\(\int \frac {\sqrt {a+b x^2} (A+B x^2)}{x^3} \, dx\) [512]

Optimal result

Integrand size = 22, antiderivative size = 84 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^3} \, dx=\frac {(A b+2 a B) \sqrt {a+b x^2}}{2 a}-\frac {A \left (a+b x^2\right )^{3/2}}{2 a x^2}-\frac {(A b+2 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{2 \sqrt {a}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {457, 79, 52, 65, 214} \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^3} \, dx=-\frac {(2 a B+A b) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{2 \sqrt {a}}+\frac {\sqrt {a+b x^2} (2 a B+A b)}{2 a}-\frac {A \left (a+b x^2\right )^{3/2}}{2 a x^2} \]

[In]

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

[Out]

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

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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 {\sqrt {a+b x} (A+B x)}{x^2} \, dx,x,x^2\right ) \\ & = -\frac {A \left (a+b x^2\right )^{3/2}}{2 a x^2}+\frac {(A b+2 a B) \text {Subst}\left (\int \frac {\sqrt {a+b x}}{x} \, dx,x,x^2\right )}{4 a} \\ & = \frac {(A b+2 a B) \sqrt {a+b x^2}}{2 a}-\frac {A \left (a+b x^2\right )^{3/2}}{2 a x^2}+\frac {1}{4} (A b+2 a B) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^2\right ) \\ & = \frac {(A b+2 a B) \sqrt {a+b x^2}}{2 a}-\frac {A \left (a+b x^2\right )^{3/2}}{2 a x^2}+\frac {(A b+2 a B) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^2}\right )}{2 b} \\ & = \frac {(A b+2 a B) \sqrt {a+b x^2}}{2 a}-\frac {A \left (a+b x^2\right )^{3/2}}{2 a x^2}-\frac {(A b+2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{2 \sqrt {a}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.77 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^3} \, dx=\frac {\sqrt {a+b x^2} \left (-A+2 B x^2\right )}{2 x^2}+\frac {(-A b-2 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{2 \sqrt {a}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.91 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.68

method	result	size
pseudoelliptic	\(-\frac {\left (A b +2 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{\sqrt {a}}\right ) x^{2}+\left (-2 x^{2} B +A \right ) \sqrt {b \,x^{2}+a}\, \sqrt {a}}{2 \sqrt {a}\, x^{2}}\)	\(57\)
risch	\(-\frac {A \sqrt {b \,x^{2}+a}}{2 x^{2}}+\sqrt {b \,x^{2}+a}\, B -\frac {\left (A b +2 B a \right ) \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{2 \sqrt {a}}\)	\(64\)
default	\(B \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )+A \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{2 a \,x^{2}}+\frac {b \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )}{2 a}\right )\)	\(106\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.68 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^3} \, dx=\left [\frac {{\left (2 \, B a + A b\right )} \sqrt {a} x^{2} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (2 \, B a x^{2} - A a\right )} \sqrt {b x^{2} + a}}{4 \, a x^{2}}, \frac {{\left (2 \, B a + A b\right )} \sqrt {-a} x^{2} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (2 \, B a x^{2} - A a\right )} \sqrt {b x^{2} + a}}{2 \, a x^{2}}\right ] \]

[In]

integrate((B*x^2+A)*(b*x^2+a)^(1/2)/x^3,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*(2*B*a*x^2 - A*a)*sqrt
(b*x^2 + a))/(a*x^2), 1/2*((2*B*a + A*b)*sqrt(-a)*x^2*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + (2*B*a*x^2 - A*a)*sqr
t(b*x^2 + a))/(a*x^2)]

Sympy [A] (verification not implemented)

Time = 11.61 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.27 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^3} \, dx=- \frac {A \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{2 x} - \frac {A b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{2 \sqrt {a}} - B \sqrt {a} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )} + \frac {B a}{\sqrt {b} x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {B \sqrt {b} x}{\sqrt {\frac {a}{b x^{2}} + 1}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.99 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^3} \, dx=-B \sqrt {a} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) - \frac {A b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{2 \, \sqrt {a}} + \sqrt {b x^{2} + a} B + \frac {\sqrt {b x^{2} + a} A b}{2 \, a} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A}{2 \, a x^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.81 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^3} \, dx=\frac {2 \, \sqrt {b x^{2} + a} B b + \frac {{\left (2 \, B a b + A b^{2}\right )} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {\sqrt {b x^{2} + a} A b}{x^{2}}}{2 \, b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.65 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.81 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^3} \, dx=B\,\sqrt {b\,x^2+a}-\frac {A\,\sqrt {b\,x^2+a}}{2\,x^2}-B\,\sqrt {a}\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )-\frac {A\,b\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{2\,\sqrt {a}} \]

[In]

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

[Out]

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