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

Optimal. Leaf size=404 \[ \frac{2 a^{5/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (5 \sqrt{a} e+21 \sqrt{b} c\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{35 \sqrt [4]{b} \sqrt{a+b x^4}}-\frac{12 a^{5/4} \sqrt [4]{b} c \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 \sqrt{a+b x^4}}-\frac{1}{2} a^{3/2} d \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )+\frac{3 a^2 f \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{16 \sqrt{b}}-\frac{\left (a+b x^4\right )^{3/2} \left (7 c-e x^2\right )}{7 x}+\frac{2}{35} x \sqrt{a+b x^4} \left (5 a e+21 b c x^2\right )+\frac{12 a \sqrt{b} c x \sqrt{a+b x^4}}{5 \left (\sqrt{a}+\sqrt{b} x^2\right )}+\frac{1}{16} a \sqrt{a+b x^4} \left (8 d+3 f x^2\right )+\frac{1}{24} \left (a+b x^4\right )^{3/2} \left (4 d+3 f x^2\right ) \]

[Out]

(12*a*Sqrt[b]*c*x*Sqrt[a + b*x^4])/(5*(Sqrt[a] + Sqrt[b]*x^2)) + (2*x*(5*a*e + 2
1*b*c*x^2)*Sqrt[a + b*x^4])/35 + (a*(8*d + 3*f*x^2)*Sqrt[a + b*x^4])/16 - ((7*c
- e*x^2)*(a + b*x^4)^(3/2))/(7*x) + ((4*d + 3*f*x^2)*(a + b*x^4)^(3/2))/24 + (3*
a^2*f*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(16*Sqrt[b]) - (a^(3/2)*d*ArcTanh[
Sqrt[a + b*x^4]/Sqrt[a]])/2 - (12*a^(5/4)*b^(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])/(5*Sqrt[a + b*x^4]) + (2*a^(5/4)*(21*Sqrt[b]*c + 5*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])/(35*b^(1/4)*Sqrt[a + b*x^4])

_______________________________________________________________________________________

Rubi [A]  time = 0.838123, antiderivative size = 404, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 14, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.467 \[ \frac{2 a^{5/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (5 \sqrt{a} e+21 \sqrt{b} c\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{35 \sqrt [4]{b} \sqrt{a+b x^4}}-\frac{12 a^{5/4} \sqrt [4]{b} c \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 \sqrt{a+b x^4}}-\frac{1}{2} a^{3/2} d \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )+\frac{3 a^2 f \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{16 \sqrt{b}}-\frac{\left (a+b x^4\right )^{3/2} \left (7 c-e x^2\right )}{7 x}+\frac{2}{35} x \sqrt{a+b x^4} \left (5 a e+21 b c x^2\right )+\frac{12 a \sqrt{b} c x \sqrt{a+b x^4}}{5 \left (\sqrt{a}+\sqrt{b} x^2\right )}+\frac{1}{16} a \sqrt{a+b x^4} \left (8 d+3 f x^2\right )+\frac{1}{24} \left (a+b x^4\right )^{3/2} \left (4 d+3 f x^2\right ) \]

Antiderivative was successfully verified.

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

[Out]

(12*a*Sqrt[b]*c*x*Sqrt[a + b*x^4])/(5*(Sqrt[a] + Sqrt[b]*x^2)) + (2*x*(5*a*e + 2
1*b*c*x^2)*Sqrt[a + b*x^4])/35 + (a*(8*d + 3*f*x^2)*Sqrt[a + b*x^4])/16 - ((7*c
- e*x^2)*(a + b*x^4)^(3/2))/(7*x) + ((4*d + 3*f*x^2)*(a + b*x^4)^(3/2))/24 + (3*
a^2*f*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(16*Sqrt[b]) - (a^(3/2)*d*ArcTanh[
Sqrt[a + b*x^4]/Sqrt[a]])/2 - (12*a^(5/4)*b^(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])/(5*Sqrt[a + b*x^4]) + (2*a^(5/4)*(21*Sqrt[b]*c + 5*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])/(35*b^(1/4)*Sqrt[a + b*x^4])

_______________________________________________________________________________________

Rubi in Sympy [A]  time = 96.5957, size = 376, normalized size = 0.93 \[ - \frac{12 a^{\frac{5}{4}} \sqrt [4]{b} c \sqrt{\frac{a + b x^{4}}{\left (\sqrt{a} + \sqrt{b} x^{2}\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x^{2}\right ) E\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}} \right )}\middle | \frac{1}{2}\right )}{5 \sqrt{a + b x^{4}}} + \frac{2 a^{\frac{5}{4}} \sqrt{\frac{a + b x^{4}}{\left (\sqrt{a} + \sqrt{b} x^{2}\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x^{2}\right ) \left (5 \sqrt{a} e + 21 \sqrt{b} c\right ) F\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}} \right )}\middle | \frac{1}{2}\right )}{35 \sqrt [4]{b} \sqrt{a + b x^{4}}} - \frac{a^{\frac{3}{2}} d \operatorname{atanh}{\left (\frac{\sqrt{a + b x^{4}}}{\sqrt{a}} \right )}}{2} + \frac{3 a^{2} f \operatorname{atanh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a + b x^{4}}} \right )}}{16 \sqrt{b}} + \frac{12 a \sqrt{b} c x \sqrt{a + b x^{4}}}{5 \left (\sqrt{a} + \sqrt{b} x^{2}\right )} + \frac{a \sqrt{a + b x^{4}} \left (8 d + 3 f x^{2}\right )}{16} + \frac{2 x \sqrt{a + b x^{4}} \left (5 a e + 21 b c x^{2}\right )}{35} + \frac{\left (a + b x^{4}\right )^{\frac{3}{2}} \left (4 d + 3 f x^{2}\right )}{24} - \frac{\left (a + b x^{4}\right )^{\frac{3}{2}} \left (7 c - e x^{2}\right )}{7 x} \]

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**2,x)

[Out]

-12*a**(5/4)*b**(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)/(5*sqrt(a + b*x**4)
) + 2*a**(5/4)*sqrt((a + b*x**4)/(sqrt(a) + sqrt(b)*x**2)**2)*(sqrt(a) + sqrt(b)
*x**2)*(5*sqrt(a)*e + 21*sqrt(b)*c)*elliptic_f(2*atan(b**(1/4)*x/a**(1/4)), 1/2)
/(35*b**(1/4)*sqrt(a + b*x**4)) - a**(3/2)*d*atanh(sqrt(a + b*x**4)/sqrt(a))/2 +
3*a**2*f*atanh(sqrt(b)*x**2/sqrt(a + b*x**4))/(16*sqrt(b)) + 12*a*sqrt(b)*c*x*s
qrt(a + b*x**4)/(5*(sqrt(a) + sqrt(b)*x**2)) + a*sqrt(a + b*x**4)*(8*d + 3*f*x**
2)/16 + 2*x*sqrt(a + b*x**4)*(5*a*e + 21*b*c*x**2)/35 + (a + b*x**4)**(3/2)*(4*d
+ 3*f*x**2)/24 - (a + b*x**4)**(3/2)*(7*c - e*x**2)/(7*x)

_______________________________________________________________________________________

Mathematica [C]  time = 0.972582, size = 328, normalized size = 0.81 \[ -\frac{1}{2} a^{3/2} d \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )-\frac{4 i a^2 e \sqrt{\frac{b x^4}{a}+1} F\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )}{7 \sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \sqrt{a+b x^4}}+\frac{3 a^2 f \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{16 \sqrt{b}}+\sqrt{a+b x^4} \left (a \left (-\frac{c}{x}+\frac{2 d}{3}+\frac{3 e x}{7}+\frac{5 f x^2}{16}\right )+b \left (\frac{c x^3}{5}+\frac{d x^4}{6}+\frac{e x^5}{7}+\frac{f x^6}{8}\right )\right )+\frac{12 i a b c \sqrt{\frac{b x^4}{a}+1} \left (E\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )-F\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )\right )}{5 \left (\frac{i \sqrt{b}}{\sqrt{a}}\right )^{3/2} \sqrt{a+b x^4}} \]

Antiderivative was successfully verified.

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

[Out]

Sqrt[a + b*x^4]*(a*((2*d)/3 - c/x + (3*e*x)/7 + (5*f*x^2)/16) + b*((c*x^3)/5 + (
d*x^4)/6 + (e*x^5)/7 + (f*x^6)/8)) + (3*a^2*f*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x
^4]])/(16*Sqrt[b]) - (a^(3/2)*d*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]])/2 + (((12*I)/5
)*a*b*c*Sqrt[1 + (b*x^4)/a]*(EllipticE[I*ArcSinh[Sqrt[(I*Sqrt[b])/Sqrt[a]]*x], -
1] - EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[b])/Sqrt[a]]*x], -1]))/(((I*Sqrt[b])/Sqrt[
a])^(3/2)*Sqrt[a + b*x^4]) - (((4*I)/7)*a^2*e*Sqrt[1 + (b*x^4)/a]*EllipticF[I*Ar
cSinh[Sqrt[(I*Sqrt[b])/Sqrt[a]]*x], -1])/(Sqrt[(I*Sqrt[b])/Sqrt[a]]*Sqrt[a + b*x
^4])

_______________________________________________________________________________________

Maple [C]  time = 0.02, size = 411, normalized size = 1. \[{\frac{{x}^{5}be}{7}\sqrt{b{x}^{4}+a}}+{\frac{3\,aex}{7}\sqrt{b{x}^{4}+a}}+{\frac{4\,e{a}^{2}}{7}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}+{\frac{3\,{a}^{2}f}{16}\ln \left ( \sqrt{b}{x}^{2}+\sqrt{b{x}^{4}+a} \right ){\frac{1}{\sqrt{b}}}}+{\frac{bf{x}^{6}}{8}\sqrt{b{x}^{4}+a}}+{\frac{5\,{x}^{2}af}{16}\sqrt{b{x}^{4}+a}}-{\frac{ac}{x}\sqrt{b{x}^{4}+a}}+{\frac{{x}^{3}bc}{5}\sqrt{b{x}^{4}+a}}+{{\frac{12\,i}{5}}c{a}^{{\frac{3}{2}}}\sqrt{b}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}-{{\frac{12\,i}{5}}c{a}^{{\frac{3}{2}}}\sqrt{b}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticE} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}-{\frac{d}{2}{a}^{{\frac{3}{2}}}\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{4}+a} \right ) } \right ) }+{\frac{bd{x}^{4}}{6}\sqrt{b{x}^{4}+a}}+{\frac{2\,ad}{3}\sqrt{b{x}^{4}+a}} \]

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^2,x)

[Out]

1/7*e*b*x^5*(b*x^4+a)^(1/2)+3/7*e*a*x*(b*x^4+a)^(1/2)+4/7*e*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*f*a^2*ln(b^(1/2)*x^2+(b*
x^4+a)^(1/2))/b^(1/2)+1/8*f*b*x^6*(b*x^4+a)^(1/2)+5/16*f*a*x^2*(b*x^4+a)^(1/2)-c
*a*(b*x^4+a)^(1/2)/x+1/5*c*b*x^3*(b*x^4+a)^(1/2)+12/5*I*c*a^(3/2)*b^(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)*EllipticF(x*(I/a^(1/2)*b^(1/2))^(1/2),I)-12/5*I*c*a^(3/2)*b^
(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)*EllipticE(x*(I/a^(1/2)*b^(1/2))^(1/2),I)-1/2*d*a
^(3/2)*ln((2*a+2*a^(1/2)*(b*x^4+a)^(1/2))/x^2)+1/6*d*b*x^4*(b*x^4+a)^(1/2)+2/3*d
*a*(b*x^4+a)^(1/2)

_______________________________________________________________________________________

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

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^2,x, algorithm="maxima")

[Out]

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

_______________________________________________________________________________________

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

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^2,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^2, x)

_______________________________________________________________________________________

Sympy [A]  time = 16.8686, size = 406, normalized size = 1. \[ \frac{a^{\frac{3}{2}} c \Gamma \left (- \frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, - \frac{1}{4} \\ \frac{3}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 x \Gamma \left (\frac{3}{4}\right )} - \frac{a^{\frac{3}{2}} d \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x^{2}} \right )}}{2} + \frac{a^{\frac{3}{2}} e x \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{1}{4} \\ \frac{5}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac{5}{4}\right )} + \frac{a^{\frac{3}{2}} f x^{2} \sqrt{1 + \frac{b x^{4}}{a}}}{4} + \frac{a^{\frac{3}{2}} f x^{2}}{16 \sqrt{1 + \frac{b x^{4}}{a}}} + \frac{\sqrt{a} b c x^{3} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac{7}{4}\right )} + \frac{\sqrt{a} b e x^{5} \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{5}{4} \\ \frac{9}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac{9}{4}\right )} + \frac{3 \sqrt{a} b f x^{6}}{16 \sqrt{1 + \frac{b x^{4}}{a}}} + \frac{a^{2} d}{2 \sqrt{b} x^{2} \sqrt{\frac{a}{b x^{4}} + 1}} + \frac{3 a^{2} f \operatorname{asinh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a}} \right )}}{16 \sqrt{b}} + \frac{a \sqrt{b} d x^{2}}{2 \sqrt{\frac{a}{b x^{4}} + 1}} + b d \left (\begin{cases} \frac{\sqrt{a} x^{4}}{4} & \text{for}\: b = 0 \\\frac{\left (a + b x^{4}\right )^{\frac{3}{2}}}{6 b} & \text{otherwise} \end{cases}\right ) + \frac{b^{2} f x^{10}}{8 \sqrt{a} \sqrt{1 + \frac{b x^{4}}{a}}} \]

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**2,x)

[Out]

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

_______________________________________________________________________________________

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

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^2,x, algorithm="giac")

[Out]

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