3.977 \(\int \frac{a+\frac{b}{x^2}}{(c+\frac{d}{x^2})^{3/2} x^5} \, dx\)

Optimal. Leaf size=68 \[ \frac{\sqrt{c+\frac{d}{x^2}} (2 b c-a d)}{d^3}+\frac{c (b c-a d)}{d^3 \sqrt{c+\frac{d}{x^2}}}-\frac{b \left (c+\frac{d}{x^2}\right )^{3/2}}{3 d^3} \]

[Out]

(c*(b*c - a*d))/(d^3*Sqrt[c + d/x^2]) + ((2*b*c - a*d)*Sqrt[c + d/x^2])/d^3 - (b*(c + d/x^2)^(3/2))/(3*d^3)

________________________________________________________________________________________

Rubi [A]  time = 0.0521264, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {446, 77} \[ \frac{\sqrt{c+\frac{d}{x^2}} (2 b c-a d)}{d^3}+\frac{c (b c-a d)}{d^3 \sqrt{c+\frac{d}{x^2}}}-\frac{b \left (c+\frac{d}{x^2}\right )^{3/2}}{3 d^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)/((c + d/x^2)^(3/2)*x^5),x]

[Out]

(c*(b*c - a*d))/(d^3*Sqrt[c + d/x^2]) + ((2*b*c - a*d)*Sqrt[c + d/x^2])/d^3 - (b*(c + d/x^2)^(3/2))/(3*d^3)

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

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

Mathematica [A]  time = 0.0226575, size = 60, normalized size = 0.88 \[ \frac{b \left (8 c^2 x^4+4 c d x^2-d^2\right )-3 a d x^2 \left (2 c x^2+d\right )}{3 d^3 x^4 \sqrt{c+\frac{d}{x^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^2)/((c + d/x^2)^(3/2)*x^5),x]

[Out]

(-3*a*d*x^2*(d + 2*c*x^2) + b*(-d^2 + 4*c*d*x^2 + 8*c^2*x^4))/(3*d^3*Sqrt[c + d/x^2]*x^4)

________________________________________________________________________________________

Maple [A]  time = 0.006, size = 69, normalized size = 1. \begin{align*} -{\frac{ \left ( 6\,acd{x}^{4}-8\,b{c}^{2}{x}^{4}+3\,a{d}^{2}{x}^{2}-4\,bcd{x}^{2}+b{d}^{2} \right ) \left ( c{x}^{2}+d \right ) }{3\,{d}^{3}{x}^{6}} \left ({\frac{c{x}^{2}+d}{{x}^{2}}} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b/x^2)/(c+d/x^2)^(3/2)/x^5,x)

[Out]

-1/3*(6*a*c*d*x^4-8*b*c^2*x^4+3*a*d^2*x^2-4*b*c*d*x^2+b*d^2)*(c*x^2+d)/((c*x^2+d)/x^2)^(3/2)/d^3/x^6

________________________________________________________________________________________

Maxima [A]  time = 0.956154, size = 109, normalized size = 1.6 \begin{align*} -\frac{1}{3} \, b{\left (\frac{{\left (c + \frac{d}{x^{2}}\right )}^{\frac{3}{2}}}{d^{3}} - \frac{6 \, \sqrt{c + \frac{d}{x^{2}}} c}{d^{3}} - \frac{3 \, c^{2}}{\sqrt{c + \frac{d}{x^{2}}} d^{3}}\right )} - a{\left (\frac{\sqrt{c + \frac{d}{x^{2}}}}{d^{2}} + \frac{c}{\sqrt{c + \frac{d}{x^{2}}} d^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)/(c+d/x^2)^(3/2)/x^5,x, algorithm="maxima")

[Out]

-1/3*b*((c + d/x^2)^(3/2)/d^3 - 6*sqrt(c + d/x^2)*c/d^3 - 3*c^2/(sqrt(c + d/x^2)*d^3)) - a*(sqrt(c + d/x^2)/d^
2 + c/(sqrt(c + d/x^2)*d^2))

________________________________________________________________________________________

Fricas [A]  time = 1.58677, size = 150, normalized size = 2.21 \begin{align*} \frac{{\left (2 \,{\left (4 \, b c^{2} - 3 \, a c d\right )} x^{4} - b d^{2} +{\left (4 \, b c d - 3 \, a d^{2}\right )} x^{2}\right )} \sqrt{\frac{c x^{2} + d}{x^{2}}}}{3 \,{\left (c d^{3} x^{4} + d^{4} x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)/(c+d/x^2)^(3/2)/x^5,x, algorithm="fricas")

[Out]

1/3*(2*(4*b*c^2 - 3*a*c*d)*x^4 - b*d^2 + (4*b*c*d - 3*a*d^2)*x^2)*sqrt((c*x^2 + d)/x^2)/(c*d^3*x^4 + d^4*x^2)

________________________________________________________________________________________

Sympy [A]  time = 7.43081, size = 61, normalized size = 0.9 \begin{align*} - \frac{b \left (c + \frac{d}{x^{2}}\right )^{\frac{3}{2}}}{3 d^{3}} - \frac{c \left (a d - b c\right )}{d^{3} \sqrt{c + \frac{d}{x^{2}}}} - \frac{\sqrt{c + \frac{d}{x^{2}}} \left (a d - 2 b c\right )}{d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x**2)/(c+d/x**2)**(3/2)/x**5,x)

[Out]

-b*(c + d/x**2)**(3/2)/(3*d**3) - c*(a*d - b*c)/(d**3*sqrt(c + d/x**2)) - sqrt(c + d/x**2)*(a*d - 2*b*c)/d**3

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a + \frac{b}{x^{2}}}{{\left (c + \frac{d}{x^{2}}\right )}^{\frac{3}{2}} x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)/(c+d/x^2)^(3/2)/x^5,x, algorithm="giac")

[Out]

integrate((a + b/x^2)/((c + d/x^2)^(3/2)*x^5), x)