3.966 \(\int \frac {a+\frac {b}{x^2}}{\sqrt {c+\frac {d}{x^2}} x^5} \, dx\)

Optimal. Leaf size=72 \[ \frac {\left (c+\frac {d}{x^2}\right )^{3/2} (2 b c-a d)}{3 d^3}-\frac {c \sqrt {c+\frac {d}{x^2}} (b c-a d)}{d^3}-\frac {b \left (c+\frac {d}{x^2}\right )^{5/2}}{5 d^3} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 72, normalized size of antiderivative = 1.00, 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} \[ \frac {\left (c+\frac {d}{x^2}\right )^{3/2} (2 b c-a d)}{3 d^3}-\frac {c \sqrt {c+\frac {d}{x^2}} (b c-a d)}{d^3}-\frac {b \left (c+\frac {d}{x^2}\right )^{5/2}}{5 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]))))

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]]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 60, normalized size = 0.83 \[ \frac {\sqrt {c+\frac {d}{x^2}} \left (b \left (-8 c^2 x^4+4 c d x^2-3 d^2\right )-5 a d x^2 \left (d-2 c x^2\right )\right )}{15 d^3 x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d/x^2]*(-5*a*d*x^2*(d - 2*c*x^2) + b*(-3*d^2 + 4*c*d*x^2 - 8*c^2*x^4)))/(15*d^3*x^4)

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fricas [A]  time = 0.48, size = 62, normalized size = 0.86 \[ -\frac {{\left (2 \, {\left (4 \, b c^{2} - 5 \, a c d\right )} x^{4} + 3 \, b d^{2} - {\left (4 \, b c d - 5 \, a d^{2}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{15 \, d^{3} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)/x^5/(c+d/x^2)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.50, size = 153, normalized size = 2.12 \[ \frac {15 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + d x^{2}}\right )}^{3} a \sqrt {c} + 20 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + d x^{2}}\right )}^{2} b c + 5 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + d x^{2}}\right )}^{2} a d + 15 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + d x^{2}}\right )} b \sqrt {c} d + 3 \, b d^{2}}{15 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + d x^{2}}\right )}^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)/x^5/(c+d/x^2)^(1/2),x, algorithm="giac")

[Out]

1/15*(15*(sqrt(c)*x^2 - sqrt(c*x^4 + d*x^2))^3*a*sqrt(c) + 20*(sqrt(c)*x^2 - sqrt(c*x^4 + d*x^2))^2*b*c + 5*(s
qrt(c)*x^2 - sqrt(c*x^4 + d*x^2))^2*a*d + 15*(sqrt(c)*x^2 - sqrt(c*x^4 + d*x^2))*b*sqrt(c)*d + 3*b*d^2)/(sqrt(
c)*x^2 - sqrt(c*x^4 + d*x^2))^5

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maple [A]  time = 0.05, size = 70, normalized size = 0.97 \[ \frac {\left (10 a c d \,x^{4}-8 b \,c^{2} x^{4}-5 a \,d^{2} x^{2}+4 b c d \,x^{2}-3 b \,d^{2}\right ) \left (c \,x^{2}+d \right )}{15 \sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, d^{3} x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b/x^2)/x^5/(c+d/x^2)^(1/2),x)

[Out]

1/15*(10*a*c*d*x^4-8*b*c^2*x^4-5*a*d^2*x^2+4*b*c*d*x^2-3*b*d^2)*(c*x^2+d)/((c*x^2+d)/x^2)^(1/2)/d^3/x^6

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maxima [A]  time = 0.49, size = 83, normalized size = 1.15 \[ -\frac {1}{15} \, b {\left (\frac {3 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}}}{d^{3}} - \frac {10 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} c}{d^{3}} + \frac {15 \, \sqrt {c + \frac {d}{x^{2}}} c^{2}}{d^{3}}\right )} - \frac {1}{3} \, a {\left (\frac {{\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}}}{d^{2}} - \frac {3 \, \sqrt {c + \frac {d}{x^{2}}} c}{d^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)/x^5/(c+d/x^2)^(1/2),x, algorithm="maxima")

[Out]

-1/15*b*(3*(c + d/x^2)^(5/2)/d^3 - 10*(c + d/x^2)^(3/2)*c/d^3 + 15*sqrt(c + d/x^2)*c^2/d^3) - 1/3*a*((c + d/x^
2)^(3/2)/d^2 - 3*sqrt(c + d/x^2)*c/d^2)

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mupad [B]  time = 4.68, size = 58, normalized size = 0.81 \[ -\frac {\sqrt {c+\frac {d}{x^2}}\,\left (8\,b\,c^2\,x^4-10\,a\,c\,d\,x^4-4\,b\,c\,d\,x^2+5\,a\,d^2\,x^2+3\,b\,d^2\right )}{15\,d^3\,x^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x^2)/(x^5*(c + d/x^2)^(1/2)),x)

[Out]

-((c + d/x^2)^(1/2)*(3*b*d^2 + 5*a*d^2*x^2 + 8*b*c^2*x^4 - 10*a*c*d*x^4 - 4*b*c*d*x^2))/(15*d^3*x^4)

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sympy [A]  time = 13.17, size = 204, normalized size = 2.83 \[ \frac {\begin {cases} \frac {- \frac {a}{2 x^{4}} - \frac {b}{3 x^{6}}}{\sqrt {c}} & \text {for}\: d = 0 \\\frac {\frac {2 a c \left (- \frac {c}{\sqrt {c + \frac {d}{x^{2}}}} - \sqrt {c + \frac {d}{x^{2}}}\right )}{d} + \frac {2 a \left (\frac {c^{2}}{\sqrt {c + \frac {d}{x^{2}}}} + 2 c \sqrt {c + \frac {d}{x^{2}}} - \frac {\left (c + \frac {d}{x^{2}}\right )^{\frac {3}{2}}}{3}\right )}{d} + \frac {2 b c \left (\frac {c^{2}}{\sqrt {c + \frac {d}{x^{2}}}} + 2 c \sqrt {c + \frac {d}{x^{2}}} - \frac {\left (c + \frac {d}{x^{2}}\right )^{\frac {3}{2}}}{3}\right )}{d^{2}} + \frac {2 b \left (- \frac {c^{3}}{\sqrt {c + \frac {d}{x^{2}}}} - 3 c^{2} \sqrt {c + \frac {d}{x^{2}}} + c \left (c + \frac {d}{x^{2}}\right )^{\frac {3}{2}} - \frac {\left (c + \frac {d}{x^{2}}\right )^{\frac {5}{2}}}{5}\right )}{d^{2}}}{d} & \text {otherwise} \end {cases}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x**2)/x**5/(c+d/x**2)**(1/2),x)

[Out]

Piecewise(((-a/(2*x**4) - b/(3*x**6))/sqrt(c), Eq(d, 0)), ((2*a*c*(-c/sqrt(c + d/x**2) - sqrt(c + d/x**2))/d +
2*a*(c**2/sqrt(c + d/x**2) + 2*c*sqrt(c + d/x**2) - (c + d/x**2)**(3/2)/3)/d + 2*b*c*(c**2/sqrt(c + d/x**2) +
2*c*sqrt(c + d/x**2) - (c + d/x**2)**(3/2)/3)/d**2 + 2*b*(-c**3/sqrt(c + d/x**2) - 3*c**2*sqrt(c + d/x**2) +
c*(c + d/x**2)**(3/2) - (c + d/x**2)**(5/2)/5)/d**2)/d, True))/2

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