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

Optimal. Leaf size=58 \[ \frac {(A b-2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{2 a^{3/2}}-\frac {A \sqrt {a+b x^2}}{2 a x^2} \]

[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

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Rubi [A]  time = 0.05, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {446, 78, 63, 208} \[ \frac {(A b-2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{2 a^{3/2}}-\frac {A \sqrt {a+b x^2}}{2 a x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 63

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 78

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] || LtQ
[p, n]))))

Rule 208

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

Rule 446

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*} \int \frac {A+B x^2}{x^3 \sqrt {a+b x^2}} \, dx &=\frac {1}{2} \operatorname {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 ) \operatorname {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 ) \operatorname {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*}

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Mathematica [A]  time = 0.03, size = 60, normalized size = 1.03 \[ \frac {1}{2} \left (-\frac {2 \left (a B-\frac {A b}{2}\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {A \sqrt {a+b x^2}}{a x^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.70, size = 124, normalized size = 2.14 \[ \left [-\frac {{\left (2 \, B a - A b\right )} \sqrt {a} x^{2} \log \left (-\frac {b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, \sqrt {b x^{2} + a} A a}{4 \, a^{2} x^{2}}, \frac {{\left (2 \, B a - A b\right )} \sqrt {-a} x^{2} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) - \sqrt {b x^{2} + a} A a}{2 \, a^{2} x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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giac [A]  time = 0.37, size = 62, normalized size = 1.07 \[ \frac {\frac {{\left (2 \, B a b - A b^{2}\right )} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} - \frac {\sqrt {b x^{2} + a} A b}{a x^{2}}}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.01, size = 79, normalized size = 1.36 \[ \frac {A b \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{2 a^{\frac {3}{2}}}-\frac {B \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{\sqrt {a}}-\frac {\sqrt {b \,x^{2}+a}\, A}{2 a \,x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.11, size = 56, normalized size = 0.97 \[ -\frac {B \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{\sqrt {a}} + \frac {A b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{2 \, a^{\frac {3}{2}}} - \frac {\sqrt {b x^{2} + a} A}{2 \, a x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 1.34, size = 60, normalized size = 1.03 \[ \frac {A\,b\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{2\,a^{3/2}}-\frac {A\,\sqrt {b\,x^2+a}}{2\,a\,x^2}-\frac {B\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{\sqrt {a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 39.42, size = 66, normalized size = 1.14 \[ - \frac {A \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{2 a x} + \frac {A b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{2 a^{\frac {3}{2}}} - \frac {B \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{\sqrt {a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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