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

Optimal. Leaf size=408 \[ \frac{2 a^{3/4} \sqrt [4]{b} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (9 \sqrt{a} e+5 \sqrt{b} c\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 \sqrt{a+b x^4}}-\frac{12 a^{5/4} \sqrt [4]{b} e \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 \sqrt{a+b x^4}}-\frac{1}{2} a^{3/2} f \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )-\frac{2 \sqrt{a+b x^4} \left (9 a e-5 b c x^2\right )}{15 x}-\frac{\left (a+b x^4\right )^{3/2} \left (5 c-3 e x^2\right )}{15 x^3}-\frac{\left (a+b x^4\right )^{3/2} \left (3 d-f x^2\right )}{6 x^2}+\frac{1}{4} \sqrt{a+b x^4} \left (2 a f+3 b d x^2\right )+\frac{3}{4} a \sqrt{b} d \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )+\frac{12 a \sqrt{b} e x \sqrt{a+b x^4}}{5 \left (\sqrt{a}+\sqrt{b} x^2\right )} \]

[Out]

(12*a*Sqrt[b]*e*x*Sqrt[a + b*x^4])/(5*(Sqrt[a] + Sqrt[b]*x^2)) - (2*(9*a*e - 5*b
*c*x^2)*Sqrt[a + b*x^4])/(15*x) + ((2*a*f + 3*b*d*x^2)*Sqrt[a + b*x^4])/4 - ((5*
c - 3*e*x^2)*(a + b*x^4)^(3/2))/(15*x^3) - ((3*d - f*x^2)*(a + b*x^4)^(3/2))/(6*
x^2) + (3*a*Sqrt[b]*d*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/4 - (a^(3/2)*f*Arc
Tanh[Sqrt[a + b*x^4]/Sqrt[a]])/2 - (12*a^(5/4)*b^(1/4)*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*Sqrt[a + b*x^4]) + (2*a^(3/4)*b^(1/4)*(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])/(15*Sqrt[a + b*x^4])

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Rubi [A]  time = 0.84761, antiderivative size = 408, 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 \[ \frac{2 a^{3/4} \sqrt [4]{b} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (9 \sqrt{a} e+5 \sqrt{b} c\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 \sqrt{a+b x^4}}-\frac{12 a^{5/4} \sqrt [4]{b} e \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 \sqrt{a+b x^4}}-\frac{1}{2} a^{3/2} f \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )-\frac{2 \sqrt{a+b x^4} \left (9 a e-5 b c x^2\right )}{15 x}-\frac{\left (a+b x^4\right )^{3/2} \left (5 c-3 e x^2\right )}{15 x^3}-\frac{\left (a+b x^4\right )^{3/2} \left (3 d-f x^2\right )}{6 x^2}+\frac{1}{4} \sqrt{a+b x^4} \left (2 a f+3 b d x^2\right )+\frac{3}{4} a \sqrt{b} d \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )+\frac{12 a \sqrt{b} e x \sqrt{a+b x^4}}{5 \left (\sqrt{a}+\sqrt{b} x^2\right )} \]

Antiderivative was successfully verified.

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

[Out]

(12*a*Sqrt[b]*e*x*Sqrt[a + b*x^4])/(5*(Sqrt[a] + Sqrt[b]*x^2)) - (2*(9*a*e - 5*b
*c*x^2)*Sqrt[a + b*x^4])/(15*x) + ((2*a*f + 3*b*d*x^2)*Sqrt[a + b*x^4])/4 - ((5*
c - 3*e*x^2)*(a + b*x^4)^(3/2))/(15*x^3) - ((3*d - f*x^2)*(a + b*x^4)^(3/2))/(6*
x^2) + (3*a*Sqrt[b]*d*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/4 - (a^(3/2)*f*Arc
Tanh[Sqrt[a + b*x^4]/Sqrt[a]])/2 - (12*a^(5/4)*b^(1/4)*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*Sqrt[a + b*x^4]) + (2*a^(3/4)*b^(1/4)*(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])/(15*Sqrt[a + b*x^4])

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Rubi in Sympy [A]  time = 91.7594, size = 381, normalized size = 0.93 \[ - \frac{12 a^{\frac{5}{4}} \sqrt [4]{b} e \sqrt{\frac{a + b x^{4}}{\left (\sqrt{a} + \sqrt{b} x^{2}\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x^{2}\right ) E\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}} \right )}\middle | \frac{1}{2}\right )}{5 \sqrt{a + b x^{4}}} + \frac{2 a^{\frac{3}{4}} \sqrt [4]{b} \sqrt{\frac{a + b x^{4}}{\left (\sqrt{a} + \sqrt{b} x^{2}\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x^{2}\right ) \left (9 \sqrt{a} e + 5 \sqrt{b} c\right ) F\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}} \right )}\middle | \frac{1}{2}\right )}{15 \sqrt{a + b x^{4}}} - \frac{a^{\frac{3}{2}} f \operatorname{atanh}{\left (\frac{\sqrt{a + b x^{4}}}{\sqrt{a}} \right )}}{2} + \frac{3 a \sqrt{b} d \operatorname{atanh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a + b x^{4}}} \right )}}{4} + \frac{12 a \sqrt{b} e x \sqrt{a + b x^{4}}}{5 \left (\sqrt{a} + \sqrt{b} x^{2}\right )} + \frac{\sqrt{a + b x^{4}} \left (4 a f + 6 b d x^{2}\right )}{8} - \frac{2 \sqrt{a + b x^{4}} \left (9 a e - 5 b c x^{2}\right )}{15 x} - \frac{\left (a + b x^{4}\right )^{\frac{3}{2}} \left (3 d - f x^{2}\right )}{6 x^{2}} - \frac{\left (a + b x^{4}\right )^{\frac{3}{2}} \left (5 c - 3 e x^{2}\right )}{15 x^{3}} \]

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**4,x)

[Out]

-12*a**(5/4)*b**(1/4)*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*sqrt(a + b*x**4)
) + 2*a**(3/4)*b**(1/4)*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)/(15*sqrt(a + b*x**4)) - a**(3/2)*f*atanh(sqrt(a + b*x**4)/sqrt(a))/2 +
3*a*sqrt(b)*d*atanh(sqrt(b)*x**2/sqrt(a + b*x**4))/4 + 12*a*sqrt(b)*e*x*sqrt(a +
b*x**4)/(5*(sqrt(a) + sqrt(b)*x**2)) + sqrt(a + b*x**4)*(4*a*f + 6*b*d*x**2)/8
- 2*sqrt(a + b*x**4)*(9*a*e - 5*b*c*x**2)/(15*x) - (a + b*x**4)**(3/2)*(3*d - f*
x**2)/(6*x**2) - (a + b*x**4)**(3/2)*(5*c - 3*e*x**2)/(15*x**3)

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Mathematica [C]  time = 1.06983, size = 327, normalized size = 0.8 \[ \frac{\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \left (-30 a^{3/2} f x^3 \sqrt{a+b x^4} \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )+\left (a+b x^4\right ) \left (b x^4 (20 c+x (15 d+2 x (6 e+5 f x)))-10 a \left (2 c+x \left (3 d+6 e x-4 f x^2\right )\right )\right )+45 a \sqrt{b} d x^3 \sqrt{a+b x^4} \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )\right )+144 a^{3/2} \sqrt{b} e x^3 \sqrt{\frac{b x^4}{a}+1} E\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )-16 a \sqrt{b} x^3 \sqrt{\frac{b x^4}{a}+1} \left (9 \sqrt{a} e+5 i \sqrt{b} c\right ) F\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )}{60 x^3 \sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \sqrt{a+b x^4}} \]

Antiderivative was successfully verified.

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

[Out]

(Sqrt[(I*Sqrt[b])/Sqrt[a]]*((a + b*x^4)*(-10*a*(2*c + x*(3*d + 6*e*x - 4*f*x^2))
+ b*x^4*(20*c + x*(15*d + 2*x*(6*e + 5*f*x)))) + 45*a*Sqrt[b]*d*x^3*Sqrt[a + b*
x^4]*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]] - 30*a^(3/2)*f*x^3*Sqrt[a + b*x^4]*A
rcTanh[Sqrt[a + b*x^4]/Sqrt[a]]) + 144*a^(3/2)*Sqrt[b]*e*x^3*Sqrt[1 + (b*x^4)/a]
*EllipticE[I*ArcSinh[Sqrt[(I*Sqrt[b])/Sqrt[a]]*x], -1] - 16*a*Sqrt[b]*((5*I)*Sqr
t[b]*c + 9*Sqrt[a]*e)*x^3*Sqrt[1 + (b*x^4)/a]*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[b
])/Sqrt[a]]*x], -1])/(60*Sqrt[(I*Sqrt[b])/Sqrt[a]]*x^3*Sqrt[a + b*x^4])

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Maple [C]  time = 0.022, size = 408, normalized size = 1. \[ -{\frac{ac}{3\,{x}^{3}}\sqrt{b{x}^{4}+a}}+{\frac{bcx}{3}\sqrt{b{x}^{4}+a}}+{\frac{4\,abc}{3}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}+{\frac{{x}^{2}bd}{4}\sqrt{b{x}^{4}+a}}+{\frac{3\,ad}{4}\sqrt{b}\ln \left ( \sqrt{b}{x}^{2}+\sqrt{b{x}^{4}+a} \right ) }-{\frac{ad}{2\,{x}^{2}}\sqrt{b{x}^{4}+a}}-{\frac{ae}{x}\sqrt{b{x}^{4}+a}}+{\frac{{x}^{3}be}{5}\sqrt{b{x}^{4}+a}}+{{\frac{12\,i}{5}}e{a}^{{\frac{3}{2}}}\sqrt{b}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}-{{\frac{12\,i}{5}}e{a}^{{\frac{3}{2}}}\sqrt{b}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticE} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}-{\frac{f}{2}{a}^{{\frac{3}{2}}}\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{4}+a} \right ) } \right ) }+{\frac{bf{x}^{4}}{6}\sqrt{b{x}^{4}+a}}+{\frac{2\,af}{3}\sqrt{b{x}^{4}+a}} \]

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^4,x)

[Out]

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

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

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^4,x, algorithm="maxima")

[Out]

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

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

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^4,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^4, x)

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Sympy [A]  time = 13.299, size = 381, normalized size = 0.93 \[ \frac{a^{\frac{3}{2}} c \Gamma \left (- \frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{4}, - \frac{1}{2} \\ \frac{1}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{3} \Gamma \left (\frac{1}{4}\right )} - \frac{a^{\frac{3}{2}} d}{2 x^{2} \sqrt{1 + \frac{b x^{4}}{a}}} + \frac{a^{\frac{3}{2}} e \Gamma \left (- \frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, - \frac{1}{4} \\ \frac{3}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 x \Gamma \left (\frac{3}{4}\right )} - \frac{a^{\frac{3}{2}} f \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x^{2}} \right )}}{2} + \frac{\sqrt{a} b c x \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{1}{4} \\ \frac{5}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac{5}{4}\right )} + \frac{\sqrt{a} b d x^{2} \sqrt{1 + \frac{b x^{4}}{a}}}{4} - \frac{\sqrt{a} b d x^{2}}{2 \sqrt{1 + \frac{b x^{4}}{a}}} + \frac{\sqrt{a} b e x^{3} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac{7}{4}\right )} + \frac{a^{2} f}{2 \sqrt{b} x^{2} \sqrt{\frac{a}{b x^{4}} + 1}} + \frac{3 a \sqrt{b} d \operatorname{asinh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a}} \right )}}{4} + \frac{a \sqrt{b} f x^{2}}{2 \sqrt{\frac{a}{b x^{4}} + 1}} + b f \left (\begin{cases} \frac{\sqrt{a} x^{4}}{4} & \text{for}\: b = 0 \\\frac{\left (a + b x^{4}\right )^{\frac{3}{2}}}{6 b} & \text{otherwise} \end{cases}\right ) \]

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**4,x)

[Out]

a**(3/2)*c*gamma(-3/4)*hyper((-3/4, -1/2), (1/4,), b*x**4*exp_polar(I*pi)/a)/(4*
x**3*gamma(1/4)) - a**(3/2)*d/(2*x**2*sqrt(1 + b*x**4/a)) + a**(3/2)*e*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)*f*asinh(sqrt(a)/(sqrt(b)*x**2))/2 + sqrt(a)*b*c*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*d*x**2*sqrt(1
+ b*x**4/a)/4 - sqrt(a)*b*d*x**2/(2*sqrt(1 + b*x**4/a)) + sqrt(a)*b*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)) + a**2
*f/(2*sqrt(b)*x**2*sqrt(a/(b*x**4) + 1)) + 3*a*sqrt(b)*d*asinh(sqrt(b)*x**2/sqrt
(a))/4 + a*sqrt(b)*f*x**2/(2*sqrt(a/(b*x**4) + 1)) + b*f*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. \[ \int \frac{{\left (b x^{4} + a\right )}^{\frac{3}{2}}{\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{4}}\,{d x} \]

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^4,x, algorithm="giac")

[Out]

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