3.522 \(\int \frac{c+d x+e x^2+f x^3}{x \sqrt{a+b x^4}} \, dx\)

Optimal. Leaf size=285 \[ \frac{\sqrt [4]{a} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (\frac{\sqrt{b} d}{\sqrt{a}}+f\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 b^{3/4} \sqrt{a+b x^4}}-\frac{\sqrt [4]{a} f \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{b^{3/4} \sqrt{a+b x^4}}-\frac{c \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{2 \sqrt{a}}+\frac{e \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{2 \sqrt{b}}+\frac{f x \sqrt{a+b x^4}}{\sqrt{b} \left (\sqrt{a}+\sqrt{b} x^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.376243, antiderivative size = 285, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 11, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.367 \[ \frac{\sqrt [4]{a} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (\frac{\sqrt{b} d}{\sqrt{a}}+f\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 b^{3/4} \sqrt{a+b x^4}}-\frac{\sqrt [4]{a} f \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{b^{3/4} \sqrt{a+b x^4}}-\frac{c \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{2 \sqrt{a}}+\frac{e \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{2 \sqrt{b}}+\frac{f x \sqrt{a+b x^4}}{\sqrt{b} \left (\sqrt{a}+\sqrt{b} x^2\right )} \]

Antiderivative was successfully verified.

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

[Out]

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

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Rubi in Sympy [A]  time = 50.6012, size = 258, normalized size = 0.91 \[ - \frac{\sqrt [4]{a} f \sqrt{\frac{a + b x^{4}}{\left (\sqrt{a} + \sqrt{b} x^{2}\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x^{2}\right ) E\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}} \right )}\middle | \frac{1}{2}\right )}{b^{\frac{3}{4}} \sqrt{a + b x^{4}}} + \frac{e \operatorname{atanh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a + b x^{4}}} \right )}}{2 \sqrt{b}} + \frac{f x \sqrt{a + b x^{4}}}{\sqrt{b} \left (\sqrt{a} + \sqrt{b} x^{2}\right )} - \frac{c \operatorname{atanh}{\left (\frac{\sqrt{a + b x^{4}}}{\sqrt{a}} \right )}}{2 \sqrt{a}} + \frac{\sqrt{\frac{a + b x^{4}}{\left (\sqrt{a} + \sqrt{b} x^{2}\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x^{2}\right ) \left (\sqrt{a} f + \sqrt{b} d\right ) F\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}} \right )}\middle | \frac{1}{2}\right )}{2 \sqrt [4]{a} b^{\frac{3}{4}} \sqrt{a + b x^{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((f*x**3+e*x**2+d*x+c)/x/(b*x**4+a)**(1/2),x)

[Out]

-a**(1/4)*f*sqrt((a + b*x**4)/(sqrt(a) + sqrt(b)*x**2)**2)*(sqrt(a) + sqrt(b)*x*
*2)*elliptic_e(2*atan(b**(1/4)*x/a**(1/4)), 1/2)/(b**(3/4)*sqrt(a + b*x**4)) + e
*atanh(sqrt(b)*x**2/sqrt(a + b*x**4))/(2*sqrt(b)) + f*x*sqrt(a + b*x**4)/(sqrt(b
)*(sqrt(a) + sqrt(b)*x**2)) - c*atanh(sqrt(a + b*x**4)/sqrt(a))/(2*sqrt(a)) + sq
rt((a + b*x**4)/(sqrt(a) + sqrt(b)*x**2)**2)*(sqrt(a) + sqrt(b)*x**2)*(sqrt(a)*f
+ sqrt(b)*d)*elliptic_f(2*atan(b**(1/4)*x/a**(1/4)), 1/2)/(2*a**(1/4)*b**(3/4)*
sqrt(a + b*x**4))

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Mathematica [C]  time = 0.918944, size = 235, normalized size = 0.82 \[ -\frac{i \sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \sqrt{a+b x^4} \left (\sqrt{a} e \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )-\sqrt{b} c \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )\right )-2 \sqrt{a} \sqrt{\frac{b x^4}{a}+1} \left (\sqrt{a} f+i \sqrt{b} d\right ) F\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )+2 a f \sqrt{\frac{b x^4}{a}+1} E\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )\right )}{2 b \sqrt{a+b x^4}} \]

Antiderivative was successfully verified.

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

[Out]

((-I/2)*Sqrt[(I*Sqrt[b])/Sqrt[a]]*(Sqrt[(I*Sqrt[b])/Sqrt[a]]*Sqrt[a + b*x^4]*(Sq
rt[a]*e*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]] - Sqrt[b]*c*ArcTanh[Sqrt[a + b*x^
4]/Sqrt[a]]) + 2*a*f*Sqrt[1 + (b*x^4)/a]*EllipticE[I*ArcSinh[Sqrt[(I*Sqrt[b])/Sq
rt[a]]*x], -1] - 2*Sqrt[a]*(I*Sqrt[b]*d + Sqrt[a]*f)*Sqrt[1 + (b*x^4)/a]*Ellipti
cF[I*ArcSinh[Sqrt[(I*Sqrt[b])/Sqrt[a]]*x], -1]))/(b*Sqrt[a + b*x^4])

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Maple [C]  time = 0.01, size = 222, normalized size = 0.8 \[{d\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}+{if\sqrt{a}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}} \left ({\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ) -{\it EllipticE} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ) \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}{\frac{1}{\sqrt{b}}}}-{\frac{c}{2}\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{4}+a} \right ) } \right ){\frac{1}{\sqrt{a}}}}+{\frac{e}{2}\ln \left ( \sqrt{b}{x}^{2}+\sqrt{b{x}^{4}+a} \right ){\frac{1}{\sqrt{b}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((f*x^3+e*x^2+d*x+c)/x/(b*x^4+a)^(1/2),x)

[Out]

d/(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)+I*f*a^(1/2)
/(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)/b^(1/2)*(EllipticF(x*(I/a^(1/2)*b^(1/2))^(1/2),I)-Ell
ipticE(x*(I/a^(1/2)*b^(1/2))^(1/2),I))-1/2*c/a^(1/2)*ln((2*a+2*a^(1/2)*(b*x^4+a)
^(1/2))/x^2)+1/2*e*ln(b^(1/2)*x^2+(b*x^4+a)^(1/2))/b^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Sympy [A]  time = 4.56604, size = 126, normalized size = 0.44 \[ \frac{e \operatorname{asinh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a}} \right )}}{2 \sqrt{b}} - \frac{c \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x^{2}} \right )}}{2 \sqrt{a}} + \frac{d x \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{4}, \frac{1}{2} \\ \frac{5}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt{a} \Gamma \left (\frac{5}{4}\right )} + \frac{f x^{3} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt{a} \Gamma \left (\frac{7}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((f*x**3+e*x**2+d*x+c)/x/(b*x**4+a)**(1/2),x)

[Out]

e*asinh(sqrt(b)*x**2/sqrt(a))/(2*sqrt(b)) - c*asinh(sqrt(a)/(sqrt(b)*x**2))/(2*s
qrt(a)) + d*x*gamma(1/4)*hyper((1/4, 1/2), (5/4,), b*x**4*exp_polar(I*pi)/a)/(4*
sqrt(a)*gamma(5/4)) + f*x**3*gamma(3/4)*hyper((1/2, 3/4), (7/4,), b*x**4*exp_pol
ar(I*pi)/a)/(4*sqrt(a)*gamma(7/4))

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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