3.156 \(\int \frac{x^2 (A+B x^2)}{(b x^2+c x^4)^{3/2}} \, dx\)

Optimal. Leaf size=64 \[ -\frac{A \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{b x^2+c x^4}}\right )}{b^{3/2}}-\frac{x (b B-A c)}{b c \sqrt{b x^2+c x^4}} \]

[Out]

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

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Rubi [A]  time = 0.138556, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {2037, 2008, 206} \[ -\frac{A \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{b x^2+c x^4}}\right )}{b^{3/2}}-\frac{x (b B-A c)}{b c \sqrt{b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]

[Out]

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

Rule 2037

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> -Si
mp[(e^(j - 1)*(b*c - a*d)*(e*x)^(m - j + 1)*(a*x^j + b*x^(j + n))^(p + 1))/(a*b*n*(p + 1)), x] - Dist[(e^j*(a*
d*(m + j*p + 1) - b*c*(m + n + p*(j + n) + 1)))/(a*b*n*(p + 1)), Int[(e*x)^(m - j)*(a*x^j + b*x^(j + n))^(p +
1), x], x] /; FreeQ[{a, b, c, d, e, j, m, n}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && Lt
Q[p, -1] && GtQ[j, 0] && LeQ[j, m] && (GtQ[e, 0] || IntegerQ[j])

Rule 2008

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

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

Mathematica [A]  time = 0.0309807, size = 73, normalized size = 1.14 \[ -\frac{x \left (\sqrt{b} (b B-A c)+A c \sqrt{b+c x^2} \tanh ^{-1}\left (\frac{\sqrt{b+c x^2}}{\sqrt{b}}\right )\right )}{b^{3/2} c \sqrt{x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]

[Out]

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

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Maple [A]  time = 0.009, size = 79, normalized size = 1.2 \begin{align*}{\frac{ \left ( c{x}^{2}+b \right ){x}^{3}}{c} \left ( A{b}^{{\frac{3}{2}}}c-A\ln \left ( 2\,{\frac{\sqrt{b}\sqrt{c{x}^{2}+b}+b}{x}} \right ) \sqrt{c{x}^{2}+b}bc-B{b}^{{\frac{5}{2}}} \right ) \left ( c{x}^{4}+b{x}^{2} \right ) ^{-{\frac{3}{2}}}{b}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x^{2} + A\right )} x^{2}}{{\left (c x^{4} + b x^{2}\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)*x^2/(c*x^4 + b*x^2)^(3/2), x)

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Fricas [A]  time = 1.44314, size = 419, normalized size = 6.55 \begin{align*} \left [\frac{{\left (A c^{2} x^{3} + A b c x\right )} \sqrt{b} \log \left (-\frac{c x^{3} + 2 \, b x - 2 \, \sqrt{c x^{4} + b x^{2}} \sqrt{b}}{x^{3}}\right ) - 2 \, \sqrt{c x^{4} + b x^{2}}{\left (B b^{2} - A b c\right )}}{2 \,{\left (b^{2} c^{2} x^{3} + b^{3} c x\right )}}, \frac{{\left (A c^{2} x^{3} + A b c x\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{c x^{4} + b x^{2}} \sqrt{-b}}{c x^{3} + b x}\right ) - \sqrt{c x^{4} + b x^{2}}{\left (B b^{2} - A b c\right )}}{b^{2} c^{2} x^{3} + b^{3} c x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")

[Out]

[1/2*((A*c^2*x^3 + A*b*c*x)*sqrt(b)*log(-(c*x^3 + 2*b*x - 2*sqrt(c*x^4 + b*x^2)*sqrt(b))/x^3) - 2*sqrt(c*x^4 +
b*x^2)*(B*b^2 - A*b*c))/(b^2*c^2*x^3 + b^3*c*x), ((A*c^2*x^3 + A*b*c*x)*sqrt(-b)*arctan(sqrt(c*x^4 + b*x^2)*s
qrt(-b)/(c*x^3 + b*x)) - sqrt(c*x^4 + b*x^2)*(B*b^2 - A*b*c))/(b^2*c^2*x^3 + b^3*c*x)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2} \left (A + B x^{2}\right )}{\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(B*x**2+A)/(c*x**4+b*x**2)**(3/2),x)

[Out]

Integral(x**2*(A + B*x**2)/(x**2*(b + c*x**2))**(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError