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

Optimal. Leaf size=76 \[ \frac{B \sqrt{a+b x^2+c x^4}}{2 c}-\frac{(b B-2 A c) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{4 c^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0639266, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {1247, 640, 621, 206} \[ \frac{B \sqrt{a+b x^2+c x^4}}{2 c}-\frac{(b B-2 A c) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{4 c^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1247

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[
Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{x \left (A+B x^2\right )}{\sqrt{a+b x^2+c x^4}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{A+B x}{\sqrt{a+b x+c x^2}} \, dx,x,x^2\right )\\ &=\frac{B \sqrt{a+b x^2+c x^4}}{2 c}+\frac{(-b B+2 A c) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x+c x^2}} \, dx,x,x^2\right )}{4 c}\\ &=\frac{B \sqrt{a+b x^2+c x^4}}{2 c}+\frac{(-b B+2 A c) \operatorname{Subst}\left (\int \frac{1}{4 c-x^2} \, dx,x,\frac{b+2 c x^2}{\sqrt{a+b x^2+c x^4}}\right )}{2 c}\\ &=\frac{B \sqrt{a+b x^2+c x^4}}{2 c}-\frac{(b B-2 A c) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{4 c^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0252939, size = 78, normalized size = 1.03 \[ \frac{1}{2} \left (\frac{(2 A c-b B) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{2 c^{3/2}}+\frac{B \sqrt{a+b x^2+c x^4}}{c}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.01, size = 93, normalized size = 1.2 \begin{align*}{\frac{B}{2\,c}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{bB}{4}\ln \left ({ \left ({\frac{b}{2}}+c{x}^{2} \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{4}+b{x}^{2}+a} \right ){c}^{-{\frac{3}{2}}}}+{\frac{A}{2}\ln \left ({ \left ({\frac{b}{2}}+c{x}^{2} \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{4}+b{x}^{2}+a} \right ){\frac{1}{\sqrt{c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(B*x^2+A)/(c*x^4+b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.50833, size = 421, normalized size = 5.54 \begin{align*} \left [\frac{4 \, \sqrt{c x^{4} + b x^{2} + a} B c -{\left (B b - 2 \, A c\right )} \sqrt{c} \log \left (-8 \, c^{2} x^{4} - 8 \, b c x^{2} - b^{2} - 4 \, \sqrt{c x^{4} + b x^{2} + a}{\left (2 \, c x^{2} + b\right )} \sqrt{c} - 4 \, a c\right )}{8 \, c^{2}}, \frac{2 \, \sqrt{c x^{4} + b x^{2} + a} B c +{\left (B b - 2 \, A c\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{4} + b x^{2} + a}{\left (2 \, c x^{2} + b\right )} \sqrt{-c}}{2 \,{\left (c^{2} x^{4} + b c x^{2} + a c\right )}}\right )}{4 \, c^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(B*x^2+A)/(c*x^4+b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \left (A + B x^{2}\right )}{\sqrt{a + b x^{2} + c x^{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.17761, size = 93, normalized size = 1.22 \begin{align*} \frac{\sqrt{c x^{4} + b x^{2} + a} B}{2 \, c} + \frac{{\left (B b - 2 \, A c\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x^{2} - \sqrt{c x^{4} + b x^{2} + a}\right )} \sqrt{c} - b \right |}\right )}{4 \, c^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(B*x^2+A)/(c*x^4+b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

1/2*sqrt(c*x^4 + b*x^2 + a)*B/c + 1/4*(B*b - 2*A*c)*log(abs(-2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*sqrt(c)
- b))/c^(3/2)