3.512 \(\int \frac{\left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^{3/2}}{x^{10}} \, dx\)

Optimal. Leaf size=405 \[ \frac{2 b^{7/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (15 \sqrt{a} e+7 \sqrt{b} c\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{105 a^{3/4} \sqrt{a+b x^4}}-\frac{4 b^{9/4} c \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 a^{3/4} \sqrt{a+b x^4}}+\frac{4 b^{5/2} c x \sqrt{a+b x^4}}{15 a \left (\sqrt{a}+\sqrt{b} x^2\right )}+\frac{1}{2} b^{3/2} f \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )-\frac{4 b^2 c \sqrt{a+b x^4}}{15 a x}-\frac{3 b^2 d \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{16 \sqrt{a}}-\frac{b \sqrt{a+b x^4} \left (\frac{224 c}{x^5}+\frac{315 d}{x^4}+\frac{480 e}{x^3}+\frac{840 f}{x^2}\right )}{1680}-\frac{1}{504} \left (a+b x^4\right )^{3/2} \left (\frac{56 c}{x^9}+\frac{63 d}{x^8}+\frac{72 e}{x^7}+\frac{84 f}{x^6}\right ) \]

[Out]

-(b*((224*c)/x^5 + (315*d)/x^4 + (480*e)/x^3 + (840*f)/x^2)*Sqrt[a + b*x^4])/168
0 - (4*b^2*c*Sqrt[a + b*x^4])/(15*a*x) + (4*b^(5/2)*c*x*Sqrt[a + b*x^4])/(15*a*(
Sqrt[a] + Sqrt[b]*x^2)) - (((56*c)/x^9 + (63*d)/x^8 + (72*e)/x^7 + (84*f)/x^6)*(
a + b*x^4)^(3/2))/504 + (b^(3/2)*f*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/2 - (
3*b^2*d*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]])/(16*Sqrt[a]) - (4*b^(9/4)*c*(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^(7/4)*(7*Sqrt[b]*c
+ 15*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])/(105*a^(3/4)*Sqrt[a + b*x^4])

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Rubi [A]  time = 0.874215, antiderivative size = 405, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 14, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.467 \[ \frac{2 b^{7/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (15 \sqrt{a} e+7 \sqrt{b} c\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{105 a^{3/4} \sqrt{a+b x^4}}-\frac{4 b^{9/4} c \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 a^{3/4} \sqrt{a+b x^4}}+\frac{4 b^{5/2} c x \sqrt{a+b x^4}}{15 a \left (\sqrt{a}+\sqrt{b} x^2\right )}+\frac{1}{2} b^{3/2} f \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )-\frac{4 b^2 c \sqrt{a+b x^4}}{15 a x}-\frac{3 b^2 d \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{16 \sqrt{a}}-\frac{b \sqrt{a+b x^4} \left (\frac{224 c}{x^5}+\frac{315 d}{x^4}+\frac{480 e}{x^3}+\frac{840 f}{x^2}\right )}{1680}-\frac{1}{504} \left (a+b x^4\right )^{3/2} \left (\frac{56 c}{x^9}+\frac{63 d}{x^8}+\frac{72 e}{x^7}+\frac{84 f}{x^6}\right ) \]

Antiderivative was successfully verified.

[In]  Int[((c + d*x + e*x^2 + f*x^3)*(a + b*x^4)^(3/2))/x^10,x]

[Out]

-(b*((224*c)/x^5 + (315*d)/x^4 + (480*e)/x^3 + (840*f)/x^2)*Sqrt[a + b*x^4])/168
0 - (4*b^2*c*Sqrt[a + b*x^4])/(15*a*x) + (4*b^(5/2)*c*x*Sqrt[a + b*x^4])/(15*a*(
Sqrt[a] + Sqrt[b]*x^2)) - (((56*c)/x^9 + (63*d)/x^8 + (72*e)/x^7 + (84*f)/x^6)*(
a + b*x^4)^(3/2))/504 + (b^(3/2)*f*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/2 - (
3*b^2*d*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]])/(16*Sqrt[a]) - (4*b^(9/4)*c*(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^(7/4)*(7*Sqrt[b]*c
+ 15*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])/(105*a^(3/4)*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**10,x)

[Out]

Timed out

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Mathematica [C]  time = 1.33716, size = 351, normalized size = 0.87 \[ \frac{-\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \left (\left (a+b x^4\right ) \left (10 a^2 \left (56 c+63 d x+72 e x^2+84 f x^3\right )+a b x^4 (1232 c+15 x (105 d+16 x (9 e+14 f x)))+1344 b^2 c x^8\right )-2520 a b^{3/2} f x^9 \sqrt{a+b x^4} \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )+945 \sqrt{a} b^2 d x^9 \sqrt{a+b x^4} \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )\right )+1344 \sqrt{a} b^{5/2} c x^9 \sqrt{\frac{b x^4}{a}+1} E\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )-192 i \sqrt{a} b^2 x^9 \sqrt{\frac{b x^4}{a}+1} \left (15 \sqrt{a} e-7 i \sqrt{b} c\right ) F\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )}{5040 a x^9 \sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \sqrt{a+b x^4}} \]

Antiderivative was successfully verified.

[In]  Integrate[((c + d*x + e*x^2 + f*x^3)*(a + b*x^4)^(3/2))/x^10,x]

[Out]

(-(Sqrt[(I*Sqrt[b])/Sqrt[a]]*((a + b*x^4)*(1344*b^2*c*x^8 + 10*a^2*(56*c + 63*d*
x + 72*e*x^2 + 84*f*x^3) + a*b*x^4*(1232*c + 15*x*(105*d + 16*x*(9*e + 14*f*x)))
) - 2520*a*b^(3/2)*f*x^9*Sqrt[a + b*x^4]*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]]
+ 945*Sqrt[a]*b^2*d*x^9*Sqrt[a + b*x^4]*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]])) + 134
4*Sqrt[a]*b^(5/2)*c*x^9*Sqrt[1 + (b*x^4)/a]*EllipticE[I*ArcSinh[Sqrt[(I*Sqrt[b])
/Sqrt[a]]*x], -1] - (192*I)*Sqrt[a]*b^2*((-7*I)*Sqrt[b]*c + 15*Sqrt[a]*e)*x^9*Sq
rt[1 + (b*x^4)/a]*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[b])/Sqrt[a]]*x], -1])/(5040*a
*Sqrt[(I*Sqrt[b])/Sqrt[a]]*x^9*Sqrt[a + b*x^4])

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Maple [C]  time = 0.03, size = 437, normalized size = 1.1 \[ -{\frac{ac}{9\,{x}^{9}}\sqrt{b{x}^{4}+a}}-{\frac{11\,bc}{45\,{x}^{5}}\sqrt{b{x}^{4}+a}}-{\frac{4\,{b}^{2}c}{15\,ax}\sqrt{b{x}^{4}+a}}+{{\frac{4\,i}{15}}c{b}^{{\frac{5}{2}}}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{a}}}{\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}-{{\frac{4\,i}{15}}c{b}^{{\frac{5}{2}}}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticE} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{a}}}{\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}-{\frac{ad}{8\,{x}^{8}}\sqrt{b{x}^{4}+a}}-{\frac{5\,bd}{16\,{x}^{4}}\sqrt{b{x}^{4}+a}}-{\frac{3\,{b}^{2}d}{16}\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{4}+a} \right ) } \right ){\frac{1}{\sqrt{a}}}}-{\frac{ae}{7\,{x}^{7}}\sqrt{b{x}^{4}+a}}-{\frac{3\,be}{7\,{x}^{3}}\sqrt{b{x}^{4}+a}}+{\frac{4\,{b}^{2}e}{7}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}+{\frac{f}{2}{b}^{{\frac{3}{2}}}\ln \left ( \sqrt{b}{x}^{2}+\sqrt{b{x}^{4}+a} \right ) }-{\frac{af}{6\,{x}^{6}}\sqrt{b{x}^{4}+a}}-{\frac{2\,fb}{3\,{x}^{2}}\sqrt{b{x}^{4}+a}} \]

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^10,x)

[Out]

-1/9*c*a*(b*x^4+a)^(1/2)/x^9-11/45*c*b*(b*x^4+a)^(1/2)/x^5-4/15*b^2*c*(b*x^4+a)^
(1/2)/a/x+4/15*I*c/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*c/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)*Ellipti
cE(x*(I/a^(1/2)*b^(1/2))^(1/2),I)-1/8*d*a/x^8*(b*x^4+a)^(1/2)-5/16*d*b/x^4*(b*x^
4+a)^(1/2)-3/16*d/a^(1/2)*b^2*ln((2*a+2*a^(1/2)*(b*x^4+a)^(1/2))/x^2)-1/7*e*a*(b
*x^4+a)^(1/2)/x^7-3/7*e*b*(b*x^4+a)^(1/2)/x^3+4/7*e*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*f*b^(3/2)*ln(b^(1/2)*x^2+(b*x^4+a
)^(1/2))-1/6*f*a/x^6*(b*x^4+a)^(1/2)-2/3*f*b/x^2*(b*x^4+a)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (b x^{4} + a\right )}^{\frac{3}{2}}{\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{10}}\,{d x} \]

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^10,x, algorithm="maxima")

[Out]

integrate((b*x^4 + a)^(3/2)*(f*x^3 + e*x^2 + d*x + c)/x^10, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{{\left (b f x^{7} + b e x^{6} + b d x^{5} + b c x^{4} + a f x^{3} + a e x^{2} + a d x + a c\right )} \sqrt{b x^{4} + a}}{x^{10}}, x\right ) \]

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^10,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^10, x)

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Sympy [A]  time = 23.0875, size = 449, normalized size = 1.11 \[ \frac{a^{\frac{3}{2}} c \Gamma \left (- \frac{9}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{9}{4}, - \frac{1}{2} \\ - \frac{5}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{9} \Gamma \left (- \frac{5}{4}\right )} + \frac{a^{\frac{3}{2}} e \Gamma \left (- \frac{7}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{7}{4}, - \frac{1}{2} \\ - \frac{3}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{7} \Gamma \left (- \frac{3}{4}\right )} + \frac{\sqrt{a} b c \Gamma \left (- \frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{5}{4}, - \frac{1}{2} \\ - \frac{1}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{5} \Gamma \left (- \frac{1}{4}\right )} + \frac{\sqrt{a} b e \Gamma \left (- \frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{4}, - \frac{1}{2} \\ \frac{1}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{3} \Gamma \left (\frac{1}{4}\right )} - \frac{\sqrt{a} b f}{2 x^{2} \sqrt{1 + \frac{b x^{4}}{a}}} - \frac{a^{2} d}{8 \sqrt{b} x^{10} \sqrt{\frac{a}{b x^{4}} + 1}} - \frac{3 a \sqrt{b} d}{16 x^{6} \sqrt{\frac{a}{b x^{4}} + 1}} - \frac{a \sqrt{b} f \sqrt{\frac{a}{b x^{4}} + 1}}{6 x^{4}} - \frac{b^{\frac{3}{2}} d \sqrt{\frac{a}{b x^{4}} + 1}}{4 x^{2}} - \frac{b^{\frac{3}{2}} d}{16 x^{2} \sqrt{\frac{a}{b x^{4}} + 1}} - \frac{b^{\frac{3}{2}} f \sqrt{\frac{a}{b x^{4}} + 1}}{6} + \frac{b^{\frac{3}{2}} f \operatorname{asinh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a}} \right )}}{2} - \frac{3 b^{2} d \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x^{2}} \right )}}{16 \sqrt{a}} - \frac{b^{2} f x^{2}}{2 \sqrt{a} \sqrt{1 + \frac{b x^{4}}{a}}} \]

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**10,x)

[Out]

a**(3/2)*c*gamma(-9/4)*hyper((-9/4, -1/2), (-5/4,), b*x**4*exp_polar(I*pi)/a)/(4
*x**9*gamma(-5/4)) + a**(3/2)*e*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*c*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*e*gamm
a(-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*f/(2*x**2*sqrt(1 + b*x**4/a)) - a**2*d/(8*sqrt(b)*x**10*sqrt(a/(b*
x**4) + 1)) - 3*a*sqrt(b)*d/(16*x**6*sqrt(a/(b*x**4) + 1)) - a*sqrt(b)*f*sqrt(a/
(b*x**4) + 1)/(6*x**4) - b**(3/2)*d*sqrt(a/(b*x**4) + 1)/(4*x**2) - b**(3/2)*d/(
16*x**2*sqrt(a/(b*x**4) + 1)) - b**(3/2)*f*sqrt(a/(b*x**4) + 1)/6 + b**(3/2)*f*a
sinh(sqrt(b)*x**2/sqrt(a))/2 - 3*b**2*d*asinh(sqrt(a)/(sqrt(b)*x**2))/(16*sqrt(a
)) - b**2*f*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. \[ \int \frac{{\left (b x^{4} + a\right )}^{\frac{3}{2}}{\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{10}}\,{d x} \]

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^10,x, algorithm="giac")

[Out]

integrate((b*x^4 + a)^(3/2)*(f*x^3 + e*x^2 + d*x + c)/x^10, x)