3.174 \(\int \frac{A+B x^2}{x^5 \sqrt{a+b x^2+c x^4}} \, dx\)

Optimal. Leaf size=124 \[ -\frac{\left (-4 a A c-4 a b B+3 A b^2\right ) \tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )}{16 a^{5/2}}+\frac{(3 A b-4 a B) \sqrt{a+b x^2+c x^4}}{8 a^2 x^2}-\frac{A \sqrt{a+b x^2+c x^4}}{4 a x^4} \]

[Out]

-(A*Sqrt[a + b*x^2 + c*x^4])/(4*a*x^4) + ((3*A*b - 4*a*B)*Sqrt[a + b*x^2 + c*x^4
])/(8*a^2*x^2) - ((3*A*b^2 - 4*a*b*B - 4*a*A*c)*ArcTanh[(2*a + b*x^2)/(2*Sqrt[a]
*Sqrt[a + b*x^2 + c*x^4])])/(16*a^(5/2))

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Rubi [A]  time = 0.382329, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185 \[ -\frac{\left (-4 a A c-4 a b B+3 A b^2\right ) \tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )}{16 a^{5/2}}+\frac{(3 A b-4 a B) \sqrt{a+b x^2+c x^4}}{8 a^2 x^2}-\frac{A \sqrt{a+b x^2+c x^4}}{4 a x^4} \]

Antiderivative was successfully verified.

[In]  Int[(A + B*x^2)/(x^5*Sqrt[a + b*x^2 + c*x^4]),x]

[Out]

-(A*Sqrt[a + b*x^2 + c*x^4])/(4*a*x^4) + ((3*A*b - 4*a*B)*Sqrt[a + b*x^2 + c*x^4
])/(8*a^2*x^2) - ((3*A*b^2 - 4*a*b*B - 4*a*A*c)*ArcTanh[(2*a + b*x^2)/(2*Sqrt[a]
*Sqrt[a + b*x^2 + c*x^4])])/(16*a^(5/2))

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Rubi in Sympy [A]  time = 34.6226, size = 114, normalized size = 0.92 \[ - \frac{A \sqrt{a + b x^{2} + c x^{4}}}{4 a x^{4}} + \frac{\left (3 A b - 4 B a\right ) \sqrt{a + b x^{2} + c x^{4}}}{8 a^{2} x^{2}} - \frac{\left (- 4 A a c + b \left (3 A b - 4 B a\right )\right ) \operatorname{atanh}{\left (\frac{2 a + b x^{2}}{2 \sqrt{a} \sqrt{a + b x^{2} + c x^{4}}} \right )}}{16 a^{\frac{5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((B*x**2+A)/x**5/(c*x**4+b*x**2+a)**(1/2),x)

[Out]

-A*sqrt(a + b*x**2 + c*x**4)/(4*a*x**4) + (3*A*b - 4*B*a)*sqrt(a + b*x**2 + c*x*
*4)/(8*a**2*x**2) - (-4*A*a*c + b*(3*A*b - 4*B*a))*atanh((2*a + b*x**2)/(2*sqrt(
a)*sqrt(a + b*x**2 + c*x**4)))/(16*a**(5/2))

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Mathematica [A]  time = 0.196443, size = 112, normalized size = 0.9 \[ \frac{\sqrt{a+b x^2+c x^4} \left (3 A b x^2-2 a \left (A+2 B x^2\right )\right )}{8 a^2 x^4}-\frac{\left (4 a A c+4 a b B-3 A b^2\right ) \left (\log \left (x^2\right )-\log \left (2 \sqrt{a} \sqrt{a+b x^2+c x^4}+2 a+b x^2\right )\right )}{16 a^{5/2}} \]

Antiderivative was successfully verified.

[In]  Integrate[(A + B*x^2)/(x^5*Sqrt[a + b*x^2 + c*x^4]),x]

[Out]

(Sqrt[a + b*x^2 + c*x^4]*(3*A*b*x^2 - 2*a*(A + 2*B*x^2)))/(8*a^2*x^4) - ((-3*A*b
^2 + 4*a*b*B + 4*a*A*c)*(Log[x^2] - Log[2*a + b*x^2 + 2*Sqrt[a]*Sqrt[a + b*x^2 +
c*x^4]]))/(16*a^(5/2))

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Maple [A]  time = 0.014, size = 194, normalized size = 1.6 \[ -{\frac{A}{4\,a{x}^{4}}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{3\,Ab}{8\,{a}^{2}{x}^{2}}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{3\,A{b}^{2}}{16}\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,a+b{x}^{2}+2\,\sqrt{a}\sqrt{c{x}^{4}+b{x}^{2}+a} \right ) } \right ){a}^{-{\frac{5}{2}}}}+{\frac{Ac}{4}\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,a+b{x}^{2}+2\,\sqrt{a}\sqrt{c{x}^{4}+b{x}^{2}+a} \right ) } \right ){a}^{-{\frac{3}{2}}}}-{\frac{B}{2\,{x}^{2}a}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{bB}{4}\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,a+b{x}^{2}+2\,\sqrt{a}\sqrt{c{x}^{4}+b{x}^{2}+a} \right ) } \right ){a}^{-{\frac{3}{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((B*x^2+A)/x^5/(c*x^4+b*x^2+a)^(1/2),x)

[Out]

-1/4*A*(c*x^4+b*x^2+a)^(1/2)/a/x^4+3/8*A*b/a^2/x^2*(c*x^4+b*x^2+a)^(1/2)-3/16*A*
b^2/a^(5/2)*ln((2*a+b*x^2+2*a^(1/2)*(c*x^4+b*x^2+a)^(1/2))/x^2)+1/4*A*c/a^(3/2)*
ln((2*a+b*x^2+2*a^(1/2)*(c*x^4+b*x^2+a)^(1/2))/x^2)-1/2*B/a/x^2*(c*x^4+b*x^2+a)^
(1/2)+1/4*B*b/a^(3/2)*ln((2*a+b*x^2+2*a^(1/2)*(c*x^4+b*x^2+a)^(1/2))/x^2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \text{Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x^2 + A)/(sqrt(c*x^4 + b*x^2 + a)*x^5),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.346938, size = 1, normalized size = 0.01 \[ \left [\frac{{\left (4 \, B a b - 3 \, A b^{2} + 4 \, A a c\right )} x^{4} \log \left (-\frac{4 \, \sqrt{c x^{4} + b x^{2} + a}{\left (a b x^{2} + 2 \, a^{2}\right )} +{\left ({\left (b^{2} + 4 \, a c\right )} x^{4} + 8 \, a b x^{2} + 8 \, a^{2}\right )} \sqrt{a}}{x^{4}}\right ) - 4 \, \sqrt{c x^{4} + b x^{2} + a}{\left ({\left (4 \, B a - 3 \, A b\right )} x^{2} + 2 \, A a\right )} \sqrt{a}}{32 \, a^{\frac{5}{2}} x^{4}}, \frac{{\left (4 \, B a b - 3 \, A b^{2} + 4 \, A a c\right )} x^{4} \arctan \left (\frac{{\left (b x^{2} + 2 \, a\right )} \sqrt{-a}}{2 \, \sqrt{c x^{4} + b x^{2} + a} a}\right ) - 2 \, \sqrt{c x^{4} + b x^{2} + a}{\left ({\left (4 \, B a - 3 \, A b\right )} x^{2} + 2 \, A a\right )} \sqrt{-a}}{16 \, \sqrt{-a} a^{2} x^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x^2 + A)/(sqrt(c*x^4 + b*x^2 + a)*x^5),x, algorithm="fricas")

[Out]

[1/32*((4*B*a*b - 3*A*b^2 + 4*A*a*c)*x^4*log(-(4*sqrt(c*x^4 + b*x^2 + a)*(a*b*x^
2 + 2*a^2) + ((b^2 + 4*a*c)*x^4 + 8*a*b*x^2 + 8*a^2)*sqrt(a))/x^4) - 4*sqrt(c*x^
4 + b*x^2 + a)*((4*B*a - 3*A*b)*x^2 + 2*A*a)*sqrt(a))/(a^(5/2)*x^4), 1/16*((4*B*
a*b - 3*A*b^2 + 4*A*a*c)*x^4*arctan(1/2*(b*x^2 + 2*a)*sqrt(-a)/(sqrt(c*x^4 + b*x
^2 + a)*a)) - 2*sqrt(c*x^4 + b*x^2 + a)*((4*B*a - 3*A*b)*x^2 + 2*A*a)*sqrt(-a))/
(sqrt(-a)*a^2*x^4)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{A + B x^{2}}{x^{5} \sqrt{a + b x^{2} + c x^{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x**2+A)/x**5/(c*x**4+b*x**2+a)**(1/2),x)

[Out]

Integral((A + B*x**2)/(x**5*sqrt(a + b*x**2 + c*x**4)), x)

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{B x^{2} + A}{\sqrt{c x^{4} + b x^{2} + a} x^{5}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x^2 + A)/(sqrt(c*x^4 + b*x^2 + a)*x^5),x, algorithm="giac")

[Out]

integrate((B*x^2 + A)/(sqrt(c*x^4 + b*x^2 + a)*x^5), x)