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

Optimal. Leaf size=50 \[ \frac{2 (A b-a B)}{a b \sqrt{a+b x}}-\frac{2 A \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{a^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0150914, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {78, 63, 208} \[ \frac{2 (A b-a B)}{a b \sqrt{a+b x}}-\frac{2 A \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{a^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x*(a + b*x)^(3/2)),x]

[Out]

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

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

Mathematica [A]  time = 0.034676, size = 50, normalized size = 1. \[ -\frac{2 A \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{a^{3/2}}-\frac{2 (a B-A b)}{a b \sqrt{a+b x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.007, size = 46, normalized size = 0.9 \begin{align*} 2\,{\frac{1}{b} \left ( -{\frac{Ab}{{a}^{3/2}}{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ) }-{\frac{-Ab+Ba}{a\sqrt{bx+a}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.41383, size = 336, normalized size = 6.72 \begin{align*} \left [\frac{{\left (A b^{2} x + A a b\right )} \sqrt{a} \log \left (\frac{b x - 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) - 2 \,{\left (B a^{2} - A a b\right )} \sqrt{b x + a}}{a^{2} b^{2} x + a^{3} b}, \frac{2 \,{\left ({\left (A b^{2} x + A a b\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) -{\left (B a^{2} - A a b\right )} \sqrt{b x + a}\right )}}{a^{2} b^{2} x + a^{3} b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 16.5693, size = 49, normalized size = 0.98 \begin{align*} \frac{2 A \operatorname{atan}{\left (\frac{\sqrt{a + b x}}{\sqrt{- a}} \right )}}{a \sqrt{- a}} - \frac{2 \left (- A b + B a\right )}{a b \sqrt{a + b x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.20671, size = 66, normalized size = 1.32 \begin{align*} \frac{2 \, A \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a} - \frac{2 \,{\left (B a - A b\right )}}{\sqrt{b x + a} a b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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