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

Optimal. Leaf size=297 \[ \frac{\left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (3 \sqrt{a} f+\sqrt{b} d\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{4 \sqrt [4]{a} b^{7/4} \sqrt{a+b x^4}}+\frac{e \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{2 b^{3/2}}+\frac{3 f x \sqrt{a+b x^4}}{2 b^{3/2} \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{3 \sqrt [4]{a} f \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 b^{7/4} \sqrt{a+b x^4}}-\frac{c+d x+e x^2+f x^3}{2 b \sqrt{a+b x^4}} \]

[Out]

-(c + d*x + e*x^2 + f*x^3)/(2*b*Sqrt[a + b*x^4]) + (3*f*x*Sqrt[a + b*x^4])/(2*b^
(3/2)*(Sqrt[a] + Sqrt[b]*x^2)) + (e*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(2*b
^(3/2)) - (3*a^(1/4)*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])/(2*b^(7/4)*Sqrt[a + b*
x^4]) + ((Sqrt[b]*d + 3*Sqrt[a]*f)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqr
t[a] + Sqrt[b]*x^2)^2]*EllipticF[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(4*a^(1/4)
*b^(7/4)*Sqrt[a + b*x^4])

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Rubi [A]  time = 0.420967, antiderivative size = 297, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267 \[ \frac{\left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (3 \sqrt{a} f+\sqrt{b} d\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{4 \sqrt [4]{a} b^{7/4} \sqrt{a+b x^4}}+\frac{e \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{2 b^{3/2}}+\frac{3 f x \sqrt{a+b x^4}}{2 b^{3/2} \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{3 \sqrt [4]{a} f \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 b^{7/4} \sqrt{a+b x^4}}-\frac{c+d x+e x^2+f x^3}{2 b \sqrt{a+b x^4}} \]

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]

-(c + d*x + e*x^2 + f*x^3)/(2*b*Sqrt[a + b*x^4]) + (3*f*x*Sqrt[a + b*x^4])/(2*b^
(3/2)*(Sqrt[a] + Sqrt[b]*x^2)) + (e*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(2*b
^(3/2)) - (3*a^(1/4)*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])/(2*b^(7/4)*Sqrt[a + b*
x^4]) + ((Sqrt[b]*d + 3*Sqrt[a]*f)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqr
t[a] + Sqrt[b]*x^2)^2]*EllipticF[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(4*a^(1/4)
*b^(7/4)*Sqrt[a + b*x^4])

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Rubi in Sympy [A]  time = 51.6553, size = 270, normalized size = 0.91 \[ - \frac{3 \sqrt [4]{a} f \sqrt{\frac{a + b x^{4}}{\left (\sqrt{a} + \sqrt{b} x^{2}\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x^{2}\right ) E\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}} \right )}\middle | \frac{1}{2}\right )}{2 b^{\frac{7}{4}} \sqrt{a + b x^{4}}} - \frac{c + d x + e x^{2} + f x^{3}}{2 b \sqrt{a + b x^{4}}} + \frac{e \operatorname{atanh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a + b x^{4}}} \right )}}{2 b^{\frac{3}{2}}} + \frac{3 f x \sqrt{a + b x^{4}}}{2 b^{\frac{3}{2}} \left (\sqrt{a} + \sqrt{b} x^{2}\right )} + \frac{\sqrt{\frac{a + b x^{4}}{\left (\sqrt{a} + \sqrt{b} x^{2}\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x^{2}\right ) \left (3 \sqrt{a} f + \sqrt{b} d\right ) F\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}} \right )}\middle | \frac{1}{2}\right )}{4 \sqrt [4]{a} b^{\frac{7}{4}} \sqrt{a + b x^{4}}} \]

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]

-3*a**(1/4)*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)/(2*b**(7/4)*sqrt(a + b*x**4))
- (c + d*x + e*x**2 + f*x**3)/(2*b*sqrt(a + b*x**4)) + e*atanh(sqrt(b)*x**2/sqr
t(a + b*x**4))/(2*b**(3/2)) + 3*f*x*sqrt(a + b*x**4)/(2*b**(3/2)*(sqrt(a) + sqrt
(b)*x**2)) + sqrt((a + b*x**4)/(sqrt(a) + sqrt(b)*x**2)**2)*(sqrt(a) + sqrt(b)*x
**2)*(3*sqrt(a)*f + sqrt(b)*d)*elliptic_f(2*atan(b**(1/4)*x/a**(1/4)), 1/2)/(4*a
**(1/4)*b**(7/4)*sqrt(a + b*x**4))

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Mathematica [C]  time = 0.541057, size = 224, normalized size = 0.75 \[ \frac{\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \left (e \sqrt{a+b x^4} \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )-\sqrt{b} (c+x (d+x (e+f x)))\right )-\sqrt{\frac{b x^4}{a}+1} \left (3 \sqrt{a} f+i \sqrt{b} d\right ) F\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )+3 \sqrt{a} f \sqrt{\frac{b x^4}{a}+1} E\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )}{2 b^{3/2} \sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \sqrt{a+b x^4}} \]

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[(I*Sqrt[b])/Sqrt[a]]*(-(Sqrt[b]*(c + x*(d + x*(e + f*x)))) + e*Sqrt[a + b*
x^4]*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]]) + 3*Sqrt[a]*f*Sqrt[1 + (b*x^4)/a]*E
llipticE[I*ArcSinh[Sqrt[(I*Sqrt[b])/Sqrt[a]]*x], -1] - (I*Sqrt[b]*d + 3*Sqrt[a]*
f)*Sqrt[1 + (b*x^4)/a]*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[b])/Sqrt[a]]*x], -1])/(2
*Sqrt[(I*Sqrt[b])/Sqrt[a]]*b^(3/2)*Sqrt[a + b*x^4])

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Maple [C]  time = 0.011, size = 331, normalized size = 1.1 \[ -{\frac{dx}{2\,b}{\frac{1}{\sqrt{ \left ({x}^{4}+{\frac{a}{b}} \right ) b}}}}+{\frac{d}{2\,b}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}-{\frac{c}{2\,b}{\frac{1}{\sqrt{b{x}^{4}+a}}}}-{\frac{e{x}^{2}}{2\,b}{\frac{1}{\sqrt{b{x}^{4}+a}}}}+{\frac{e}{2}\ln \left ( \sqrt{b}{x}^{2}+\sqrt{b{x}^{4}+a} \right ){b}^{-{\frac{3}{2}}}}-{\frac{f{x}^{3}}{2\,b}{\frac{1}{\sqrt{ \left ({x}^{4}+{\frac{a}{b}} \right ) b}}}}+{{\frac{3\,i}{2}}f\sqrt{a}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){b}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}-{{\frac{3\,i}{2}}f\sqrt{a}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticE} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){b}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}} \]

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]

-1/2*d/b*x/((x^4+a/b)*b)^(1/2)+1/2*d/b/(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*c/b/(b*x^4+a)^(1/2)-1/2*e*x^2/b/(b*x^4+a)^(1/2)+1/
2*e/b^(3/2)*ln(b^(1/2)*x^2+(b*x^4+a)^(1/2))-1/2*f/b*x^3/((x^4+a/b)*b)^(1/2)+3/2*
I*f/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)-3/2*I*f/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)

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-1/2*c/(sqrt(b*x^4 + a)*b) + integrate((f*x^6 + e*x^5 + d*x^4)/(b*x^4 + a)^(3/2)
, x)

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

integral((f*x^6 + e*x^5 + d*x^4 + c*x^3)/(b*x^4 + a)^(3/2), x)

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Sympy [A]  time = 31.075, size = 156, normalized size = 0.53 \[ c \left (\begin{cases} - \frac{1}{2 b \sqrt{a + b x^{4}}} & \text{for}\: b \neq 0 \\\frac{x^{4}}{4 a^{\frac{3}{2}}} & \text{otherwise} \end{cases}\right ) + e \left (\frac{\operatorname{asinh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a}} \right )}}{2 b^{\frac{3}{2}}} - \frac{x^{2}}{2 \sqrt{a} b \sqrt{1 + \frac{b x^{4}}{a}}}\right ) + \frac{d x^{5} \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{5}{4}, \frac{3}{2} \\ \frac{9}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac{3}{2}} \Gamma \left (\frac{9}{4}\right )} + \frac{f x^{7} \Gamma \left (\frac{7}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{3}{2}, \frac{7}{4} \\ \frac{11}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac{3}{2}} \Gamma \left (\frac{11}{4}\right )} \]

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]

c*Piecewise((-1/(2*b*sqrt(a + b*x**4)), Ne(b, 0)), (x**4/(4*a**(3/2)), True)) +
e*(asinh(sqrt(b)*x**2/sqrt(a))/(2*b**(3/2)) - x**2/(2*sqrt(a)*b*sqrt(1 + b*x**4/
a))) + d*x**5*gamma(5/4)*hyper((5/4, 3/2), (9/4,), b*x**4*exp_polar(I*pi)/a)/(4*
a**(3/2)*gamma(9/4)) + f*x**7*gamma(7/4)*hyper((3/2, 7/4), (11/4,), b*x**4*exp_p
olar(I*pi)/a)/(4*a**(3/2)*gamma(11/4))

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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