3.510 \(\int \frac{\left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^{3/2}}{x^8} \, dx\)

Optimal. Leaf size=412 \[ \frac{2 b^{5/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (21 \sqrt{a} e+5 \sqrt{b} c\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{35 \sqrt [4]{a} \sqrt{a+b x^4}}+\frac{1}{2} b^{3/2} d \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )+\frac{12 b^{3/2} e x \sqrt{a+b x^4}}{5 \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{12 \sqrt [4]{a} b^{5/4} e \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 \sqrt{a+b x^4}}-\frac{1}{420} \left (a+b x^4\right )^{3/2} \left (\frac{60 c}{x^7}+\frac{70 d}{x^6}+\frac{84 e}{x^5}+\frac{105 f}{x^4}\right )-\frac{2 b \sqrt{a+b x^4} \left (5 c-21 e x^2\right )}{35 x^3}-\frac{b \sqrt{a+b x^4} \left (2 d-3 f x^2\right )}{4 x^2}-\frac{12 b e \sqrt{a+b x^4}}{5 x}-\frac{3}{4} \sqrt{a} b f \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right ) \]

[Out]

(-12*b*e*Sqrt[a + b*x^4])/(5*x) + (12*b^(3/2)*e*x*Sqrt[a + b*x^4])/(5*(Sqrt[a] +
Sqrt[b]*x^2)) - (2*b*(5*c - 21*e*x^2)*Sqrt[a + b*x^4])/(35*x^3) - (b*(2*d - 3*f
*x^2)*Sqrt[a + b*x^4])/(4*x^2) - (((60*c)/x^7 + (70*d)/x^6 + (84*e)/x^5 + (105*f
)/x^4)*(a + b*x^4)^(3/2))/420 + (b^(3/2)*d*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]
])/2 - (3*Sqrt[a]*b*f*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]])/4 - (12*a^(1/4)*b^(5/4)*
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*b^(5/4)*(5*Sqrt[b]
*c + 21*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])/(35*a^(1/4)*Sqrt[a + b*x^4
])

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Rubi [A]  time = 0.913646, antiderivative size = 412, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 16, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.533 \[ \frac{2 b^{5/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (21 \sqrt{a} e+5 \sqrt{b} c\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{35 \sqrt [4]{a} \sqrt{a+b x^4}}+\frac{1}{2} b^{3/2} d \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )+\frac{12 b^{3/2} e x \sqrt{a+b x^4}}{5 \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{12 \sqrt [4]{a} b^{5/4} e \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 \sqrt{a+b x^4}}-\frac{1}{420} \left (a+b x^4\right )^{3/2} \left (\frac{60 c}{x^7}+\frac{70 d}{x^6}+\frac{84 e}{x^5}+\frac{105 f}{x^4}\right )-\frac{2 b \sqrt{a+b x^4} \left (5 c-21 e x^2\right )}{35 x^3}-\frac{b \sqrt{a+b x^4} \left (2 d-3 f x^2\right )}{4 x^2}-\frac{12 b e \sqrt{a+b x^4}}{5 x}-\frac{3}{4} \sqrt{a} b f \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right ) \]

Antiderivative was successfully verified.

[In]  Int[((c + d*x + e*x^2 + f*x^3)*(a + b*x^4)^(3/2))/x^8,x]

[Out]

(-12*b*e*Sqrt[a + b*x^4])/(5*x) + (12*b^(3/2)*e*x*Sqrt[a + b*x^4])/(5*(Sqrt[a] +
Sqrt[b]*x^2)) - (2*b*(5*c - 21*e*x^2)*Sqrt[a + b*x^4])/(35*x^3) - (b*(2*d - 3*f
*x^2)*Sqrt[a + b*x^4])/(4*x^2) - (((60*c)/x^7 + (70*d)/x^6 + (84*e)/x^5 + (105*f
)/x^4)*(a + b*x^4)^(3/2))/420 + (b^(3/2)*d*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]
])/2 - (3*Sqrt[a]*b*f*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]])/4 - (12*a^(1/4)*b^(5/4)*
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*b^(5/4)*(5*Sqrt[b]
*c + 21*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])/(35*a^(1/4)*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**8,x)

[Out]

Timed out

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Mathematica [C]  time = 1.12642, size = 330, normalized size = 0.8 \[ \frac{-\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \left (-210 b^{3/2} d x^7 \sqrt{a+b x^4} \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )+\left (a+b x^4\right ) \left (a (60 c+7 x (10 d+3 x (4 e+5 f x)))+2 b x^4 (90 c+7 x (20 d+3 x (14 e-5 f x)))\right )+315 \sqrt{a} b f x^7 \sqrt{a+b x^4} \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )\right )-48 b^{3/2} x^7 \sqrt{\frac{b x^4}{a}+1} \left (21 \sqrt{a} e+5 i \sqrt{b} c\right ) F\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )+1008 \sqrt{a} b^{3/2} e x^7 \sqrt{\frac{b x^4}{a}+1} E\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )}{420 x^7 \sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \sqrt{a+b x^4}} \]

Antiderivative was successfully verified.

[In]  Integrate[((c + d*x + e*x^2 + f*x^3)*(a + b*x^4)^(3/2))/x^8,x]

[Out]

(-(Sqrt[(I*Sqrt[b])/Sqrt[a]]*((a + b*x^4)*(2*b*x^4*(90*c + 7*x*(20*d + 3*x*(14*e
- 5*f*x))) + a*(60*c + 7*x*(10*d + 3*x*(4*e + 5*f*x)))) - 210*b^(3/2)*d*x^7*Sqr
t[a + b*x^4]*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]] + 315*Sqrt[a]*b*f*x^7*Sqrt[a
+ b*x^4]*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]])) + 1008*Sqrt[a]*b^(3/2)*e*x^7*Sqrt[1
+ (b*x^4)/a]*EllipticE[I*ArcSinh[Sqrt[(I*Sqrt[b])/Sqrt[a]]*x], -1] - 48*b^(3/2)
*((5*I)*Sqrt[b]*c + 21*Sqrt[a]*e)*x^7*Sqrt[1 + (b*x^4)/a]*EllipticF[I*ArcSinh[Sq
rt[(I*Sqrt[b])/Sqrt[a]]*x], -1])/(420*Sqrt[(I*Sqrt[b])/Sqrt[a]]*x^7*Sqrt[a + b*x
^4])

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Maple [C]  time = 0.026, size = 411, normalized size = 1. \[ -{\frac{ac}{7\,{x}^{7}}\sqrt{b{x}^{4}+a}}-{\frac{3\,bc}{7\,{x}^{3}}\sqrt{b{x}^{4}+a}}+{\frac{4\,{b}^{2}c}{7}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}+{\frac{d}{2}{b}^{{\frac{3}{2}}}\ln \left ( \sqrt{b}{x}^{2}+\sqrt{b{x}^{4}+a} \right ) }-{\frac{ad}{6\,{x}^{6}}\sqrt{b{x}^{4}+a}}-{\frac{2\,bd}{3\,{x}^{2}}\sqrt{b{x}^{4}+a}}-{\frac{ae}{5\,{x}^{5}}\sqrt{b{x}^{4}+a}}-{\frac{7\,be}{5\,x}\sqrt{b{x}^{4}+a}}+{{\frac{12\,i}{5}}e{b}^{{\frac{3}{2}}}\sqrt{a}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}-{{\frac{12\,i}{5}}e{b}^{{\frac{3}{2}}}\sqrt{a}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticE} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}+{\frac{fb}{2}\sqrt{b{x}^{4}+a}}-{\frac{af}{4\,{x}^{4}}\sqrt{b{x}^{4}+a}}-{\frac{3\,fb}{4}\sqrt{a}\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{4}+a} \right ) } \right ) } \]

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^8,x)

[Out]

-1/7*c*a*(b*x^4+a)^(1/2)/x^7-3/7*c*b*(b*x^4+a)^(1/2)/x^3+4/7*c*b^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)+1/2*d*b^(3/2)*ln(b^(1/2)*x
^2+(b*x^4+a)^(1/2))-1/6*d*a/x^6*(b*x^4+a)^(1/2)-2/3*d*b/x^2*(b*x^4+a)^(1/2)-1/5*
e*a*(b*x^4+a)^(1/2)/x^5-7/5*b*e*(b*x^4+a)^(1/2)/x+12/5*I*e*b^(3/2)*a^(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*b^(3/2)*a
^(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*
b*(b*x^4+a)^(1/2)-1/4*f*a/x^4*(b*x^4+a)^(1/2)-3/4*f*a^(1/2)*b*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. \[ \int \frac{{\left (b x^{4} + a\right )}^{\frac{3}{2}}{\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{8}}\,{d x} \]

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^8,x, algorithm="maxima")

[Out]

integrate((b*x^4 + a)^(3/2)*(f*x^3 + e*x^2 + d*x + c)/x^8, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{{\left (b f x^{7} + b e x^{6} + b d x^{5} + b c x^{4} + a f x^{3} + a e x^{2} + a d x + a c\right )} \sqrt{b x^{4} + a}}{x^{8}}, x\right ) \]

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^8,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^8, x)

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Sympy [A]  time = 16.3071, size = 415, normalized size = 1.01 \[ \frac{a^{\frac{3}{2}} c \Gamma \left (- \frac{7}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{7}{4}, - \frac{1}{2} \\ - \frac{3}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{7} \Gamma \left (- \frac{3}{4}\right )} + \frac{a^{\frac{3}{2}} e \Gamma \left (- \frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{5}{4}, - \frac{1}{2} \\ - \frac{1}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{5} \Gamma \left (- \frac{1}{4}\right )} + \frac{\sqrt{a} b c \Gamma \left (- \frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{4}, - \frac{1}{2} \\ \frac{1}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{3} \Gamma \left (\frac{1}{4}\right )} - \frac{\sqrt{a} b d}{2 x^{2} \sqrt{1 + \frac{b x^{4}}{a}}} + \frac{\sqrt{a} b e \Gamma \left (- \frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, - \frac{1}{4} \\ \frac{3}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 x \Gamma \left (\frac{3}{4}\right )} - \frac{3 \sqrt{a} b f \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x^{2}} \right )}}{4} - \frac{a \sqrt{b} d \sqrt{\frac{a}{b x^{4}} + 1}}{6 x^{4}} - \frac{a \sqrt{b} f \sqrt{\frac{a}{b x^{4}} + 1}}{4 x^{2}} + \frac{a \sqrt{b} f}{2 x^{2} \sqrt{\frac{a}{b x^{4}} + 1}} - \frac{b^{\frac{3}{2}} d \sqrt{\frac{a}{b x^{4}} + 1}}{6} + \frac{b^{\frac{3}{2}} d \operatorname{asinh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a}} \right )}}{2} + \frac{b^{\frac{3}{2}} f x^{2}}{2 \sqrt{\frac{a}{b x^{4}} + 1}} - \frac{b^{2} d x^{2}}{2 \sqrt{a} \sqrt{1 + \frac{b x^{4}}{a}}} \]

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**8,x)

[Out]

a**(3/2)*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)) + a**(3/2)*e*gamma(-5/4)*hyper((-5/4, -1/2), (-1/4,), b*x**4*
exp_polar(I*pi)/a)/(4*x**5*gamma(-1/4)) + sqrt(a)*b*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)) - sqrt(a)*b*d/(2*x**
2*sqrt(1 + b*x**4/a)) + sqrt(a)*b*e*gamma(-1/4)*hyper((-1/2, -1/4), (3/4,), b*x*
*4*exp_polar(I*pi)/a)/(4*x*gamma(3/4)) - 3*sqrt(a)*b*f*asinh(sqrt(a)/(sqrt(b)*x*
*2))/4 - a*sqrt(b)*d*sqrt(a/(b*x**4) + 1)/(6*x**4) - a*sqrt(b)*f*sqrt(a/(b*x**4)
+ 1)/(4*x**2) + a*sqrt(b)*f/(2*x**2*sqrt(a/(b*x**4) + 1)) - b**(3/2)*d*sqrt(a/(
b*x**4) + 1)/6 + b**(3/2)*d*asinh(sqrt(b)*x**2/sqrt(a))/2 + b**(3/2)*f*x**2/(2*s
qrt(a/(b*x**4) + 1)) - b**2*d*x**2/(2*sqrt(a)*sqrt(1 + b*x**4/a))

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (b x^{4} + a\right )}^{\frac{3}{2}}{\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{8}}\,{d x} \]

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^8,x, algorithm="giac")

[Out]

integrate((b*x^4 + a)^(3/2)*(f*x^3 + e*x^2 + d*x + c)/x^8, x)