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

Optimal. Leaf size=57 \[ -\frac{2 \sqrt{b x+c x^2} (3 b B-2 A c)}{3 b^2 x}-\frac{2 A \sqrt{b x+c x^2}}{3 b x^2} \]

[Out]

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

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Rubi [A]  time = 0.042887, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {792, 650} \[ -\frac{2 \sqrt{b x+c x^2} (3 b B-2 A c)}{3 b^2 x}-\frac{2 A \sqrt{b x+c x^2}}{3 b x^2} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^2*Sqrt[b*x + c*x^2]),x]

[Out]

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

Rule 792

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d*g - e*f)*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/((2*c*d - b*e)*(m + p + 1)), x] + Dist[(m*(g*(c*d - b*e)
+ c*e*f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p,
x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((L
tQ[m, -1] &&  !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0]

Rule 650

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^m*(a +
b*x + c*x^2)^(p + 1))/((p + 1)*(2*c*d - b*e)), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] &&
EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p + 2, 0]

Rubi steps

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

Mathematica [A]  time = 0.0208504, size = 35, normalized size = 0.61 \[ -\frac{2 \sqrt{x (b+c x)} (A (b-2 c x)+3 b B x)}{3 b^2 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.004, size = 39, normalized size = 0.7 \begin{align*} -{\frac{ \left ( 2\,cx+2\,b \right ) \left ( -2\,Acx+3\,bBx+Ab \right ) }{3\,x{b}^{2}}{\frac{1}{\sqrt{c{x}^{2}+bx}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.96217, size = 81, normalized size = 1.42 \begin{align*} -\frac{2 \, \sqrt{c x^{2} + b x}{\left (A b +{\left (3 \, B b - 2 \, A c\right )} x\right )}}{3 \, b^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A + B x}{x^{2} \sqrt{x \left (b + c x\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + B*x)/(x**2*sqrt(x*(b + c*x))), x)

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Giac [A]  time = 1.15844, size = 103, normalized size = 1.81 \begin{align*} \frac{2 \,{\left (3 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} B + 3 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} A \sqrt{c} + A b\right )}}{3 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(3*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B + 3*(sqrt(c)*x - sqrt(c*x^2 + b*x))*A*sqrt(c) + A*b)/(sqrt(c)*x - s
qrt(c*x^2 + b*x))^3