\(\int \frac {A+B x^2}{x^2 (a+b x^2)^{3/2}} \, dx\) [577]

Optimal result

Integrand size = 22, antiderivative size = 47 \[ \int \frac {A+B x^2}{x^2 \left (a+b x^2\right )^{3/2}} \, dx=-\frac {A}{a x \sqrt {a+b x^2}}-\frac {(2 A b-a B) x}{a^2 \sqrt {a+b x^2}} \]

[Out]

-A/a/x/(b*x^2+a)^(1/2)-(2*A*b-B*a)*x/a^2/(b*x^2+a)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {464, 197} \[ \int \frac {A+B x^2}{x^2 \left (a+b x^2\right )^{3/2}} \, dx=-\frac {x (2 A b-a B)}{a^2 \sqrt {a+b x^2}}-\frac {A}{a x \sqrt {a+b x^2}} \]

[In]

Int[(A + B*x^2)/(x^2*(a + b*x^2)^(3/2)),x]

[Out]

-(A/(a*x*Sqrt[a + b*x^2])) - ((2*A*b - a*B)*x)/(a^2*Sqrt[a + b*x^2])

Rule 197

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 464

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*} \text {integral}& = -\frac {A}{a x \sqrt {a+b x^2}}-\frac {(2 A b-a B) \int \frac {1}{\left (a+b x^2\right )^{3/2}} \, dx}{a} \\ & = -\frac {A}{a x \sqrt {a+b x^2}}-\frac {(2 A b-a B) x}{a^2 \sqrt {a+b x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.77 \[ \int \frac {A+B x^2}{x^2 \left (a+b x^2\right )^{3/2}} \, dx=\frac {-a A-2 A b x^2+a B x^2}{a^2 x \sqrt {a+b x^2}} \]

[In]

Integrate[(A + B*x^2)/(x^2*(a + b*x^2)^(3/2)),x]

[Out]

(-(a*A) - 2*A*b*x^2 + a*B*x^2)/(a^2*x*Sqrt[a + b*x^2])

Maple [A] (verified)

Time = 2.84 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.77

method	result	size
gosper	\(-\frac {2 A b \,x^{2}-B a \,x^{2}+A a}{x \sqrt {b \,x^{2}+a}\, a^{2}}\)	\(36\)
trager	\(-\frac {2 A b \,x^{2}-B a \,x^{2}+A a}{x \sqrt {b \,x^{2}+a}\, a^{2}}\)	\(36\)
pseudoelliptic	\(-\frac {\left (-x^{2} B +A \right ) a +2 A b \,x^{2}}{\sqrt {b \,x^{2}+a}\, x \,a^{2}}\)	\(36\)
risch	\(-\frac {A \sqrt {b \,x^{2}+a}}{a^{2} x}-\frac {x \left (A b -B a \right )}{\sqrt {b \,x^{2}+a}\, a^{2}}\)	\(43\)
default	\(\frac {B x}{a \sqrt {b \,x^{2}+a}}+A \left (-\frac {1}{a x \sqrt {b \,x^{2}+a}}-\frac {2 b x}{a^{2} \sqrt {b \,x^{2}+a}}\right )\)	\(53\)

[In]

int((B*x^2+A)/x^2/(b*x^2+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-(2*A*b*x^2-B*a*x^2+A*a)/x/(b*x^2+a)^(1/2)/a^2

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.91 \[ \int \frac {A+B x^2}{x^2 \left (a+b x^2\right )^{3/2}} \, dx=\frac {{\left ({\left (B a - 2 \, A b\right )} x^{2} - A a\right )} \sqrt {b x^{2} + a}}{a^{2} b x^{3} + a^{3} x} \]

[In]

integrate((B*x^2+A)/x^2/(b*x^2+a)^(3/2),x, algorithm="fricas")

[Out]

((B*a - 2*A*b)*x^2 - A*a)*sqrt(b*x^2 + a)/(a^2*b*x^3 + a^3*x)

Sympy [A] (verification not implemented)

Time = 2.49 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.45 \[ \int \frac {A+B x^2}{x^2 \left (a+b x^2\right )^{3/2}} \, dx=A \left (- \frac {1}{a \sqrt {b} x^{2} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {2 \sqrt {b}}{a^{2} \sqrt {\frac {a}{b x^{2}} + 1}}\right ) + \frac {B x}{a^{\frac {3}{2}} \sqrt {1 + \frac {b x^{2}}{a}}} \]

[In]

integrate((B*x**2+A)/x**2/(b*x**2+a)**(3/2),x)

[Out]

A*(-1/(a*sqrt(b)*x**2*sqrt(a/(b*x**2) + 1)) - 2*sqrt(b)/(a**2*sqrt(a/(b*x**2) + 1))) + B*x/(a**(3/2)*sqrt(1 +
b*x**2/a))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.09 \[ \int \frac {A+B x^2}{x^2 \left (a+b x^2\right )^{3/2}} \, dx=\frac {B x}{\sqrt {b x^{2} + a} a} - \frac {2 \, A b x}{\sqrt {b x^{2} + a} a^{2}} - \frac {A}{\sqrt {b x^{2} + a} a x} \]

[In]

integrate((B*x^2+A)/x^2/(b*x^2+a)^(3/2),x, algorithm="maxima")

[Out]

B*x/(sqrt(b*x^2 + a)*a) - 2*A*b*x/(sqrt(b*x^2 + a)*a^2) - A/(sqrt(b*x^2 + a)*a*x)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.21 \[ \int \frac {A+B x^2}{x^2 \left (a+b x^2\right )^{3/2}} \, dx=\frac {2 \, A \sqrt {b}}{{\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )} a} + \frac {{\left (B a - A b\right )} x}{\sqrt {b x^{2} + a} a^{2}} \]

[In]

integrate((B*x^2+A)/x^2/(b*x^2+a)^(3/2),x, algorithm="giac")

[Out]

2*A*sqrt(b)/(((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)*a) + (B*a - A*b)*x/(sqrt(b*x^2 + a)*a^2)

Mupad [B] (verification not implemented)

Time = 5.28 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.98 \[ \int \frac {A+B x^2}{x^2 \left (a+b x^2\right )^{3/2}} \, dx=-\frac {\sqrt {b\,x^2+a}\,\left (\frac {A}{a}-x^2\,\left (\frac {B}{a}-\frac {2\,A\,b}{a^2}\right )\right )}{b\,x^3+a\,x} \]

[In]

int((A + B*x^2)/(x^2*(a + b*x^2)^(3/2)),x)

[Out]

-((a + b*x^2)^(1/2)*(A/a - x^2*(B/a - (2*A*b)/a^2)))/(a*x + b*x^3)