3.493 \(\int \frac{\left (c+d x+e x^2+f x^3\right ) \sqrt{a+b x^4}}{x^6} \, dx\)

Optimal. Leaf size=360 \[ \frac{b^{3/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (5 \sqrt{a} e+3 \sqrt{b} c\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 a^{3/4} \sqrt{a+b x^4}}-\frac{2 b^{5/4} c \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 a^{3/4} \sqrt{a+b x^4}}+\frac{2 b^{3/2} c x \sqrt{a+b x^4}}{5 a \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{1}{60} \sqrt{a+b x^4} \left (\frac{12 c}{x^5}+\frac{15 d}{x^4}+\frac{20 e}{x^3}+\frac{30 f}{x^2}\right )-\frac{2 b c \sqrt{a+b x^4}}{5 a x}-\frac{b d \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{4 \sqrt{a}}+\frac{1}{2} \sqrt{b} f \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right ) \]

[Out]

-(((12*c)/x^5 + (15*d)/x^4 + (20*e)/x^3 + (30*f)/x^2)*Sqrt[a + b*x^4])/60 - (2*b
*c*Sqrt[a + b*x^4])/(5*a*x) + (2*b^(3/2)*c*x*Sqrt[a + b*x^4])/(5*a*(Sqrt[a] + Sq
rt[b]*x^2)) + (Sqrt[b]*f*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/2 - (b*d*ArcTan
h[Sqrt[a + b*x^4]/Sqrt[a]])/(4*Sqrt[a]) - (2*b^(5/4)*c*(Sqrt[a] + Sqrt[b]*x^2)*S
qrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticE[2*ArcTan[(b^(1/4)*x)/a^(1/4
)], 1/2])/(5*a^(3/4)*Sqrt[a + b*x^4]) + (b^(3/4)*(3*Sqrt[b]*c + 5*Sqrt[a]*e)*(Sq
rt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticF[2*Arc
Tan[(b^(1/4)*x)/a^(1/4)], 1/2])/(15*a^(3/4)*Sqrt[a + b*x^4])

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Rubi [A]  time = 0.818206, antiderivative size = 360, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 14, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.467 \[ \frac{b^{3/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (5 \sqrt{a} e+3 \sqrt{b} c\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 a^{3/4} \sqrt{a+b x^4}}-\frac{2 b^{5/4} c \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 a^{3/4} \sqrt{a+b x^4}}+\frac{2 b^{3/2} c x \sqrt{a+b x^4}}{5 a \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{1}{60} \sqrt{a+b x^4} \left (\frac{12 c}{x^5}+\frac{15 d}{x^4}+\frac{20 e}{x^3}+\frac{30 f}{x^2}\right )-\frac{2 b c \sqrt{a+b x^4}}{5 a x}-\frac{b d \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{4 \sqrt{a}}+\frac{1}{2} \sqrt{b} f \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right ) \]

Antiderivative was successfully verified.

[In]  Int[((c + d*x + e*x^2 + f*x^3)*Sqrt[a + b*x^4])/x^6,x]

[Out]

-(((12*c)/x^5 + (15*d)/x^4 + (20*e)/x^3 + (30*f)/x^2)*Sqrt[a + b*x^4])/60 - (2*b
*c*Sqrt[a + b*x^4])/(5*a*x) + (2*b^(3/2)*c*x*Sqrt[a + b*x^4])/(5*a*(Sqrt[a] + Sq
rt[b]*x^2)) + (Sqrt[b]*f*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/2 - (b*d*ArcTan
h[Sqrt[a + b*x^4]/Sqrt[a]])/(4*Sqrt[a]) - (2*b^(5/4)*c*(Sqrt[a] + Sqrt[b]*x^2)*S
qrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticE[2*ArcTan[(b^(1/4)*x)/a^(1/4
)], 1/2])/(5*a^(3/4)*Sqrt[a + b*x^4]) + (b^(3/4)*(3*Sqrt[b]*c + 5*Sqrt[a]*e)*(Sq
rt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticF[2*Arc
Tan[(b^(1/4)*x)/a^(1/4)], 1/2])/(15*a^(3/4)*Sqrt[a + b*x^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((f*x**3+e*x**2+d*x+c)*(b*x**4+a)**(1/2)/x**6,x)

[Out]

Timed out

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Mathematica [C]  time = 1.16428, size = 314, normalized size = 0.87 \[ \frac{24 \sqrt{a} b^{3/2} c x^5 \sqrt{\frac{b x^4}{a}+1} E\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )-\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \left (\left (a+b x^4\right ) \left (12 a c+5 a x \left (3 d+4 e x+6 f x^2\right )+24 b c x^4\right )+15 \sqrt{a} b d x^5 \sqrt{a+b x^4} \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )-30 a \sqrt{b} f x^5 \sqrt{a+b x^4} \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )\right )-8 i \sqrt{a} b x^5 \sqrt{\frac{b x^4}{a}+1} \left (5 \sqrt{a} e-3 i \sqrt{b} c\right ) F\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )}{60 a x^5 \sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \sqrt{a+b x^4}} \]

Antiderivative was successfully verified.

[In]  Integrate[((c + d*x + e*x^2 + f*x^3)*Sqrt[a + b*x^4])/x^6,x]

[Out]

(-(Sqrt[(I*Sqrt[b])/Sqrt[a]]*((a + b*x^4)*(12*a*c + 24*b*c*x^4 + 5*a*x*(3*d + 4*
e*x + 6*f*x^2)) - 30*a*Sqrt[b]*f*x^5*Sqrt[a + b*x^4]*ArcTanh[(Sqrt[b]*x^2)/Sqrt[
a + b*x^4]] + 15*Sqrt[a]*b*d*x^5*Sqrt[a + b*x^4]*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]
])) + 24*Sqrt[a]*b^(3/2)*c*x^5*Sqrt[1 + (b*x^4)/a]*EllipticE[I*ArcSinh[Sqrt[(I*S
qrt[b])/Sqrt[a]]*x], -1] - (8*I)*Sqrt[a]*b*((-3*I)*Sqrt[b]*c + 5*Sqrt[a]*e)*x^5*
Sqrt[1 + (b*x^4)/a]*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[b])/Sqrt[a]]*x], -1])/(60*a
*Sqrt[(I*Sqrt[b])/Sqrt[a]]*x^5*Sqrt[a + b*x^4])

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Maple [C]  time = 0.027, size = 404, normalized size = 1.1 \[ -{\frac{c}{5\,{x}^{5}}\sqrt{b{x}^{4}+a}}-{\frac{2\,bc}{5\,ax}\sqrt{b{x}^{4}+a}}+{{\frac{2\,i}{5}}c{b}^{{\frac{3}{2}}}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{a}}}{\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}-{{\frac{2\,i}{5}}c{b}^{{\frac{3}{2}}}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticE} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{a}}}{\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}-{\frac{d}{4\,a{x}^{4}} \left ( b{x}^{4}+a \right ) ^{{\frac{3}{2}}}}-{\frac{bd}{4}\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{4}+a} \right ) } \right ){\frac{1}{\sqrt{a}}}}+{\frac{bd}{4\,a}\sqrt{b{x}^{4}+a}}-{\frac{e}{3\,{x}^{3}}\sqrt{b{x}^{4}+a}}+{\frac{2\,be}{3}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}-{\frac{f}{2\,a{x}^{2}} \left ( b{x}^{4}+a \right ) ^{{\frac{3}{2}}}}+{\frac{bf{x}^{2}}{2\,a}\sqrt{b{x}^{4}+a}}+{\frac{f}{2}\sqrt{b}\ln \left ( \sqrt{b}{x}^{2}+\sqrt{b{x}^{4}+a} \right ) } \]

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^6,x)

[Out]

-1/5*c/x^5*(b*x^4+a)^(1/2)-2/5*b*c*(b*x^4+a)^(1/2)/a/x+2/5*I*c/a^(1/2)*b^(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)-2/5*I*c/a^(1/2
)*b^(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)-1/4
*d/a/x^4*(b*x^4+a)^(3/2)-1/4*d*b/a^(1/2)*ln((2*a+2*a^(1/2)*(b*x^4+a)^(1/2))/x^2)
+1/4*d*b/a*(b*x^4+a)^(1/2)-1/3*e/x^3*(b*x^4+a)^(1/2)+2/3*e*b/(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*f/a/x^2*(b*x^4+a)^(3/2)+1/2*
f*b/a*x^2*(b*x^4+a)^(1/2)+1/2*f*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. \[ \int \frac{\sqrt{b x^{4} + a}{\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{6}}\,{d x} \]

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^6,x, algorithm="maxima")

[Out]

integrate(sqrt(b*x^4 + a)*(f*x^3 + e*x^2 + d*x + c)/x^6, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{\sqrt{b x^{4} + a}{\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{6}}, x\right ) \]

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^6,x, algorithm="fricas")

[Out]

integral(sqrt(b*x^4 + a)*(f*x^3 + e*x^2 + d*x + c)/x^6, x)

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Sympy [A]  time = 8.72219, size = 216, normalized size = 0.6 \[ \frac{\sqrt{a} c \Gamma \left (- \frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{5}{4}, - \frac{1}{2} \\ - \frac{1}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{5} \Gamma \left (- \frac{1}{4}\right )} + \frac{\sqrt{a} e \Gamma \left (- \frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{4}, - \frac{1}{2} \\ \frac{1}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{3} \Gamma \left (\frac{1}{4}\right )} - \frac{\sqrt{a} f}{2 x^{2} \sqrt{1 + \frac{b x^{4}}{a}}} - \frac{\sqrt{b} d \sqrt{\frac{a}{b x^{4}} + 1}}{4 x^{2}} + \frac{\sqrt{b} f \operatorname{asinh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a}} \right )}}{2} - \frac{b d \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x^{2}} \right )}}{4 \sqrt{a}} - \frac{b f x^{2}}{2 \sqrt{a} \sqrt{1 + \frac{b x^{4}}{a}}} \]

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**6,x)

[Out]

sqrt(a)*c*gamma(-5/4)*hyper((-5/4, -1/2), (-1/4,), b*x**4*exp_polar(I*pi)/a)/(4*
x**5*gamma(-1/4)) + sqrt(a)*e*gamma(-3/4)*hyper((-3/4, -1/2), (1/4,), b*x**4*exp
_polar(I*pi)/a)/(4*x**3*gamma(1/4)) - sqrt(a)*f/(2*x**2*sqrt(1 + b*x**4/a)) - sq
rt(b)*d*sqrt(a/(b*x**4) + 1)/(4*x**2) + sqrt(b)*f*asinh(sqrt(b)*x**2/sqrt(a))/2
- b*d*asinh(sqrt(a)/(sqrt(b)*x**2))/(4*sqrt(a)) - b*f*x**2/(2*sqrt(a)*sqrt(1 + b
*x**4/a))

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{\sqrt{b x^{4} + a}{\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{6}}\,{d x} \]

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^6,x, algorithm="giac")

[Out]

integrate(sqrt(b*x^4 + a)*(f*x^3 + e*x^2 + d*x + c)/x^6, x)