\(\int \frac {A+B x^2}{x^3 \sqrt {a+b x^2}} \, dx\) [563]

Optimal result

Integrand size = 22, antiderivative size = 58 \[ \int \frac {A+B x^2}{x^3 \sqrt {a+b x^2}} \, dx=-\frac {A \sqrt {a+b x^2}}{2 a x^2}+\frac {(A b-2 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{2 a^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {457, 79, 65, 214} \[ \int \frac {A+B x^2}{x^3 \sqrt {a+b x^2}} \, dx=\frac {(A b-2 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{2 a^{3/2}}-\frac {A \sqrt {a+b x^2}}{2 a x^2} \]

[In]

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

[Out]

-1/2*(A*Sqrt[a + b*x^2])/(a*x^2) + ((A*b - 2*a*B)*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(2*a^(3/2))

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 457

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00 \[ \int \frac {A+B x^2}{x^3 \sqrt {a+b x^2}} \, dx=-\frac {A \sqrt {a+b x^2}}{2 a x^2}+\frac {(A b-2 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{2 a^{3/2}} \]

[In]

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

[Out]

-1/2*(A*Sqrt[a + b*x^2])/(a*x^2) + ((A*b - 2*a*B)*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(2*a^(3/2))

Maple [A] (verified)

Time = 2.83 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.81

method	result	size
pseudoelliptic	\(\frac {\left (A b -2 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{\sqrt {a}}\right )}{2 a^{\frac {3}{2}}}-\frac {A \sqrt {b \,x^{2}+a}}{2 a \,x^{2}}\)	\(47\)
risch	\(-\frac {A \sqrt {b \,x^{2}+a}}{2 a \,x^{2}}+\frac {\left (A b -2 B a \right ) \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{2 a^{\frac {3}{2}}}\)	\(56\)
default	\(-\frac {B \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{\sqrt {a}}+A \left (-\frac {\sqrt {b \,x^{2}+a}}{2 a \,x^{2}}+\frac {b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{2 a^{\frac {3}{2}}}\right )\)	\(80\)

[In]

int((B*x^2+A)/x^3/(b*x^2+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 124, normalized size of antiderivative = 2.14 \[ \int \frac {A+B x^2}{x^3 \sqrt {a+b x^2}} \, dx=\left [-\frac {{\left (2 \, B a - A b\right )} \sqrt {a} x^{2} \log \left (-\frac {b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, \sqrt {b x^{2} + a} A a}{4 \, a^{2} x^{2}}, \frac {{\left (2 \, B a - A b\right )} \sqrt {-a} x^{2} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) - \sqrt {b x^{2} + a} A a}{2 \, a^{2} x^{2}}\right ] \]

[In]

integrate((B*x^2+A)/x^3/(b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

[-1/4*((2*B*a - A*b)*sqrt(a)*x^2*log(-(b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*sqrt(b*x^2 + a)*A*a)/
(a^2*x^2), 1/2*((2*B*a - A*b)*sqrt(-a)*x^2*arctan(sqrt(-a)/sqrt(b*x^2 + a)) - sqrt(b*x^2 + a)*A*a)/(a^2*x^2)]

Sympy [A] (verification not implemented)

Time = 5.77 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.14 \[ \int \frac {A+B x^2}{x^3 \sqrt {a+b x^2}} \, dx=- \frac {A \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{2 a x} + \frac {A b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{2 a^{\frac {3}{2}}} - \frac {B \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{\sqrt {a}} \]

[In]

integrate((B*x**2+A)/x**3/(b*x**2+a)**(1/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.97 \[ \int \frac {A+B x^2}{x^3 \sqrt {a+b x^2}} \, dx=-\frac {B \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{\sqrt {a}} + \frac {A b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{2 \, a^{\frac {3}{2}}} - \frac {\sqrt {b x^{2} + a} A}{2 \, a x^{2}} \]

[In]

integrate((B*x^2+A)/x^3/(b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.07 \[ \int \frac {A+B x^2}{x^3 \sqrt {a+b x^2}} \, dx=\frac {\frac {{\left (2 \, B a b - A b^{2}\right )} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} - \frac {\sqrt {b x^{2} + a} A b}{a x^{2}}}{2 \, b} \]

[In]

integrate((B*x^2+A)/x^3/(b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.83 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.03 \[ \int \frac {A+B x^2}{x^3 \sqrt {a+b x^2}} \, dx=\frac {A\,b\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{2\,a^{3/2}}-\frac {A\,\sqrt {b\,x^2+a}}{2\,a\,x^2}-\frac {B\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{\sqrt {a}} \]

[In]

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

[Out]

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