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

Optimal. Leaf size=382 \[ \frac{2 a^{7/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (7 \sqrt{a} e+15 \sqrt{b} c\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{105 b^{3/4} \sqrt{a+b x^4}}-\frac{4 a^{9/4} e \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 b^{3/4} \sqrt{a+b x^4}}+\frac{3 a^2 d \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{16 \sqrt{b}}+\frac{4 a^2 e x \sqrt{a+b x^4}}{15 \sqrt{b} \left (\sqrt{a}+\sqrt{b} x^2\right )}+\frac{1}{63} x \left (a+b x^4\right )^{3/2} \left (9 c+7 e x^2\right )+\frac{2}{105} a x \sqrt{a+b x^4} \left (15 c+7 e x^2\right )+\frac{1}{8} d x^2 \left (a+b x^4\right )^{3/2}+\frac{3}{16} a d x^2 \sqrt{a+b x^4}+\frac{f \left (a+b x^4\right )^{5/2}}{10 b} \]

[Out]

(3*a*d*x^2*Sqrt[a + b*x^4])/16 + (4*a^2*e*x*Sqrt[a + b*x^4])/(15*Sqrt[b]*(Sqrt[a
] + Sqrt[b]*x^2)) + (2*a*x*(15*c + 7*e*x^2)*Sqrt[a + b*x^4])/105 + (d*x^2*(a + b
*x^4)^(3/2))/8 + (x*(9*c + 7*e*x^2)*(a + b*x^4)^(3/2))/63 + (f*(a + b*x^4)^(5/2)
)/(10*b) + (3*a^2*d*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(16*Sqrt[b]) - (4*a^
(9/4)*e*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*Elli
pticE[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(15*b^(3/4)*Sqrt[a + b*x^4]) + (2*a^(
7/4)*(15*Sqrt[b]*c + 7*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])/(105*b^(3/4
)*Sqrt[a + b*x^4])

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Rubi [A]  time = 0.639411, antiderivative size = 382, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 10, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.37 \[ \frac{2 a^{7/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (7 \sqrt{a} e+15 \sqrt{b} c\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{105 b^{3/4} \sqrt{a+b x^4}}-\frac{4 a^{9/4} e \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 b^{3/4} \sqrt{a+b x^4}}+\frac{3 a^2 d \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{16 \sqrt{b}}+\frac{4 a^2 e x \sqrt{a+b x^4}}{15 \sqrt{b} \left (\sqrt{a}+\sqrt{b} x^2\right )}+\frac{1}{63} x \left (a+b x^4\right )^{3/2} \left (9 c+7 e x^2\right )+\frac{2}{105} a x \sqrt{a+b x^4} \left (15 c+7 e x^2\right )+\frac{1}{8} d x^2 \left (a+b x^4\right )^{3/2}+\frac{3}{16} a d x^2 \sqrt{a+b x^4}+\frac{f \left (a+b x^4\right )^{5/2}}{10 b} \]

Antiderivative was successfully verified.

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

[Out]

(3*a*d*x^2*Sqrt[a + b*x^4])/16 + (4*a^2*e*x*Sqrt[a + b*x^4])/(15*Sqrt[b]*(Sqrt[a
] + Sqrt[b]*x^2)) + (2*a*x*(15*c + 7*e*x^2)*Sqrt[a + b*x^4])/105 + (d*x^2*(a + b
*x^4)^(3/2))/8 + (x*(9*c + 7*e*x^2)*(a + b*x^4)^(3/2))/63 + (f*(a + b*x^4)^(5/2)
)/(10*b) + (3*a^2*d*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(16*Sqrt[b]) - (4*a^
(9/4)*e*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*Elli
pticE[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(15*b^(3/4)*Sqrt[a + b*x^4]) + (2*a^(
7/4)*(15*Sqrt[b]*c + 7*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])/(105*b^(3/4
)*Sqrt[a + b*x^4])

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Rubi in Sympy [A]  time = 61.3009, size = 357, normalized size = 0.93 \[ - \frac{4 a^{\frac{9}{4}} e \sqrt{\frac{a + b x^{4}}{\left (\sqrt{a} + \sqrt{b} x^{2}\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x^{2}\right ) E\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}} \right )}\middle | \frac{1}{2}\right )}{15 b^{\frac{3}{4}} \sqrt{a + b x^{4}}} + \frac{2 a^{\frac{7}{4}} \sqrt{\frac{a + b x^{4}}{\left (\sqrt{a} + \sqrt{b} x^{2}\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x^{2}\right ) \left (7 \sqrt{a} e + 15 \sqrt{b} c\right ) F\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}} \right )}\middle | \frac{1}{2}\right )}{105 b^{\frac{3}{4}} \sqrt{a + b x^{4}}} + \frac{3 a^{2} d \operatorname{atanh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a + b x^{4}}} \right )}}{16 \sqrt{b}} + \frac{4 a^{2} e x \sqrt{a + b x^{4}}}{15 \sqrt{b} \left (\sqrt{a} + \sqrt{b} x^{2}\right )} + \frac{3 a d x^{2} \sqrt{a + b x^{4}}}{16} + \frac{2 a x \sqrt{a + b x^{4}} \left (45 c + 21 e x^{2}\right )}{315} + \frac{d x^{2} \left (a + b x^{4}\right )^{\frac{3}{2}}}{8} + \frac{x \left (a + b x^{4}\right )^{\frac{3}{2}} \left (9 c + 7 e x^{2}\right )}{63} + \frac{f \left (a + b x^{4}\right )^{\frac{5}{2}}}{10 b} \]

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)**(3/2),x)

[Out]

-4*a**(9/4)*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)/(15*b**(3/4)*sqrt(a + b*x**4)
) + 2*a**(7/4)*sqrt((a + b*x**4)/(sqrt(a) + sqrt(b)*x**2)**2)*(sqrt(a) + sqrt(b)
*x**2)*(7*sqrt(a)*e + 15*sqrt(b)*c)*elliptic_f(2*atan(b**(1/4)*x/a**(1/4)), 1/2)
/(105*b**(3/4)*sqrt(a + b*x**4)) + 3*a**2*d*atanh(sqrt(b)*x**2/sqrt(a + b*x**4))
/(16*sqrt(b)) + 4*a**2*e*x*sqrt(a + b*x**4)/(15*sqrt(b)*(sqrt(a) + sqrt(b)*x**2)
) + 3*a*d*x**2*sqrt(a + b*x**4)/16 + 2*a*x*sqrt(a + b*x**4)*(45*c + 21*e*x**2)/3
15 + d*x**2*(a + b*x**4)**(3/2)/8 + x*(a + b*x**4)**(3/2)*(9*c + 7*e*x**2)/63 +
f*(a + b*x**4)**(5/2)/(10*b)

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Mathematica [C]  time = 0.796908, size = 294, normalized size = 0.77 \[ \frac{1344 a^{5/2} \sqrt{b} e \sqrt{\frac{b x^4}{a}+1} E\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )+\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \left (\left (a+b x^4\right ) \left (504 a^2 f+a b x (2160 c+7 x (225 d+16 x (11 e+9 f x)))+2 b^2 x^5 \left (360 c+7 x \left (45 d+40 e x+36 f x^2\right )\right )\right )+945 a^2 \sqrt{b} d \sqrt{a+b x^4} \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )\right )-192 a^2 \sqrt{b} \sqrt{\frac{b x^4}{a}+1} \left (7 \sqrt{a} e+15 i \sqrt{b} c\right ) F\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )}{5040 b \sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \sqrt{a+b x^4}} \]

Antiderivative was successfully verified.

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

[Out]

(Sqrt[(I*Sqrt[b])/Sqrt[a]]*((a + b*x^4)*(504*a^2*f + 2*b^2*x^5*(360*c + 7*x*(45*
d + 40*e*x + 36*f*x^2)) + a*b*x*(2160*c + 7*x*(225*d + 16*x*(11*e + 9*f*x)))) +
945*a^2*Sqrt[b]*d*Sqrt[a + b*x^4]*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]]) + 1344
*a^(5/2)*Sqrt[b]*e*Sqrt[1 + (b*x^4)/a]*EllipticE[I*ArcSinh[Sqrt[(I*Sqrt[b])/Sqrt
[a]]*x], -1] - 192*a^2*Sqrt[b]*((15*I)*Sqrt[b]*c + 7*Sqrt[a]*e)*Sqrt[1 + (b*x^4)
/a]*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[b])/Sqrt[a]]*x], -1])/(5040*Sqrt[(I*Sqrt[b]
)/Sqrt[a]]*b*Sqrt[a + b*x^4])

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Maple [C]  time = 0.011, size = 368, normalized size = 1. \[{\frac{bc{x}^{5}}{7}\sqrt{b{x}^{4}+a}}+{\frac{3\,acx}{7}\sqrt{b{x}^{4}+a}}+{\frac{4\,{a}^{2}c}{7}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}+{\frac{3\,{a}^{2}d}{16}\ln \left ( \sqrt{b}{x}^{2}+\sqrt{b{x}^{4}+a} \right ){\frac{1}{\sqrt{b}}}}+{\frac{bd{x}^{6}}{8}\sqrt{b{x}^{4}+a}}+{\frac{5\,ad{x}^{2}}{16}\sqrt{b{x}^{4}+a}}+{\frac{be{x}^{7}}{9}\sqrt{b{x}^{4}+a}}+{\frac{11\,ae{x}^{3}}{45}\sqrt{b{x}^{4}+a}}+{{\frac{4\,i}{15}}e{a}^{{\frac{5}{2}}}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}{\frac{1}{\sqrt{b}}}}-{{\frac{4\,i}{15}}e{a}^{{\frac{5}{2}}}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticE} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}{\frac{1}{\sqrt{b}}}}+{\frac{f}{10\,b} \left ( b{x}^{4}+a \right ) ^{{\frac{5}{2}}}} \]

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)^(3/2),x)

[Out]

1/7*c*b*x^5*(b*x^4+a)^(1/2)+3/7*c*a*x*(b*x^4+a)^(1/2)+4/7*c*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)+3/16*d*a^2*ln(b^(1/2)*x^2+(b*
x^4+a)^(1/2))/b^(1/2)+1/8*d*b*x^6*(b*x^4+a)^(1/2)+5/16*a*d*x^2*(b*x^4+a)^(1/2)+1
/9*e*b*x^7*(b*x^4+a)^(1/2)+11/45*e*a*x^3*(b*x^4+a)^(1/2)+4/15*I*e*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)/b^(1/2)*EllipticF(x*(I/a^(1/2)*b^(1/2))^(1/2),I)-4/15*I*e*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)/b^(1/2)*EllipticE(x*(I/a^(1/2)*b^(1/2))^(1/2),I)
+1/10*f*(b*x^4+a)^(5/2)/b

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left ({\left (b f x^{7} + b e x^{6} + b d x^{5} + b c x^{4} + a f x^{3} + a e x^{2} + a d x + a c\right )} \sqrt{b x^{4} + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

integral((b*f*x^7 + b*e*x^6 + b*d*x^5 + b*c*x^4 + a*f*x^3 + a*e*x^2 + a*d*x + a*
c)*sqrt(b*x^4 + a), x)

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Sympy [A]  time = 13.8238, size = 394, normalized size = 1.03 \[ \frac{a^{\frac{3}{2}} c x \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{1}{4} \\ \frac{5}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac{5}{4}\right )} + \frac{a^{\frac{3}{2}} d x^{2} \sqrt{1 + \frac{b x^{4}}{a}}}{4} + \frac{a^{\frac{3}{2}} d x^{2}}{16 \sqrt{1 + \frac{b x^{4}}{a}}} + \frac{a^{\frac{3}{2}} e x^{3} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac{7}{4}\right )} + \frac{\sqrt{a} b c x^{5} \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{5}{4} \\ \frac{9}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac{9}{4}\right )} + \frac{3 \sqrt{a} b d x^{6}}{16 \sqrt{1 + \frac{b x^{4}}{a}}} + \frac{\sqrt{a} b e x^{7} \Gamma \left (\frac{7}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{7}{4} \\ \frac{11}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac{11}{4}\right )} + \frac{3 a^{2} d \operatorname{asinh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a}} \right )}}{16 \sqrt{b}} + a f \left (\begin{cases} \frac{\sqrt{a} x^{4}}{4} & \text{for}\: b = 0 \\\frac{\left (a + b x^{4}\right )^{\frac{3}{2}}}{6 b} & \text{otherwise} \end{cases}\right ) + b f \left (\begin{cases} - \frac{a^{2} \sqrt{a + b x^{4}}}{15 b^{2}} + \frac{a x^{4} \sqrt{a + b x^{4}}}{30 b} + \frac{x^{8} \sqrt{a + b x^{4}}}{10} & \text{for}\: b \neq 0 \\\frac{\sqrt{a} x^{8}}{8} & \text{otherwise} \end{cases}\right ) + \frac{b^{2} d x^{10}}{8 \sqrt{a} \sqrt{1 + \frac{b x^{4}}{a}}} \]

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)**(3/2),x)

[Out]

a**(3/2)*c*x*gamma(1/4)*hyper((-1/2, 1/4), (5/4,), b*x**4*exp_polar(I*pi)/a)/(4*
gamma(5/4)) + a**(3/2)*d*x**2*sqrt(1 + b*x**4/a)/4 + a**(3/2)*d*x**2/(16*sqrt(1
+ b*x**4/a)) + a**(3/2)*e*x**3*gamma(3/4)*hyper((-1/2, 3/4), (7/4,), b*x**4*exp_
polar(I*pi)/a)/(4*gamma(7/4)) + sqrt(a)*b*c*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)*b*d*x**6/(16*sqrt(1
+ b*x**4/a)) + sqrt(a)*b*e*x**7*gamma(7/4)*hyper((-1/2, 7/4), (11/4,), b*x**4*ex
p_polar(I*pi)/a)/(4*gamma(11/4)) + 3*a**2*d*asinh(sqrt(b)*x**2/sqrt(a))/(16*sqrt
(b)) + a*f*Piecewise((sqrt(a)*x**4/4, Eq(b, 0)), ((a + b*x**4)**(3/2)/(6*b), Tru
e)) + b*f*Piecewise((-a**2*sqrt(a + b*x**4)/(15*b**2) + a*x**4*sqrt(a + b*x**4)/
(30*b) + x**8*sqrt(a + b*x**4)/10, Ne(b, 0)), (sqrt(a)*x**8/8, True)) + b**2*d*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. \[ \int{\left (b x^{4} + a\right )}^{\frac{3}{2}}{\left (f x^{3} + e x^{2} + d x + c\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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