3.1083 \(\int \frac{x^{9/2}}{\left (a+b x^2+c x^4\right )^3} \, dx\)

Optimal. Leaf size=533 \[ -\frac{3 x^{3/2} \left (-4 a c+5 b^2+8 b c x^2\right )}{16 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}+\frac{x^{3/2} \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac{3 \sqrt [4]{c} \left (4 b \sqrt{b^2-4 a c}+20 a c+11 b^2\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{16\ 2^{3/4} \left (b^2-4 a c\right )^{5/2} \sqrt [4]{-\sqrt{b^2-4 a c}-b}}+\frac{3 \sqrt [4]{c} \left (-4 b \sqrt{b^2-4 a c}+20 a c+11 b^2\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{16\ 2^{3/4} \left (b^2-4 a c\right )^{5/2} \sqrt [4]{\sqrt{b^2-4 a c}-b}}+\frac{3 \sqrt [4]{c} \left (4 b \sqrt{b^2-4 a c}+20 a c+11 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{16\ 2^{3/4} \left (b^2-4 a c\right )^{5/2} \sqrt [4]{-\sqrt{b^2-4 a c}-b}}-\frac{3 \sqrt [4]{c} \left (-4 b \sqrt{b^2-4 a c}+20 a c+11 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{16\ 2^{3/4} \left (b^2-4 a c\right )^{5/2} \sqrt [4]{\sqrt{b^2-4 a c}-b}} \]

[Out]

(x^(3/2)*(2*a + b*x^2))/(4*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)^2) - (3*x^(3/2)*(5*
b^2 - 4*a*c + 8*b*c*x^2))/(16*(b^2 - 4*a*c)^2*(a + b*x^2 + c*x^4)) - (3*c^(1/4)*
(11*b^2 + 20*a*c + 4*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^(3/4)*(b^2 - 4*a*c)^(5/2)*(-b - Sqrt[b^2 - 4*a
*c])^(1/4)) + (3*c^(1/4)*(11*b^2 + 20*a*c - 4*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^(3/4)*(b^2 - 4*a*c)^(
5/2)*(-b + Sqrt[b^2 - 4*a*c])^(1/4)) + (3*c^(1/4)*(11*b^2 + 20*a*c + 4*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^(3/4)*(b^2 - 4*a*c)^(5/2)*(-b - Sqrt[b^2 - 4*a*c])^(1/4)) - (3*c^(1/4)*(11*
b^2 + 20*a*c - 4*b*Sqrt[b^2 - 4*a*c])*ArcTanh[(2^(1/4)*c^(1/4)*Sqrt[x])/(-b + Sq
rt[b^2 - 4*a*c])^(1/4)])/(16*2^(3/4)*(b^2 - 4*a*c)^(5/2)*(-b + Sqrt[b^2 - 4*a*c]
)^(1/4))

_______________________________________________________________________________________

Rubi [A]  time = 2.70076, 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 \[ -\frac{3 x^{3/2} \left (-4 a c+5 b^2+8 b c x^2\right )}{16 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}+\frac{x^{3/2} \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac{3 \sqrt [4]{c} \left (4 b \sqrt{b^2-4 a c}+20 a c+11 b^2\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{16\ 2^{3/4} \left (b^2-4 a c\right )^{5/2} \sqrt [4]{-\sqrt{b^2-4 a c}-b}}+\frac{3 \sqrt [4]{c} \left (-4 b \sqrt{b^2-4 a c}+20 a c+11 b^2\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{16\ 2^{3/4} \left (b^2-4 a c\right )^{5/2} \sqrt [4]{\sqrt{b^2-4 a c}-b}}+\frac{3 \sqrt [4]{c} \left (4 b \sqrt{b^2-4 a c}+20 a c+11 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{16\ 2^{3/4} \left (b^2-4 a c\right )^{5/2} \sqrt [4]{-\sqrt{b^2-4 a c}-b}}-\frac{3 \sqrt [4]{c} \left (-4 b \sqrt{b^2-4 a c}+20 a c+11 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{16\ 2^{3/4} \left (b^2-4 a c\right )^{5/2} \sqrt [4]{\sqrt{b^2-4 a c}-b}} \]

Antiderivative was successfully verified.

[In]  Int[x^(9/2)/(a + b*x^2 + c*x^4)^3,x]

[Out]

(x^(3/2)*(2*a + b*x^2))/(4*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)^2) - (3*x^(3/2)*(5*
b^2 - 4*a*c + 8*b*c*x^2))/(16*(b^2 - 4*a*c)^2*(a + b*x^2 + c*x^4)) - (3*c^(1/4)*
(11*b^2 + 20*a*c + 4*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^(3/4)*(b^2 - 4*a*c)^(5/2)*(-b - Sqrt[b^2 - 4*a
*c])^(1/4)) + (3*c^(1/4)*(11*b^2 + 20*a*c - 4*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^(3/4)*(b^2 - 4*a*c)^(
5/2)*(-b + Sqrt[b^2 - 4*a*c])^(1/4)) + (3*c^(1/4)*(11*b^2 + 20*a*c + 4*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^(3/4)*(b^2 - 4*a*c)^(5/2)*(-b - Sqrt[b^2 - 4*a*c])^(1/4)) - (3*c^(1/4)*(11*
b^2 + 20*a*c - 4*b*Sqrt[b^2 - 4*a*c])*ArcTanh[(2^(1/4)*c^(1/4)*Sqrt[x])/(-b + Sq
rt[b^2 - 4*a*c])^(1/4)])/(16*2^(3/4)*(b^2 - 4*a*c)^(5/2)*(-b + Sqrt[b^2 - 4*a*c]
)^(1/4))

_______________________________________________________________________________________

Rubi in Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \[ \text{Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate(x**(9/2)/(c*x**4+b*x**2+a)**3,x)

[Out]

Timed out

_______________________________________________________________________________________

Mathematica [C]  time = 0.380375, size = 176, normalized size = 0.33 \[ \frac{-3 \text{RootSum}\left [\text{$\#$1}^8 c+\text{$\#$1}^4 b+a\&,\frac{8 \text{$\#$1}^4 b c \log \left (\sqrt{x}-\text{$\#$1}\right )-20 a c \log \left (\sqrt{x}-\text{$\#$1}\right )-7 b^2 \log \left (\sqrt{x}-\text{$\#$1}\right )}{2 \text{$\#$1}^5 c+\text{$\#$1} b}\&\right ]-\frac{12 x^{3/2} \left (-4 a c+5 b^2+8 b c x^2\right )}{a+b x^2+c x^4}+\frac{16 x^{3/2} \left (b^2-4 a c\right ) \left (2 a+b x^2\right )}{\left (a+b x^2+c x^4\right )^2}}{64 \left (b^2-4 a c\right )^2} \]

Antiderivative was successfully verified.

[In]  Integrate[x^(9/2)/(a + b*x^2 + c*x^4)^3,x]

[Out]

((16*(b^2 - 4*a*c)*x^(3/2)*(2*a + b*x^2))/(a + b*x^2 + c*x^4)^2 - (12*x^(3/2)*(5
*b^2 - 4*a*c + 8*b*c*x^2))/(a + b*x^2 + c*x^4) - 3*RootSum[a + b*#1^4 + c*#1^8 &
, (-7*b^2*Log[Sqrt[x] - #1] - 20*a*c*Log[Sqrt[x] - #1] + 8*b*c*Log[Sqrt[x] - #1
]*#1^4)/(b*#1 + 2*c*#1^5) & ])/(64*(b^2 - 4*a*c)^2)

_______________________________________________________________________________________

Maple [C]  time = 0.045, size = 244, normalized size = 0.5 \[ 2\,{\frac{1}{ \left ( c{x}^{4}+b{x}^{2}+a \right ) ^{2}} \left ( -1/32\,{\frac{a \left ( 20\,ac+7\,{b}^{2} \right ){x}^{3/2}}{16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4}}}-1/32\,{\frac{b \left ( 28\,ac+11\,{b}^{2} \right ){x}^{7/2}}{16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4}}}+{\frac{ \left ( 12\,ac-39\,{b}^{2} \right ) c{x}^{11/2}}{512\,{a}^{2}{c}^{2}-256\,a{b}^{2}c+32\,{b}^{4}}}-3/4\,{\frac{b{c}^{2}{x}^{15/2}}{16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4}}} \right ) }+{\frac{3}{64}\sum _{{\it \_R}={\it RootOf} \left ({{\it \_Z}}^{8}c+{{\it \_Z}}^{4}b+a \right ) }{\frac{-8\,bc{{\it \_R}}^{6}+ \left ( 20\,ac+7\,{b}^{2} \right ){{\it \_R}}^{2}}{ \left ( 16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4} \right ) \left ( 2\,{{\it \_R}}^{7}c+{{\it \_R}}^{3}b \right ) }\ln \left ( \sqrt{x}-{\it \_R} \right ) }} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int(x^(9/2)/(c*x^4+b*x^2+a)^3,x)

[Out]

2*(-1/32*a*(20*a*c+7*b^2)/(16*a^2*c^2-8*a*b^2*c+b^4)*x^(3/2)-1/32*b*(28*a*c+11*b
^2)/(16*a^2*c^2-8*a*b^2*c+b^4)*x^(7/2)+3/32*(4*a*c-13*b^2)*c/(16*a^2*c^2-8*a*b^2
*c+b^4)*x^(11/2)-3/4*b*c^2/(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*sum((-8*b*c*_R^6+(20*a*c+7*b^2)*_R^2)/(16*a^2*c^2-8*a*b^2*c+b^4)/(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. \[ -\frac{24 \, b c^{2} x^{\frac{15}{2}} + 3 \,{\left (13 \, b^{2} c - 4 \, a c^{2}\right )} x^{\frac{11}{2}} +{\left (11 \, b^{3} + 28 \, a b c\right )} x^{\frac{7}{2}} +{\left (7 \, a b^{2} + 20 \, a^{2} c\right )} x^{\frac{3}{2}}}{16 \,{\left ({\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} x^{8} + 2 \,{\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} x^{6} + a^{2} b^{4} - 8 \, a^{3} b^{2} c + 16 \, a^{4} c^{2} +{\left (b^{6} - 6 \, a b^{4} c + 32 \, a^{3} c^{3}\right )} x^{4} + 2 \,{\left (a b^{5} - 8 \, a^{2} b^{3} c + 16 \, a^{3} b c^{2}\right )} x^{2}\right )}} - \int \frac{3 \,{\left (8 \, b c x^{\frac{5}{2}} -{\left (7 \, b^{2} + 20 \, a c\right )} \sqrt{x}\right )}}{32 \,{\left (a b^{4} - 8 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} +{\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{4} +{\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x^{2}\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(x^(9/2)/(c*x^4 + b*x^2 + a)^3,x, algorithm="maxima")

[Out]

-1/16*(24*b*c^2*x^(15/2) + 3*(13*b^2*c - 4*a*c^2)*x^(11/2) + (11*b^3 + 28*a*b*c)
*x^(7/2) + (7*a*b^2 + 20*a^2*c)*x^(3/2))/((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) - integrate(3/32*(8*b*c*x^(5/2) - (7*b^2 + 20*a*c)*sqrt(x))/(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)

_______________________________________________________________________________________

Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \[ \text{Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(x^(9/2)/(c*x^4 + b*x^2 + a)^3,x, algorithm="fricas")

[Out]

Timed out

_______________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \[ \text{Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(x**(9/2)/(c*x**4+b*x**2+a)**3,x)

[Out]

Timed out

_______________________________________________________________________________________

GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{x^{\frac{9}{2}}}{{\left (c x^{4} + b x^{2} + a\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(x^(9/2)/(c*x^4 + b*x^2 + a)^3,x, algorithm="giac")

[Out]

integrate(x^(9/2)/(c*x^4 + b*x^2 + a)^3, x)