left (b^2-4 a c\right )^{5/2} \left (-b-\sqrt{b^2-4 a c}\right )^{3/4}}-\frac{c^{3/4} \left (41 b^2+28 a c-36 b \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 )}{16 \sqrt [4]{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 = 0.443112, size = 177, normalized size = 0.33 \[ -\frac{\text{RootSum}\left [\text{$\#$1}^4 b+\text{$\#$1}^8 c+a\& ,\frac{72 \text{$\#$1}^4 b c \log \left (\sqrt{x}-\text{$\#$1}\right )-28 a c \log \left (\sqrt{x}-\text{$\#$1}\right )-5 b^2 \log \left (\sqrt{x}-\text{$\#$1}\right )}{\text{$\#$1}^3 b+2 \text{$\#$1}^7 c}\& \right ]+\frac{4 \sqrt{x} \left (28 a^2 c+a \left (5 b^2+36 b c x^2-4 c^2 x^4\right )+b x^2 \left (9 b^2+37 b c x^2+24 c^2 x^4\right )\right )}{\left (a+b x^2+c x^4\right )^2}}{64 \left (b^2-4 a c\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(7/2)/(a + b*x^2 + c*x^4)^3,x]

[Out]

-((4*Sqrt[x]*(28*a^2*c + a*(5*b^2 + 36*b*c*x^2 - 4*c^2*x^4) + b*x^2*(9*b^2 + 37*b*c*x^2 + 24*c^2*x^4)))/(a + b
*x^2 + c*x^4)^2 + RootSum[a + b*#1^4 + c*#1^8 & , (-5*b^2*Log[Sqrt[x] - #1] - 28*a*c*Log[Sqrt[x] - #1] + 72*b*
c*Log[Sqrt[x] - #1]*#1^4)/(b*#1^3 + 2*c*#1^7) & ])/(64*(b^2 - 4*a*c)^2)

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Maple [C]  time = 0.27, size = 237, normalized size = 0.4 \begin{align*} 2\,{\frac{1}{ \left ( c{x}^{4}+b{x}^{2}+a \right ) ^{2}} \left ( -1/32\,{\frac{a \left ( 28\,ac+5\,{b}^{2} \right ) \sqrt{x}}{16\,{a}^{2}{c}^{2}-8\,ac{b}^{2}+{b}^{4}}}-{\frac{9\,b \left ( 4\,ac+{b}^{2} \right ){x}^{5/2}}{512\,{a}^{2}{c}^{2}-256\,ac{b}^{2}+32\,{b}^{4}}}+1/32\,{\frac{c \left ( 4\,ac-37\,{b}^{2} \right ){x}^{9/2}}{16\,{a}^{2}{c}^{2}-8\,ac{b}^{2}+{b}^{4}}}-3/4\,{\frac{b{c}^{2}{x}^{13/2}}{16\,{a}^{2}{c}^{2}-8\,ac{b}^{2}+{b}^{4}}} \right ) }+{\frac{1}{1024\,{a}^{2}{c}^{2}-512\,ac{b}^{2}+64\,{b}^{4}}\sum _{{\it \_R}={\it RootOf} \left ({{\it \_Z}}^{8}c+{{\it \_Z}}^{4}b+a \right ) }{\frac{-72\,{{\it \_R}}^{4}bc+28\,ac+5\,{b}^{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^(7/2)/(c*x^4+b*x^2+a)^3,x)

[Out]

2*(-1/32*a*(28*a*c+5*b^2)/(16*a^2*c^2-8*a*b^2*c+b^4)*x^(1/2)-9/32*b*(4*a*c+b^2)/(16*a^2*c^2-8*a*b^2*c+b^4)*x^(
5/2)+1/32*c*(4*a*c-37*b^2)/(16*a^2*c^2-8*a*b^2*c+b^4)*x^(9/2)-3/4*b*c^2/(16*a^2*c^2-8*a*b^2*c+b^4)*x^(13/2))/(
c*x^4+b*x^2+a)^2+1/64/(16*a^2*c^2-8*a*b^2*c+b^4)*sum((-72*_R^4*b*c+28*a*c+5*b^2)/(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{{\left (5 \, b^{2} c^{2} + 28 \, a c^{3}\right )} x^{\frac{17}{2}} + 2 \,{\left (5 \, b^{3} c + 16 \, a b c^{2}\right )} x^{\frac{13}{2}} +{\left (5 \, b^{4} + a b^{2} c + 60 \, a^{2} c^{2}\right )} x^{\frac{9}{2}} +{\left (a b^{3} + 20 \, a^{2} b c\right )} x^{\frac{5}{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{{\left (5 \, b^{2} c + 28 \, a c^{2}\right )} x^{\frac{7}{2}} + 5 \,{\left (b^{3} + 20 \, a b c\right )} x^{\frac{3}{2}}}{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^(7/2)/(c*x^4+b*x^2+a)^3,x, algorithm="maxima")

[Out]

1/16*((5*b^2*c^2 + 28*a*c^3)*x^(17/2) + 2*(5*b^3*c + 16*a*b*c^2)*x^(13/2) + (5*b^4 + a*b^2*c + 60*a^2*c^2)*x^(
9/2) + (a*b^3 + 20*a^2*b*c)*x^(5/2))/((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 + 1
6*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(1/32*((5*b^2*c + 28*a*c^2)*x^(7/2) + 5*(b^3 + 20*a*b*c)*x^(3
/2))/(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)

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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^(7/2)/(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**(7/2)/(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^(7/2)/(c*x^4+b*x^2+a)^3,x, algorithm="giac")

[Out]

Timed out