3.523 \(\int \frac{c+d x+e x^2+f x^3}{x^2 \sqrt{a+b x^4}} \, dx\)

Optimal. Leaf size=309 \[ \frac{\left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (\sqrt{a} e+\sqrt{b} c\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 a^{3/4} \sqrt [4]{b} \sqrt{a+b x^4}}-\frac{\sqrt [4]{b} c \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{a^{3/4} \sqrt{a+b x^4}}-\frac{c \sqrt{a+b x^4}}{a x}+\frac{\sqrt{b} c x \sqrt{a+b x^4}}{a \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{d \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{2 \sqrt{a}}+\frac{f \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{2 \sqrt{b}} \]

[Out]

-((c*Sqrt[a + b*x^4])/(a*x)) + (Sqrt[b]*c*x*Sqrt[a + b*x^4])/(a*(Sqrt[a] + Sqrt[
b]*x^2)) + (f*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(2*Sqrt[b]) - (d*ArcTanh[S
qrt[a + b*x^4]/Sqrt[a]])/(2*Sqrt[a]) - (b^(1/4)*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])/(a^(3/4)*Sqrt[a + b*x^4]) + ((Sqrt[b]*c + 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])/(2*a^(3/4)*b^(1/4)*Sqrt[a + b*x^4])

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Rubi [A]  time = 0.564272, antiderivative size = 309, 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 \[ \frac{\left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (\sqrt{a} e+\sqrt{b} c\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 a^{3/4} \sqrt [4]{b} \sqrt{a+b x^4}}-\frac{\sqrt [4]{b} c \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{a^{3/4} \sqrt{a+b x^4}}-\frac{c \sqrt{a+b x^4}}{a x}+\frac{\sqrt{b} c x \sqrt{a+b x^4}}{a \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{d \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{2 \sqrt{a}}+\frac{f \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{2 \sqrt{b}} \]

Antiderivative was successfully verified.

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

[Out]

-((c*Sqrt[a + b*x^4])/(a*x)) + (Sqrt[b]*c*x*Sqrt[a + b*x^4])/(a*(Sqrt[a] + Sqrt[
b]*x^2)) + (f*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(2*Sqrt[b]) - (d*ArcTanh[S
qrt[a + b*x^4]/Sqrt[a]])/(2*Sqrt[a]) - (b^(1/4)*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])/(a^(3/4)*Sqrt[a + b*x^4]) + ((Sqrt[b]*c + 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])/(2*a^(3/4)*b^(1/4)*Sqrt[a + b*x^4])

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Rubi in Sympy [A]  time = 62.9062, size = 275, normalized size = 0.89 \[ \frac{f \operatorname{atanh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a + b x^{4}}} \right )}}{2 \sqrt{b}} + \frac{\sqrt{b} c x \sqrt{a + b x^{4}}}{a \left (\sqrt{a} + \sqrt{b} x^{2}\right )} - \frac{c \sqrt{a + b x^{4}}}{a x} - \frac{d \operatorname{atanh}{\left (\frac{\sqrt{a + b x^{4}}}{\sqrt{a}} \right )}}{2 \sqrt{a}} - \frac{\sqrt [4]{b} c \sqrt{\frac{a + b x^{4}}{\left (\sqrt{a} + \sqrt{b} x^{2}\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x^{2}\right ) E\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}} \right )}\middle | \frac{1}{2}\right )}{a^{\frac{3}{4}} \sqrt{a + b x^{4}}} + \frac{\sqrt{\frac{a + b x^{4}}{\left (\sqrt{a} + \sqrt{b} x^{2}\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x^{2}\right ) \left (\sqrt{a} e + \sqrt{b} c\right ) F\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}} \right )}\middle | \frac{1}{2}\right )}{2 a^{\frac{3}{4}} \sqrt [4]{b} \sqrt{a + b x^{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((f*x**3+e*x**2+d*x+c)/x**2/(b*x**4+a)**(1/2),x)

[Out]

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

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Mathematica [C]  time = 3.77512, size = 250, normalized size = 0.81 \[ \frac{1}{2} \left (-\frac{2 c \sqrt{a+b x^4}}{a x}-\frac{d \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{\sqrt{a}}+\frac{f \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{\sqrt{b}}\right )-\frac{\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \sqrt{\frac{b x^4}{a}+1} \left (\sqrt{a} e-i \sqrt{b} c\right ) F\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )}{\sqrt{b} \sqrt{a+b x^4}}-\frac{i c \sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \sqrt{\frac{b x^4}{a}+1} E\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )}{\sqrt{a+b x^4}} \]

Antiderivative was successfully verified.

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

[Out]

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

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Maple [C]  time = 0.012, size = 299, normalized size = 1. \[{e\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}+{\frac{f}{2}\ln \left ( \sqrt{b}{x}^{2}+\sqrt{b{x}^{4}+a} \right ){\frac{1}{\sqrt{b}}}}-{\frac{c}{ax}\sqrt{b{x}^{4}+a}}+{ic\sqrt{b}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{a}}}{\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}-{ic\sqrt{b}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticE} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{a}}}{\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}-{\frac{d}{2}\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{4}+a} \right ) } \right ){\frac{1}{\sqrt{a}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((f*x^3+e*x^2+d*x+c)/x^2/(b*x^4+a)^(1/2),x)

[Out]

e/(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*ln(b^
(1/2)*x^2+(b*x^4+a)^(1/2))/b^(1/2)-c*(b*x^4+a)^(1/2)/a/x+I*c*b^(1/2)/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)-I*c*b^(1/2)/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*d/a^(
1/2)*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. \[ \int \frac{f x^{3} + e x^{2} + d x + c}{\sqrt{b x^{4} + a} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Sympy [A]  time = 4.66349, size = 128, normalized size = 0.41 \[ \frac{f \operatorname{asinh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a}} \right )}}{2 \sqrt{b}} + \frac{c \Gamma \left (- \frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{4}, \frac{1}{2} \\ \frac{3}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt{a} x \Gamma \left (\frac{3}{4}\right )} - \frac{d \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x^{2}} \right )}}{2 \sqrt{a}} + \frac{e x \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{4}, \frac{1}{2} \\ \frac{5}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt{a} \Gamma \left (\frac{5}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((f*x**3+e*x**2+d*x+c)/x**2/(b*x**4+a)**(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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