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

Optimal. Leaf size=72 \[ \frac{A b \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{2 a^{3/2}}-\frac{A \sqrt{a+b x^2}}{2 a x^2}-\frac{B \sqrt{a+b x^2}}{a x} \]

[Out]

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

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Rubi [A]  time = 0.0531302, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {835, 807, 266, 63, 208} \[ \frac{A b \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{2 a^{3/2}}-\frac{A \sqrt{a+b x^2}}{2 a x^2}-\frac{B \sqrt{a+b x^2}}{a x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 835

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((e*f - d*g)
*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

Rule 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&& EqQ[Simplify[m + 2*p + 3], 0]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/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}{x^3 \sqrt{a+b x^2}} \, dx &=-\frac{A \sqrt{a+b x^2}}{2 a x^2}-\frac{\int \frac{-2 a B+A b x}{x^2 \sqrt{a+b x^2}} \, dx}{2 a}\\ &=-\frac{A \sqrt{a+b x^2}}{2 a x^2}-\frac{B \sqrt{a+b x^2}}{a x}-\frac{(A b) \int \frac{1}{x \sqrt{a+b x^2}} \, dx}{2 a}\\ &=-\frac{A \sqrt{a+b x^2}}{2 a x^2}-\frac{B \sqrt{a+b x^2}}{a x}-\frac{(A b) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^2\right )}{4 a}\\ &=-\frac{A \sqrt{a+b x^2}}{2 a x^2}-\frac{B \sqrt{a+b x^2}}{a x}-\frac{A \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^2}\right )}{2 a}\\ &=-\frac{A \sqrt{a+b x^2}}{2 a x^2}-\frac{B \sqrt{a+b x^2}}{a x}+\frac{A b \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{2 a^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.101697, size = 63, normalized size = 0.88 \[ \frac{\sqrt{a+b x^2} \left (\frac{A b \tanh ^{-1}\left (\sqrt{\frac{b x^2}{a}+1}\right )}{\sqrt{\frac{b x^2}{a}+1}}-\frac{a (A+2 B x)}{x^2}\right )}{2 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.008, size = 68, normalized size = 0.9 \begin{align*} -{\frac{A}{2\,a{x}^{2}}\sqrt{b{x}^{2}+a}}+{\frac{Ab}{2}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ){a}^{-{\frac{3}{2}}}}-{\frac{B}{ax}\sqrt{b{x}^{2}+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.56395, size = 306, normalized size = 4.25 \begin{align*} \left [\frac{A \sqrt{a} b 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 + A a\right )} \sqrt{b x^{2} + a}}{4 \, a^{2} x^{2}}, -\frac{A \sqrt{-a} b x^{2} \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) +{\left (2 \, B a x + A a\right )} \sqrt{b x^{2} + a}}{2 \, a^{2} x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 3.15571, size = 66, normalized size = 0.92 \begin{align*} - \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 \sqrt{b} \sqrt{\frac{a}{b x^{2}} + 1}}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Giac [B]  time = 1.26266, size = 197, normalized size = 2.74 \begin{align*} -\frac{A b \arctan \left (-\frac{\sqrt{b} x - \sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a} + \frac{{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{3} A b + 2 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} B a \sqrt{b} +{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )} A a b - 2 \, B a^{2} \sqrt{b}}{{\left ({\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} - a\right )}^{2} a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-A*b*arctan(-(sqrt(b)*x - sqrt(b*x^2 + a))/sqrt(-a))/(sqrt(-a)*a) + ((sqrt(b)*x - sqrt(b*x^2 + a))^3*A*b + 2*(
sqrt(b)*x - sqrt(b*x^2 + a))^2*B*a*sqrt(b) + (sqrt(b)*x - sqrt(b*x^2 + a))*A*a*b - 2*B*a^2*sqrt(b))/(((sqrt(b)
*x - sqrt(b*x^2 + a))^2 - a)^2*a)