3.538 \(\int \frac{c+d x+e x^2+f x^3}{x^4 \left (a+b x^4\right )^{3/2}} \, dx\)

Optimal. Leaf size=387 \[ -\frac{\sqrt [4]{b} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (5 \sqrt{b} c-9 \sqrt{a} e\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{12 a^{9/4} \sqrt{a+b x^4}}-\frac{3 \sqrt [4]{b} e \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 a^{7/4} \sqrt{a+b x^4}}-\frac{f \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{2 a^{3/2}}-\frac{x \left (b c+b d x+b e x^2+b f x^3\right )}{2 a^2 \sqrt{a+b x^4}}-\frac{c \sqrt{a+b x^4}}{3 a^2 x^3}-\frac{d \sqrt{a+b x^4}}{2 a^2 x^2}-\frac{e \sqrt{a+b x^4}}{a^2 x}+\frac{3 \sqrt{b} e x \sqrt{a+b x^4}}{2 a^2 \left (\sqrt{a}+\sqrt{b} x^2\right )}+\frac{f \sqrt{a+b x^4}}{2 a^2} \]

[Out]

-(x*(b*c + b*d*x + b*e*x^2 + b*f*x^3))/(2*a^2*Sqrt[a + b*x^4]) + (f*Sqrt[a + b*x
^4])/(2*a^2) - (c*Sqrt[a + b*x^4])/(3*a^2*x^3) - (d*Sqrt[a + b*x^4])/(2*a^2*x^2)
- (e*Sqrt[a + b*x^4])/(a^2*x) + (3*Sqrt[b]*e*x*Sqrt[a + b*x^4])/(2*a^2*(Sqrt[a]
+ Sqrt[b]*x^2)) - (f*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]])/(2*a^(3/2)) - (3*b^(1/4)
*e*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticE
[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(2*a^(7/4)*Sqrt[a + b*x^4]) - (b^(1/4)*(5*
Sqrt[b]*c - 9*Sqrt[a]*e)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqr
t[b]*x^2)^2]*EllipticF[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(12*a^(9/4)*Sqrt[a +
b*x^4])

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Rubi [A]  time = 1.04488, antiderivative size = 387, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 13, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.433 \[ -\frac{\sqrt [4]{b} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (5 \sqrt{b} c-9 \sqrt{a} e\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{12 a^{9/4} \sqrt{a+b x^4}}-\frac{3 \sqrt [4]{b} e \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 a^{7/4} \sqrt{a+b x^4}}-\frac{f \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{2 a^{3/2}}-\frac{x \left (b c+b d x+b e x^2+b f x^3\right )}{2 a^2 \sqrt{a+b x^4}}-\frac{c \sqrt{a+b x^4}}{3 a^2 x^3}-\frac{d \sqrt{a+b x^4}}{2 a^2 x^2}-\frac{e \sqrt{a+b x^4}}{a^2 x}+\frac{3 \sqrt{b} e x \sqrt{a+b x^4}}{2 a^2 \left (\sqrt{a}+\sqrt{b} x^2\right )}+\frac{f \sqrt{a+b x^4}}{2 a^2} \]

Antiderivative was successfully verified.

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

[Out]

-(x*(b*c + b*d*x + b*e*x^2 + b*f*x^3))/(2*a^2*Sqrt[a + b*x^4]) + (f*Sqrt[a + b*x
^4])/(2*a^2) - (c*Sqrt[a + b*x^4])/(3*a^2*x^3) - (d*Sqrt[a + b*x^4])/(2*a^2*x^2)
- (e*Sqrt[a + b*x^4])/(a^2*x) + (3*Sqrt[b]*e*x*Sqrt[a + b*x^4])/(2*a^2*(Sqrt[a]
+ Sqrt[b]*x^2)) - (f*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]])/(2*a^(3/2)) - (3*b^(1/4)
*e*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticE
[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(2*a^(7/4)*Sqrt[a + b*x^4]) - (b^(1/4)*(5*
Sqrt[b]*c - 9*Sqrt[a]*e)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqr
t[b]*x^2)^2]*EllipticF[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(12*a^(9/4)*Sqrt[a +
b*x^4])

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Rubi in Sympy [A]  time = 15.818, size = 32, normalized size = 0.08 \[ \frac{x \left (\frac{c}{x^{4}} + \frac{d}{x^{3}} + \frac{e}{x^{2}} + \frac{f}{x}\right )}{2 a \sqrt{a + b x^{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

x*(c/x**4 + d/x**3 + e/x**2 + f/x)/(2*a*sqrt(a + b*x**4))

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Mathematica [C]  time = 0.719223, size = 267, normalized size = 0.69 \[ \frac{-\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \left (3 \sqrt{a} f x^3 \sqrt{a+b x^4} \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )+2 a c+3 a x (d+x (2 e-f x))+b x^4 \left (5 c+6 d x+9 e x^2\right )\right )-\sqrt{b} x^3 \sqrt{\frac{b x^4}{a}+1} \left (9 \sqrt{a} e-5 i \sqrt{b} c\right ) F\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )+9 \sqrt{a} \sqrt{b} e x^3 \sqrt{\frac{b x^4}{a}+1} E\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )}{6 a^2 x^3 \sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \sqrt{a+b x^4}} \]

Antiderivative was successfully verified.

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

[Out]

(-(Sqrt[(I*Sqrt[b])/Sqrt[a]]*(2*a*c + b*x^4*(5*c + 6*d*x + 9*e*x^2) + 3*a*x*(d +
x*(2*e - f*x)) + 3*Sqrt[a]*f*x^3*Sqrt[a + b*x^4]*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a
]])) + 9*Sqrt[a]*Sqrt[b]*e*x^3*Sqrt[1 + (b*x^4)/a]*EllipticE[I*ArcSinh[Sqrt[(I*S
qrt[b])/Sqrt[a]]*x], -1] - Sqrt[b]*((-5*I)*Sqrt[b]*c + 9*Sqrt[a]*e)*x^3*Sqrt[1 +
(b*x^4)/a]*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[b])/Sqrt[a]]*x], -1])/(6*a^2*Sqrt[(
I*Sqrt[b])/Sqrt[a]]*x^3*Sqrt[a + b*x^4])

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Maple [C]  time = 0.014, size = 383, normalized size = 1. \[ -{\frac{bcx}{2\,{a}^{2}}{\frac{1}{\sqrt{ \left ({x}^{4}+{\frac{a}{b}} \right ) b}}}}-{\frac{c}{3\,{a}^{2}{x}^{3}}\sqrt{b{x}^{4}+a}}-{\frac{5\,bc}{6\,{a}^{2}}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}-{\frac{d \left ( 2\,b{x}^{4}+a \right ) }{2\,{x}^{2}{a}^{2}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}-{\frac{{x}^{3}be}{2\,{a}^{2}}{\frac{1}{\sqrt{ \left ({x}^{4}+{\frac{a}{b}} \right ) b}}}}-{\frac{e}{{a}^{2}x}\sqrt{b{x}^{4}+a}}+{{\frac{3\,i}{2}}e\sqrt{b}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){a}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}-{{\frac{3\,i}{2}}e\sqrt{b}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticE} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){a}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}+{\frac{f}{2\,a}{\frac{1}{\sqrt{b{x}^{4}+a}}}}-{\frac{f}{2}\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{4}+a} \right ) } \right ){a}^{-{\frac{3}{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-1/2*c*b/a^2*x/((x^4+a/b)*b)^(1/2)-1/3*c*(b*x^4+a)^(1/2)/a^2/x^3-5/6*c*b/a^2/(I/
a^(1/2)*b^(1/2))^(1/2)*(1-I/a^(1/2)*b^(1/2)*x^2)^(1/2)*(1+I/a^(1/2)*b^(1/2)*x^2)
^(1/2)/(b*x^4+a)^(1/2)*EllipticF(x*(I/a^(1/2)*b^(1/2))^(1/2),I)-1/2*d/x^2*(2*b*x
^4+a)/(b*x^4+a)^(1/2)/a^2-1/2*e*b/a^2*x^3/((x^4+a/b)*b)^(1/2)-e*(b*x^4+a)^(1/2)/
a^2/x+3/2*I*e*b^(1/2)/a^(3/2)/(I/a^(1/2)*b^(1/2))^(1/2)*(1-I/a^(1/2)*b^(1/2)*x^2
)^(1/2)*(1+I/a^(1/2)*b^(1/2)*x^2)^(1/2)/(b*x^4+a)^(1/2)*EllipticF(x*(I/a^(1/2)*b
^(1/2))^(1/2),I)-3/2*I*e*b^(1/2)/a^(3/2)/(I/a^(1/2)*b^(1/2))^(1/2)*(1-I/a^(1/2)*
b^(1/2)*x^2)^(1/2)*(1+I/a^(1/2)*b^(1/2)*x^2)^(1/2)/(b*x^4+a)^(1/2)*EllipticE(x*(
I/a^(1/2)*b^(1/2))^(1/2),I)+1/2*f/a/(b*x^4+a)^(1/2)-1/2*f/a^(3/2)*ln((2*a+2*a^(1
/2)*(b*x^4+a)^(1/2))/x^2)

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

integrate((f*x^3 + e*x^2 + d*x + c)/((b*x^4 + a)^(3/2)*x^4), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{f x^{3} + e x^{2} + d x + c}{{\left (b x^{8} + a x^{4}\right )} \sqrt{b x^{4} + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

integral((f*x^3 + e*x^2 + d*x + c)/((b*x^8 + a*x^4)*sqrt(b*x^4 + a)), x)

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Sympy [A]  time = 114.883, size = 321, normalized size = 0.83 \[ d \left (- \frac{1}{2 a \sqrt{b} x^{4} \sqrt{\frac{a}{b x^{4}} + 1}} - \frac{\sqrt{b}}{a^{2} \sqrt{\frac{a}{b x^{4}} + 1}}\right ) + f \left (\frac{2 a^{3} \sqrt{1 + \frac{b x^{4}}{a}}}{4 a^{\frac{9}{2}} + 4 a^{\frac{7}{2}} b x^{4}} + \frac{a^{3} \log{\left (\frac{b x^{4}}{a} \right )}}{4 a^{\frac{9}{2}} + 4 a^{\frac{7}{2}} b x^{4}} - \frac{2 a^{3} \log{\left (\sqrt{1 + \frac{b x^{4}}{a}} + 1 \right )}}{4 a^{\frac{9}{2}} + 4 a^{\frac{7}{2}} b x^{4}} + \frac{a^{2} b x^{4} \log{\left (\frac{b x^{4}}{a} \right )}}{4 a^{\frac{9}{2}} + 4 a^{\frac{7}{2}} b x^{4}} - \frac{2 a^{2} b x^{4} \log{\left (\sqrt{1 + \frac{b x^{4}}{a}} + 1 \right )}}{4 a^{\frac{9}{2}} + 4 a^{\frac{7}{2}} b x^{4}}\right ) + \frac{c \Gamma \left (- \frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{4}, \frac{3}{2} \\ \frac{1}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac{3}{2}} x^{3} \Gamma \left (\frac{1}{4}\right )} + \frac{e \Gamma \left (- \frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{4}, \frac{3}{2} \\ \frac{3}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac{3}{2}} x \Gamma \left (\frac{3}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

d*(-1/(2*a*sqrt(b)*x**4*sqrt(a/(b*x**4) + 1)) - sqrt(b)/(a**2*sqrt(a/(b*x**4) +
1))) + f*(2*a**3*sqrt(1 + b*x**4/a)/(4*a**(9/2) + 4*a**(7/2)*b*x**4) + a**3*log(
b*x**4/a)/(4*a**(9/2) + 4*a**(7/2)*b*x**4) - 2*a**3*log(sqrt(1 + b*x**4/a) + 1)/
(4*a**(9/2) + 4*a**(7/2)*b*x**4) + a**2*b*x**4*log(b*x**4/a)/(4*a**(9/2) + 4*a**
(7/2)*b*x**4) - 2*a**2*b*x**4*log(sqrt(1 + b*x**4/a) + 1)/(4*a**(9/2) + 4*a**(7/
2)*b*x**4)) + c*gamma(-3/4)*hyper((-3/4, 3/2), (1/4,), b*x**4*exp_polar(I*pi)/a)
/(4*a**(3/2)*x**3*gamma(1/4)) + e*gamma(-1/4)*hyper((-1/4, 3/2), (3/4,), b*x**4*
exp_polar(I*pi)/a)/(4*a**(3/2)*x*gamma(3/4))

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

integrate((f*x^3 + e*x^2 + d*x + c)/((b*x^4 + a)^(3/2)*x^4), x)