3.1081 \(\int \frac{x^{13/2}}{\left (a+b x^2+c x^4\right )^3} \, dx\)

Optimal. Leaf size=569 \[ \frac{x^{7/2} \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}+\frac{x^{3/2} \left (x^2 \left (28 a c+5 b^2\right )+24 a b\right )}{16 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}+\frac{\left (\sqrt{b^2-4 a c} \left (28 a c+5 b^2\right )+172 a b c+5 b^3\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{32\ 2^{3/4} c^{3/4} \left (b^2-4 a c\right )^{5/2} \sqrt [4]{-\sqrt{b^2-4 a c}-b}}+\frac{\left (-\frac{172 a b c+5 b^3}{\sqrt{b^2-4 a c}}+28 a c+5 b^2\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{32\ 2^{3/4} c^{3/4} \left (b^2-4 a c\right )^2 \sqrt [4]{\sqrt{b^2-4 a c}-b}}-\frac{\left (\sqrt{b^2-4 a c} \left (28 a c+5 b^2\right )+172 a b c+5 b^3\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{32\ 2^{3/4} c^{3/4} \left (b^2-4 a c\right )^{5/2} \sqrt [4]{-\sqrt{b^2-4 a c}-b}}-\frac{\left (-\frac{172 a b c+5 b^3}{\sqrt{b^2-4 a c}}+28 a c+5 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{32\ 2^{3/4} c^{3/4} \left (b^2-4 a c\right )^2 \sqrt [4]{\sqrt{b^2-4 a c}-b}} \]

[Out]

(x^(7/2)*(2*a + b*x^2))/(4*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)^2) + (x^(3/2)*(24*a
*b + (5*b^2 + 28*a*c)*x^2))/(16*(b^2 - 4*a*c)^2*(a + b*x^2 + c*x^4)) + ((5*b^3 +
172*a*b*c + Sqrt[b^2 - 4*a*c]*(5*b^2 + 28*a*c))*ArcTan[(2^(1/4)*c^(1/4)*Sqrt[x]
)/(-b - Sqrt[b^2 - 4*a*c])^(1/4)])/(32*2^(3/4)*c^(3/4)*(b^2 - 4*a*c)^(5/2)*(-b -
Sqrt[b^2 - 4*a*c])^(1/4)) + ((5*b^2 + 28*a*c - (5*b^3 + 172*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)*c^(3/4)*(b^2 - 4*a*c)^2*(-b + Sqrt[b^2 - 4*a*c])^(1/4)) - ((5*b^3 + 172*a*b
*c + Sqrt[b^2 - 4*a*c]*(5*b^2 + 28*a*c))*ArcTanh[(2^(1/4)*c^(1/4)*Sqrt[x])/(-b -
Sqrt[b^2 - 4*a*c])^(1/4)])/(32*2^(3/4)*c^(3/4)*(b^2 - 4*a*c)^(5/2)*(-b - Sqrt[b
^2 - 4*a*c])^(1/4)) - ((5*b^2 + 28*a*c - (5*b^3 + 172*a*b*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)])/(32*2^(3/4)*c
^(3/4)*(b^2 - 4*a*c)^2*(-b + Sqrt[b^2 - 4*a*c])^(1/4))

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Rubi [A]  time = 3.78662, antiderivative size = 569, 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 \[ \frac{x^{7/2} \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}+\frac{x^{3/2} \left (x^2 \left (28 a c+5 b^2\right )+24 a b\right )}{16 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}+\frac{\left (\sqrt{b^2-4 a c} \left (28 a c+5 b^2\right )+172 a b c+5 b^3\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{32\ 2^{3/4} c^{3/4} \left (b^2-4 a c\right )^{5/2} \sqrt [4]{-\sqrt{b^2-4 a c}-b}}+\frac{\left (-\frac{172 a b c+5 b^3}{\sqrt{b^2-4 a c}}+28 a c+5 b^2\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{32\ 2^{3/4} c^{3/4} \left (b^2-4 a c\right )^2 \sqrt [4]{\sqrt{b^2-4 a c}-b}}-\frac{\left (\sqrt{b^2-4 a c} \left (28 a c+5 b^2\right )+172 a b c+5 b^3\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{32\ 2^{3/4} c^{3/4} \left (b^2-4 a c\right )^{5/2} \sqrt [4]{-\sqrt{b^2-4 a c}-b}}-\frac{\left (-\frac{172 a b c+5 b^3}{\sqrt{b^2-4 a c}}+28 a c+5 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{32\ 2^{3/4} c^{3/4} \left (b^2-4 a c\right )^2 \sqrt [4]{\sqrt{b^2-4 a c}-b}} \]

Antiderivative was successfully verified.

[In]  Int[x^(13/2)/(a + b*x^2 + c*x^4)^3,x]

[Out]

(x^(7/2)*(2*a + b*x^2))/(4*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)^2) + (x^(3/2)*(24*a
*b + (5*b^2 + 28*a*c)*x^2))/(16*(b^2 - 4*a*c)^2*(a + b*x^2 + c*x^4)) + ((5*b^3 +
172*a*b*c + Sqrt[b^2 - 4*a*c]*(5*b^2 + 28*a*c))*ArcTan[(2^(1/4)*c^(1/4)*Sqrt[x]
)/(-b - Sqrt[b^2 - 4*a*c])^(1/4)])/(32*2^(3/4)*c^(3/4)*(b^2 - 4*a*c)^(5/2)*(-b -
Sqrt[b^2 - 4*a*c])^(1/4)) + ((5*b^2 + 28*a*c - (5*b^3 + 172*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)*c^(3/4)*(b^2 - 4*a*c)^2*(-b + Sqrt[b^2 - 4*a*c])^(1/4)) - ((5*b^3 + 172*a*b
*c + Sqrt[b^2 - 4*a*c]*(5*b^2 + 28*a*c))*ArcTanh[(2^(1/4)*c^(1/4)*Sqrt[x])/(-b -
Sqrt[b^2 - 4*a*c])^(1/4)])/(32*2^(3/4)*c^(3/4)*(b^2 - 4*a*c)^(5/2)*(-b - Sqrt[b
^2 - 4*a*c])^(1/4)) - ((5*b^2 + 28*a*c - (5*b^3 + 172*a*b*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)])/(32*2^(3/4)*c
^(3/4)*(b^2 - 4*a*c)^2*(-b + Sqrt[b^2 - 4*a*c])^(1/4))

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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**(13/2)/(c*x**4+b*x**2+a)**3,x)

[Out]

Timed out

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Mathematica [C]  time = 0.618805, size = 216, normalized size = 0.38 \[ \frac{c \left (a+b x^2+c x^4\right )^2 \text{RootSum}\left [\text{$\#$1}^8 c+\text{$\#$1}^4 b+a\&,\frac{28 \text{$\#$1}^4 a c \log \left (\sqrt{x}-\text{$\#$1}\right )+5 \text{$\#$1}^4 b^2 \log \left (\sqrt{x}-\text{$\#$1}\right )-72 a b \log \left (\sqrt{x}-\text{$\#$1}\right )}{2 \text{$\#$1}^5 c+\text{$\#$1} b}\&\right ]-16 x^{3/2} \left (b^2-4 a c\right ) \left (a \left (b-2 c x^2\right )+b^2 x^2\right )+4 x^{3/2} \left (8 a b c+28 a c^2 x^2+4 b^3+5 b^2 c x^2\right ) \left (a+b x^2+c x^4\right )}{64 c \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )^2} \]

Antiderivative was successfully verified.

[In]  Integrate[x^(13/2)/(a + b*x^2 + c*x^4)^3,x]

[Out]

(4*x^(3/2)*(4*b^3 + 8*a*b*c + 5*b^2*c*x^2 + 28*a*c^2*x^2)*(a + b*x^2 + c*x^4) -
16*(b^2 - 4*a*c)*x^(3/2)*(b^2*x^2 + a*(b - 2*c*x^2)) + c*(a + b*x^2 + c*x^4)^2*R
ootSum[a + b*#1^4 + c*#1^8 & , (-72*a*b*Log[Sqrt[x] - #1] + 5*b^2*Log[Sqrt[x] -
#1]*#1^4 + 28*a*c*Log[Sqrt[x] - #1]*#1^4)/(b*#1 + 2*c*#1^5) & ])/(64*c*(b^2 - 4*
a*c)^2*(a + b*x^2 + c*x^4)^2)

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Maple [C]  time = 0.076, size = 242, normalized size = 0.4 \[ 2\,{\frac{1}{ \left ( c{x}^{4}+b{x}^{2}+a \right ) ^{2}} \left ( 3/4\,{\frac{{a}^{2}b{x}^{3/2}}{16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4}}}-1/32\,{\frac{a \left ( 4\,ac-37\,{b}^{2} \right ){x}^{7/2}}{16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4}}}+{\frac{9\,b \left ( 4\,ac+{b}^{2} \right ){x}^{11/2}}{512\,{a}^{2}{c}^{2}-256\,a{b}^{2}c+32\,{b}^{4}}}+1/32\,{\frac{c \left ( 28\,ac+5\,{b}^{2} \right ){x}^{15/2}}{16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4}}} \right ) }+{\frac{1}{64}\sum _{{\it \_R}={\it RootOf} \left ({{\it \_Z}}^{8}c+{{\it \_Z}}^{4}b+a \right ) }{\frac{ \left ( 28\,ac+5\,{b}^{2} \right ){{\it \_R}}^{6}-72\,{{\it \_R}}^{2}ab}{ \left ( 16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4} \right ) \left ( 2\,{{\it \_R}}^{7}c+{{\it \_R}}^{3}b \right ) }\ln \left ( \sqrt{x}-{\it \_R} \right ) }} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int(x^(13/2)/(c*x^4+b*x^2+a)^3,x)

[Out]

2*(3/4*a^2*b/(16*a^2*c^2-8*a*b^2*c+b^4)*x^(3/2)-1/32*a*(4*a*c-37*b^2)/(16*a^2*c^
2-8*a*b^2*c+b^4)*x^(7/2)+9/32*b*(4*a*c+b^2)/(16*a^2*c^2-8*a*b^2*c+b^4)*x^(11/2)+
1/32*c*(28*a*c+5*b^2)/(16*a^2*c^2-8*a*b^2*c+b^4)*x^(15/2))/(c*x^4+b*x^2+a)^2+1/6
4*sum(((28*a*c+5*b^2)*_R^6-72*_R^2*a*b)/(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))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \frac{{\left (5 \, b^{2} c + 28 \, a c^{2}\right )} x^{\frac{15}{2}} + 9 \,{\left (b^{3} + 4 \, a b c\right )} x^{\frac{11}{2}} + 24 \, a^{2} b x^{\frac{3}{2}} +{\left (37 \, a b^{2} - 4 \, a^{2} c\right )} x^{\frac{7}{2}}}{16 \,{\left ({\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} x^{8} + 2 \,{\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} x^{6} + a^{2} b^{4} - 8 \, a^{3} b^{2} c + 16 \, a^{4} c^{2} +{\left (b^{6} - 6 \, a b^{4} c + 32 \, a^{3} c^{3}\right )} x^{4} + 2 \,{\left (a b^{5} - 8 \, a^{2} b^{3} c + 16 \, a^{3} b c^{2}\right )} x^{2}\right )}} + \int \frac{{\left (5 \, b^{2} + 28 \, a c\right )} x^{\frac{5}{2}} - 72 \, a b \sqrt{x}}{32 \,{\left (a b^{4} - 8 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} +{\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{4} +{\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x^{2}\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(x^(13/2)/(c*x^4 + b*x^2 + a)^3,x, algorithm="maxima")

[Out]

1/16*((5*b^2*c + 28*a*c^2)*x^(15/2) + 9*(b^3 + 4*a*b*c)*x^(11/2) + 24*a^2*b*x^(3
/2) + (37*a*b^2 - 4*a^2*c)*x^(7/2))/((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(1/32*((5*b^2 + 28*a*c)*x^(5/2) - 72*a*b*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)

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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^(13/2)/(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. \[ \text{Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(x**(13/2)/(c*x**4+b*x**2+a)**3,x)

[Out]

Timed out

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{x^{\frac{13}{2}}}{{\left (c x^{4} + b x^{2} + a\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(x^(13/2)/(c*x^4 + b*x^2 + a)^3,x, algorithm="giac")

[Out]

integrate(x^(13/2)/(c*x^4 + b*x^2 + a)^3, x)