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

Optimal. Leaf size=406 \[ \frac{2 a^{5/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (5 \sqrt{a} f+21 \sqrt{b} d\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{35 \sqrt [4]{b} \sqrt{a+b x^4}}-\frac{12 a^{5/4} \sqrt [4]{b} d \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 \sqrt{a+b x^4}}-\frac{1}{2} a^{3/2} e \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )-\frac{\left (a+b x^4\right )^{3/2} \left (3 c-e x^2\right )}{6 x^2}+\frac{1}{4} \sqrt{a+b x^4} \left (2 a e+3 b c x^2\right )+\frac{3}{4} a \sqrt{b} c \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )-\frac{\left (a+b x^4\right )^{3/2} \left (7 d-f x^2\right )}{7 x}+\frac{2}{35} x \sqrt{a+b x^4} \left (5 a f+21 b d x^2\right )+\frac{12 a \sqrt{b} d x \sqrt{a+b x^4}}{5 \left (\sqrt{a}+\sqrt{b} x^2\right )} \]

[Out]

(12*a*Sqrt[b]*d*x*Sqrt[a + b*x^4])/(5*(Sqrt[a] + Sqrt[b]*x^2)) + ((2*a*e + 3*b*c
*x^2)*Sqrt[a + b*x^4])/4 + (2*x*(5*a*f + 21*b*d*x^2)*Sqrt[a + b*x^4])/35 - ((3*c
- e*x^2)*(a + b*x^4)^(3/2))/(6*x^2) - ((7*d - f*x^2)*(a + b*x^4)^(3/2))/(7*x) +
(3*a*Sqrt[b]*c*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/4 - (a^(3/2)*e*ArcTanh[S
qrt[a + b*x^4]/Sqrt[a]])/2 - (12*a^(5/4)*b^(1/4)*d*(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]*d + 5*Sqrt[a]*f)*(Sqrt[a] + S
qrt[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])

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Rubi [A]  time = 0.826172, antiderivative size = 406, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 15, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5 \[ \frac{2 a^{5/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (5 \sqrt{a} f+21 \sqrt{b} d\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{35 \sqrt [4]{b} \sqrt{a+b x^4}}-\frac{12 a^{5/4} \sqrt [4]{b} d \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 \sqrt{a+b x^4}}-\frac{1}{2} a^{3/2} e \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )-\frac{\left (a+b x^4\right )^{3/2} \left (3 c-e x^2\right )}{6 x^2}+\frac{1}{4} \sqrt{a+b x^4} \left (2 a e+3 b c x^2\right )+\frac{3}{4} a \sqrt{b} c \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )-\frac{\left (a+b x^4\right )^{3/2} \left (7 d-f x^2\right )}{7 x}+\frac{2}{35} x \sqrt{a+b x^4} \left (5 a f+21 b d x^2\right )+\frac{12 a \sqrt{b} d x \sqrt{a+b x^4}}{5 \left (\sqrt{a}+\sqrt{b} x^2\right )} \]

Antiderivative was successfully verified.

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

[Out]

(12*a*Sqrt[b]*d*x*Sqrt[a + b*x^4])/(5*(Sqrt[a] + Sqrt[b]*x^2)) + ((2*a*e + 3*b*c
*x^2)*Sqrt[a + b*x^4])/4 + (2*x*(5*a*f + 21*b*d*x^2)*Sqrt[a + b*x^4])/35 - ((3*c
- e*x^2)*(a + b*x^4)^(3/2))/(6*x^2) - ((7*d - f*x^2)*(a + b*x^4)^(3/2))/(7*x) +
(3*a*Sqrt[b]*c*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/4 - (a^(3/2)*e*ArcTanh[S
qrt[a + b*x^4]/Sqrt[a]])/2 - (12*a^(5/4)*b^(1/4)*d*(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]*d + 5*Sqrt[a]*f)*(Sqrt[a] + S
qrt[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])

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Rubi in Sympy [A]  time = 93.4706, size = 377, normalized size = 0.93 \[ - \frac{12 a^{\frac{5}{4}} \sqrt [4]{b} d \sqrt{\frac{a + b x^{4}}{\left (\sqrt{a} + \sqrt{b} x^{2}\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x^{2}\right ) E\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}} \right )}\middle | \frac{1}{2}\right )}{5 \sqrt{a + b x^{4}}} + \frac{2 a^{\frac{5}{4}} \sqrt{\frac{a + b x^{4}}{\left (\sqrt{a} + \sqrt{b} x^{2}\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x^{2}\right ) \left (5 \sqrt{a} f + 21 \sqrt{b} d\right ) F\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}} \right )}\middle | \frac{1}{2}\right )}{35 \sqrt [4]{b} \sqrt{a + b x^{4}}} - \frac{a^{\frac{3}{2}} e \operatorname{atanh}{\left (\frac{\sqrt{a + b x^{4}}}{\sqrt{a}} \right )}}{2} + \frac{3 a \sqrt{b} c \operatorname{atanh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a + b x^{4}}} \right )}}{4} + \frac{12 a \sqrt{b} d x \sqrt{a + b x^{4}}}{5 \left (\sqrt{a} + \sqrt{b} x^{2}\right )} + \frac{2 x \sqrt{a + b x^{4}} \left (5 a f + 21 b d x^{2}\right )}{35} + \frac{\sqrt{a + b x^{4}} \left (4 a e + 6 b c x^{2}\right )}{8} - \frac{\left (a + b x^{4}\right )^{\frac{3}{2}} \left (7 d - f x^{2}\right )}{7 x} - \frac{\left (a + b x^{4}\right )^{\frac{3}{2}} \left (3 c - e x^{2}\right )}{6 x^{2}} \]

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**3,x)

[Out]

-12*a**(5/4)*b**(1/4)*d*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)*f + 21*sqrt(b)*d)*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)*e*atanh(sqrt(a + b*x**4)/sqrt(a))/2 +
3*a*sqrt(b)*c*atanh(sqrt(b)*x**2/sqrt(a + b*x**4))/4 + 12*a*sqrt(b)*d*x*sqrt(a
+ b*x**4)/(5*(sqrt(a) + sqrt(b)*x**2)) + 2*x*sqrt(a + b*x**4)*(5*a*f + 21*b*d*x*
*2)/35 + sqrt(a + b*x**4)*(4*a*e + 6*b*c*x**2)/8 - (a + b*x**4)**(3/2)*(7*d - f*
x**2)/(7*x) - (a + b*x**4)**(3/2)*(3*c - e*x**2)/(6*x**2)

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Mathematica [C]  time = 1.15303, size = 326, normalized size = 0.8 \[ \frac{\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \left (-210 a^{3/2} e x^2 \sqrt{a+b x^4} \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )+\left (a+b x^4\right ) \left (-210 a c+20 a x (x (14 e+9 f x)-21 d)+b x^4 \left (105 c+84 d x+70 e x^2+60 f x^3\right )\right )+315 a \sqrt{b} c x^2 \sqrt{a+b x^4} \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )\right )-48 i a^{3/2} x^2 \sqrt{\frac{b x^4}{a}+1} \left (5 \sqrt{a} f-21 i \sqrt{b} d\right ) F\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )+1008 a^{3/2} \sqrt{b} d x^2 \sqrt{\frac{b x^4}{a}+1} E\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )}{420 x^2 \sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \sqrt{a+b x^4}} \]

Antiderivative was successfully verified.

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

[Out]

(Sqrt[(I*Sqrt[b])/Sqrt[a]]*((a + b*x^4)*(-210*a*c + b*x^4*(105*c + 84*d*x + 70*e
*x^2 + 60*f*x^3) + 20*a*x*(-21*d + x*(14*e + 9*f*x))) + 315*a*Sqrt[b]*c*x^2*Sqrt
[a + b*x^4]*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]] - 210*a^(3/2)*e*x^2*Sqrt[a +
b*x^4]*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]]) + 1008*a^(3/2)*Sqrt[b]*d*x^2*Sqrt[1 + (
b*x^4)/a]*EllipticE[I*ArcSinh[Sqrt[(I*Sqrt[b])/Sqrt[a]]*x], -1] - (48*I)*a^(3/2)
*((-21*I)*Sqrt[b]*d + 5*Sqrt[a]*f)*x^2*Sqrt[1 + (b*x^4)/a]*EllipticF[I*ArcSinh[S
qrt[(I*Sqrt[b])/Sqrt[a]]*x], -1])/(420*Sqrt[(I*Sqrt[b])/Sqrt[a]]*x^2*Sqrt[a + b*
x^4])

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Maple [C]  time = 0.027, size = 409, normalized size = 1. \[{\frac{bf{x}^{5}}{7}\sqrt{b{x}^{4}+a}}+{\frac{3\,afx}{7}\sqrt{b{x}^{4}+a}}+{\frac{4\,{a}^{2}f}{7}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}+{\frac{{x}^{2}bc}{4}\sqrt{b{x}^{4}+a}}+{\frac{3\,ac}{4}\sqrt{b}\ln \left ( \sqrt{b}{x}^{2}+\sqrt{b{x}^{4}+a} \right ) }-{\frac{ac}{2\,{x}^{2}}\sqrt{b{x}^{4}+a}}-{\frac{ad}{x}\sqrt{b{x}^{4}+a}}+{\frac{{x}^{3}bd}{5}\sqrt{b{x}^{4}+a}}+{{\frac{12\,i}{5}}d{a}^{{\frac{3}{2}}}\sqrt{b}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}-{{\frac{12\,i}{5}}d{a}^{{\frac{3}{2}}}\sqrt{b}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticE} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}-{\frac{e}{2}{a}^{{\frac{3}{2}}}\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{4}+a} \right ) } \right ) }+{\frac{{x}^{4}be}{6}\sqrt{b{x}^{4}+a}}+{\frac{2\,ae}{3}\sqrt{b{x}^{4}+a}} \]

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^3,x)

[Out]

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

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

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^3,x, algorithm="maxima")

[Out]

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

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

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^3,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^3, x)

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Sympy [A]  time = 13.3853, size = 377, normalized size = 0.93 \[ - \frac{a^{\frac{3}{2}} c}{2 x^{2} \sqrt{1 + \frac{b x^{4}}{a}}} + \frac{a^{\frac{3}{2}} d \Gamma \left (- \frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, - \frac{1}{4} \\ \frac{3}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 x \Gamma \left (\frac{3}{4}\right )} - \frac{a^{\frac{3}{2}} e \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x^{2}} \right )}}{2} + \frac{a^{\frac{3}{2}} f x \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{1}{4} \\ \frac{5}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac{5}{4}\right )} + \frac{\sqrt{a} b c x^{2} \sqrt{1 + \frac{b x^{4}}{a}}}{4} - \frac{\sqrt{a} b c x^{2}}{2 \sqrt{1 + \frac{b x^{4}}{a}}} + \frac{\sqrt{a} b d x^{3} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac{7}{4}\right )} + \frac{\sqrt{a} b f x^{5} \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{5}{4} \\ \frac{9}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac{9}{4}\right )} + \frac{a^{2} e}{2 \sqrt{b} x^{2} \sqrt{\frac{a}{b x^{4}} + 1}} + \frac{3 a \sqrt{b} c \operatorname{asinh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a}} \right )}}{4} + \frac{a \sqrt{b} e x^{2}}{2 \sqrt{\frac{a}{b x^{4}} + 1}} + b e \left (\begin{cases} \frac{\sqrt{a} x^{4}}{4} & \text{for}\: b = 0 \\\frac{\left (a + b x^{4}\right )^{\frac{3}{2}}}{6 b} & \text{otherwise} \end{cases}\right ) \]

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**3,x)

[Out]

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

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

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^3,x, algorithm="giac")

[Out]

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