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

Optimal. Leaf size=95 \[ -\frac{3 A \sqrt{a+b x^2}}{2 a^2 x^2}+\frac{3 A b \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{2 a^{5/2}}-\frac{2 B \sqrt{a+b x^2}}{a^2 x}+\frac{A+B x}{a x^2 \sqrt{a+b x^2}} \]

[Out]

(A + B*x)/(a*x^2*Sqrt[a + b*x^2]) - (3*A*Sqrt[a + b*x^2])/(2*a^2*x^2) - (2*B*Sqrt[a + b*x^2])/(a^2*x) + (3*A*b
*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(2*a^(5/2))

________________________________________________________________________________________

Rubi [A]  time = 0.0783209, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {823, 835, 807, 266, 63, 208} \[ -\frac{3 A \sqrt{a+b x^2}}{2 a^2 x^2}+\frac{3 A b \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{2 a^{5/2}}-\frac{2 B \sqrt{a+b x^2}}{a^2 x}+\frac{A+B x}{a x^2 \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(A + B*x)/(a*x^2*Sqrt[a + b*x^2]) - (3*A*Sqrt[a + b*x^2])/(2*a^2*x^2) - (2*B*Sqrt[a + b*x^2])/(a^2*x) + (3*A*b
*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(2*a^(5/2))

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 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 \left (a+b x^2\right )^{3/2}} \, dx &=\frac{A+B x}{a x^2 \sqrt{a+b x^2}}-\frac{\int \frac{-3 a A b-2 a b B x}{x^3 \sqrt{a+b x^2}} \, dx}{a^2 b}\\ &=\frac{A+B x}{a x^2 \sqrt{a+b x^2}}-\frac{3 A \sqrt{a+b x^2}}{2 a^2 x^2}+\frac{\int \frac{4 a^2 b B-3 a A b^2 x}{x^2 \sqrt{a+b x^2}} \, dx}{2 a^3 b}\\ &=\frac{A+B x}{a x^2 \sqrt{a+b x^2}}-\frac{3 A \sqrt{a+b x^2}}{2 a^2 x^2}-\frac{2 B \sqrt{a+b x^2}}{a^2 x}-\frac{(3 A b) \int \frac{1}{x \sqrt{a+b x^2}} \, dx}{2 a^2}\\ &=\frac{A+B x}{a x^2 \sqrt{a+b x^2}}-\frac{3 A \sqrt{a+b x^2}}{2 a^2 x^2}-\frac{2 B \sqrt{a+b x^2}}{a^2 x}-\frac{(3 A b) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^2\right )}{4 a^2}\\ &=\frac{A+B x}{a x^2 \sqrt{a+b x^2}}-\frac{3 A \sqrt{a+b x^2}}{2 a^2 x^2}-\frac{2 B \sqrt{a+b x^2}}{a^2 x}-\frac{(3 A) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^2}\right )}{2 a^2}\\ &=\frac{A+B x}{a x^2 \sqrt{a+b x^2}}-\frac{3 A \sqrt{a+b x^2}}{2 a^2 x^2}-\frac{2 B \sqrt{a+b x^2}}{a^2 x}+\frac{3 A b \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{2 a^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.15961, size = 75, normalized size = 0.79 \[ \frac{3 A b \sqrt{\frac{b x^2}{a}+1} \tanh ^{-1}\left (\sqrt{\frac{b x^2}{a}+1}\right )-\frac{a (A+2 B x)}{x^2}-b (3 A+4 B x)}{2 a^2 \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((a*(A + 2*B*x))/x^2) - b*(3*A + 4*B*x) + 3*A*b*Sqrt[1 + (b*x^2)/a]*ArcTanh[Sqrt[1 + (b*x^2)/a]])/(2*a^2*Sqr
t[a + b*x^2])

________________________________________________________________________________________

Maple [A]  time = 0.008, size = 101, normalized size = 1.1 \begin{align*} -{\frac{A}{2\,a{x}^{2}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}-{\frac{3\,Ab}{2\,{a}^{2}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}+{\frac{3\,Ab}{2}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ){a}^{-{\frac{5}{2}}}}-{\frac{B}{ax}{\frac{1}{\sqrt{b{x}^{2}+a}}}}-2\,{\frac{bBx}{{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^3/(b*x^2+a)^(3/2),x)

[Out]

-1/2*A/a/x^2/(b*x^2+a)^(1/2)-3/2*A*b/a^2/(b*x^2+a)^(1/2)+3/2*A*b/a^(5/2)*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x)
-B/a/x/(b*x^2+a)^(1/2)-2*B*b/a^2*x/(b*x^2+a)^(1/2)

________________________________________________________________________________________

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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.87817, size = 474, normalized size = 4.99 \begin{align*} \left [\frac{3 \,{\left (A b^{2} x^{4} + A a b x^{2}\right )} \sqrt{a} \log \left (-\frac{b x^{2} + 2 \, \sqrt{b x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) - 2 \,{\left (4 \, B a b x^{3} + 3 \, A a b x^{2} + 2 \, B a^{2} x + A a^{2}\right )} \sqrt{b x^{2} + a}}{4 \,{\left (a^{3} b x^{4} + a^{4} x^{2}\right )}}, -\frac{3 \,{\left (A b^{2} x^{4} + A a b x^{2}\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) +{\left (4 \, B a b x^{3} + 3 \, A a b x^{2} + 2 \, B a^{2} x + A a^{2}\right )} \sqrt{b x^{2} + a}}{2 \,{\left (a^{3} b x^{4} + a^{4} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(3*(A*b^2*x^4 + A*a*b*x^2)*sqrt(a)*log(-(b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*(4*B*a*b*x^3 +
3*A*a*b*x^2 + 2*B*a^2*x + A*a^2)*sqrt(b*x^2 + a))/(a^3*b*x^4 + a^4*x^2), -1/2*(3*(A*b^2*x^4 + A*a*b*x^2)*sqrt
(-a)*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + (4*B*a*b*x^3 + 3*A*a*b*x^2 + 2*B*a^2*x + A*a^2)*sqrt(b*x^2 + a))/(a^3*
b*x^4 + a^4*x^2)]

________________________________________________________________________________________

Sympy [A]  time = 8.60718, size = 124, normalized size = 1.31 \begin{align*} A \left (- \frac{1}{2 a \sqrt{b} x^{3} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{3 \sqrt{b}}{2 a^{2} x \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{3 b \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )}}{2 a^{\frac{5}{2}}}\right ) + B \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*(-1/(2*a*sqrt(b)*x**3*sqrt(a/(b*x**2) + 1)) - 3*sqrt(b)/(2*a**2*x*sqrt(a/(b*x**2) + 1)) + 3*b*asinh(sqrt(a)/
(sqrt(b)*x))/(2*a**(5/2))) + B*(-1/(a*sqrt(b)*x**2*sqrt(a/(b*x**2) + 1)) - 2*sqrt(b)/(a**2*sqrt(a/(b*x**2) + 1
)))

________________________________________________________________________________________

Giac [B]  time = 1.25495, size = 231, normalized size = 2.43 \begin{align*} -\frac{\frac{B b x}{a^{2}} + \frac{A b}{a^{2}}}{\sqrt{b x^{2} + a}} - \frac{3 \, A b \arctan \left (-\frac{\sqrt{b} x - \sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{2}} + \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^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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