3.514 \(\int \frac{\left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^{3/2}}{x^{12}} \, dx\)

Optimal. Leaf size=424 \[ -\frac{2 b^{9/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (15 \sqrt{b} c-77 \sqrt{a} e\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{1155 a^{5/4} \sqrt{a+b x^4}}-\frac{4 b^{9/4} e \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 a^{3/4} \sqrt{a+b x^4}}+\frac{4 b^{5/2} e x \sqrt{a+b x^4}}{15 a \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{4 b^2 c \sqrt{a+b x^4}}{77 a x^3}-\frac{b^2 d \sqrt{a+b x^4}}{10 a x^2}-\frac{4 b^2 e \sqrt{a+b x^4}}{15 a x}-\frac{3 b^2 f \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{16 \sqrt{a}}-\frac{b \sqrt{a+b x^4} \left (\frac{1440 c}{x^7}+\frac{1848 d}{x^6}+\frac{2464 e}{x^5}+\frac{3465 f}{x^4}\right )}{18480}-\frac{\left (a+b x^4\right )^{3/2} \left (\frac{360 c}{x^{11}}+\frac{396 d}{x^{10}}+\frac{440 e}{x^9}+\frac{495 f}{x^8}\right )}{3960} \]

[Out]

-(b*((1440*c)/x^7 + (1848*d)/x^6 + (2464*e)/x^5 + (3465*f)/x^4)*Sqrt[a + b*x^4])
/18480 - (4*b^2*c*Sqrt[a + b*x^4])/(77*a*x^3) - (b^2*d*Sqrt[a + b*x^4])/(10*a*x^
2) - (4*b^2*e*Sqrt[a + b*x^4])/(15*a*x) + (4*b^(5/2)*e*x*Sqrt[a + b*x^4])/(15*a*
(Sqrt[a] + Sqrt[b]*x^2)) - (((360*c)/x^11 + (396*d)/x^10 + (440*e)/x^9 + (495*f)
/x^8)*(a + b*x^4)^(3/2))/3960 - (3*b^2*f*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]])/(16*S
qrt[a]) - (4*b^(9/4)*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])/(15*a^(3/4)*Sqrt[a + b
*x^4]) - (2*b^(9/4)*(15*Sqrt[b]*c - 77*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])/(1155*a^(5/4)*Sqrt[a + b*x^4])

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Rubi [A]  time = 0.98442, antiderivative size = 424, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 12, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4 \[ -\frac{2 b^{9/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (15 \sqrt{b} c-77 \sqrt{a} e\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{1155 a^{5/4} \sqrt{a+b x^4}}-\frac{4 b^{9/4} e \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 a^{3/4} \sqrt{a+b x^4}}+\frac{4 b^{5/2} e x \sqrt{a+b x^4}}{15 a \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{4 b^2 c \sqrt{a+b x^4}}{77 a x^3}-\frac{b^2 d \sqrt{a+b x^4}}{10 a x^2}-\frac{4 b^2 e \sqrt{a+b x^4}}{15 a x}-\frac{3 b^2 f \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{16 \sqrt{a}}-\frac{b \sqrt{a+b x^4} \left (\frac{1440 c}{x^7}+\frac{1848 d}{x^6}+\frac{2464 e}{x^5}+\frac{3465 f}{x^4}\right )}{18480}-\frac{\left (a+b x^4\right )^{3/2} \left (\frac{360 c}{x^{11}}+\frac{396 d}{x^{10}}+\frac{440 e}{x^9}+\frac{495 f}{x^8}\right )}{3960} \]

Antiderivative was successfully verified.

[In]  Int[((c + d*x + e*x^2 + f*x^3)*(a + b*x^4)^(3/2))/x^12,x]

[Out]

-(b*((1440*c)/x^7 + (1848*d)/x^6 + (2464*e)/x^5 + (3465*f)/x^4)*Sqrt[a + b*x^4])
/18480 - (4*b^2*c*Sqrt[a + b*x^4])/(77*a*x^3) - (b^2*d*Sqrt[a + b*x^4])/(10*a*x^
2) - (4*b^2*e*Sqrt[a + b*x^4])/(15*a*x) + (4*b^(5/2)*e*x*Sqrt[a + b*x^4])/(15*a*
(Sqrt[a] + Sqrt[b]*x^2)) - (((360*c)/x^11 + (396*d)/x^10 + (440*e)/x^9 + (495*f)
/x^8)*(a + b*x^4)^(3/2))/3960 - (3*b^2*f*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]])/(16*S
qrt[a]) - (4*b^(9/4)*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])/(15*a^(3/4)*Sqrt[a + b
*x^4]) - (2*b^(9/4)*(15*Sqrt[b]*c - 77*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])/(1155*a^(5/4)*Sqrt[a + b*x^4])

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Rubi in Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \[ \text{Timed out} \]

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**12,x)

[Out]

Timed out

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Mathematica [C]  time = 0.994866, size = 317, normalized size = 0.75 \[ \frac{-\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \left (\left (a+b x^4\right ) \left (14 a^2 (360 c+11 x (36 d+5 x (8 e+9 f x)))+a b x^4 \left (9360 c+77 x \left (144 d+176 e x+225 f x^2\right )\right )+24 b^2 x^8 (120 c+77 x (3 d+8 e x))\right )+10395 \sqrt{a} b^2 f x^{11} \sqrt{a+b x^4} \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )\right )-192 b^{5/2} x^{11} \sqrt{\frac{b x^4}{a}+1} \left (77 \sqrt{a} e-15 i \sqrt{b} c\right ) F\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )+14784 \sqrt{a} b^{5/2} e x^{11} \sqrt{\frac{b x^4}{a}+1} E\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )}{55440 a x^{11} \sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \sqrt{a+b x^4}} \]

Antiderivative was successfully verified.

[In]  Integrate[((c + d*x + e*x^2 + f*x^3)*(a + b*x^4)^(3/2))/x^12,x]

[Out]

(-(Sqrt[(I*Sqrt[b])/Sqrt[a]]*((a + b*x^4)*(24*b^2*x^8*(120*c + 77*x*(3*d + 8*e*x
)) + a*b*x^4*(9360*c + 77*x*(144*d + 176*e*x + 225*f*x^2)) + 14*a^2*(360*c + 11*
x*(36*d + 5*x*(8*e + 9*f*x)))) + 10395*Sqrt[a]*b^2*f*x^11*Sqrt[a + b*x^4]*ArcTan
h[Sqrt[a + b*x^4]/Sqrt[a]])) + 14784*Sqrt[a]*b^(5/2)*e*x^11*Sqrt[1 + (b*x^4)/a]*
EllipticE[I*ArcSinh[Sqrt[(I*Sqrt[b])/Sqrt[a]]*x], -1] - 192*b^(5/2)*((-15*I)*Sqr
t[b]*c + 77*Sqrt[a]*e)*x^11*Sqrt[1 + (b*x^4)/a]*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt
[b])/Sqrt[a]]*x], -1])/(55440*a*Sqrt[(I*Sqrt[b])/Sqrt[a]]*x^11*Sqrt[a + b*x^4])

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Maple [C]  time = 0.031, size = 441, normalized size = 1. \[ -{\frac{ac}{11\,{x}^{11}}\sqrt{b{x}^{4}+a}}-{\frac{13\,bc}{77\,{x}^{7}}\sqrt{b{x}^{4}+a}}-{\frac{4\,{b}^{2}c}{77\,a{x}^{3}}\sqrt{b{x}^{4}+a}}-{\frac{4\,{b}^{3}c}{77\,a}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}-{\frac{d \left ({b}^{2}{x}^{8}+2\,ab{x}^{4}+{a}^{2} \right ) }{10\,{x}^{10}a}\sqrt{b{x}^{4}+a}}-{\frac{ae}{9\,{x}^{9}}\sqrt{b{x}^{4}+a}}-{\frac{11\,be}{45\,{x}^{5}}\sqrt{b{x}^{4}+a}}-{\frac{4\,{b}^{2}e}{15\,ax}\sqrt{b{x}^{4}+a}}+{{\frac{4\,i}{15}}e{b}^{{\frac{5}{2}}}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{a}}}{\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}-{{\frac{4\,i}{15}}e{b}^{{\frac{5}{2}}}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticE} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{a}}}{\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}-{\frac{af}{8\,{x}^{8}}\sqrt{b{x}^{4}+a}}-{\frac{5\,fb}{16\,{x}^{4}}\sqrt{b{x}^{4}+a}}-{\frac{3\,f{b}^{2}}{16}\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{4}+a} \right ) } \right ){\frac{1}{\sqrt{a}}}} \]

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^12,x)

[Out]

-1/11*c*a*(b*x^4+a)^(1/2)/x^11-13/77*c*b*(b*x^4+a)^(1/2)/x^7-4/77*b^2*c*(b*x^4+a
)^(1/2)/a/x^3-4/77*c/a*b^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*d*(b*x^4+a)^(1/2)/x^10/a*(b^2*x^8+2*a*b*x^4+a^2)-1/9*e*a*(b*x
^4+a)^(1/2)/x^9-11/45*e*b*(b*x^4+a)^(1/2)/x^5-4/15*b^2*e*(b*x^4+a)^(1/2)/a/x+4/1
5*I*e/a^(1/2)*b^(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)*EllipticF(x*(I/a^(1/2)*b^(1/2))^
(1/2),I)-4/15*I*e/a^(1/2)*b^(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)*EllipticE(x*(I/a^(1/
2)*b^(1/2))^(1/2),I)-1/8*f*a/x^8*(b*x^4+a)^(1/2)-5/16*f*b/x^4*(b*x^4+a)^(1/2)-3/
16*f/a^(1/2)*b^2*ln((2*a+2*a^(1/2)*(b*x^4+a)^(1/2))/x^2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (b x^{4} + a\right )}^{\frac{3}{2}}{\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{12}}\,{d x} \]

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^12,x, algorithm="maxima")

[Out]

integrate((b*x^4 + a)^(3/2)*(f*x^3 + e*x^2 + d*x + c)/x^12, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{{\left (b f x^{7} + b e x^{6} + b d x^{5} + b c x^{4} + a f x^{3} + a e x^{2} + a d x + a c\right )} \sqrt{b x^{4} + a}}{x^{12}}, x\right ) \]

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^12,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^12, x)

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Sympy [A]  time = 27.6757, size = 401, normalized size = 0.95 \[ \frac{a^{\frac{3}{2}} c \Gamma \left (- \frac{11}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{11}{4}, - \frac{1}{2} \\ - \frac{7}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{11} \Gamma \left (- \frac{7}{4}\right )} + \frac{a^{\frac{3}{2}} e \Gamma \left (- \frac{9}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{9}{4}, - \frac{1}{2} \\ - \frac{5}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{9} \Gamma \left (- \frac{5}{4}\right )} + \frac{\sqrt{a} b c \Gamma \left (- \frac{7}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{7}{4}, - \frac{1}{2} \\ - \frac{3}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{7} \Gamma \left (- \frac{3}{4}\right )} + \frac{\sqrt{a} b e \Gamma \left (- \frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{5}{4}, - \frac{1}{2} \\ - \frac{1}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{5} \Gamma \left (- \frac{1}{4}\right )} - \frac{a^{2} f}{8 \sqrt{b} x^{10} \sqrt{\frac{a}{b x^{4}} + 1}} - \frac{a \sqrt{b} d \sqrt{\frac{a}{b x^{4}} + 1}}{10 x^{8}} - \frac{3 a \sqrt{b} f}{16 x^{6} \sqrt{\frac{a}{b x^{4}} + 1}} - \frac{b^{\frac{3}{2}} d \sqrt{\frac{a}{b x^{4}} + 1}}{5 x^{4}} - \frac{b^{\frac{3}{2}} f \sqrt{\frac{a}{b x^{4}} + 1}}{4 x^{2}} - \frac{b^{\frac{3}{2}} f}{16 x^{2} \sqrt{\frac{a}{b x^{4}} + 1}} - \frac{b^{\frac{5}{2}} d \sqrt{\frac{a}{b x^{4}} + 1}}{10 a} - \frac{3 b^{2} f \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x^{2}} \right )}}{16 \sqrt{a}} \]

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**12,x)

[Out]

a**(3/2)*c*gamma(-11/4)*hyper((-11/4, -1/2), (-7/4,), b*x**4*exp_polar(I*pi)/a)/
(4*x**11*gamma(-7/4)) + a**(3/2)*e*gamma(-9/4)*hyper((-9/4, -1/2), (-5/4,), b*x*
*4*exp_polar(I*pi)/a)/(4*x**9*gamma(-5/4)) + sqrt(a)*b*c*gamma(-7/4)*hyper((-7/4
, -1/2), (-3/4,), b*x**4*exp_polar(I*pi)/a)/(4*x**7*gamma(-3/4)) + sqrt(a)*b*e*g
amma(-5/4)*hyper((-5/4, -1/2), (-1/4,), b*x**4*exp_polar(I*pi)/a)/(4*x**5*gamma(
-1/4)) - a**2*f/(8*sqrt(b)*x**10*sqrt(a/(b*x**4) + 1)) - a*sqrt(b)*d*sqrt(a/(b*x
**4) + 1)/(10*x**8) - 3*a*sqrt(b)*f/(16*x**6*sqrt(a/(b*x**4) + 1)) - b**(3/2)*d*
sqrt(a/(b*x**4) + 1)/(5*x**4) - b**(3/2)*f*sqrt(a/(b*x**4) + 1)/(4*x**2) - b**(3
/2)*f/(16*x**2*sqrt(a/(b*x**4) + 1)) - b**(5/2)*d*sqrt(a/(b*x**4) + 1)/(10*a) -
3*b**2*f*asinh(sqrt(a)/(sqrt(b)*x**2))/(16*sqrt(a))

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (b x^{4} + a\right )}^{\frac{3}{2}}{\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{12}}\,{d x} \]

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^12,x, algorithm="giac")

[Out]

integrate((b*x^4 + a)^(3/2)*(f*x^3 + e*x^2 + d*x + c)/x^12, x)