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

Optimal. Leaf size=403 \[ \frac{2 a^{7/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (7 \sqrt{a} f+15 \sqrt{b} d\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{105 b^{3/4} \sqrt{a+b x^4}}-\frac{4 a^{9/4} f \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 b^{3/4} \sqrt{a+b x^4}}-\frac{1}{2} a^{3/2} c \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )+\frac{3 a^2 e \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{16 \sqrt{b}}+\frac{4 a^2 f x \sqrt{a+b x^4}}{15 \sqrt{b} \left (\sqrt{a}+\sqrt{b} x^2\right )}+\frac{1}{16} a \sqrt{a+b x^4} \left (8 c+3 e x^2\right )+\frac{1}{24} \left (a+b x^4\right )^{3/2} \left (4 c+3 e x^2\right )+\frac{2}{105} a x \sqrt{a+b x^4} \left (15 d+7 f x^2\right )+\frac{1}{63} x \left (a+b x^4\right )^{3/2} \left (9 d+7 f x^2\right ) \]

[Out]

(4*a^2*f*x*Sqrt[a + b*x^4])/(15*Sqrt[b]*(Sqrt[a] + Sqrt[b]*x^2)) + (a*(8*c + 3*e
*x^2)*Sqrt[a + b*x^4])/16 + (2*a*x*(15*d + 7*f*x^2)*Sqrt[a + b*x^4])/105 + ((4*c
+ 3*e*x^2)*(a + b*x^4)^(3/2))/24 + (x*(9*d + 7*f*x^2)*(a + b*x^4)^(3/2))/63 + (
3*a^2*e*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(16*Sqrt[b]) - (a^(3/2)*c*ArcTan
h[Sqrt[a + b*x^4]/Sqrt[a]])/2 - (4*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^(3/4)*Sqrt[a + b*x^4]) + (2*a^(7/4)*(15*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^(3/4)*Sqrt[a + b*x^4])

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Rubi [A]  time = 0.910761, antiderivative size = 403, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 13, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.433 \[ \frac{2 a^{7/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (7 \sqrt{a} f+15 \sqrt{b} d\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{105 b^{3/4} \sqrt{a+b x^4}}-\frac{4 a^{9/4} f \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 b^{3/4} \sqrt{a+b x^4}}-\frac{1}{2} a^{3/2} c \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )+\frac{3 a^2 e \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{16 \sqrt{b}}+\frac{4 a^2 f x \sqrt{a+b x^4}}{15 \sqrt{b} \left (\sqrt{a}+\sqrt{b} x^2\right )}+\frac{1}{16} a \sqrt{a+b x^4} \left (8 c+3 e x^2\right )+\frac{1}{24} \left (a+b x^4\right )^{3/2} \left (4 c+3 e x^2\right )+\frac{2}{105} a x \sqrt{a+b x^4} \left (15 d+7 f x^2\right )+\frac{1}{63} x \left (a+b x^4\right )^{3/2} \left (9 d+7 f x^2\right ) \]

Antiderivative was successfully verified.

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

[Out]

(4*a^2*f*x*Sqrt[a + b*x^4])/(15*Sqrt[b]*(Sqrt[a] + Sqrt[b]*x^2)) + (a*(8*c + 3*e
*x^2)*Sqrt[a + b*x^4])/16 + (2*a*x*(15*d + 7*f*x^2)*Sqrt[a + b*x^4])/105 + ((4*c
+ 3*e*x^2)*(a + b*x^4)^(3/2))/24 + (x*(9*d + 7*f*x^2)*(a + b*x^4)^(3/2))/63 + (
3*a^2*e*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(16*Sqrt[b]) - (a^(3/2)*c*ArcTan
h[Sqrt[a + b*x^4]/Sqrt[a]])/2 - (4*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^(3/4)*Sqrt[a + b*x^4]) + (2*a^(7/4)*(15*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^(3/4)*Sqrt[a + b*x^4])

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Rubi in Sympy [A]  time = 90.4774, size = 376, normalized size = 0.93 \[ - \frac{4 a^{\frac{9}{4}} f \sqrt{\frac{a + b x^{4}}{\left (\sqrt{a} + \sqrt{b} x^{2}\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x^{2}\right ) E\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}} \right )}\middle | \frac{1}{2}\right )}{15 b^{\frac{3}{4}} \sqrt{a + b x^{4}}} + \frac{2 a^{\frac{7}{4}} \sqrt{\frac{a + b x^{4}}{\left (\sqrt{a} + \sqrt{b} x^{2}\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x^{2}\right ) \left (7 \sqrt{a} f + 15 \sqrt{b} d\right ) F\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}} \right )}\middle | \frac{1}{2}\right )}{105 b^{\frac{3}{4}} \sqrt{a + b x^{4}}} - \frac{a^{\frac{3}{2}} c \operatorname{atanh}{\left (\frac{\sqrt{a + b x^{4}}}{\sqrt{a}} \right )}}{2} + \frac{3 a^{2} e \operatorname{atanh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a + b x^{4}}} \right )}}{16 \sqrt{b}} + \frac{4 a^{2} f x \sqrt{a + b x^{4}}}{15 \sqrt{b} \left (\sqrt{a} + \sqrt{b} x^{2}\right )} + \frac{2 a x \sqrt{a + b x^{4}} \left (45 d + 21 f x^{2}\right )}{315} + \frac{a \sqrt{a + b x^{4}} \left (8 c + 3 e x^{2}\right )}{16} + \frac{x \left (a + b x^{4}\right )^{\frac{3}{2}} \left (9 d + 7 f x^{2}\right )}{63} + \frac{\left (a + b x^{4}\right )^{\frac{3}{2}} \left (4 c + 3 e x^{2}\right )}{24} \]

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,x)

[Out]

-4*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**(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)*f + 15*sqrt(b)*d)*elliptic_f(2*atan(b**(1/4)*x/a**(1/4)), 1/2)
/(105*b**(3/4)*sqrt(a + b*x**4)) - a**(3/2)*c*atanh(sqrt(a + b*x**4)/sqrt(a))/2
+ 3*a**2*e*atanh(sqrt(b)*x**2/sqrt(a + b*x**4))/(16*sqrt(b)) + 4*a**2*f*x*sqrt(a
+ b*x**4)/(15*sqrt(b)*(sqrt(a) + sqrt(b)*x**2)) + 2*a*x*sqrt(a + b*x**4)*(45*d
+ 21*f*x**2)/315 + a*sqrt(a + b*x**4)*(8*c + 3*e*x**2)/16 + x*(a + b*x**4)**(3/2
)*(9*d + 7*f*x**2)/63 + (a + b*x**4)**(3/2)*(4*c + 3*e*x**2)/24

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Mathematica [C]  time = 0.895006, size = 319, normalized size = 0.79 \[ -\frac{1}{2} a^{3/2} c \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )-\frac{4 i a^2 d \sqrt{\frac{b x^4}{a}+1} F\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )}{7 \sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \sqrt{a+b x^4}}+\frac{3 a^2 e \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{16 \sqrt{b}}+\frac{4 i a^2 f \sqrt{\frac{b x^4}{a}+1} \left (E\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )-F\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )\right )}{15 \left (\frac{i \sqrt{b}}{\sqrt{a}}\right )^{3/2} \sqrt{a+b x^4}}+\frac{\sqrt{a+b x^4} \left (a (3360 c+x (2160 d+7 x (225 e+176 f x)))+10 b x^4 (84 c+x (72 d+7 x (9 e+8 f x)))\right )}{5040} \]

Antiderivative was successfully verified.

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

[Out]

(Sqrt[a + b*x^4]*(10*b*x^4*(84*c + x*(72*d + 7*x*(9*e + 8*f*x))) + a*(3360*c + x
*(2160*d + 7*x*(225*e + 176*f*x)))))/5040 + (3*a^2*e*ArcTanh[(Sqrt[b]*x^2)/Sqrt[
a + b*x^4]])/(16*Sqrt[b]) - (a^(3/2)*c*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]])/2 + (((
4*I)/15)*a^2*f*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*d*Sqrt[1 + (b*x^4)/a]*Ellipti
cF[I*ArcSinh[Sqrt[(I*Sqrt[b])/Sqrt[a]]*x], -1])/(Sqrt[(I*Sqrt[b])/Sqrt[a]]*Sqrt[
a + b*x^4])

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Maple [C]  time = 0.025, size = 411, normalized size = 1. \[{\frac{bd{x}^{5}}{7}\sqrt{b{x}^{4}+a}}+{\frac{3\,adx}{7}\sqrt{b{x}^{4}+a}}+{\frac{4\,{a}^{2}d}{7}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}+{\frac{bf{x}^{7}}{9}\sqrt{b{x}^{4}+a}}+{\frac{11\,{x}^{3}af}{45}\sqrt{b{x}^{4}+a}}+{{\frac{4\,i}{15}}f{a}^{{\frac{5}{2}}}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}{\frac{1}{\sqrt{b}}}}-{{\frac{4\,i}{15}}f{a}^{{\frac{5}{2}}}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticE} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}{\frac{1}{\sqrt{b}}}}-{\frac{c}{2}{a}^{{\frac{3}{2}}}\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{4}+a} \right ) } \right ) }+{\frac{bc{x}^{4}}{6}\sqrt{b{x}^{4}+a}}+{\frac{2\,ac}{3}\sqrt{b{x}^{4}+a}}+{\frac{3\,e{a}^{2}}{16}\ln \left ( \sqrt{b}{x}^{2}+\sqrt{b{x}^{4}+a} \right ){\frac{1}{\sqrt{b}}}}+{\frac{be{x}^{6}}{8}\sqrt{b{x}^{4}+a}}+{\frac{5\,ae{x}^{2}}{16}\sqrt{b{x}^{4}+a}} \]

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,x)

[Out]

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

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

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,x, algorithm="maxima")

[Out]

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

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

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,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, x)

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Sympy [A]  time = 17.0931, size = 405, normalized size = 1. \[ - \frac{a^{\frac{3}{2}} c \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x^{2}} \right )}}{2} + \frac{a^{\frac{3}{2}} d x \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{1}{4} \\ \frac{5}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac{5}{4}\right )} + \frac{a^{\frac{3}{2}} e x^{2} \sqrt{1 + \frac{b x^{4}}{a}}}{4} + \frac{a^{\frac{3}{2}} e x^{2}}{16 \sqrt{1 + \frac{b x^{4}}{a}}} + \frac{a^{\frac{3}{2}} f x^{3} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac{7}{4}\right )} + \frac{\sqrt{a} b d x^{5} \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{5}{4} \\ \frac{9}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac{9}{4}\right )} + \frac{3 \sqrt{a} b e x^{6}}{16 \sqrt{1 + \frac{b x^{4}}{a}}} + \frac{\sqrt{a} b f x^{7} \Gamma \left (\frac{7}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{7}{4} \\ \frac{11}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac{11}{4}\right )} + \frac{a^{2} c}{2 \sqrt{b} x^{2} \sqrt{\frac{a}{b x^{4}} + 1}} + \frac{3 a^{2} e \operatorname{asinh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a}} \right )}}{16 \sqrt{b}} + \frac{a \sqrt{b} c x^{2}}{2 \sqrt{\frac{a}{b x^{4}} + 1}} + b c \left (\begin{cases} \frac{\sqrt{a} x^{4}}{4} & \text{for}\: b = 0 \\\frac{\left (a + b x^{4}\right )^{\frac{3}{2}}}{6 b} & \text{otherwise} \end{cases}\right ) + \frac{b^{2} e x^{10}}{8 \sqrt{a} \sqrt{1 + \frac{b x^{4}}{a}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-a**(3/2)*c*asinh(sqrt(a)/(sqrt(b)*x**2))/2 + a**(3/2)*d*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)*e*x**2*sqrt
(1 + b*x**4/a)/4 + a**(3/2)*e*x**2/(16*sqrt(1 + b*x**4/a)) + a**(3/2)*f*x**3*gam
ma(3/4)*hyper((-1/2, 3/4), (7/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(7/4)) + sq
rt(a)*b*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)*b*e*x**6/(16*sqrt(1 + b*x**4/a)) + sqrt(a)*b*f*x**7*ga
mma(7/4)*hyper((-1/2, 7/4), (11/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(11/4)) +
a**2*c/(2*sqrt(b)*x**2*sqrt(a/(b*x**4) + 1)) + 3*a**2*e*asinh(sqrt(b)*x**2/sqrt
(a))/(16*sqrt(b)) + a*sqrt(b)*c*x**2/(2*sqrt(a/(b*x**4) + 1)) + b*c*Piecewise((s
qrt(a)*x**4/4, Eq(b, 0)), ((a + b*x**4)**(3/2)/(6*b), True)) + b**2*e*x**10/(8*s
qrt(a)*sqrt(1 + b*x**4/a))

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

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,x, algorithm="giac")

[Out]

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