3.513 \(\int \frac{\left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^{3/2}}{x^{11}} \, dx\)

Optimal. Leaf size=399 \[ \frac{2 b^{7/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (15 \sqrt{a} f+7 \sqrt{b} d\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{105 a^{3/4} \sqrt{a+b x^4}}-\frac{4 b^{9/4} d \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 a^{3/4} \sqrt{a+b x^4}}+\frac{4 b^{5/2} d x \sqrt{a+b x^4}}{15 a \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{b^2 c \sqrt{a+b x^4}}{10 a x^2}-\frac{4 b^2 d \sqrt{a+b x^4}}{15 a x}-\frac{3 b^2 e \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{16 \sqrt{a}}-\frac{b \sqrt{a+b x^4} \left (\frac{168 c}{x^6}+\frac{224 d}{x^5}+\frac{315 e}{x^4}+\frac{480 f}{x^3}\right )}{1680}-\frac{\left (a+b x^4\right )^{3/2} \left (\frac{252 c}{x^{10}}+\frac{280 d}{x^9}+\frac{315 e}{x^8}+\frac{360 f}{x^7}\right )}{2520} \]

[Out]

-(b*((168*c)/x^6 + (224*d)/x^5 + (315*e)/x^4 + (480*f)/x^3)*Sqrt[a + b*x^4])/168
0 - (b^2*c*Sqrt[a + b*x^4])/(10*a*x^2) - (4*b^2*d*Sqrt[a + b*x^4])/(15*a*x) + (4
*b^(5/2)*d*x*Sqrt[a + b*x^4])/(15*a*(Sqrt[a] + Sqrt[b]*x^2)) - (((252*c)/x^10 +
(280*d)/x^9 + (315*e)/x^8 + (360*f)/x^7)*(a + b*x^4)^(3/2))/2520 - (3*b^2*e*ArcT
anh[Sqrt[a + b*x^4]/Sqrt[a]])/(16*Sqrt[a]) - (4*b^(9/4)*d*(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]*d + 15*Sqrt[a]
*f)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*Elliptic
F[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 = 0.864589, antiderivative size = 399, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 12, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4 \[ \frac{2 b^{7/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (15 \sqrt{a} f+7 \sqrt{b} d\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{105 a^{3/4} \sqrt{a+b x^4}}-\frac{4 b^{9/4} d \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 a^{3/4} \sqrt{a+b x^4}}+\frac{4 b^{5/2} d x \sqrt{a+b x^4}}{15 a \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{b^2 c \sqrt{a+b x^4}}{10 a x^2}-\frac{4 b^2 d \sqrt{a+b x^4}}{15 a x}-\frac{3 b^2 e \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{16 \sqrt{a}}-\frac{b \sqrt{a+b x^4} \left (\frac{168 c}{x^6}+\frac{224 d}{x^5}+\frac{315 e}{x^4}+\frac{480 f}{x^3}\right )}{1680}-\frac{\left (a+b x^4\right )^{3/2} \left (\frac{252 c}{x^{10}}+\frac{280 d}{x^9}+\frac{315 e}{x^8}+\frac{360 f}{x^7}\right )}{2520} \]

Antiderivative was successfully verified.

[In]  Int[((c + d*x + e*x^2 + f*x^3)*(a + b*x^4)^(3/2))/x^11,x]

[Out]

-(b*((168*c)/x^6 + (224*d)/x^5 + (315*e)/x^4 + (480*f)/x^3)*Sqrt[a + b*x^4])/168
0 - (b^2*c*Sqrt[a + b*x^4])/(10*a*x^2) - (4*b^2*d*Sqrt[a + b*x^4])/(15*a*x) + (4
*b^(5/2)*d*x*Sqrt[a + b*x^4])/(15*a*(Sqrt[a] + Sqrt[b]*x^2)) - (((252*c)/x^10 +
(280*d)/x^9 + (315*e)/x^8 + (360*f)/x^7)*(a + b*x^4)^(3/2))/2520 - (3*b^2*e*ArcT
anh[Sqrt[a + b*x^4]/Sqrt[a]])/(16*Sqrt[a]) - (4*b^(9/4)*d*(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]*d + 15*Sqrt[a]
*f)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*Elliptic
F[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**11,x)

[Out]

Timed out

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Mathematica [C]  time = 1.06158, size = 314, normalized size = 0.79 \[ \frac{-\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \left (\left (a+b x^4\right ) \left (a^2 (504 c+10 x (56 d+9 x (7 e+8 f x)))+a b x^4 (1008 c+x (1232 d+45 x (35 e+48 f x)))+168 b^2 x^8 (3 c+8 d x)\right )+945 \sqrt{a} b^2 e x^{10} \sqrt{a+b x^4} \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )\right )+1344 \sqrt{a} b^{5/2} d x^{10} \sqrt{\frac{b x^4}{a}+1} E\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )-192 i \sqrt{a} b^2 x^{10} \sqrt{\frac{b x^4}{a}+1} \left (15 \sqrt{a} f-7 i \sqrt{b} d\right ) F\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )}{5040 a x^{10} \sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \sqrt{a+b x^4}} \]

Antiderivative was successfully verified.

[In]  Integrate[((c + d*x + e*x^2 + f*x^3)*(a + b*x^4)^(3/2))/x^11,x]

[Out]

(-(Sqrt[(I*Sqrt[b])/Sqrt[a]]*((a + b*x^4)*(168*b^2*x^8*(3*c + 8*d*x) + a^2*(504*
c + 10*x*(56*d + 9*x*(7*e + 8*f*x))) + a*b*x^4*(1008*c + x*(1232*d + 45*x*(35*e
+ 48*f*x)))) + 945*Sqrt[a]*b^2*e*x^10*Sqrt[a + b*x^4]*ArcTanh[Sqrt[a + b*x^4]/Sq
rt[a]])) + 1344*Sqrt[a]*b^(5/2)*d*x^10*Sqrt[1 + (b*x^4)/a]*EllipticE[I*ArcSinh[S
qrt[(I*Sqrt[b])/Sqrt[a]]*x], -1] - (192*I)*Sqrt[a]*b^2*((-7*I)*Sqrt[b]*d + 15*Sq
rt[a]*f)*x^10*Sqrt[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^10*Sqrt[a + b*x^4])

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Maple [C]  time = 0.026, size = 417, normalized size = 1.1 \[ -{\frac{c \left ({b}^{2}{x}^{8}+2\,ab{x}^{4}+{a}^{2} \right ) }{10\,{x}^{10}a}\sqrt{b{x}^{4}+a}}-{\frac{ad}{9\,{x}^{9}}\sqrt{b{x}^{4}+a}}-{\frac{11\,bd}{45\,{x}^{5}}\sqrt{b{x}^{4}+a}}-{\frac{4\,{b}^{2}d}{15\,ax}\sqrt{b{x}^{4}+a}}+{{\frac{4\,i}{15}}d{b}^{{\frac{5}{2}}}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{a}}}{\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}-{{\frac{4\,i}{15}}d{b}^{{\frac{5}{2}}}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticE} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{a}}}{\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}-{\frac{ae}{8\,{x}^{8}}\sqrt{b{x}^{4}+a}}-{\frac{5\,be}{16\,{x}^{4}}\sqrt{b{x}^{4}+a}}-{\frac{3\,{b}^{2}e}{16}\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{4}+a} \right ) } \right ){\frac{1}{\sqrt{a}}}}-{\frac{af}{7\,{x}^{7}}\sqrt{b{x}^{4}+a}}-{\frac{3\,fb}{7\,{x}^{3}}\sqrt{b{x}^{4}+a}}+{\frac{4\,f{b}^{2}}{7}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}} \]

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^11,x)

[Out]

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

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

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^11,x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{{\left (b f x^{7} + b e x^{6} + b d x^{5} + b c x^{4} + a f x^{3} + a e x^{2} + a d x + a c\right )} \sqrt{b x^{4} + a}}{x^{11}}, x\right ) \]

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^11,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^11, x)

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Sympy [A]  time = 24.6385, size = 398, normalized size = 1. \[ \frac{a^{\frac{3}{2}} d \Gamma \left (- \frac{9}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{9}{4}, - \frac{1}{2} \\ - \frac{5}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{9} \Gamma \left (- \frac{5}{4}\right )} + \frac{a^{\frac{3}{2}} f \Gamma \left (- \frac{7}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{7}{4}, - \frac{1}{2} \\ - \frac{3}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{7} \Gamma \left (- \frac{3}{4}\right )} + \frac{\sqrt{a} b d \Gamma \left (- \frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{5}{4}, - \frac{1}{2} \\ - \frac{1}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{5} \Gamma \left (- \frac{1}{4}\right )} + \frac{\sqrt{a} b f \Gamma \left (- \frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{4}, - \frac{1}{2} \\ \frac{1}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{3} \Gamma \left (\frac{1}{4}\right )} - \frac{a^{2} e}{8 \sqrt{b} x^{10} \sqrt{\frac{a}{b x^{4}} + 1}} - \frac{a \sqrt{b} c \sqrt{\frac{a}{b x^{4}} + 1}}{10 x^{8}} - \frac{3 a \sqrt{b} e}{16 x^{6} \sqrt{\frac{a}{b x^{4}} + 1}} - \frac{b^{\frac{3}{2}} c \sqrt{\frac{a}{b x^{4}} + 1}}{5 x^{4}} - \frac{b^{\frac{3}{2}} e \sqrt{\frac{a}{b x^{4}} + 1}}{4 x^{2}} - \frac{b^{\frac{3}{2}} e}{16 x^{2} \sqrt{\frac{a}{b x^{4}} + 1}} - \frac{b^{\frac{5}{2}} c \sqrt{\frac{a}{b x^{4}} + 1}}{10 a} - \frac{3 b^{2} e \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x^{2}} \right )}}{16 \sqrt{a}} \]

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**11,x)

[Out]

a**(3/2)*d*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)*f*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*d*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*f*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)
) - a**2*e/(8*sqrt(b)*x**10*sqrt(a/(b*x**4) + 1)) - a*sqrt(b)*c*sqrt(a/(b*x**4)
+ 1)/(10*x**8) - 3*a*sqrt(b)*e/(16*x**6*sqrt(a/(b*x**4) + 1)) - b**(3/2)*c*sqrt(
a/(b*x**4) + 1)/(5*x**4) - b**(3/2)*e*sqrt(a/(b*x**4) + 1)/(4*x**2) - b**(3/2)*e
/(16*x**2*sqrt(a/(b*x**4) + 1)) - b**(5/2)*c*sqrt(a/(b*x**4) + 1)/(10*a) - 3*b**
2*e*asinh(sqrt(a)/(sqrt(b)*x**2))/(16*sqrt(a))

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (b x^{4} + a\right )}^{\frac{3}{2}}{\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{11}}\,{d x} \]

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^11,x, algorithm="giac")

[Out]

integrate((b*x^4 + a)^(3/2)*(f*x^3 + e*x^2 + d*x + c)/x^11, x)