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

Optimal. Leaf size=49 \[ \frac{(A b-2 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{a^{3/2}}-\frac{A \sqrt{a+b x}}{a x} \]

[Out]

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

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Rubi [A]  time = 0.0218117, antiderivative size = 49, 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} \[ \frac{(A b-2 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{a^{3/2}}-\frac{A \sqrt{a+b x}}{a x} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^2*Sqrt[a + b*x]),x]

[Out]

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

Mathematica [A]  time = 0.025058, size = 49, normalized size = 1. \[ \frac{(A b-2 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{a^{3/2}}-\frac{A \sqrt{a+b x}}{a x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.007, size = 42, normalized size = 0.9 \begin{align*}{(Ab-2\,Ba){\it Artanh} \left ({\sqrt{bx+a}{\frac{1}{\sqrt{a}}}} \right ){a}^{-{\frac{3}{2}}}}-{\frac{A}{ax}\sqrt{bx+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^2/(b*x+a)^(1/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.71236, size = 270, normalized size = 5.51 \begin{align*} \left [-\frac{{\left (2 \, B a - A b\right )} \sqrt{a} x \log \left (\frac{b x + 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) + 2 \, \sqrt{b x + a} A a}{2 \, a^{2} x}, \frac{{\left (2 \, B a - A b\right )} \sqrt{-a} x \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) - \sqrt{b x + a} A a}{a^{2} x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 17.0983, size = 82, normalized size = 1.67 \begin{align*} - \frac{A \sqrt{b} \sqrt{\frac{a}{b x} + 1}}{a \sqrt{x}} + \frac{A b \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} \sqrt{x}} \right )}}{a^{\frac{3}{2}}} + \frac{2 B \operatorname{atan}{\left (\frac{1}{\sqrt{- \frac{1}{a}} \sqrt{a + b x}} \right )}}{a \sqrt{- \frac{1}{a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.13294, size = 78, normalized size = 1.59 \begin{align*} -\frac{\frac{\sqrt{b x + a} A b}{a x} - \frac{{\left (2 \, B a b - A b^{2}\right )} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a}}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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