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

Optimal. Leaf size=70 \[ -\frac{2 A \sqrt{a+b x^2}}{a^2 x}-\frac{B \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{a^{3/2}}+\frac{A+B x}{a x \sqrt{a+b x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0570752, antiderivative size = 70, 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 = {823, 807, 266, 63, 208} \[ -\frac{2 A \sqrt{a+b x^2}}{a^2 x}-\frac{B \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{a^{3/2}}+\frac{A+B x}{a x \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[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^2 \left (a+b x^2\right )^{3/2}} \, dx &=\frac{A+B x}{a x \sqrt{a+b x^2}}-\frac{\int \frac{-2 a A b-a b B x}{x^2 \sqrt{a+b x^2}} \, dx}{a^2 b}\\ &=\frac{A+B x}{a x \sqrt{a+b x^2}}-\frac{2 A \sqrt{a+b x^2}}{a^2 x}+\frac{B \int \frac{1}{x \sqrt{a+b x^2}} \, dx}{a}\\ &=\frac{A+B x}{a x \sqrt{a+b x^2}}-\frac{2 A \sqrt{a+b x^2}}{a^2 x}+\frac{B \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^2\right )}{2 a}\\ &=\frac{A+B x}{a x \sqrt{a+b x^2}}-\frac{2 A \sqrt{a+b x^2}}{a^2 x}+\frac{B \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 x}{a x \sqrt{a+b x^2}}-\frac{2 A \sqrt{a+b x^2}}{a^2 x}-\frac{B \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{a^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0376964, size = 72, normalized size = 1.03 \[ -\frac{a (A-B x)+\sqrt{a} B x \sqrt{a+b x^2} \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )+2 A b x^2}{a^2 x \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.008, size = 80, normalized size = 1.1 \begin{align*}{\frac{B}{a}{\frac{1}{\sqrt{b{x}^{2}+a}}}}-{B\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ){a}^{-{\frac{3}{2}}}}-{\frac{A}{ax}{\frac{1}{\sqrt{b{x}^{2}+a}}}}-2\,{\frac{Abx}{{a}^{2}\sqrt{b{x}^{2}+a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.88477, size = 381, normalized size = 5.44 \begin{align*} \left [\frac{{\left (B b x^{3} + B a x\right )} \sqrt{a} \log \left (-\frac{b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) - 2 \,{\left (2 \, A b x^{2} - B a x + A a\right )} \sqrt{b x^{2} + a}}{2 \,{\left (a^{2} b x^{3} + a^{3} x\right )}}, \frac{{\left (B b x^{3} + B a x\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) -{\left (2 \, A b x^{2} - B a x + A a\right )} \sqrt{b x^{2} + a}}{a^{2} b x^{3} + a^{3} x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 8.08795, size = 235, normalized size = 3.36 \begin{align*} A \left (- \frac{1}{a \sqrt{b} x^{2} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{2 \sqrt{b}}{a^{2} \sqrt{\frac{a}{b x^{2}} + 1}}\right ) + B \left (\frac{2 a^{3} \sqrt{1 + \frac{b x^{2}}{a}}}{2 a^{\frac{9}{2}} + 2 a^{\frac{7}{2}} b x^{2}} + \frac{a^{3} \log{\left (\frac{b x^{2}}{a} \right )}}{2 a^{\frac{9}{2}} + 2 a^{\frac{7}{2}} b x^{2}} - \frac{2 a^{3} \log{\left (\sqrt{1 + \frac{b x^{2}}{a}} + 1 \right )}}{2 a^{\frac{9}{2}} + 2 a^{\frac{7}{2}} b x^{2}} + \frac{a^{2} b x^{2} \log{\left (\frac{b x^{2}}{a} \right )}}{2 a^{\frac{9}{2}} + 2 a^{\frac{7}{2}} b x^{2}} - \frac{2 a^{2} b x^{2} \log{\left (\sqrt{1 + \frac{b x^{2}}{a}} + 1 \right )}}{2 a^{\frac{9}{2}} + 2 a^{\frac{7}{2}} b x^{2}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*(-1/(a*sqrt(b)*x**2*sqrt(a/(b*x**2) + 1)) - 2*sqrt(b)/(a**2*sqrt(a/(b*x**2) + 1))) + B*(2*a**3*sqrt(1 + b*x*
*2/a)/(2*a**(9/2) + 2*a**(7/2)*b*x**2) + a**3*log(b*x**2/a)/(2*a**(9/2) + 2*a**(7/2)*b*x**2) - 2*a**3*log(sqrt
(1 + b*x**2/a) + 1)/(2*a**(9/2) + 2*a**(7/2)*b*x**2) + a**2*b*x**2*log(b*x**2/a)/(2*a**(9/2) + 2*a**(7/2)*b*x*
*2) - 2*a**2*b*x**2*log(sqrt(1 + b*x**2/a) + 1)/(2*a**(9/2) + 2*a**(7/2)*b*x**2))

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Giac [A]  time = 1.22262, size = 130, normalized size = 1.86 \begin{align*} -\frac{\frac{A b x}{a^{2}} - \frac{B}{a}}{\sqrt{b x^{2} + a}} + \frac{2 \, B \arctan \left (-\frac{\sqrt{b} x - \sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a} + \frac{2 \, A \sqrt{b}}{{\left ({\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} - a\right )} a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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