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

Optimal. Leaf size=45 \[ \frac {a x}{c \sqrt {c+\frac {d}{x^2}}}-\frac {b c-2 a d}{c^2 x \sqrt {c+\frac {d}{x^2}}} \]

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Rubi [A]  time = 0.03, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {375, 453, 191} \begin {gather*} \frac {a x}{c \sqrt {c+\frac {d}{x^2}}}-\frac {b c-2 a d}{c^2 x \sqrt {c+\frac {d}{x^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)/(c + d/x^2)^(3/2),x]

[Out]

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

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 375

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> -Subst[Int[((a + b/x^n)^p*(c +
d/x^n)^q)/x^2, x], x, 1/x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && ILtQ[n, 0]

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 33, normalized size = 0.73 \begin {gather*} \frac {a c x^2+2 a d-b c}{c^2 x \sqrt {c+\frac {d}{x^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.08, size = 40, normalized size = 0.89 \begin {gather*} \frac {x \sqrt {c+\frac {d}{x^2}} \left (a c x^2+2 a d-b c\right )}{c^2 \left (c x^2+d\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.48, size = 47, normalized size = 1.04 \begin {gather*} \frac {{\left (a c x^{3} - {\left (b c - 2 \, a d\right )} x\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{c^{3} x^{2} + c^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep)]Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is
real):Check [sign(t_nostep),sign(t_nostep+sqrt(d)/c*sign(t_nostep))]sym2poly/r2sym(const gen & e,const index_
m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.04, size = 43, normalized size = 0.96 \begin {gather*} \frac {\left (a \,x^{2} c +2 a d -b c \right ) \left (c \,x^{2}+d \right )}{\left (\frac {c \,x^{2}+d}{x^{2}}\right )^{\frac {3}{2}} c^{2} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.68, size = 53, normalized size = 1.18 \begin {gather*} a {\left (\frac {\sqrt {c + \frac {d}{x^{2}}} x}{c^{2}} + \frac {d}{\sqrt {c + \frac {d}{x^{2}}} c^{2} x}\right )} - \frac {b}{\sqrt {c + \frac {d}{x^{2}}} c x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.90, size = 38, normalized size = 0.84 \begin {gather*} \frac {\left (c\,x^2+d\right )\,\left (a\,c\,x^2+2\,a\,d-b\,c\right )}{c^2\,x^3\,{\left (c+\frac {d}{x^2}\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x^2)/(c + d/x^2)^(3/2),x)

[Out]

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

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sympy [A]  time = 7.51, size = 65, normalized size = 1.44 \begin {gather*} a \left (\frac {x^{2}}{c \sqrt {d} \sqrt {\frac {c x^{2}}{d} + 1}} + \frac {2 \sqrt {d}}{c^{2} \sqrt {\frac {c x^{2}}{d} + 1}}\right ) - \frac {b}{c \sqrt {d} \sqrt {\frac {c x^{2}}{d} + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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