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

Optimal. Leaf size=345 \[ \frac{a^{3/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (3 \sqrt{a} f+5 \sqrt{b} d\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 b^{3/4} \sqrt{a+b x^4}}-\frac{2 a^{5/4} f \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 b^{3/4} \sqrt{a+b x^4}}+\frac{1}{4} \sqrt{a+b x^4} \left (2 c+e x^2\right )-\frac{1}{2} \sqrt{a} c \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )+\frac{1}{15} x \sqrt{a+b x^4} \left (5 d+3 f x^2\right )+\frac{a e \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{4 \sqrt{b}}+\frac{2 a f x \sqrt{a+b x^4}}{5 \sqrt{b} \left (\sqrt{a}+\sqrt{b} x^2\right )} \]

[Out]

(2*a*f*x*Sqrt[a + b*x^4])/(5*Sqrt[b]*(Sqrt[a] + Sqrt[b]*x^2)) + ((2*c + e*x^2)*S
qrt[a + b*x^4])/4 + (x*(5*d + 3*f*x^2)*Sqrt[a + b*x^4])/15 + (a*e*ArcTanh[(Sqrt[
b]*x^2)/Sqrt[a + b*x^4]])/(4*Sqrt[b]) - (Sqrt[a]*c*ArcTanh[Sqrt[a + b*x^4]/Sqrt[
a]])/2 - (2*a^(5/4)*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])/(5*b^(3/4)*Sqrt[a + b*x
^4]) + (a^(3/4)*(5*Sqrt[b]*d + 3*Sqrt[a]*f)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*
x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticF[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(
15*b^(3/4)*Sqrt[a + b*x^4])

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Rubi [A]  time = 0.6867, antiderivative size = 345, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 13, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.433 \[ \frac{a^{3/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (3 \sqrt{a} f+5 \sqrt{b} d\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 b^{3/4} \sqrt{a+b x^4}}-\frac{2 a^{5/4} f \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 b^{3/4} \sqrt{a+b x^4}}+\frac{1}{4} \sqrt{a+b x^4} \left (2 c+e x^2\right )-\frac{1}{2} \sqrt{a} c \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )+\frac{1}{15} x \sqrt{a+b x^4} \left (5 d+3 f x^2\right )+\frac{a e \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{4 \sqrt{b}}+\frac{2 a f x \sqrt{a+b x^4}}{5 \sqrt{b} \left (\sqrt{a}+\sqrt{b} x^2\right )} \]

Antiderivative was successfully verified.

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

[Out]

(2*a*f*x*Sqrt[a + b*x^4])/(5*Sqrt[b]*(Sqrt[a] + Sqrt[b]*x^2)) + ((2*c + e*x^2)*S
qrt[a + b*x^4])/4 + (x*(5*d + 3*f*x^2)*Sqrt[a + b*x^4])/15 + (a*e*ArcTanh[(Sqrt[
b]*x^2)/Sqrt[a + b*x^4]])/(4*Sqrt[b]) - (Sqrt[a]*c*ArcTanh[Sqrt[a + b*x^4]/Sqrt[
a]])/2 - (2*a^(5/4)*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])/(5*b^(3/4)*Sqrt[a + b*x
^4]) + (a^(3/4)*(5*Sqrt[b]*d + 3*Sqrt[a]*f)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*
x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticF[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(
15*b^(3/4)*Sqrt[a + b*x^4])

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Rubi in Sympy [A]  time = 69.0769, size = 316, normalized size = 0.92 \[ - \frac{2 a^{\frac{5}{4}} f \sqrt{\frac{a + b x^{4}}{\left (\sqrt{a} + \sqrt{b} x^{2}\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x^{2}\right ) E\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}} \right )}\middle | \frac{1}{2}\right )}{5 b^{\frac{3}{4}} \sqrt{a + b x^{4}}} + \frac{a^{\frac{3}{4}} \sqrt{\frac{a + b x^{4}}{\left (\sqrt{a} + \sqrt{b} x^{2}\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x^{2}\right ) \left (3 \sqrt{a} f + 5 \sqrt{b} d\right ) F\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}} \right )}\middle | \frac{1}{2}\right )}{15 b^{\frac{3}{4}} \sqrt{a + b x^{4}}} - \frac{\sqrt{a} c \operatorname{atanh}{\left (\frac{\sqrt{a + b x^{4}}}{\sqrt{a}} \right )}}{2} + \frac{a e \operatorname{atanh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a + b x^{4}}} \right )}}{4 \sqrt{b}} + \frac{2 a f x \sqrt{a + b x^{4}}}{5 \sqrt{b} \left (\sqrt{a} + \sqrt{b} x^{2}\right )} + \frac{x \sqrt{a + b x^{4}} \left (5 d + 3 f x^{2}\right )}{15} + \frac{\sqrt{a + b x^{4}} \left (2 c + e x^{2}\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-2*a**(5/4)*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)/(5*b**(3/4)*sqrt(a + b*x**4))
+ a**(3/4)*sqrt((a + b*x**4)/(sqrt(a) + sqrt(b)*x**2)**2)*(sqrt(a) + sqrt(b)*x*
*2)*(3*sqrt(a)*f + 5*sqrt(b)*d)*elliptic_f(2*atan(b**(1/4)*x/a**(1/4)), 1/2)/(15
*b**(3/4)*sqrt(a + b*x**4)) - sqrt(a)*c*atanh(sqrt(a + b*x**4)/sqrt(a))/2 + a*e*
atanh(sqrt(b)*x**2/sqrt(a + b*x**4))/(4*sqrt(b)) + 2*a*f*x*sqrt(a + b*x**4)/(5*s
qrt(b)*(sqrt(a) + sqrt(b)*x**2)) + x*sqrt(a + b*x**4)*(5*d + 3*f*x**2)/15 + sqrt
(a + b*x**4)*(2*c + e*x**2)/4

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Mathematica [C]  time = 2.35789, size = 280, normalized size = 0.81 \[ \frac{24 a^{3/2} f \sqrt{\frac{b x^4}{a}+1} E\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )+\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \left (\sqrt{b} \left (\left (a+b x^4\right ) (30 c+x (20 d+3 x (5 e+4 f x)))-30 \sqrt{a} c \sqrt{a+b x^4} \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )\right )+15 a e \sqrt{a+b x^4} \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )\right )-8 a \sqrt{\frac{b x^4}{a}+1} \left (3 \sqrt{a} f+5 i \sqrt{b} d\right ) F\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )}{60 \sqrt{b} \sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \sqrt{a+b x^4}} \]

Antiderivative was successfully verified.

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

[Out]

(Sqrt[(I*Sqrt[b])/Sqrt[a]]*(15*a*e*Sqrt[a + b*x^4]*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a
+ b*x^4]] + Sqrt[b]*((a + b*x^4)*(30*c + x*(20*d + 3*x*(5*e + 4*f*x))) - 30*Sqrt
[a]*c*Sqrt[a + b*x^4]*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]])) + 24*a^(3/2)*f*Sqrt[1 +
(b*x^4)/a]*EllipticE[I*ArcSinh[Sqrt[(I*Sqrt[b])/Sqrt[a]]*x], -1] - 8*a*((5*I)*S
qrt[b]*d + 3*Sqrt[a]*f)*Sqrt[1 + (b*x^4)/a]*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[b])
/Sqrt[a]]*x], -1])/(60*Sqrt[(I*Sqrt[b])/Sqrt[a]]*Sqrt[b]*Sqrt[a + b*x^4])

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Maple [C]  time = 0.019, size = 339, normalized size = 1. \[{\frac{dx}{3}\sqrt{b{x}^{4}+a}}+{\frac{2\,ad}{3}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}+{\frac{f{x}^{3}}{5}\sqrt{b{x}^{4}+a}}+{{\frac{2\,i}{5}}f{a}^{{\frac{3}{2}}}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}{\frac{1}{\sqrt{b}}}}-{{\frac{2\,i}{5}}f{a}^{{\frac{3}{2}}}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticE} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}{\frac{1}{\sqrt{b}}}}+{\frac{c}{2}\sqrt{b{x}^{4}+a}}-{\frac{c}{2}\sqrt{a}\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{4}+a} \right ) } \right ) }+{\frac{e{x}^{2}}{4}\sqrt{b{x}^{4}+a}}+{\frac{ae}{4}\ln \left ( \sqrt{b}{x}^{2}+\sqrt{b{x}^{4}+a} \right ){\frac{1}{\sqrt{b}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Sympy [A]  time = 7.37201, size = 204, normalized size = 0.59 \[ - \frac{\sqrt{a} c \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x^{2}} \right )}}{2} + \frac{\sqrt{a} d x \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{1}{4} \\ \frac{5}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac{5}{4}\right )} + \frac{\sqrt{a} e x^{2} \sqrt{1 + \frac{b x^{4}}{a}}}{4} + \frac{\sqrt{a} f x^{3} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac{7}{4}\right )} + \frac{a c}{2 \sqrt{b} x^{2} \sqrt{\frac{a}{b x^{4}} + 1}} + \frac{a e \operatorname{asinh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a}} \right )}}{4 \sqrt{b}} + \frac{\sqrt{b} c x^{2}}{2 \sqrt{\frac{a}{b x^{4}} + 1}} \]

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)**(1/2)/x,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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