3.933 \(\int \frac{(a+\frac{b}{x^2}) \sqrt{c+\frac{d}{x^2}}}{x^3} \, dx\)

Optimal. Leaf size=46 \[ \frac{\left (c+\frac{d}{x^2}\right )^{3/2} (b c-a d)}{3 d^2}-\frac{b \left (c+\frac{d}{x^2}\right )^{5/2}}{5 d^2} \]

[Out]

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

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Rubi [A]  time = 0.0373817, antiderivative size = 46, 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 = {444, 43} \[ \frac{\left (c+\frac{d}{x^2}\right )^{3/2} (b c-a d)}{3 d^2}-\frac{b \left (c+\frac{d}{x^2}\right )^{5/2}}{5 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0162534, size = 47, normalized size = 1.02 \[ -\frac{\sqrt{c+\frac{d}{x^2}} \left (c x^2+d\right ) \left (5 a d x^2-2 b c x^2+3 b d\right )}{15 d^2 x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[c + d/x^2]*(d + c*x^2)*(3*b*d - 2*b*c*x^2 + 5*a*d*x^2))/(15*d^2*x^4)

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Maple [A]  time = 0.006, size = 48, normalized size = 1. \begin{align*} -{\frac{ \left ( 5\,ad{x}^{2}-2\,bc{x}^{2}+3\,bd \right ) \left ( c{x}^{2}+d \right ) }{15\,{d}^{2}{x}^{4}}\sqrt{{\frac{c{x}^{2}+d}{{x}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b/x^2)*(c+d/x^2)^(1/2)/x^3,x)

[Out]

-1/15*((c*x^2+d)/x^2)^(1/2)*(5*a*d*x^2-2*b*c*x^2+3*b*d)*(c*x^2+d)/d^2/x^4

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Maxima [A]  time = 0.940565, size = 66, normalized size = 1.43 \begin{align*} -\frac{1}{15} \, b{\left (\frac{3 \,{\left (c + \frac{d}{x^{2}}\right )}^{\frac{5}{2}}}{d^{2}} - \frac{5 \,{\left (c + \frac{d}{x^{2}}\right )}^{\frac{3}{2}} c}{d^{2}}\right )} - \frac{a{\left (c + \frac{d}{x^{2}}\right )}^{\frac{3}{2}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)*(c+d/x^2)^(1/2)/x^3,x, algorithm="maxima")

[Out]

-1/15*b*(3*(c + d/x^2)^(5/2)/d^2 - 5*(c + d/x^2)^(3/2)*c/d^2) - 1/3*a*(c + d/x^2)^(3/2)/d

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Fricas [A]  time = 1.32251, size = 132, normalized size = 2.87 \begin{align*} \frac{{\left ({\left (2 \, b c^{2} - 5 \, a c d\right )} x^{4} - 3 \, b d^{2} -{\left (b c d + 5 \, a d^{2}\right )} x^{2}\right )} \sqrt{\frac{c x^{2} + d}{x^{2}}}}{15 \, d^{2} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)*(c+d/x^2)^(1/2)/x^3,x, algorithm="fricas")

[Out]

1/15*((2*b*c^2 - 5*a*c*d)*x^4 - 3*b*d^2 - (b*c*d + 5*a*d^2)*x^2)*sqrt((c*x^2 + d)/x^2)/(d^2*x^4)

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Sympy [A]  time = 3.30036, size = 58, normalized size = 1.26 \begin{align*} - \frac{a \left (\begin{cases} \frac{\sqrt{c}}{x^{2}} & \text{for}\: d = 0 \\\frac{2 \left (c + \frac{d}{x^{2}}\right )^{\frac{3}{2}}}{3 d} & \text{otherwise} \end{cases}\right )}{2} - \frac{b \left (- \frac{c \left (c + \frac{d}{x^{2}}\right )^{\frac{3}{2}}}{3} + \frac{\left (c + \frac{d}{x^{2}}\right )^{\frac{5}{2}}}{5}\right )}{d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x**2)*(c+d/x**2)**(1/2)/x**3,x)

[Out]

-a*Piecewise((sqrt(c)/x**2, Eq(d, 0)), (2*(c + d/x**2)**(3/2)/(3*d), True))/2 - b*(-c*(c + d/x**2)**(3/2)/3 +
(c + d/x**2)**(5/2)/5)/d**2

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Giac [B]  time = 2.19649, size = 338, normalized size = 7.35 \begin{align*} \frac{2 \,{\left (15 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{8} a c^{\frac{3}{2}} \mathrm{sgn}\left (x\right ) + 30 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{6} b c^{\frac{5}{2}} \mathrm{sgn}\left (x\right ) - 30 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{6} a c^{\frac{3}{2}} d \mathrm{sgn}\left (x\right ) + 10 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{4} b c^{\frac{5}{2}} d \mathrm{sgn}\left (x\right ) + 20 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{4} a c^{\frac{3}{2}} d^{2} \mathrm{sgn}\left (x\right ) + 10 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{2} b c^{\frac{5}{2}} d^{2} \mathrm{sgn}\left (x\right ) - 10 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{2} a c^{\frac{3}{2}} d^{3} \mathrm{sgn}\left (x\right ) - 2 \, b c^{\frac{5}{2}} d^{3} \mathrm{sgn}\left (x\right ) + 5 \, a c^{\frac{3}{2}} d^{4} \mathrm{sgn}\left (x\right )\right )}}{15 \,{\left ({\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{2} - d\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)*(c+d/x^2)^(1/2)/x^3,x, algorithm="giac")

[Out]

2/15*(15*(sqrt(c)*x - sqrt(c*x^2 + d))^8*a*c^(3/2)*sgn(x) + 30*(sqrt(c)*x - sqrt(c*x^2 + d))^6*b*c^(5/2)*sgn(x
) - 30*(sqrt(c)*x - sqrt(c*x^2 + d))^6*a*c^(3/2)*d*sgn(x) + 10*(sqrt(c)*x - sqrt(c*x^2 + d))^4*b*c^(5/2)*d*sgn
(x) + 20*(sqrt(c)*x - sqrt(c*x^2 + d))^4*a*c^(3/2)*d^2*sgn(x) + 10*(sqrt(c)*x - sqrt(c*x^2 + d))^2*b*c^(5/2)*d
^2*sgn(x) - 10*(sqrt(c)*x - sqrt(c*x^2 + d))^2*a*c^(3/2)*d^3*sgn(x) - 2*b*c^(5/2)*d^3*sgn(x) + 5*a*c^(3/2)*d^4
*sgn(x))/((sqrt(c)*x - sqrt(c*x^2 + d))^2 - d)^5