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

Optimal. Leaf size=142 \[ \frac{\sqrt{b x^2+c x^4} (2 b B-3 A c)}{2 b^2 c x^3}-\frac{(2 b B-3 A c) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{b x^2+c x^4}}\right )}{2 b^{5/2}}-\frac{2 b B-3 A c}{3 b c x \sqrt{b x^2+c x^4}}-\frac{B}{3 c x \sqrt{b x^2+c x^4}} \]

[Out]

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

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Rubi [A]  time = 0.0887823, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {1145, 2006, 2025, 2008, 206} \[ \frac{\sqrt{b x^2+c x^4} (2 b B-3 A c)}{2 b^2 c x^3}-\frac{(2 b B-3 A c) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{b x^2+c x^4}}\right )}{2 b^{5/2}}-\frac{2 b B-3 A c}{3 b c x \sqrt{b x^2+c x^4}}-\frac{B}{3 c x \sqrt{b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1145

Int[((d_) + (e_.)*(x_)^2)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(e*(b*x^2 + c*x^4)^(p + 1))/(c
*(4*p + 3)*x), x] - Dist[(b*e*(2*p + 1) - c*d*(4*p + 3))/(c*(4*p + 3)), Int[(b*x^2 + c*x^4)^p, x], x] /; FreeQ
[{b, c, d, e, p}, x] &&  !IntegerQ[p] && NeQ[4*p + 3, 0] && NeQ[b*e*(2*p + 1) - c*d*(4*p + 3), 0]

Rule 2006

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(a*x^j + b*x^n)^(p + 1)/(a*(n - j)*(p + 1)*
x^(j - 1)), x] + Dist[(n*p + n - j + 1)/(a*(n - j)*(p + 1)), Int[(a*x^j + b*x^n)^(p + 1)/x^j, x], x] /; FreeQ[
{a, b}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && LtQ[p, -1]

Rule 2025

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(j - 1)*(c*x)^(m - j +
1)*(a*x^j + b*x^n)^(p + 1))/(a*(m + j*p + 1)), x] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*p + 1)
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j,
n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[m + j*p + 1, 0]

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{A+B x^2}{\left (b x^2+c x^4\right )^{3/2}} \, dx &=-\frac{B}{3 c x \sqrt{b x^2+c x^4}}+\frac{(-2 b B+3 A c) \int \frac{1}{\left (b x^2+c x^4\right )^{3/2}} \, dx}{3 c}\\ &=-\frac{B}{3 c x \sqrt{b x^2+c x^4}}-\frac{2 b B-3 A c}{3 b c x \sqrt{b x^2+c x^4}}+\frac{(-2 b B+3 A c) \int \frac{1}{x^2 \sqrt{b x^2+c x^4}} \, dx}{b c}\\ &=-\frac{B}{3 c x \sqrt{b x^2+c x^4}}-\frac{2 b B-3 A c}{3 b c x \sqrt{b x^2+c x^4}}+\frac{(2 b B-3 A c) \sqrt{b x^2+c x^4}}{2 b^2 c x^3}+\frac{(2 b B-3 A c) \int \frac{1}{\sqrt{b x^2+c x^4}} \, dx}{2 b^2}\\ &=-\frac{B}{3 c x \sqrt{b x^2+c x^4}}-\frac{2 b B-3 A c}{3 b c x \sqrt{b x^2+c x^4}}+\frac{(2 b B-3 A c) \sqrt{b x^2+c x^4}}{2 b^2 c x^3}-\frac{(2 b B-3 A c) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{b x^2+c x^4}}\right )}{2 b^2}\\ &=-\frac{B}{3 c x \sqrt{b x^2+c x^4}}-\frac{2 b B-3 A c}{3 b c x \sqrt{b x^2+c x^4}}+\frac{(2 b B-3 A c) \sqrt{b x^2+c x^4}}{2 b^2 c x^3}-\frac{(2 b B-3 A c) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{b x^2+c x^4}}\right )}{2 b^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0240062, size = 61, normalized size = 0.43 \[ \frac{x^2 (2 b B-3 A c) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{c x^2}{b}+1\right )-A b}{2 b^2 x \sqrt{x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(A*b) + (2*b*B - 3*A*c)*x^2*Hypergeometric2F1[-1/2, 1, 1/2, 1 + (c*x^2)/b])/(2*b^2*x*Sqrt[x^2*(b + c*x^2)])

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Maple [A]  time = 0.01, size = 129, normalized size = 0.9 \begin{align*} -{\frac{ \left ( c{x}^{2}+b \right ) x}{2} \left ( 3\,A{b}^{3/2}{x}^{2}c-3\,A\ln \left ( 2\,{\frac{\sqrt{b}\sqrt{c{x}^{2}+b}+b}{x}} \right ) \sqrt{c{x}^{2}+b}{x}^{2}bc-2\,B{b}^{5/2}{x}^{2}+2\,B\ln \left ( 2\,{\frac{\sqrt{b}\sqrt{c{x}^{2}+b}+b}{x}} \right ) \sqrt{c{x}^{2}+b}{x}^{2}{b}^{2}+A{b}^{{\frac{5}{2}}} \right ) \left ( c{x}^{4}+b{x}^{2} \right ) ^{-{\frac{3}{2}}}{b}^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*x*(c*x^2+b)*(3*A*b^(3/2)*x^2*c-3*A*ln(2*(b^(1/2)*(c*x^2+b)^(1/2)+b)/x)*(c*x^2+b)^(1/2)*x^2*b*c-2*B*b^(5/2
)*x^2+2*B*ln(2*(b^(1/2)*(c*x^2+b)^(1/2)+b)/x)*(c*x^2+b)^(1/2)*x^2*b^2+A*b^(5/2))/(c*x^4+b*x^2)^(3/2)/b^(7/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B x^{2} + A}{{\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((B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.48199, size = 549, normalized size = 3.87 \begin{align*} \left [-\frac{{\left ({\left (2 \, B b c - 3 \, A c^{2}\right )} x^{5} +{\left (2 \, B b^{2} - 3 \, A b c\right )} x^{3}\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 (A b^{2} -{\left (2 \, B b^{2} - 3 \, A b c\right )} x^{2}\right )}}{4 \,{\left (b^{3} c x^{5} + b^{4} x^{3}\right )}}, \frac{{\left ({\left (2 \, B b c - 3 \, A c^{2}\right )} x^{5} +{\left (2 \, B b^{2} - 3 \, A b c\right )} x^{3}\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 (A b^{2} -{\left (2 \, B b^{2} - 3 \, A b c\right )} x^{2}\right )}}{2 \,{\left (b^{3} c x^{5} + b^{4} x^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x