3.116 \(\int \frac{A+B x}{x^3 \sqrt{b x+c x^2}} \, dx\)

Optimal. Leaf size=90 \[ \frac{4 c \sqrt{b x+c x^2} (5 b B-4 A c)}{15 b^3 x}-\frac{2 \sqrt{b x+c x^2} (5 b B-4 A c)}{15 b^2 x^2}-\frac{2 A \sqrt{b x+c x^2}}{5 b x^3} \]

[Out]

(-2*A*Sqrt[b*x + c*x^2])/(5*b*x^3) - (2*(5*b*B - 4*A*c)*Sqrt[b*x + c*x^2])/(15*b^2*x^2) + (4*c*(5*b*B - 4*A*c)
*Sqrt[b*x + c*x^2])/(15*b^3*x)

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Rubi [A]  time = 0.0813891, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {792, 658, 650} \[ \frac{4 c \sqrt{b x+c x^2} (5 b B-4 A c)}{15 b^3 x}-\frac{2 \sqrt{b x+c x^2} (5 b B-4 A c)}{15 b^2 x^2}-\frac{2 A \sqrt{b x+c x^2}}{5 b x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A*Sqrt[b*x + c*x^2])/(5*b*x^3) - (2*(5*b*B - 4*A*c)*Sqrt[b*x + c*x^2])/(15*b^2*x^2) + (4*c*(5*b*B - 4*A*c)
*Sqrt[b*x + c*x^2])/(15*b^3*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 658

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))/((m + p + 1)*(2*c*d - b*e)), x] + Dist[(c*Simplify[m + 2*p + 2])/((m + p + 1)*(2*c*d -
b*e)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], 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] && ILtQ[Simplify[m + 2*p + 2], 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^3 \sqrt{b x+c x^2}} \, dx &=-\frac{2 A \sqrt{b x+c x^2}}{5 b x^3}+\frac{\left (2 \left (-3 (-b B+A c)+\frac{1}{2} (-b B+2 A c)\right )\right ) \int \frac{1}{x^2 \sqrt{b x+c x^2}} \, dx}{5 b}\\ &=-\frac{2 A \sqrt{b x+c x^2}}{5 b x^3}-\frac{2 (5 b B-4 A c) \sqrt{b x+c x^2}}{15 b^2 x^2}-\frac{(2 c (5 b B-4 A c)) \int \frac{1}{x \sqrt{b x+c x^2}} \, dx}{15 b^2}\\ &=-\frac{2 A \sqrt{b x+c x^2}}{5 b x^3}-\frac{2 (5 b B-4 A c) \sqrt{b x+c x^2}}{15 b^2 x^2}+\frac{4 c (5 b B-4 A c) \sqrt{b x+c x^2}}{15 b^3 x}\\ \end{align*}

Mathematica [A]  time = 0.0300194, size = 54, normalized size = 0.6 \[ -\frac{2 \sqrt{x (b+c x)} \left (A \left (3 b^2-4 b c x+8 c^2 x^2\right )+5 b B x (b-2 c x)\right )}{15 b^3 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[x*(b + c*x)]*(5*b*B*x*(b - 2*c*x) + A*(3*b^2 - 4*b*c*x + 8*c^2*x^2)))/(15*b^3*x^3)

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Maple [A]  time = 0.004, size = 62, normalized size = 0.7 \begin{align*} -{\frac{ \left ( 2\,cx+2\,b \right ) \left ( 8\,A{c}^{2}{x}^{2}-10\,B{x}^{2}bc-4\,Abcx+5\,{b}^{2}Bx+3\,A{b}^{2} \right ) }{15\,{x}^{2}{b}^{3}}{\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^3/(c*x^2+b*x)^(1/2),x)

[Out]

-2/15*(c*x+b)*(8*A*c^2*x^2-10*B*b*c*x^2-4*A*b*c*x+5*B*b^2*x+3*A*b^2)/x^2/b^3/(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^3/(c*x^2+b*x)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.00558, size = 131, normalized size = 1.46 \begin{align*} -\frac{2 \,{\left (3 \, A b^{2} - 2 \,{\left (5 \, B b c - 4 \, A c^{2}\right )} x^{2} +{\left (5 \, B b^{2} - 4 \, A b c\right )} x\right )} \sqrt{c x^{2} + b x}}{15 \, b^{3} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(3*A*b^2 - 2*(5*B*b*c - 4*A*c^2)*x^2 + (5*B*b^2 - 4*A*b*c)*x)*sqrt(c*x^2 + b*x)/(b^3*x^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A + B x}{x^{3} \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**3/(c*x**2+b*x)**(1/2),x)

[Out]

Integral((A + B*x)/(x**3*sqrt(x*(b + c*x))), x)

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Giac [A]  time = 1.15484, size = 180, normalized size = 2. \begin{align*} \frac{2 \,{\left (15 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{3} B \sqrt{c} + 5 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} B b + 20 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} A c + 15 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} A b \sqrt{c} + 3 \, A b^{2}\right )}}{15 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(15*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*B*sqrt(c) + 5*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*b + 20*(sqrt(c)*x
- sqrt(c*x^2 + b*x))^2*A*c + 15*(sqrt(c)*x - sqrt(c*x^2 + b*x))*A*b*sqrt(c) + 3*A*b^2)/(sqrt(c)*x - sqrt(c*x^
2 + b*x))^5