\(\int \frac {A+B x^2}{x \sqrt {a+b x^2}} \, dx\) [561]

Optimal result

Integrand size = 22, antiderivative size = 43 \[ \int \frac {A+B x^2}{x \sqrt {a+b x^2}} \, dx=\frac {B \sqrt {a+b x^2}}{b}-\frac {A \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{\sqrt {a}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 43, 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, 81, 65, 214} \[ \int \frac {A+B x^2}{x \sqrt {a+b x^2}} \, dx=\frac {B \sqrt {a+b x^2}}{b}-\frac {A \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{\sqrt {a}} \]

[In]

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

[Out]

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

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 81

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

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00 \[ \int \frac {A+B x^2}{x \sqrt {a+b x^2}} \, dx=\frac {B \sqrt {a+b x^2}}{b}-\frac {A \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{\sqrt {a}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.84 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.95

method	result	size
pseudoelliptic	\(\frac {-A b \,\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{\sqrt {a}}\right )+\sqrt {b \,x^{2}+a}\, B \sqrt {a}}{b \sqrt {a}}\)	\(41\)
default	\(\frac {B \sqrt {b \,x^{2}+a}}{b}-\frac {A \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{\sqrt {a}}\)	\(45\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 102, normalized size of antiderivative = 2.37 \[ \int \frac {A+B x^2}{x \sqrt {a+b x^2}} \, dx=\left [\frac {A \sqrt {a} b \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, \sqrt {b x^{2} + a} B a}{2 \, a b}, \frac {A \sqrt {-a} b \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + \sqrt {b x^{2} + a} B a}{a b}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 1.67 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.65 \[ \int \frac {A+B x^2}{x \sqrt {a+b x^2}} \, dx=\frac {A \left (\begin {cases} \frac {2 \operatorname {atan}{\left (\frac {\sqrt {a + b x^{2}}}{\sqrt {- a}} \right )}}{\sqrt {- a}} & \text {for}\: b \neq 0 \\- \frac {\log {\left (\frac {1}{x^{2}} \right )}}{\sqrt {a}} & \text {otherwise} \end {cases}\right )}{2} - \frac {B \left (\begin {cases} - \frac {x^{2}}{\sqrt {a}} & \text {for}\: b = 0 \\- \frac {2 \sqrt {a + b x^{2}}}{b} & \text {otherwise} \end {cases}\right )}{2} \]

[In]

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

[Out]

A*Piecewise((2*atan(sqrt(a + b*x**2)/sqrt(-a))/sqrt(-a), Ne(b, 0)), (-log(x**(-2))/sqrt(a), True))/2 - B*Piece
wise((-x**2/sqrt(a), Eq(b, 0)), (-2*sqrt(a + b*x**2)/b, True))/2

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.77 \[ \int \frac {A+B x^2}{x \sqrt {a+b x^2}} \, dx=-\frac {A \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{\sqrt {a}} + \frac {\sqrt {b x^{2} + a} B}{b} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.47 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.88 \[ \int \frac {A+B x^2}{x \sqrt {a+b x^2}} \, dx=\frac {A \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + \frac {\sqrt {b x^{2} + a} B}{b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.52 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int \frac {A+B x^2}{x \sqrt {a+b x^2}} \, dx=\frac {B\,\sqrt {b\,x^2+a}}{b}-\frac {A\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{\sqrt {a}} \]

[In]

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

[Out]

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