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

Optimal. Leaf size=452 \[ -\frac{2 a^{11/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (77 \sqrt{a} f+65 \sqrt{b} d\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5005 b^{7/4} \sqrt{a+b x^4}}+\frac{4 a^{13/4} f \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{65 b^{7/4} \sqrt{a+b x^4}}-\frac{a^3 e \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{32 b^{3/2}}-\frac{4 a^3 f x \sqrt{a+b x^4}}{65 b^{3/2} \left (\sqrt{a}+\sqrt{b} x^2\right )}+\frac{4 a^2 d x \sqrt{a+b x^4}}{77 b}-\frac{a^2 e x^2 \sqrt{a+b x^4}}{32 b}+\frac{4 a^2 f x^3 \sqrt{a+b x^4}}{195 b}+\frac{\left (a+b x^4\right )^{5/2} \left (6 c+5 e x^2\right )}{60 b}+\frac{1}{143} x^5 \left (a+b x^4\right )^{3/2} \left (13 d+11 f x^2\right )+\frac{2 a x^5 \sqrt{a+b x^4} \left (117 d+77 f x^2\right )}{3003}-\frac{a e x^2 \left (a+b x^4\right )^{3/2}}{48 b} \]

[Out]

(4*a^2*d*x*Sqrt[a + b*x^4])/(77*b) - (a^2*e*x^2*Sqrt[a + b*x^4])/(32*b) + (4*a^2
*f*x^3*Sqrt[a + b*x^4])/(195*b) - (4*a^3*f*x*Sqrt[a + b*x^4])/(65*b^(3/2)*(Sqrt[
a] + Sqrt[b]*x^2)) + (2*a*x^5*(117*d + 77*f*x^2)*Sqrt[a + b*x^4])/3003 - (a*e*x^
2*(a + b*x^4)^(3/2))/(48*b) + (x^5*(13*d + 11*f*x^2)*(a + b*x^4)^(3/2))/143 + ((
6*c + 5*e*x^2)*(a + b*x^4)^(5/2))/(60*b) - (a^3*e*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a +
b*x^4]])/(32*b^(3/2)) + (4*a^(13/4)*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])/(65*b^
(7/4)*Sqrt[a + b*x^4]) - (2*a^(11/4)*(65*Sqrt[b]*d + 77*Sqrt[a]*f)*(Sqrt[a] + Sq
rt[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])/(5005*b^(7/4)*Sqrt[a + b*x^4])

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Rubi [A]  time = 1.19028, antiderivative size = 452, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 11, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.367 \[ -\frac{2 a^{11/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (77 \sqrt{a} f+65 \sqrt{b} d\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5005 b^{7/4} \sqrt{a+b x^4}}+\frac{4 a^{13/4} f \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{65 b^{7/4} \sqrt{a+b x^4}}-\frac{a^3 e \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{32 b^{3/2}}-\frac{4 a^3 f x \sqrt{a+b x^4}}{65 b^{3/2} \left (\sqrt{a}+\sqrt{b} x^2\right )}+\frac{4 a^2 d x \sqrt{a+b x^4}}{77 b}-\frac{a^2 e x^2 \sqrt{a+b x^4}}{32 b}+\frac{4 a^2 f x^3 \sqrt{a+b x^4}}{195 b}+\frac{\left (a+b x^4\right )^{5/2} \left (6 c+5 e x^2\right )}{60 b}+\frac{1}{143} x^5 \left (a+b x^4\right )^{3/2} \left (13 d+11 f x^2\right )+\frac{2 a x^5 \sqrt{a+b x^4} \left (117 d+77 f x^2\right )}{3003}-\frac{a e x^2 \left (a+b x^4\right )^{3/2}}{48 b} \]

Antiderivative was successfully verified.

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

[Out]

(4*a^2*d*x*Sqrt[a + b*x^4])/(77*b) - (a^2*e*x^2*Sqrt[a + b*x^4])/(32*b) + (4*a^2
*f*x^3*Sqrt[a + b*x^4])/(195*b) - (4*a^3*f*x*Sqrt[a + b*x^4])/(65*b^(3/2)*(Sqrt[
a] + Sqrt[b]*x^2)) + (2*a*x^5*(117*d + 77*f*x^2)*Sqrt[a + b*x^4])/3003 - (a*e*x^
2*(a + b*x^4)^(3/2))/(48*b) + (x^5*(13*d + 11*f*x^2)*(a + b*x^4)^(3/2))/143 + ((
6*c + 5*e*x^2)*(a + b*x^4)^(5/2))/(60*b) - (a^3*e*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a +
b*x^4]])/(32*b^(3/2)) + (4*a^(13/4)*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])/(65*b^
(7/4)*Sqrt[a + b*x^4]) - (2*a^(11/4)*(65*Sqrt[b]*d + 77*Sqrt[a]*f)*(Sqrt[a] + Sq
rt[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])/(5005*b^(7/4)*Sqrt[a + b*x^4])

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Rubi in Sympy [A]  time = 108.96, size = 418, normalized size = 0.92 \[ \frac{4 a^{\frac{13}{4}} f \sqrt{\frac{a + b x^{4}}{\left (\sqrt{a} + \sqrt{b} x^{2}\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x^{2}\right ) E\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}} \right )}\middle | \frac{1}{2}\right )}{65 b^{\frac{7}{4}} \sqrt{a + b x^{4}}} - \frac{2 a^{\frac{11}{4}} \sqrt{\frac{a + b x^{4}}{\left (\sqrt{a} + \sqrt{b} x^{2}\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x^{2}\right ) \left (77 \sqrt{a} f + 65 \sqrt{b} d\right ) F\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}} \right )}\middle | \frac{1}{2}\right )}{5005 b^{\frac{7}{4}} \sqrt{a + b x^{4}}} - \frac{a^{3} e \operatorname{atanh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a + b x^{4}}} \right )}}{32 b^{\frac{3}{2}}} - \frac{4 a^{3} f x \sqrt{a + b x^{4}}}{65 b^{\frac{3}{2}} \left (\sqrt{a} + \sqrt{b} x^{2}\right )} + \frac{4 a^{2} d x \sqrt{a + b x^{4}}}{77 b} - \frac{a^{2} e x^{2} \sqrt{a + b x^{4}}}{32 b} + \frac{4 a^{2} f x^{3} \sqrt{a + b x^{4}}}{195 b} + \frac{2 a x^{5} \sqrt{a + b x^{4}} \left (117 d + 77 f x^{2}\right )}{3003} - \frac{a e x^{2} \left (a + b x^{4}\right )^{\frac{3}{2}}}{48 b} + \frac{x^{5} \left (a + b x^{4}\right )^{\frac{3}{2}} \left (13 d + 11 f x^{2}\right )}{143} + \frac{\left (a + b x^{4}\right )^{\frac{5}{2}} \left (6 c + 5 e x^{2}\right )}{60 b} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

4*a**(13/4)*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)/(65*b**(7/4)*sqrt(a + b*x**4)
) - 2*a**(11/4)*sqrt((a + b*x**4)/(sqrt(a) + sqrt(b)*x**2)**2)*(sqrt(a) + sqrt(b
)*x**2)*(77*sqrt(a)*f + 65*sqrt(b)*d)*elliptic_f(2*atan(b**(1/4)*x/a**(1/4)), 1/
2)/(5005*b**(7/4)*sqrt(a + b*x**4)) - a**3*e*atanh(sqrt(b)*x**2/sqrt(a + b*x**4)
)/(32*b**(3/2)) - 4*a**3*f*x*sqrt(a + b*x**4)/(65*b**(3/2)*(sqrt(a) + sqrt(b)*x*
*2)) + 4*a**2*d*x*sqrt(a + b*x**4)/(77*b) - a**2*e*x**2*sqrt(a + b*x**4)/(32*b)
+ 4*a**2*f*x**3*sqrt(a + b*x**4)/(195*b) + 2*a*x**5*sqrt(a + b*x**4)*(117*d + 77
*f*x**2)/3003 - a*e*x**2*(a + b*x**4)**(3/2)/(48*b) + x**5*(a + b*x**4)**(3/2)*(
13*d + 11*f*x**2)/143 + (a + b*x**4)**(5/2)*(6*c + 5*e*x**2)/(60*b)

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Mathematica [C]  time = 6.13172, size = 431, normalized size = 0.95 \[ \sqrt{a+b x^4} \left (\frac{a^2 c}{10 b}+\frac{4 a^2 d x}{77 b}+\frac{a^2 e x^2}{32 b}+\frac{4 a^2 f x^3}{195 b}+\frac{1}{5} a c x^4+\frac{13}{77} a d x^5+\frac{7}{48} a e x^6+\frac{5}{39} a f x^7+\frac{1}{10} b c x^8+\frac{1}{11} b d x^9+\frac{1}{12} b e x^{10}+\frac{1}{13} b f x^{11}\right )-\frac{a^3 \left (-\frac{4160 i d \sqrt{1-\frac{i \sqrt{b} x^2}{\sqrt{a}}} \sqrt{1+\frac{i \sqrt{b} x^2}{\sqrt{a}}} F\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )}{\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \sqrt{a+b x^4}}+\frac{5005 e \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{2 \sqrt{b}}+\frac{4928 \sqrt{a} f \sqrt{1-\frac{i \sqrt{b} x^2}{\sqrt{a}}} \sqrt{1+\frac{i \sqrt{b} x^2}{\sqrt{a}}} \left (E\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )-F\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )\right )}{\sqrt{b} \sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \sqrt{a+b x^4}}\right )}{80080 b} \]

Antiderivative was successfully verified.

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

[Out]

Sqrt[a + b*x^4]*((a^2*c)/(10*b) + (4*a^2*d*x)/(77*b) + (a^2*e*x^2)/(32*b) + (4*a
^2*f*x^3)/(195*b) + (a*c*x^4)/5 + (13*a*d*x^5)/77 + (7*a*e*x^6)/48 + (5*a*f*x^7)
/39 + (b*c*x^8)/10 + (b*d*x^9)/11 + (b*e*x^10)/12 + (b*f*x^11)/13) - (a^3*((5005
*e*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(2*Sqrt[b]) + (4928*Sqrt[a]*f*Sqrt[1
- (I*Sqrt[b]*x^2)/Sqrt[a]]*Sqrt[1 + (I*Sqrt[b]*x^2)/Sqrt[a]]*(EllipticE[I*ArcSin
h[Sqrt[(I*Sqrt[b])/Sqrt[a]]*x], -1] - EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[b])/Sqrt[
a]]*x], -1]))/(Sqrt[(I*Sqrt[b])/Sqrt[a]]*Sqrt[b]*Sqrt[a + b*x^4]) - ((4160*I)*d*
Sqrt[1 - (I*Sqrt[b]*x^2)/Sqrt[a]]*Sqrt[1 + (I*Sqrt[b]*x^2)/Sqrt[a]]*EllipticF[I*
ArcSinh[Sqrt[(I*Sqrt[b])/Sqrt[a]]*x], -1])/(Sqrt[(I*Sqrt[b])/Sqrt[a]]*Sqrt[a + b
*x^4])))/(80080*b)

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Maple [C]  time = 0.013, size = 434, normalized size = 1. \[{\frac{bd{x}^{9}}{11}\sqrt{b{x}^{4}+a}}+{\frac{13\,ad{x}^{5}}{77}\sqrt{b{x}^{4}+a}}+{\frac{4\,{a}^{2}dx}{77\,b}\sqrt{b{x}^{4}+a}}-{\frac{4\,{a}^{3}d}{77\,b}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}+{\frac{c}{10\,b} \left ( b{x}^{4}+a \right ) ^{{\frac{5}{2}}}}+{\frac{{a}^{2}e{x}^{2}}{32\,b}\sqrt{b{x}^{4}+a}}-{\frac{{a}^{3}e}{32}\ln \left ( \sqrt{b}{x}^{2}+\sqrt{b{x}^{4}+a} \right ){b}^{-{\frac{3}{2}}}}+{\frac{be{x}^{10}}{12}\sqrt{b{x}^{4}+a}}+{\frac{7\,ae{x}^{6}}{48}\sqrt{b{x}^{4}+a}}+{\frac{bf{x}^{11}}{13}\sqrt{b{x}^{4}+a}}+{\frac{5\,af{x}^{7}}{39}\sqrt{b{x}^{4}+a}}+{\frac{4\,{a}^{2}f{x}^{3}}{195\,b}\sqrt{b{x}^{4}+a}}-{{\frac{4\,i}{65}}f{a}^{{\frac{7}{2}}}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){b}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}+{{\frac{4\,i}{65}}f{a}^{{\frac{7}{2}}}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticE} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){b}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

1/11*d*b*x^9*(b*x^4+a)^(1/2)+13/77*d*a*x^5*(b*x^4+a)^(1/2)+4/77*a^2*d*x*(b*x^4+a
)^(1/2)/b-4/77*d/b*a^3/(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)+1/10*c/b*(b*x^4+a)^(5/2)+1/32*a^2*e*x^2*(b*x^4+a)^(1/2)/b-1/32*e*a^3/b
^(3/2)*ln(b^(1/2)*x^2+(b*x^4+a)^(1/2))+1/12*e*b*x^10*(b*x^4+a)^(1/2)+7/48*e*a*x^
6*(b*x^4+a)^(1/2)+1/13*f*b*x^11*(b*x^4+a)^(1/2)+5/39*f*a*x^7*(b*x^4+a)^(1/2)+4/1
95*a^2*f*x^3*(b*x^4+a)^(1/2)/b-4/65*I*f/b^(3/2)*a^(7/2)/(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)+4/65*I*f/b^(3/2)*a^(7/2)/(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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \frac{{\left (b x^{4} + a\right )}^{\frac{5}{2}} c}{10 \, b} + \int{\left (b f x^{10} + b e x^{9} + b d x^{8} + a f x^{6} + a e x^{5} + a d x^{4}\right )} \sqrt{b x^{4} + a}\,{d x} \]

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]

1/10*(b*x^4 + a)^(5/2)*c/b + integrate((b*f*x^10 + b*e*x^9 + b*d*x^8 + a*f*x^6 +
a*e*x^5 + a*d*x^4)*sqrt(b*x^4 + a), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left ({\left (b f x^{10} + b e x^{9} + b d x^{8} + b c x^{7} + a f x^{6} + a e x^{5} + a d x^{4} + a c x^{3}\right )} \sqrt{b x^{4} + a}, x\right ) \]

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^3,x, algorithm="fricas")

[Out]

integral((b*f*x^10 + b*e*x^9 + b*d*x^8 + b*c*x^7 + a*f*x^6 + a*e*x^5 + a*d*x^4 +
a*c*x^3)*sqrt(b*x^4 + a), x)

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Sympy [A]  time = 23.8878, size = 398, normalized size = 0.88 \[ \frac{a^{\frac{5}{2}} e x^{2}}{32 b \sqrt{1 + \frac{b x^{4}}{a}}} + \frac{a^{\frac{3}{2}} d x^{5} \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{5}{4} \\ \frac{9}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac{9}{4}\right )} + \frac{17 a^{\frac{3}{2}} e x^{6}}{96 \sqrt{1 + \frac{b x^{4}}{a}}} + \frac{a^{\frac{3}{2}} f x^{7} \Gamma \left (\frac{7}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{7}{4} \\ \frac{11}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac{11}{4}\right )} + \frac{\sqrt{a} b d x^{9} \Gamma \left (\frac{9}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{9}{4} \\ \frac{13}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac{13}{4}\right )} + \frac{11 \sqrt{a} b e x^{10}}{48 \sqrt{1 + \frac{b x^{4}}{a}}} + \frac{\sqrt{a} b f x^{11} \Gamma \left (\frac{11}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{11}{4} \\ \frac{15}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac{15}{4}\right )} - \frac{a^{3} e \operatorname{asinh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a}} \right )}}{32 b^{\frac{3}{2}}} + a c \left (\begin{cases} \frac{\sqrt{a} x^{4}}{4} & \text{for}\: b = 0 \\\frac{\left (a + b x^{4}\right )^{\frac{3}{2}}}{6 b} & \text{otherwise} \end{cases}\right ) + b c \left (\begin{cases} - \frac{a^{2} \sqrt{a + b x^{4}}}{15 b^{2}} + \frac{a x^{4} \sqrt{a + b x^{4}}}{30 b} + \frac{x^{8} \sqrt{a + b x^{4}}}{10} & \text{for}\: b \neq 0 \\\frac{\sqrt{a} x^{8}}{8} & \text{otherwise} \end{cases}\right ) + \frac{b^{2} e x^{14}}{12 \sqrt{a} \sqrt{1 + \frac{b x^{4}}{a}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

a**(5/2)*e*x**2/(32*b*sqrt(1 + b*x**4/a)) + a**(3/2)*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)) + 17*a**(3/2)*e*x**6/
(96*sqrt(1 + b*x**4/a)) + a**(3/2)*f*x**7*gamma(7/4)*hyper((-1/2, 7/4), (11/4,),
b*x**4*exp_polar(I*pi)/a)/(4*gamma(11/4)) + sqrt(a)*b*d*x**9*gamma(9/4)*hyper((
-1/2, 9/4), (13/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(13/4)) + 11*sqrt(a)*b*e*
x**10/(48*sqrt(1 + b*x**4/a)) + sqrt(a)*b*f*x**11*gamma(11/4)*hyper((-1/2, 11/4)
, (15/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(15/4)) - a**3*e*asinh(sqrt(b)*x**2
/sqrt(a))/(32*b**(3/2)) + a*c*Piecewise((sqrt(a)*x**4/4, Eq(b, 0)), ((a + b*x**4
)**(3/2)/(6*b), True)) + b*c*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*e*x**14/(12*sqrt(a)*sqrt(1 + b*x**4/a))

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

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)