3.93 \(\int \frac{\left (A+B x^2\right ) \sqrt{b x^2+c x^4}}{x} \, dx\)

Optimal. Leaf size=100 \[ -\frac{b (b B-4 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )}{8 c^{3/2}}-\frac{\sqrt{b x^2+c x^4} (b B-4 A c)}{8 c}+\frac{B \left (b x^2+c x^4\right )^{3/2}}{4 c x^2} \]

[Out]

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

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Rubi [A]  time = 0.381558, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192 \[ -\frac{b (b B-4 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )}{8 c^{3/2}}-\frac{\sqrt{b x^2+c x^4} (b B-4 A c)}{8 c}+\frac{B \left (b x^2+c x^4\right )^{3/2}}{4 c x^2} \]

Antiderivative was successfully verified.

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

[Out]

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

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Rubi in Sympy [A]  time = 23.4429, size = 87, normalized size = 0.87 \[ \frac{B \left (b x^{2} + c x^{4}\right )^{\frac{3}{2}}}{4 c x^{2}} + \frac{b \left (4 A c - B b\right ) \operatorname{atanh}{\left (\frac{\sqrt{c} x^{2}}{\sqrt{b x^{2} + c x^{4}}} \right )}}{8 c^{\frac{3}{2}}} + \frac{\left (4 A c - B b\right ) \sqrt{b x^{2} + c x^{4}}}{8 c} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Mathematica [A]  time = 0.165027, size = 99, normalized size = 0.99 \[ \frac{x \left (\sqrt{c} x \left (b+c x^2\right ) \left (4 A c+b B+2 B c x^2\right )+b \sqrt{b+c x^2} (4 A c-b B) \log \left (\sqrt{c} \sqrt{b+c x^2}+c x\right )\right )}{8 c^{3/2} \sqrt{x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

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

[Out]

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

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Maple [A]  time = 0.01, size = 127, normalized size = 1.3 \[{\frac{1}{8\,x}\sqrt{c{x}^{4}+b{x}^{2}} \left ( 2\,Bx \left ( c{x}^{2}+b \right ) ^{3/2}{c}^{3/2}+4\,Ax\sqrt{c{x}^{2}+b}{c}^{5/2}-Bbx\sqrt{c{x}^{2}+b}{c}^{{\frac{3}{2}}}+4\,Ab\ln \left ( \sqrt{c}x+\sqrt{c{x}^{2}+b} \right ){c}^{2}-B{b}^{2}\ln \left ( \sqrt{c}x+\sqrt{c{x}^{2}+b} \right ) c \right ){\frac{1}{\sqrt{c{x}^{2}+b}}}{c}^{-{\frac{5}{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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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(sqrt(c*x^4 + b*x^2)*(B*x^2 + A)/x,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.240051, size = 1, normalized size = 0.01 \[ \left [-\frac{{\left (B b^{2} - 4 \, A b c\right )} \sqrt{c} \log \left (-{\left (2 \, c x^{2} + b\right )} \sqrt{c} - 2 \, \sqrt{c x^{4} + b x^{2}} c\right ) - 2 \,{\left (2 \, B c^{2} x^{2} + B b c + 4 \, A c^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{16 \, c^{2}}, \frac{{\left (B b^{2} - 4 \, A b c\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{-c} x^{2}}{\sqrt{c x^{4} + b x^{2}}}\right ) +{\left (2 \, B c^{2} x^{2} + B b c + 4 \, A c^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{8 \, c^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{\sqrt{x^{2} \left (b + c x^{2}\right )} \left (A + B x^{2}\right )}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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GIAC/XCAS [A]  time = 0.219777, size = 142, normalized size = 1.42 \[ \frac{1}{8} \,{\left (2 \, B x^{2}{\rm sign}\left (x\right ) + \frac{B b c{\rm sign}\left (x\right ) + 4 \, A c^{2}{\rm sign}\left (x\right )}{c^{2}}\right )} \sqrt{c x^{2} + b} x + \frac{{\left (B b^{2}{\rm sign}\left (x\right ) - 4 \, A b c{\rm sign}\left (x\right )\right )}{\rm ln}\left ({\left | -\sqrt{c} x + \sqrt{c x^{2} + b} \right |}\right )}{8 \, c^{\frac{3}{2}}} - \frac{{\left (B b^{2}{\rm ln}\left (\sqrt{b}\right ) - 4 \, A b c{\rm ln}\left (\sqrt{b}\right )\right )}{\rm sign}\left (x\right )}{8 \, c^{\frac{3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

1/8*(2*B*x^2*sign(x) + (B*b*c*sign(x) + 4*A*c^2*sign(x))/c^2)*sqrt(c*x^2 + b)*x
+ 1/8*(B*b^2*sign(x) - 4*A*b*c*sign(x))*ln(abs(-sqrt(c)*x + sqrt(c*x^2 + b)))/c^
(3/2) - 1/8*(B*b^2*ln(sqrt(b)) - 4*A*b*c*ln(sqrt(b)))*sign(x)/c^(3/2)