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

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

[Out]

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

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Rubi [A]  time = 0.0216724, 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 = {80, 63, 205} \[ \frac{2 (A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{\sqrt{a} b^{3/2}}+\frac{2 B \sqrt{x}}{b} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(Sqrt[x]*(a + b*x)),x]

[Out]

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

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

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 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

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

Mathematica [A]  time = 0.0267192, size = 49, normalized size = 1. \[ \frac{2 (A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{\sqrt{a} b^{3/2}}+\frac{2 B \sqrt{x}}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.006, size = 53, normalized size = 1.1 \begin{align*} 2\,{\frac{B\sqrt{x}}{b}}+2\,{\frac{A}{\sqrt{ab}}\arctan \left ({\frac{b\sqrt{x}}{\sqrt{ab}}} \right ) }-2\,{\frac{Ba}{b\sqrt{ab}}\arctan \left ({\frac{b\sqrt{x}}{\sqrt{ab}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)/(b*x+a)/x^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.43824, size = 250, normalized size = 5.1 \begin{align*} \left [\frac{2 \, B a b \sqrt{x} +{\left (B a - A b\right )} \sqrt{-a b} \log \left (\frac{b x - a - 2 \, \sqrt{-a b} \sqrt{x}}{b x + a}\right )}{a b^{2}}, \frac{2 \,{\left (B a b \sqrt{x} +{\left (B a - A b\right )} \sqrt{a b} \arctan \left (\frac{\sqrt{a b}}{b \sqrt{x}}\right )\right )}}{a b^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 2.38755, size = 218, normalized size = 4.45 \begin{align*} \begin{cases} \tilde{\infty } \left (- \frac{2 A}{\sqrt{x}} + 2 B \sqrt{x}\right ) & \text{for}\: a = 0 \wedge b = 0 \\\frac{- \frac{2 A}{\sqrt{x}} + 2 B \sqrt{x}}{b} & \text{for}\: a = 0 \\\frac{2 A \sqrt{x} + \frac{2 B x^{\frac{3}{2}}}{3}}{a} & \text{for}\: b = 0 \\- \frac{i A \log{\left (- i \sqrt{a} \sqrt{\frac{1}{b}} + \sqrt{x} \right )}}{\sqrt{a} b \sqrt{\frac{1}{b}}} + \frac{i A \log{\left (i \sqrt{a} \sqrt{\frac{1}{b}} + \sqrt{x} \right )}}{\sqrt{a} b \sqrt{\frac{1}{b}}} + \frac{i B \sqrt{a} \log{\left (- i \sqrt{a} \sqrt{\frac{1}{b}} + \sqrt{x} \right )}}{b^{2} \sqrt{\frac{1}{b}}} - \frac{i B \sqrt{a} \log{\left (i \sqrt{a} \sqrt{\frac{1}{b}} + \sqrt{x} \right )}}{b^{2} \sqrt{\frac{1}{b}}} + \frac{2 B \sqrt{x}}{b} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*(-2*A/sqrt(x) + 2*B*sqrt(x)), Eq(a, 0) & Eq(b, 0)), ((-2*A/sqrt(x) + 2*B*sqrt(x))/b, Eq(a, 0)),
((2*A*sqrt(x) + 2*B*x**(3/2)/3)/a, Eq(b, 0)), (-I*A*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(sqrt(a)*b*sqrt(1/b))
+ I*A*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(sqrt(a)*b*sqrt(1/b)) + I*B*sqrt(a)*log(-I*sqrt(a)*sqrt(1/b) + sqrt(
x))/(b**2*sqrt(1/b)) - I*B*sqrt(a)*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(b**2*sqrt(1/b)) + 2*B*sqrt(x)/b, True))

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Giac [A]  time = 1.14286, size = 53, normalized size = 1.08 \begin{align*} \frac{2 \, B \sqrt{x}}{b} - \frac{2 \,{\left (B a - A b\right )} \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{\sqrt{a b} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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