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

Optimal. Leaf size=308 \[ \frac{\sqrt [4]{a} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (3 \sqrt{b} c-\sqrt{a} e\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{6 b^{5/4} \sqrt{a+b x^4}}-\frac{\sqrt [4]{a} c \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{b^{3/4} \sqrt{a+b x^4}}-\frac{a f \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{4 b^{3/2}}+\frac{c x \sqrt{a+b x^4}}{\sqrt{b} \left (\sqrt{a}+\sqrt{b} x^2\right )}+\frac{\sqrt{a+b x^4} \left (2 d+f x^2\right )}{4 b}+\frac{e x \sqrt{a+b x^4}}{3 b} \]

[Out]

(e*x*Sqrt[a + b*x^4])/(3*b) + (c*x*Sqrt[a + b*x^4])/(Sqrt[b]*(Sqrt[a] + Sqrt[b]*
x^2)) + ((2*d + f*x^2)*Sqrt[a + b*x^4])/(4*b) - (a*f*ArcTanh[(Sqrt[b]*x^2)/Sqrt[
a + b*x^4]])/(4*b^(3/2)) - (a^(1/4)*c*(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)*(3*Sqrt[b]*c - Sqrt[a]*e)*(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])/(6*b^(5/4)*Sqrt[a + b*x^4])

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Rubi [A]  time = 0.58183, antiderivative size = 308, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3 \[ \frac{\sqrt [4]{a} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (3 \sqrt{b} c-\sqrt{a} e\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{6 b^{5/4} \sqrt{a+b x^4}}-\frac{\sqrt [4]{a} c \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{b^{3/4} \sqrt{a+b x^4}}-\frac{a f \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{4 b^{3/2}}+\frac{c x \sqrt{a+b x^4}}{\sqrt{b} \left (\sqrt{a}+\sqrt{b} x^2\right )}+\frac{\sqrt{a+b x^4} \left (2 d+f x^2\right )}{4 b}+\frac{e x \sqrt{a+b x^4}}{3 b} \]

Antiderivative was successfully verified.

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

[Out]

(e*x*Sqrt[a + b*x^4])/(3*b) + (c*x*Sqrt[a + b*x^4])/(Sqrt[b]*(Sqrt[a] + Sqrt[b]*
x^2)) + ((2*d + f*x^2)*Sqrt[a + b*x^4])/(4*b) - (a*f*ArcTanh[(Sqrt[b]*x^2)/Sqrt[
a + b*x^4]])/(4*b^(3/2)) - (a^(1/4)*c*(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)*(3*Sqrt[b]*c - Sqrt[a]*e)*(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])/(6*b^(5/4)*Sqrt[a + b*x^4])

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Rubi in Sympy [A]  time = 61.0822, size = 275, normalized size = 0.89 \[ - \frac{\sqrt [4]{a} c \sqrt{\frac{a + b x^{4}}{\left (\sqrt{a} + \sqrt{b} x^{2}\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x^{2}\right ) E\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}} \right )}\middle | \frac{1}{2}\right )}{b^{\frac{3}{4}} \sqrt{a + b x^{4}}} - \frac{\sqrt [4]{a} \sqrt{\frac{a + b x^{4}}{\left (\sqrt{a} + \sqrt{b} x^{2}\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x^{2}\right ) \left (\sqrt{a} e - 3 \sqrt{b} c\right ) F\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}} \right )}\middle | \frac{1}{2}\right )}{6 b^{\frac{5}{4}} \sqrt{a + b x^{4}}} - \frac{a f \operatorname{atanh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a + b x^{4}}} \right )}}{4 b^{\frac{3}{2}}} + \frac{e x \sqrt{a + b x^{4}}}{3 b} + \frac{\sqrt{a + b x^{4}} \left (2 d + f x^{2}\right )}{4 b} + \frac{c x \sqrt{a + b x^{4}}}{\sqrt{b} \left (\sqrt{a} + \sqrt{b} x^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-a**(1/4)*c*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)) - a
**(1/4)*sqrt((a + b*x**4)/(sqrt(a) + sqrt(b)*x**2)**2)*(sqrt(a) + sqrt(b)*x**2)*
(sqrt(a)*e - 3*sqrt(b)*c)*elliptic_f(2*atan(b**(1/4)*x/a**(1/4)), 1/2)/(6*b**(5/
4)*sqrt(a + b*x**4)) - a*f*atanh(sqrt(b)*x**2/sqrt(a + b*x**4))/(4*b**(3/2)) + e
*x*sqrt(a + b*x**4)/(3*b) + sqrt(a + b*x**4)*(2*d + f*x**2)/(4*b) + c*x*sqrt(a +
b*x**4)/(sqrt(b)*(sqrt(a) + sqrt(b)*x**2))

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Mathematica [C]  time = 0.658796, size = 245, normalized size = 0.8 \[ \frac{4 i \sqrt{a} \sqrt{b} \sqrt{\frac{b x^4}{a}+1} \left (\sqrt{a} e+3 i \sqrt{b} c\right ) F\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )+12 \sqrt{a} b c \sqrt{\frac{b x^4}{a}+1} E\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )+\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \left (\sqrt{b} \left (a+b x^4\right ) \left (6 d+4 e x+3 f x^2\right )-3 a f \sqrt{a+b x^4} \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )\right )}{12 b^{3/2} \sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \sqrt{a+b x^4}} \]

Antiderivative was successfully verified.

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

[Out]

(Sqrt[(I*Sqrt[b])/Sqrt[a]]*(Sqrt[b]*(6*d + 4*e*x + 3*f*x^2)*(a + b*x^4) - 3*a*f*
Sqrt[a + b*x^4]*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]]) + 12*Sqrt[a]*b*c*Sqrt[1
+ (b*x^4)/a]*EllipticE[I*ArcSinh[Sqrt[(I*Sqrt[b])/Sqrt[a]]*x], -1] + (4*I)*Sqrt[
a]*Sqrt[b]*((3*I)*Sqrt[b]*c + Sqrt[a]*e)*Sqrt[1 + (b*x^4)/a]*EllipticF[I*ArcSinh
[Sqrt[(I*Sqrt[b])/Sqrt[a]]*x], -1])/(12*Sqrt[(I*Sqrt[b])/Sqrt[a]]*b^(3/2)*Sqrt[a
+ b*x^4])

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Maple [C]  time = 0.01, size = 248, normalized size = 0.8 \[{\frac{d}{2\,b}\sqrt{b{x}^{4}+a}}+{ic\sqrt{a}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}} \left ({\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ) -{\it EllipticE} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ) \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}{\frac{1}{\sqrt{b}}}}+{\frac{ex}{3\,b}\sqrt{b{x}^{4}+a}}-{\frac{ae}{3\,b}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}+{\frac{f{x}^{2}}{4\,b}\sqrt{b{x}^{4}+a}}-{\frac{af}{4}\ln \left ( \sqrt{b}{x}^{2}+\sqrt{b{x}^{4}+a} \right ){b}^{-{\frac{3}{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{f x^{5} + e x^{4} + d x^{3} + c x^{2}}{\sqrt{b x^{4} + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

integral((f*x^5 + e*x^4 + d*x^3 + c*x^2)/sqrt(b*x^4 + a), x)

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Sympy [A]  time = 6.53185, size = 156, normalized size = 0.51 \[ \frac{\sqrt{a} f x^{2} \sqrt{1 + \frac{b x^{4}}{a}}}{4 b} - \frac{a f \operatorname{asinh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a}} \right )}}{4 b^{\frac{3}{2}}} + d \left (\begin{cases} \frac{x^{4}}{4 \sqrt{a}} & \text{for}\: b = 0 \\\frac{\sqrt{a + b x^{4}}}{2 b} & \text{otherwise} \end{cases}\right ) + \frac{c x^{3} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt{a} \Gamma \left (\frac{7}{4}\right )} + \frac{e x^{5} \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{5}{4} \\ \frac{9}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt{a} \Gamma \left (\frac{9}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

sqrt(a)*f*x**2*sqrt(1 + b*x**4/a)/(4*b) - a*f*asinh(sqrt(b)*x**2/sqrt(a))/(4*b**
(3/2)) + d*Piecewise((x**4/(4*sqrt(a)), Eq(b, 0)), (sqrt(a + b*x**4)/(2*b), True
)) + c*x**3*gamma(3/4)*hyper((1/2, 3/4), (7/4,), b*x**4*exp_polar(I*pi)/a)/(4*sq
rt(a)*gamma(7/4)) + e*x**5*gamma(5/4)*hyper((1/2, 5/4), (9/4,), b*x**4*exp_polar
(I*pi)/a)/(4*sqrt(a)*gamma(9/4))

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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