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

Optimal. Leaf size=137 \[ -\frac{3 \sqrt{b x^2+c x^4} (4 b B-5 A c)}{8 b^3 x^3}+\frac{4 b B-5 A c}{4 b^2 x \sqrt{b x^2+c x^4}}+\frac{3 c (4 b B-5 A c) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{b x^2+c x^4}}\right )}{8 b^{7/2}}-\frac{A}{4 b x^3 \sqrt{b x^2+c x^4}} \]

[Out]

-A/(4*b*x^3*Sqrt[b*x^2 + c*x^4]) + (4*b*B - 5*A*c)/(4*b^2*x*Sqrt[b*x^2 + c*x^4]) - (3*(4*b*B - 5*A*c)*Sqrt[b*x
^2 + c*x^4])/(8*b^3*x^3) + (3*c*(4*b*B - 5*A*c)*ArcTanh[(Sqrt[b]*x)/Sqrt[b*x^2 + c*x^4]])/(8*b^(7/2))

________________________________________________________________________________________

Rubi [A]  time = 0.191907, antiderivative size = 137, 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, Rules used = {2038, 2006, 2025, 2008, 206} \[ -\frac{3 \sqrt{b x^2+c x^4} (4 b B-5 A c)}{8 b^3 x^3}+\frac{4 b B-5 A c}{4 b^2 x \sqrt{b x^2+c x^4}}+\frac{3 c (4 b B-5 A c) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{b x^2+c x^4}}\right )}{8 b^{7/2}}-\frac{A}{4 b x^3 \sqrt{b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-A/(4*b*x^3*Sqrt[b*x^2 + c*x^4]) + (4*b*B - 5*A*c)/(4*b^2*x*Sqrt[b*x^2 + c*x^4]) - (3*(4*b*B - 5*A*c)*Sqrt[b*x
^2 + c*x^4])/(8*b^3*x^3) + (3*c*(4*b*B - 5*A*c)*ArcTanh[(Sqrt[b]*x)/Sqrt[b*x^2 + c*x^4]])/(8*b^(7/2))

Rule 2038

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Sim
p[(c*e^(j - 1)*(e*x)^(m - j + 1)*(a*x^j + b*x^(j + n))^(p + 1))/(a*(m + j*p + 1)), x] + Dist[(a*d*(m + j*p + 1
) - b*c*(m + n + p*(j + n) + 1))/(a*e^n*(m + j*p + 1)), Int[(e*x)^(m + n)*(a*x^j + b*x^(j + n))^p, x], x] /; F
reeQ[{a, b, c, d, e, j, p}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && (LtQ[m
+ j*p, -1] || (IntegersQ[m - 1/2, p - 1/2] && LtQ[p, 0] && LtQ[m, -(n*p) - 1])) && (GtQ[e, 0] || IntegersQ[j,
n]) && NeQ[m + j*p + 1, 0] && NeQ[m - n + j*p + 1, 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}{x^2 \left (b x^2+c x^4\right )^{3/2}} \, dx &=-\frac{A}{4 b x^3 \sqrt{b x^2+c x^4}}-\frac{(-4 b B+5 A c) \int \frac{1}{\left (b x^2+c x^4\right )^{3/2}} \, dx}{4 b}\\ &=-\frac{A}{4 b x^3 \sqrt{b x^2+c x^4}}+\frac{4 b B-5 A c}{4 b^2 x \sqrt{b x^2+c x^4}}+\frac{(3 (4 b B-5 A c)) \int \frac{1}{x^2 \sqrt{b x^2+c x^4}} \, dx}{4 b^2}\\ &=-\frac{A}{4 b x^3 \sqrt{b x^2+c x^4}}+\frac{4 b B-5 A c}{4 b^2 x \sqrt{b x^2+c x^4}}-\frac{3 (4 b B-5 A c) \sqrt{b x^2+c x^4}}{8 b^3 x^3}-\frac{(3 c (4 b B-5 A c)) \int \frac{1}{\sqrt{b x^2+c x^4}} \, dx}{8 b^3}\\ &=-\frac{A}{4 b x^3 \sqrt{b x^2+c x^4}}+\frac{4 b B-5 A c}{4 b^2 x \sqrt{b x^2+c x^4}}-\frac{3 (4 b B-5 A c) \sqrt{b x^2+c x^4}}{8 b^3 x^3}+\frac{(3 c (4 b B-5 A c)) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{b x^2+c x^4}}\right )}{8 b^3}\\ &=-\frac{A}{4 b x^3 \sqrt{b x^2+c x^4}}+\frac{4 b B-5 A c}{4 b^2 x \sqrt{b x^2+c x^4}}-\frac{3 (4 b B-5 A c) \sqrt{b x^2+c x^4}}{8 b^3 x^3}+\frac{3 c (4 b B-5 A c) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{b x^2+c x^4}}\right )}{8 b^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0266065, size = 64, normalized size = 0.47 \[ \frac{c x^4 (5 A c-4 b B) \, _2F_1\left (-\frac{1}{2},2;\frac{1}{2};\frac{c x^2}{b}+1\right )-A b^2}{4 b^3 x^3 \sqrt{x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(A*b^2) + c*(-4*b*B + 5*A*c)*x^4*Hypergeometric2F1[-1/2, 2, 1/2, 1 + (c*x^2)/b])/(4*b^3*x^3*Sqrt[x^2*(b + c*
x^2)])

________________________________________________________________________________________

Maple [A]  time = 0.01, size = 157, normalized size = 1.2 \begin{align*} -{\frac{c{x}^{2}+b}{8\,x} \left ( 12\,B{b}^{5/2}{x}^{4}c-15\,A{b}^{3/2}{x}^{4}{c}^{2}+15\,A\ln \left ( 2\,{\frac{\sqrt{b}\sqrt{c{x}^{2}+b}+b}{x}} \right ) \sqrt{c{x}^{2}+b}{x}^{4}b{c}^{2}+4\,B{b}^{7/2}{x}^{2}-12\,B\ln \left ( 2\,{\frac{\sqrt{b}\sqrt{c{x}^{2}+b}+b}{x}} \right ) \sqrt{c{x}^{2}+b}{x}^{4}{b}^{2}c-5\,A{b}^{5/2}{x}^{2}c+2\,A{b}^{7/2} \right ) \left ( c{x}^{4}+b{x}^{2} \right ) ^{-{\frac{3}{2}}}{b}^{-{\frac{9}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(c*x^2+b)*(12*B*b^(5/2)*x^4*c-15*A*b^(3/2)*x^4*c^2+15*A*ln(2*(b^(1/2)*(c*x^2+b)^(1/2)+b)/x)*(c*x^2+b)^(1/
2)*x^4*b*c^2+4*B*b^(7/2)*x^2-12*B*ln(2*(b^(1/2)*(c*x^2+b)^(1/2)+b)/x)*(c*x^2+b)^(1/2)*x^4*b^2*c-5*A*b^(5/2)*x^
2*c+2*A*b^(7/2))/x/(c*x^4+b*x^2)^(3/2)/b^(9/2)

________________________________________________________________________________________

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}} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^2/(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^2), x)

________________________________________________________________________________________

Fricas [A]  time = 1.4119, size = 671, normalized size = 4.9 \begin{align*} \left [-\frac{3 \,{\left ({\left (4 \, B b c^{2} - 5 \, A c^{3}\right )} x^{7} +{\left (4 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{5}\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 \,{\left (3 \,{\left (4 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{4} + 2 \, A b^{3} +{\left (4 \, B b^{3} - 5 \, A b^{2} c\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{16 \,{\left (b^{4} c x^{7} + b^{5} x^{5}\right )}}, -\frac{3 \,{\left ({\left (4 \, B b c^{2} - 5 \, A c^{3}\right )} x^{7} +{\left (4 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{5}\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{c x^{4} + b x^{2}} \sqrt{-b}}{c x^{3} + b x}\right ) +{\left (3 \,{\left (4 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{4} + 2 \, A b^{3} +{\left (4 \, B b^{3} - 5 \, A b^{2} c\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{8 \,{\left (b^{4} c x^{7} + b^{5} x^{5}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*(3*((4*B*b*c^2 - 5*A*c^3)*x^7 + (4*B*b^2*c - 5*A*b*c^2)*x^5)*sqrt(b)*log(-(c*x^3 + 2*b*x - 2*sqrt(c*x^4
+ b*x^2)*sqrt(b))/x^3) + 2*(3*(4*B*b^2*c - 5*A*b*c^2)*x^4 + 2*A*b^3 + (4*B*b^3 - 5*A*b^2*c)*x^2)*sqrt(c*x^4 +
b*x^2))/(b^4*c*x^7 + b^5*x^5), -1/8*(3*((4*B*b*c^2 - 5*A*c^3)*x^7 + (4*B*b^2*c - 5*A*b*c^2)*x^5)*sqrt(-b)*arc
tan(sqrt(c*x^4 + b*x^2)*sqrt(-b)/(c*x^3 + b*x)) + (3*(4*B*b^2*c - 5*A*b*c^2)*x^4 + 2*A*b^3 + (4*B*b^3 - 5*A*b^
2*c)*x^2)*sqrt(c*x^4 + b*x^2))/(b^4*c*x^7 + b^5*x^5)]

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Giac [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}} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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