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

Optimal. Leaf size=323 \[ -\frac{\sqrt [4]{b} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (\sqrt{b} c-3 \sqrt{a} e\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{6 a^{5/4} \sqrt{a+b x^4}}-\frac{\sqrt [4]{b} e \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{a^{3/4} \sqrt{a+b x^4}}-\frac{c \sqrt{a+b x^4}}{3 a x^3}-\frac{d \sqrt{a+b x^4}}{2 a x^2}-\frac{e \sqrt{a+b x^4}}{a x}+\frac{\sqrt{b} e x \sqrt{a+b x^4}}{a \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{f \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{2 \sqrt{a}} \]

[Out]

-(c*Sqrt[a + b*x^4])/(3*a*x^3) - (d*Sqrt[a + b*x^4])/(2*a*x^2) - (e*Sqrt[a + b*x
^4])/(a*x) + (Sqrt[b]*e*x*Sqrt[a + b*x^4])/(a*(Sqrt[a] + Sqrt[b]*x^2)) - (f*ArcT
anh[Sqrt[a + b*x^4]/Sqrt[a]])/(2*Sqrt[a]) - (b^(1/4)*e*(Sqrt[a] + Sqrt[b]*x^2)*S
qrt[(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]) - (b^(1/4)*(Sqrt[b]*c - 3*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])/(6*a^(5/4)*Sqrt[a + b*x^4])

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Rubi [A]  time = 0.654556, antiderivative size = 323, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 10, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333 \[ -\frac{\sqrt [4]{b} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (\sqrt{b} c-3 \sqrt{a} e\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{6 a^{5/4} \sqrt{a+b x^4}}-\frac{\sqrt [4]{b} e \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{a^{3/4} \sqrt{a+b x^4}}-\frac{c \sqrt{a+b x^4}}{3 a x^3}-\frac{d \sqrt{a+b x^4}}{2 a x^2}-\frac{e \sqrt{a+b x^4}}{a x}+\frac{\sqrt{b} e x \sqrt{a+b x^4}}{a \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{f \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{2 \sqrt{a}} \]

Antiderivative was successfully verified.

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

[Out]

-(c*Sqrt[a + b*x^4])/(3*a*x^3) - (d*Sqrt[a + b*x^4])/(2*a*x^2) - (e*Sqrt[a + b*x
^4])/(a*x) + (Sqrt[b]*e*x*Sqrt[a + b*x^4])/(a*(Sqrt[a] + Sqrt[b]*x^2)) - (f*ArcT
anh[Sqrt[a + b*x^4]/Sqrt[a]])/(2*Sqrt[a]) - (b^(1/4)*e*(Sqrt[a] + Sqrt[b]*x^2)*S
qrt[(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]) - (b^(1/4)*(Sqrt[b]*c - 3*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])/(6*a^(5/4)*Sqrt[a + b*x^4])

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Rubi in Sympy [A]  time = 74.6426, size = 286, normalized size = 0.89 \[ \frac{\sqrt{b} e x \sqrt{a + b x^{4}}}{a \left (\sqrt{a} + \sqrt{b} x^{2}\right )} - \frac{c \sqrt{a + b x^{4}}}{3 a x^{3}} - \frac{d \sqrt{a + b x^{4}}}{2 a x^{2}} - \frac{e \sqrt{a + b x^{4}}}{a x} - \frac{f \operatorname{atanh}{\left (\frac{\sqrt{a + b x^{4}}}{\sqrt{a}} \right )}}{2 \sqrt{a}} - \frac{\sqrt [4]{b} e \sqrt{\frac{a + b x^{4}}{\left (\sqrt{a} + \sqrt{b} x^{2}\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x^{2}\right ) E\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}} \right )}\middle | \frac{1}{2}\right )}{a^{\frac{3}{4}} \sqrt{a + b x^{4}}} + \frac{\sqrt [4]{b} \sqrt{\frac{a + b x^{4}}{\left (\sqrt{a} + \sqrt{b} x^{2}\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x^{2}\right ) \left (3 \sqrt{a} e - \sqrt{b} c\right ) F\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}} \right )}\middle | \frac{1}{2}\right )}{6 a^{\frac{5}{4}} \sqrt{a + b x^{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

sqrt(b)*e*x*sqrt(a + b*x**4)/(a*(sqrt(a) + sqrt(b)*x**2)) - c*sqrt(a + b*x**4)/(
3*a*x**3) - d*sqrt(a + b*x**4)/(2*a*x**2) - e*sqrt(a + b*x**4)/(a*x) - f*atanh(s
qrt(a + b*x**4)/sqrt(a))/(2*sqrt(a)) - b**(1/4)*e*sqrt((a + b*x**4)/(sqrt(a) + s
qrt(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)) + b**(1/4)*sqrt((a + b*x**4)/(sqrt(a) + sqrt(
b)*x**2)**2)*(sqrt(a) + sqrt(b)*x**2)*(3*sqrt(a)*e - sqrt(b)*c)*elliptic_f(2*ata
n(b**(1/4)*x/a**(1/4)), 1/2)/(6*a**(5/4)*sqrt(a + b*x**4))

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Mathematica [C]  time = 0.65337, size = 249, normalized size = 0.77 \[ \frac{-\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \left (\left (a+b x^4\right ) (2 c+3 x (d+2 e x))+3 \sqrt{a} f x^3 \sqrt{a+b x^4} \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )\right )-2 \sqrt{b} x^3 \sqrt{\frac{b x^4}{a}+1} \left (3 \sqrt{a} e-i \sqrt{b} c\right ) F\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )+6 \sqrt{a} \sqrt{b} e x^3 \sqrt{\frac{b x^4}{a}+1} E\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )}{6 a x^3 \sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \sqrt{a+b x^4}} \]

Antiderivative was successfully verified.

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

[Out]

(-(Sqrt[(I*Sqrt[b])/Sqrt[a]]*((a + b*x^4)*(2*c + 3*x*(d + 2*e*x)) + 3*Sqrt[a]*f*
x^3*Sqrt[a + b*x^4]*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]])) + 6*Sqrt[a]*Sqrt[b]*e*x^3
*Sqrt[1 + (b*x^4)/a]*EllipticE[I*ArcSinh[Sqrt[(I*Sqrt[b])/Sqrt[a]]*x], -1] - 2*S
qrt[b]*((-I)*Sqrt[b]*c + 3*Sqrt[a]*e)*x^3*Sqrt[1 + (b*x^4)/a]*EllipticF[I*ArcSin
h[Sqrt[(I*Sqrt[b])/Sqrt[a]]*x], -1])/(6*a*Sqrt[(I*Sqrt[b])/Sqrt[a]]*x^3*Sqrt[a +
b*x^4])

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Maple [C]  time = 0.014, size = 316, normalized size = 1. \[ -{\frac{c}{3\,a{x}^{3}}\sqrt{b{x}^{4}+a}}-{\frac{bc}{3\,a}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}-{\frac{d}{2\,a{x}^{2}}\sqrt{b{x}^{4}+a}}-{\frac{e}{ax}\sqrt{b{x}^{4}+a}}+{ie\sqrt{b}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{a}}}{\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}-{ie\sqrt{b}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticE} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{a}}}{\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}-{\frac{f}{2}\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{4}+a} \right ) } \right ){\frac{1}{\sqrt{a}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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^4),x, algorithm="maxima")

[Out]

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

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

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^4),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 4.87073, size = 131, normalized size = 0.41 \[ - \frac{\sqrt{b} d \sqrt{\frac{a}{b x^{4}} + 1}}{2 a} + \frac{c \Gamma \left (- \frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{4}, \frac{1}{2} \\ \frac{1}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt{a} x^{3} \Gamma \left (\frac{1}{4}\right )} + \frac{e \Gamma \left (- \frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{4}, \frac{1}{2} \\ \frac{3}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt{a} x \Gamma \left (\frac{3}{4}\right )} - \frac{f \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x^{2}} \right )}}{2 \sqrt{a}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

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^4),x, algorithm="giac")

[Out]

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