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

Optimal. Leaf size=66 \[ \frac{B \sqrt{b x^2+c x^4}}{2 c}-\frac{(b B-2 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )}{2 c^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.123152, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {2034, 640, 620, 206} \[ \frac{B \sqrt{b x^2+c x^4}}{2 c}-\frac{(b B-2 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )}{2 c^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2034

Int[(x_)^(m_.)*((b_.)*(x_)^(k_.) + (a_.)*(x_)^(j_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n
, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a*x^Simplify[j/n] + b*x^Simplify[k/n])^p*(c + d*x)^q, x], x, x^n], x]
/; FreeQ[{a, b, c, d, j, k, m, n, p, q}, x] &&  !IntegerQ[p] && NeQ[k, j] && IntegerQ[Simplify[j/n]] && Integ
erQ[Simplify[k/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]

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 620

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

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

Mathematica [A]  time = 0.0434992, size = 81, normalized size = 1.23 \[ \frac{x \left (B \sqrt{c} x \left (b+c x^2\right )-\sqrt{b+c x^2} (b B-2 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b+c x^2}}\right )\right )}{2 c^{3/2} \sqrt{x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.006, size = 88, normalized size = 1.3 \begin{align*}{\frac{x}{2}\sqrt{c{x}^{2}+b} \left ( B{c}^{{\frac{3}{2}}}\sqrt{c{x}^{2}+b}x+2\,A\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+b} \right ){c}^{2}-B\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+b} \right ) bc \right ){\frac{1}{\sqrt{c{x}^{4}+b{x}^{2}}}}{c}^{-{\frac{5}{2}}}} \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)^(1/2),x)

[Out]

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

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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)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.1539, size = 302, normalized size = 4.58 \begin{align*} \left [\frac{2 \, \sqrt{c x^{4} + b x^{2}} B c -{\left (B b - 2 \, A c\right )} \sqrt{c} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt{c x^{4} + b x^{2}} \sqrt{c}\right )}{4 \, c^{2}}, \frac{\sqrt{c x^{4} + b x^{2}} B c +{\left (B b - 2 \, A c\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{4} + b x^{2}} \sqrt{-c}}{c x^{2} + b}\right )}{2 \, 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)^(1/2),x, algorithm="fricas")

[Out]

[1/4*(2*sqrt(c*x^4 + b*x^2)*B*c - (B*b - 2*A*c)*sqrt(c)*log(-2*c*x^2 - b - 2*sqrt(c*x^4 + b*x^2)*sqrt(c)))/c^2
, 1/2*(sqrt(c*x^4 + b*x^2)*B*c + (B*b - 2*A*c)*sqrt(-c)*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-c)/(c*x^2 + b)))/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{x^{2} \left (b + c x^{2}\right )}}\, 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)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.23461, size = 90, normalized size = 1.36 \begin{align*} \frac{\sqrt{c x^{4} + b x^{2}} 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}}\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)^(1/2),x, algorithm="giac")

[Out]

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