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

Optimal. Leaf size=43 \[ \frac {\sqrt {c+\frac {d}{x^2}} (b c-a d)}{d^2}-\frac {b \left (c+\frac {d}{x^2}\right )^{3/2}}{3 d^2} \]

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Rubi [A]  time = 0.04, antiderivative size = 43, 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 = {444, 43} \begin {gather*} \frac {\sqrt {c+\frac {d}{x^2}} (b c-a d)}{d^2}-\frac {b \left (c+\frac {d}{x^2}\right )^{3/2}}{3 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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]

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 39, normalized size = 0.91 \begin {gather*} -\frac {\sqrt {c+\frac {d}{x^2}} \left (3 a d x^2+b \left (d-2 c x^2\right )\right )}{3 d^2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(Sqrt[c + d/x^2]*(3*a*d*x^2 + b*(d - 2*c*x^2)))/(d^2*x^2)

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IntegrateAlgebraic [A]  time = 0.05, size = 44, normalized size = 1.02 \begin {gather*} \frac {\sqrt {\frac {c x^2+d}{x^2}} \left (-3 a d x^2+2 b c x^2-b d\right )}{3 d^2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.41, size = 39, normalized size = 0.91 \begin {gather*} \frac {{\left ({\left (2 \, b c - 3 \, a d\right )} x^{2} - b d\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{3 \, d^{2} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.41, size = 88, normalized size = 2.05 \begin {gather*} \frac {3 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + d x^{2}}\right )}^{2} a + 3 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + d x^{2}}\right )} b \sqrt {c} + b d}{3 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + d x^{2}}\right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 47, normalized size = 1.09 \begin {gather*} -\frac {\left (3 a d \,x^{2}-2 b c \,x^{2}+b d \right ) \left (c \,x^{2}+d \right )}{3 \sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, d^{2} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.53, size = 48, normalized size = 1.12 \begin {gather*} -\frac {1}{3} \, b {\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 )} - \frac {a \sqrt {c + \frac {d}{x^{2}}}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.56, size = 35, normalized size = 0.81 \begin {gather*} -\frac {\sqrt {c+\frac {d}{x^2}}\,\left (b\,d+3\,a\,d\,x^2-2\,b\,c\,x^2\right )}{3\,d^2\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 7.05, size = 138, normalized size = 3.21 \begin {gather*} \frac {\begin {cases} \frac {- \frac {a}{x^{2}} - \frac {b}{2 x^{4}}}{\sqrt {c}} & \text {for}\: d = 0 \\\frac {\frac {2 a c}{\sqrt {c + \frac {d}{x^{2}}}} + 2 a \left (- \frac {c}{\sqrt {c + \frac {d}{x^{2}}}} - \sqrt {c + \frac {d}{x^{2}}}\right ) + \frac {2 b c \left (- \frac {c}{\sqrt {c + \frac {d}{x^{2}}}} - \sqrt {c + \frac {d}{x^{2}}}\right )}{d} + \frac {2 b \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}}{d} & \text {otherwise} \end {cases}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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