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

Optimal. Leaf size=361 \[ -\frac{a^{3/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (9 \sqrt{a} e+5 \sqrt{b} c\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{30 b^{7/4} \sqrt{a+b x^4}}+\frac{3 a^{5/4} e \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 b^{7/4} \sqrt{a+b x^4}}-\frac{a d \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{4 b^{3/2}}-\frac{3 a e x \sqrt{a+b x^4}}{5 b^{3/2} \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{\sqrt{a+b x^4} \left (4 a f-3 b d x^2\right )}{12 b^2}+\frac{c x \sqrt{a+b x^4}}{3 b}+\frac{e x^3 \sqrt{a+b x^4}}{5 b}+\frac{f x^4 \sqrt{a+b x^4}}{6 b} \]

[Out]

(c*x*Sqrt[a + b*x^4])/(3*b) + (e*x^3*Sqrt[a + b*x^4])/(5*b) + (f*x^4*Sqrt[a + b*
x^4])/(6*b) - (3*a*e*x*Sqrt[a + b*x^4])/(5*b^(3/2)*(Sqrt[a] + Sqrt[b]*x^2)) - ((
4*a*f - 3*b*d*x^2)*Sqrt[a + b*x^4])/(12*b^2) - (a*d*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a
+ b*x^4]])/(4*b^(3/2)) + (3*a^(5/4)*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])/(5*b^(
7/4)*Sqrt[a + b*x^4]) - (a^(3/4)*(5*Sqrt[b]*c + 9*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])/(30*b^(7/4)*Sqrt[a + b*x^4])

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Rubi [A]  time = 0.810429, antiderivative size = 361, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 10, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333 \[ -\frac{a^{3/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (9 \sqrt{a} e+5 \sqrt{b} c\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{30 b^{7/4} \sqrt{a+b x^4}}+\frac{3 a^{5/4} e \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 b^{7/4} \sqrt{a+b x^4}}-\frac{a d \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{4 b^{3/2}}-\frac{3 a e x \sqrt{a+b x^4}}{5 b^{3/2} \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{\sqrt{a+b x^4} \left (4 a f-3 b d x^2\right )}{12 b^2}+\frac{c x \sqrt{a+b x^4}}{3 b}+\frac{e x^3 \sqrt{a+b x^4}}{5 b}+\frac{f x^4 \sqrt{a+b x^4}}{6 b} \]

Antiderivative was successfully verified.

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

[Out]

(c*x*Sqrt[a + b*x^4])/(3*b) + (e*x^3*Sqrt[a + b*x^4])/(5*b) + (f*x^4*Sqrt[a + b*
x^4])/(6*b) - (3*a*e*x*Sqrt[a + b*x^4])/(5*b^(3/2)*(Sqrt[a] + Sqrt[b]*x^2)) - ((
4*a*f - 3*b*d*x^2)*Sqrt[a + b*x^4])/(12*b^2) - (a*d*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a
+ b*x^4]])/(4*b^(3/2)) + (3*a^(5/4)*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])/(5*b^(
7/4)*Sqrt[a + b*x^4]) - (a^(3/4)*(5*Sqrt[b]*c + 9*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])/(30*b^(7/4)*Sqrt[a + b*x^4])

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Rubi in Sympy [A]  time = 85.7179, size = 330, normalized size = 0.91 \[ \frac{3 a^{\frac{5}{4}} e \sqrt{\frac{a + b x^{4}}{\left (\sqrt{a} + \sqrt{b} x^{2}\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x^{2}\right ) E\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}} \right )}\middle | \frac{1}{2}\right )}{5 b^{\frac{7}{4}} \sqrt{a + b x^{4}}} - \frac{a^{\frac{3}{4}} \sqrt{\frac{a + b x^{4}}{\left (\sqrt{a} + \sqrt{b} x^{2}\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x^{2}\right ) \left (9 \sqrt{a} e + 5 \sqrt{b} c\right ) F\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}} \right )}\middle | \frac{1}{2}\right )}{30 b^{\frac{7}{4}} \sqrt{a + b x^{4}}} - \frac{a d \operatorname{atanh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a + b x^{4}}} \right )}}{4 b^{\frac{3}{2}}} - \frac{3 a e x \sqrt{a + b x^{4}}}{5 b^{\frac{3}{2}} \left (\sqrt{a} + \sqrt{b} x^{2}\right )} + \frac{c x \sqrt{a + b x^{4}}}{3 b} + \frac{e x^{3} \sqrt{a + b x^{4}}}{5 b} + \frac{f x^{4} \sqrt{a + b x^{4}}}{6 b} - \frac{\sqrt{a + b x^{4}} \left (4 a f - 3 b d x^{2}\right )}{12 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

3*a**(5/4)*e*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**(7/4)*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)*(9*sqrt(a)*e + 5*sqrt(b)*c)*elliptic_f(2*atan(b**(1/4)*x/a**(1/4)), 1/2)/(30*
b**(7/4)*sqrt(a + b*x**4)) - a*d*atanh(sqrt(b)*x**2/sqrt(a + b*x**4))/(4*b**(3/2
)) - 3*a*e*x*sqrt(a + b*x**4)/(5*b**(3/2)*(sqrt(a) + sqrt(b)*x**2)) + c*x*sqrt(a
+ b*x**4)/(3*b) + e*x**3*sqrt(a + b*x**4)/(5*b) + f*x**4*sqrt(a + b*x**4)/(6*b)
- sqrt(a + b*x**4)*(4*a*f - 3*b*d*x**2)/(12*b**2)

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Mathematica [C]  time = 1.01655, size = 259, normalized size = 0.72 \[ \frac{-36 a^{3/2} \sqrt{b} e \sqrt{\frac{b x^4}{a}+1} E\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )+\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \left (-\left (a+b x^4\right ) (20 a f-b x (20 c+x (15 d+2 x (6 e+5 f x))))-15 a \sqrt{b} d \sqrt{a+b x^4} \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )\right )+4 a \sqrt{b} \sqrt{\frac{b x^4}{a}+1} \left (9 \sqrt{a} e+5 i \sqrt{b} c\right ) F\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )}{60 b^2 \sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \sqrt{a+b x^4}} \]

Antiderivative was successfully verified.

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

[Out]

(Sqrt[(I*Sqrt[b])/Sqrt[a]]*(-((a + b*x^4)*(20*a*f - b*x*(20*c + x*(15*d + 2*x*(6
*e + 5*f*x))))) - 15*a*Sqrt[b]*d*Sqrt[a + b*x^4]*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a +
b*x^4]]) - 36*a^(3/2)*Sqrt[b]*e*Sqrt[1 + (b*x^4)/a]*EllipticE[I*ArcSinh[Sqrt[(I*
Sqrt[b])/Sqrt[a]]*x], -1] + 4*a*Sqrt[b]*((5*I)*Sqrt[b]*c + 9*Sqrt[a]*e)*Sqrt[1 +
(b*x^4)/a]*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[b])/Sqrt[a]]*x], -1])/(60*Sqrt[(I*S
qrt[b])/Sqrt[a]]*b^2*Sqrt[a + b*x^4])

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Maple [C]  time = 0.022, size = 335, normalized size = 0.9 \[{\frac{cx}{3\,b}\sqrt{b{x}^{4}+a}}-{\frac{ac}{3\,b}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}+{\frac{d{x}^{2}}{4\,b}\sqrt{b{x}^{4}+a}}-{\frac{ad}{4}\ln \left ( \sqrt{b}{x}^{2}+\sqrt{b{x}^{4}+a} \right ){b}^{-{\frac{3}{2}}}}+{\frac{e{x}^{3}}{5\,b}\sqrt{b{x}^{4}+a}}-{{\frac{3\,i}{5}}e{a}^{{\frac{3}{2}}}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){b}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}+{{\frac{3\,i}{5}}e{a}^{{\frac{3}{2}}}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticE} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){b}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}-{\frac{f \left ( -b{x}^{4}+2\,a \right ) }{6\,{b}^{2}}\sqrt{b{x}^{4}+a}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{f x^{7} + e x^{6} + d x^{5} + c x^{4}}{\sqrt{b x^{4} + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

integral((f*x^7 + e*x^6 + d*x^5 + c*x^4)/sqrt(b*x^4 + a), x)

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Sympy [A]  time = 8.13287, size = 177, normalized size = 0.49 \[ \frac{\sqrt{a} d x^{2} \sqrt{1 + \frac{b x^{4}}{a}}}{4 b} - \frac{a d \operatorname{asinh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a}} \right )}}{4 b^{\frac{3}{2}}} + f \left (\begin{cases} - \frac{a \sqrt{a + b x^{4}}}{3 b^{2}} + \frac{x^{4} \sqrt{a + b x^{4}}}{6 b} & \text{for}\: b \neq 0 \\\frac{x^{8}}{8 \sqrt{a}} & \text{otherwise} \end{cases}\right ) + \frac{c x^{5} \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{5}{4} \\ \frac{9}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt{a} \Gamma \left (\frac{9}{4}\right )} + \frac{e x^{7} \Gamma \left (\frac{7}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{7}{4} \\ \frac{11}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt{a} \Gamma \left (\frac{11}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

sqrt(a)*d*x**2*sqrt(1 + b*x**4/a)/(4*b) - a*d*asinh(sqrt(b)*x**2/sqrt(a))/(4*b**
(3/2)) + f*Piecewise((-a*sqrt(a + b*x**4)/(3*b**2) + x**4*sqrt(a + b*x**4)/(6*b)
, Ne(b, 0)), (x**8/(8*sqrt(a)), True)) + c*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)) + e*x**7*gamma(7/4)*hyper(
(1/2, 7/4), (11/4,), b*x**4*exp_polar(I*pi)/a)/(4*sqrt(a)*gamma(11/4))

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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