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

Optimal. Leaf size=53 \[ \frac{A b-a B}{a b \sqrt{a+b x^2}}-\frac{A \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{a^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0387111, antiderivative size = 53, normalized size of antiderivative = 1., 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} \[ \frac{A b-a B}{a b \sqrt{a+b x^2}}-\frac{A \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{a^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]]

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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.0326145, size = 53, normalized size = 1. \[ \frac{A b-a B}{a b \sqrt{a+b x^2}}-\frac{A \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{a^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.006, size = 60, normalized size = 1.1 \begin{align*} -{\frac{B}{b}{\frac{1}{\sqrt{b{x}^{2}+a}}}}+{\frac{A}{a}{\frac{1}{\sqrt{b{x}^{2}+a}}}}-{A\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ){a}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.61111, size = 365, normalized size = 6.89 \begin{align*} \left [\frac{{\left (A b^{2} x^{2} + A a b\right )} \sqrt{a} \log \left (-\frac{b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) - 2 \,{\left (B a^{2} - A a b\right )} \sqrt{b x^{2} + a}}{2 \,{\left (a^{2} b^{2} x^{2} + a^{3} b\right )}}, \frac{{\left (A b^{2} x^{2} + A a b\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) -{\left (B a^{2} - A a b\right )} \sqrt{b x^{2} + a}}{a^{2} b^{2} x^{2} + a^{3} b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((A*b^2*x^2 + A*a*b)*sqrt(a)*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*(B*a^2 - A*a*b)*sqrt
(b*x^2 + a))/(a^2*b^2*x^2 + a^3*b), ((A*b^2*x^2 + A*a*b)*sqrt(-a)*arctan(sqrt(-a)/sqrt(b*x^2 + a)) - (B*a^2 -
A*a*b)*sqrt(b*x^2 + a))/(a^2*b^2*x^2 + a^3*b)]

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Sympy [A]  time = 8.91535, size = 48, normalized size = 0.91 \begin{align*} \frac{A \operatorname{atan}{\left (\frac{\sqrt{a + b x^{2}}}{\sqrt{- a}} \right )}}{a \sqrt{- a}} - \frac{- A b + B a}{a b \sqrt{a + b x^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.11949, size = 70, normalized size = 1.32 \begin{align*} \frac{A \arctan \left (\frac{\sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a} - \frac{B a - A b}{\sqrt{b x^{2} + a} a b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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