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

Optimal. Leaf size=394 \[ -\frac{a^{7/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (7 \sqrt{a} f+5 \sqrt{b} d\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{105 b^{7/4} \sqrt{a+b x^4}}+\frac{2 a^{9/4} f \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 b^{7/4} \sqrt{a+b x^4}}-\frac{a^2 e \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{16 b^{3/2}}-\frac{2 a^2 f x \sqrt{a+b x^4}}{15 b^{3/2} \left (\sqrt{a}+\sqrt{b} x^2\right )}+\frac{\left (a+b x^4\right )^{3/2} \left (4 c+3 e x^2\right )}{24 b}+\frac{1}{63} x^5 \sqrt{a+b x^4} \left (9 d+7 f x^2\right )+\frac{2 a d x \sqrt{a+b x^4}}{21 b}-\frac{a e x^2 \sqrt{a+b x^4}}{16 b}+\frac{2 a f x^3 \sqrt{a+b x^4}}{45 b} \]

[Out]

(2*a*d*x*Sqrt[a + b*x^4])/(21*b) - (a*e*x^2*Sqrt[a + b*x^4])/(16*b) + (2*a*f*x^3
*Sqrt[a + b*x^4])/(45*b) - (2*a^2*f*x*Sqrt[a + b*x^4])/(15*b^(3/2)*(Sqrt[a] + Sq
rt[b]*x^2)) + (x^5*(9*d + 7*f*x^2)*Sqrt[a + b*x^4])/63 + ((4*c + 3*e*x^2)*(a + b
*x^4)^(3/2))/(24*b) - (a^2*e*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(16*b^(3/2)
) + (2*a^(9/4)*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])/(15*b^(7/4)*Sqrt[a + b*x^4])
- (a^(7/4)*(5*Sqrt[b]*d + 7*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])/(105*
b^(7/4)*Sqrt[a + b*x^4])

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Rubi [A]  time = 0.978458, antiderivative size = 394, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 11, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.367 \[ -\frac{a^{7/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (7 \sqrt{a} f+5 \sqrt{b} d\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{105 b^{7/4} \sqrt{a+b x^4}}+\frac{2 a^{9/4} f \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 b^{7/4} \sqrt{a+b x^4}}-\frac{a^2 e \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{16 b^{3/2}}-\frac{2 a^2 f x \sqrt{a+b x^4}}{15 b^{3/2} \left (\sqrt{a}+\sqrt{b} x^2\right )}+\frac{\left (a+b x^4\right )^{3/2} \left (4 c+3 e x^2\right )}{24 b}+\frac{1}{63} x^5 \sqrt{a+b x^4} \left (9 d+7 f x^2\right )+\frac{2 a d x \sqrt{a+b x^4}}{21 b}-\frac{a e x^2 \sqrt{a+b x^4}}{16 b}+\frac{2 a f x^3 \sqrt{a+b x^4}}{45 b} \]

Antiderivative was successfully verified.

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

[Out]

(2*a*d*x*Sqrt[a + b*x^4])/(21*b) - (a*e*x^2*Sqrt[a + b*x^4])/(16*b) + (2*a*f*x^3
*Sqrt[a + b*x^4])/(45*b) - (2*a^2*f*x*Sqrt[a + b*x^4])/(15*b^(3/2)*(Sqrt[a] + Sq
rt[b]*x^2)) + (x^5*(9*d + 7*f*x^2)*Sqrt[a + b*x^4])/63 + ((4*c + 3*e*x^2)*(a + b
*x^4)^(3/2))/(24*b) - (a^2*e*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(16*b^(3/2)
) + (2*a^(9/4)*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])/(15*b^(7/4)*Sqrt[a + b*x^4])
- (a^(7/4)*(5*Sqrt[b]*d + 7*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])/(105*
b^(7/4)*Sqrt[a + b*x^4])

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Rubi in Sympy [A]  time = 94.692, size = 362, normalized size = 0.92 \[ \frac{2 a^{\frac{9}{4}} f \sqrt{\frac{a + b x^{4}}{\left (\sqrt{a} + \sqrt{b} x^{2}\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x^{2}\right ) E\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}} \right )}\middle | \frac{1}{2}\right )}{15 b^{\frac{7}{4}} \sqrt{a + b x^{4}}} - \frac{a^{\frac{7}{4}} \sqrt{\frac{a + b x^{4}}{\left (\sqrt{a} + \sqrt{b} x^{2}\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x^{2}\right ) \left (7 \sqrt{a} f + 5 \sqrt{b} d\right ) F\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}} \right )}\middle | \frac{1}{2}\right )}{105 b^{\frac{7}{4}} \sqrt{a + b x^{4}}} - \frac{a^{2} e \operatorname{atanh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a + b x^{4}}} \right )}}{16 b^{\frac{3}{2}}} - \frac{2 a^{2} f x \sqrt{a + b x^{4}}}{15 b^{\frac{3}{2}} \left (\sqrt{a} + \sqrt{b} x^{2}\right )} + \frac{2 a d x \sqrt{a + b x^{4}}}{21 b} - \frac{a e x^{2} \sqrt{a + b x^{4}}}{16 b} + \frac{2 a f x^{3} \sqrt{a + b x^{4}}}{45 b} + \frac{x^{5} \sqrt{a + b x^{4}} \left (9 d + 7 f x^{2}\right )}{63} + \frac{\left (a + b x^{4}\right )^{\frac{3}{2}} \left (4 c + 3 e x^{2}\right )}{24 b} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

2*a**(9/4)*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)/(15*b**(7/4)*sqrt(a + b*x**4))
- a**(7/4)*sqrt((a + b*x**4)/(sqrt(a) + sqrt(b)*x**2)**2)*(sqrt(a) + sqrt(b)*x*
*2)*(7*sqrt(a)*f + 5*sqrt(b)*d)*elliptic_f(2*atan(b**(1/4)*x/a**(1/4)), 1/2)/(10
5*b**(7/4)*sqrt(a + b*x**4)) - a**2*e*atanh(sqrt(b)*x**2/sqrt(a + b*x**4))/(16*b
**(3/2)) - 2*a**2*f*x*sqrt(a + b*x**4)/(15*b**(3/2)*(sqrt(a) + sqrt(b)*x**2)) +
2*a*d*x*sqrt(a + b*x**4)/(21*b) - a*e*x**2*sqrt(a + b*x**4)/(16*b) + 2*a*f*x**3*
sqrt(a + b*x**4)/(45*b) + x**5*sqrt(a + b*x**4)*(9*d + 7*f*x**2)/63 + (a + b*x**
4)**(3/2)*(4*c + 3*e*x**2)/(24*b)

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Mathematica [C]  time = 0.833208, size = 275, normalized size = 0.7 \[ \frac{-672 a^{5/2} f \sqrt{\frac{b x^4}{a}+1} E\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )+\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \left (\sqrt{b} \left (a+b x^4\right ) \left (a (840 c+x (480 d+7 x (45 e+32 f x)))+10 b x^4 (84 c+x (72 d+7 x (9 e+8 f x)))\right )-315 a^2 e \sqrt{a+b x^4} \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )\right )+96 a^2 \sqrt{\frac{b x^4}{a}+1} \left (7 \sqrt{a} f+5 i \sqrt{b} d\right ) F\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )}{5040 b^{3/2} \sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \sqrt{a+b x^4}} \]

Antiderivative was successfully verified.

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

[Out]

(Sqrt[(I*Sqrt[b])/Sqrt[a]]*(Sqrt[b]*(a + b*x^4)*(10*b*x^4*(84*c + x*(72*d + 7*x*
(9*e + 8*f*x))) + a*(840*c + x*(480*d + 7*x*(45*e + 32*f*x)))) - 315*a^2*e*Sqrt[
a + b*x^4]*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]]) - 672*a^(5/2)*f*Sqrt[1 + (b*x
^4)/a]*EllipticE[I*ArcSinh[Sqrt[(I*Sqrt[b])/Sqrt[a]]*x], -1] + 96*a^2*((5*I)*Sqr
t[b]*d + 7*Sqrt[a]*f)*Sqrt[1 + (b*x^4)/a]*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[b])/S
qrt[a]]*x], -1])/(5040*Sqrt[(I*Sqrt[b])/Sqrt[a]]*b^(3/2)*Sqrt[a + b*x^4])

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Maple [C]  time = 0.013, size = 380, normalized size = 1. \[{\frac{{x}^{5}d}{7}\sqrt{b{x}^{4}+a}}+{\frac{2\,adx}{21\,b}\sqrt{b{x}^{4}+a}}-{\frac{2\,{a}^{2}d}{21\,b}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}+{\frac{c}{6\,b} \left ( b{x}^{4}+a \right ) ^{{\frac{3}{2}}}}+{\frac{e{x}^{2}}{8\,b} \left ( b{x}^{4}+a \right ) ^{{\frac{3}{2}}}}-{\frac{ae{x}^{2}}{16\,b}\sqrt{b{x}^{4}+a}}-{\frac{e{a}^{2}}{16}\ln \left ( \sqrt{b}{x}^{2}+\sqrt{b{x}^{4}+a} \right ){b}^{-{\frac{3}{2}}}}+{\frac{f{x}^{7}}{9}\sqrt{b{x}^{4}+a}}+{\frac{2\,{x}^{3}af}{45\,b}\sqrt{b{x}^{4}+a}}-{{\frac{2\,i}{15}}f{a}^{{\frac{5}{2}}}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){b}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}+{{\frac{2\,i}{15}}f{a}^{{\frac{5}{2}}}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticE} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){b}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

1/7*x^5*d*(b*x^4+a)^(1/2)+2/21*a*d*x*(b*x^4+a)^(1/2)/b-2/21*d/b*a^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)+1/6*c/b*(b*x^4+a)^(3/2)+1
/8*e*x^2*(b*x^4+a)^(3/2)/b-1/16*a*e*x^2*(b*x^4+a)^(1/2)/b-1/16*e*a^2/b^(3/2)*ln(
b^(1/2)*x^2+(b*x^4+a)^(1/2))+1/9*f*x^7*(b*x^4+a)^(1/2)+2/45*a*f*x^3*(b*x^4+a)^(1
/2)/b-2/15*I*f/b^(3/2)*a^(5/2)/(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)+2/15*I*f/b^(3/2)*a^(5/2)/(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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \frac{{\left (b x^{4} + a\right )}^{\frac{3}{2}} c}{6 \, b} + \int{\left (f x^{6} + e x^{5} + d x^{4}\right )} \sqrt{b x^{4} + a}\,{d x} \]

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^3,x, algorithm="maxima")

[Out]

1/6*(b*x^4 + a)^(3/2)*c/b + integrate((f*x^6 + e*x^5 + d*x^4)*sqrt(b*x^4 + a), x
)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left ({\left (f x^{6} + e x^{5} + d x^{4} + c x^{3}\right )} \sqrt{b x^{4} + a}, x\right ) \]

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^3,x, algorithm="fricas")

[Out]

integral((f*x^6 + e*x^5 + d*x^4 + c*x^3)*sqrt(b*x^4 + a), x)

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Sympy [A]  time = 8.80098, size = 212, normalized size = 0.54 \[ \frac{a^{\frac{3}{2}} e x^{2}}{16 b \sqrt{1 + \frac{b x^{4}}{a}}} + \frac{\sqrt{a} d x^{5} \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{5}{4} \\ \frac{9}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac{9}{4}\right )} + \frac{3 \sqrt{a} e x^{6}}{16 \sqrt{1 + \frac{b x^{4}}{a}}} + \frac{\sqrt{a} f x^{7} \Gamma \left (\frac{7}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{7}{4} \\ \frac{11}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac{11}{4}\right )} - \frac{a^{2} e \operatorname{asinh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a}} \right )}}{16 b^{\frac{3}{2}}} + c \left (\begin{cases} \frac{\sqrt{a} x^{4}}{4} & \text{for}\: b = 0 \\\frac{\left (a + b x^{4}\right )^{\frac{3}{2}}}{6 b} & \text{otherwise} \end{cases}\right ) + \frac{b e x^{10}}{8 \sqrt{a} \sqrt{1 + \frac{b x^{4}}{a}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

a**(3/2)*e*x**2/(16*b*sqrt(1 + b*x**4/a)) + sqrt(a)*d*x**5*gamma(5/4)*hyper((-1/
2, 5/4), (9/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(9/4)) + 3*sqrt(a)*e*x**6/(16
*sqrt(1 + b*x**4/a)) + sqrt(a)*f*x**7*gamma(7/4)*hyper((-1/2, 7/4), (11/4,), b*x
**4*exp_polar(I*pi)/a)/(4*gamma(11/4)) - a**2*e*asinh(sqrt(b)*x**2/sqrt(a))/(16*
b**(3/2)) + c*Piecewise((sqrt(a)*x**4/4, Eq(b, 0)), ((a + b*x**4)**(3/2)/(6*b),
True)) + b*e*x**10/(8*sqrt(a)*sqrt(1 + b*x**4/a))

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

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^3,x, algorithm="giac")

[Out]

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