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

Optimal. Leaf size=46 \[ \frac{\left (c+\frac{d}{x^2}\right )^{5/2} (b c-a d)}{5 d^2}-\frac{b \left (c+\frac{d}{x^2}\right )^{7/2}}{7 d^2} \]

[Out]

((b*c - a*d)*(c + d/x^2)^(5/2))/(5*d^2) - (b*(c + d/x^2)^(7/2))/(7*d^2)

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Rubi [A]  time = 0.0364029, 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} \[ \frac{\left (c+\frac{d}{x^2}\right )^{5/2} (b c-a d)}{5 d^2}-\frac{b \left (c+\frac{d}{x^2}\right )^{7/2}}{7 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c - a*d)*(c + d/x^2)^(5/2))/(5*d^2) - (b*(c + d/x^2)^(7/2))/(7*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 ) \left (c+\frac{d}{x^2}\right )^{3/2}}{x^3} \, dx &=-\left (\frac{1}{2} \operatorname{Subst}\left (\int (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{(-b c+a d) (c+d x)^{3/2}}{d}+\frac{b (c+d x)^{5/2}}{d}\right ) \, dx,x,\frac{1}{x^2}\right )\right )\\ &=\frac{(b c-a d) \left (c+\frac{d}{x^2}\right )^{5/2}}{5 d^2}-\frac{b \left (c+\frac{d}{x^2}\right )^{7/2}}{7 d^2}\\ \end{align*}

Mathematica [A]  time = 0.0183914, size = 49, normalized size = 1.07 \[ -\frac{\sqrt{c+\frac{d}{x^2}} \left (c x^2+d\right )^2 \left (7 a d x^2-2 b c x^2+5 b d\right )}{35 d^2 x^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[c + d/x^2]*(d + c*x^2)^2*(5*b*d - 2*b*c*x^2 + 7*a*d*x^2))/(35*d^2*x^6)

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Maple [A]  time = 0.006, size = 48, normalized size = 1. \begin{align*} -{\frac{ \left ( 7\,ad{x}^{2}-2\,bc{x}^{2}+5\,bd \right ) \left ( c{x}^{2}+d \right ) }{35\,{d}^{2}{x}^{4}} \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^3,x)

[Out]

-1/35*((c*x^2+d)/x^2)^(3/2)*(7*a*d*x^2-2*b*c*x^2+5*b*d)*(c*x^2+d)/d^2/x^4

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Maxima [A]  time = 0.947076, size = 66, normalized size = 1.43 \begin{align*} -\frac{a{\left (c + \frac{d}{x^{2}}\right )}^{\frac{5}{2}}}{5 \, d} - \frac{1}{35} \,{\left (\frac{5 \,{\left (c + \frac{d}{x^{2}}\right )}^{\frac{7}{2}}}{d^{2}} - \frac{7 \,{\left (c + \frac{d}{x^{2}}\right )}^{\frac{5}{2}} c}{d^{2}}\right )} b \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^3,x, algorithm="maxima")

[Out]

-1/5*a*(c + d/x^2)^(5/2)/d - 1/35*(5*(c + d/x^2)^(7/2)/d^2 - 7*(c + d/x^2)^(5/2)*c/d^2)*b

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Fricas [B]  time = 1.37909, size = 180, normalized size = 3.91 \begin{align*} \frac{{\left ({\left (2 \, b c^{3} - 7 \, a c^{2} d\right )} x^{6} -{\left (b c^{2} d + 14 \, a c d^{2}\right )} x^{4} - 5 \, b d^{3} -{\left (8 \, b c d^{2} + 7 \, a d^{3}\right )} x^{2}\right )} \sqrt{\frac{c x^{2} + d}{x^{2}}}}{35 \, d^{2} x^{6}} \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^3,x, algorithm="fricas")

[Out]

1/35*((2*b*c^3 - 7*a*c^2*d)*x^6 - (b*c^2*d + 14*a*c*d^2)*x^4 - 5*b*d^3 - (8*b*c*d^2 + 7*a*d^3)*x^2)*sqrt((c*x^
2 + d)/x^2)/(d^2*x^6)

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Sympy [A]  time = 10.1932, size = 138, normalized size = 3. \begin{align*} - \frac{a c \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{a \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} - \frac{b c \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}} - \frac{b \left (\frac{c^{2} \left (c + \frac{d}{x^{2}}\right )^{\frac{3}{2}}}{3} - \frac{2 c \left (c + \frac{d}{x^{2}}\right )^{\frac{5}{2}}}{5} + \frac{\left (c + \frac{d}{x^{2}}\right )^{\frac{7}{2}}}{7}\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)**(3/2)/x**3,x)

[Out]

-a*c*Piecewise((sqrt(c)/x**2, Eq(d, 0)), (2*(c + d/x**2)**(3/2)/(3*d), True))/2 - a*(-c*(c + d/x**2)**(3/2)/3
+ (c + d/x**2)**(5/2)/5)/d - b*c*(-c*(c + d/x**2)**(3/2)/3 + (c + d/x**2)**(5/2)/5)/d**2 - b*(c**2*(c + d/x**2
)**(3/2)/3 - 2*c*(c + d/x**2)**(5/2)/5 + (c + d/x**2)**(7/2)/7)/d**2

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Giac [B]  time = 5.43887, size = 500, normalized size = 10.87 \begin{align*} \frac{2 \,{\left (35 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{12} a c^{\frac{5}{2}} \mathrm{sgn}\left (x\right ) + 70 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{10} b c^{\frac{7}{2}} \mathrm{sgn}\left (x\right ) - 70 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{10} a c^{\frac{5}{2}} d \mathrm{sgn}\left (x\right ) + 70 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{8} b c^{\frac{7}{2}} d \mathrm{sgn}\left (x\right ) + 105 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{8} a c^{\frac{5}{2}} d^{2} \mathrm{sgn}\left (x\right ) + 140 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{6} b c^{\frac{7}{2}} d^{2} \mathrm{sgn}\left (x\right ) - 140 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{6} a c^{\frac{5}{2}} d^{3} \mathrm{sgn}\left (x\right ) + 28 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{4} b c^{\frac{7}{2}} d^{3} \mathrm{sgn}\left (x\right ) + 77 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{4} a c^{\frac{5}{2}} d^{4} \mathrm{sgn}\left (x\right ) + 14 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{2} b c^{\frac{7}{2}} d^{4} \mathrm{sgn}\left (x\right ) - 14 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{2} a c^{\frac{5}{2}} d^{5} \mathrm{sgn}\left (x\right ) - 2 \, b c^{\frac{7}{2}} d^{5} \mathrm{sgn}\left (x\right ) + 7 \, a c^{\frac{5}{2}} d^{6} \mathrm{sgn}\left (x\right )\right )}}{35 \,{\left ({\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{2} - d\right )}^{7}} \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^3,x, algorithm="giac")

[Out]

2/35*(35*(sqrt(c)*x - sqrt(c*x^2 + d))^12*a*c^(5/2)*sgn(x) + 70*(sqrt(c)*x - sqrt(c*x^2 + d))^10*b*c^(7/2)*sgn
(x) - 70*(sqrt(c)*x - sqrt(c*x^2 + d))^10*a*c^(5/2)*d*sgn(x) + 70*(sqrt(c)*x - sqrt(c*x^2 + d))^8*b*c^(7/2)*d*
sgn(x) + 105*(sqrt(c)*x - sqrt(c*x^2 + d))^8*a*c^(5/2)*d^2*sgn(x) + 140*(sqrt(c)*x - sqrt(c*x^2 + d))^6*b*c^(7
/2)*d^2*sgn(x) - 140*(sqrt(c)*x - sqrt(c*x^2 + d))^6*a*c^(5/2)*d^3*sgn(x) + 28*(sqrt(c)*x - sqrt(c*x^2 + d))^4
*b*c^(7/2)*d^3*sgn(x) + 77*(sqrt(c)*x - sqrt(c*x^2 + d))^4*a*c^(5/2)*d^4*sgn(x) + 14*(sqrt(c)*x - sqrt(c*x^2 +
d))^2*b*c^(7/2)*d^4*sgn(x) - 14*(sqrt(c)*x - sqrt(c*x^2 + d))^2*a*c^(5/2)*d^5*sgn(x) - 2*b*c^(7/2)*d^5*sgn(x)
+ 7*a*c^(5/2)*d^6*sgn(x))/((sqrt(c)*x - sqrt(c*x^2 + d))^2 - d)^7