3.75 \(\int \frac{(A+B x) \sqrt{b x+c x^2}}{x^5} \, dx\)

Optimal. Leaf size=90 \[ \frac{4 c \left (b x+c x^2\right )^{3/2} (7 b B-4 A c)}{105 b^3 x^3}-\frac{2 \left (b x+c x^2\right )^{3/2} (7 b B-4 A c)}{35 b^2 x^4}-\frac{2 A \left (b x+c x^2\right )^{3/2}}{7 b x^5} \]

[Out]

(-2*A*(b*x + c*x^2)^(3/2))/(7*b*x^5) - (2*(7*b*B - 4*A*c)*(b*x + c*x^2)^(3/2))/(35*b^2*x^4) + (4*c*(7*b*B - 4*
A*c)*(b*x + c*x^2)^(3/2))/(105*b^3*x^3)

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Rubi [A]  time = 0.0870264, 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} \[ \frac{4 c \left (b x+c x^2\right )^{3/2} (7 b B-4 A c)}{105 b^3 x^3}-\frac{2 \left (b x+c x^2\right )^{3/2} (7 b B-4 A c)}{35 b^2 x^4}-\frac{2 A \left (b x+c x^2\right )^{3/2}}{7 b x^5} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*Sqrt[b*x + c*x^2])/x^5,x]

[Out]

(-2*A*(b*x + c*x^2)^(3/2))/(7*b*x^5) - (2*(7*b*B - 4*A*c)*(b*x + c*x^2)^(3/2))/(35*b^2*x^4) + (4*c*(7*b*B - 4*
A*c)*(b*x + c*x^2)^(3/2))/(105*b^3*x^3)

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) \sqrt{b x+c x^2}}{x^5} \, dx &=-\frac{2 A \left (b x+c x^2\right )^{3/2}}{7 b x^5}+\frac{\left (2 \left (-5 (-b B+A c)+\frac{3}{2} (-b B+2 A c)\right )\right ) \int \frac{\sqrt{b x+c x^2}}{x^4} \, dx}{7 b}\\ &=-\frac{2 A \left (b x+c x^2\right )^{3/2}}{7 b x^5}-\frac{2 (7 b B-4 A c) \left (b x+c x^2\right )^{3/2}}{35 b^2 x^4}-\frac{(2 c (7 b B-4 A c)) \int \frac{\sqrt{b x+c x^2}}{x^3} \, dx}{35 b^2}\\ &=-\frac{2 A \left (b x+c x^2\right )^{3/2}}{7 b x^5}-\frac{2 (7 b B-4 A c) \left (b x+c x^2\right )^{3/2}}{35 b^2 x^4}+\frac{4 c (7 b B-4 A c) \left (b x+c x^2\right )^{3/2}}{105 b^3 x^3}\\ \end{align*}

Mathematica [A]  time = 0.0334338, size = 56, normalized size = 0.62 \[ -\frac{2 (x (b+c x))^{3/2} \left (A \left (15 b^2-12 b c x+8 c^2 x^2\right )+7 b B x (3 b-2 c x)\right )}{105 b^3 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*Sqrt[b*x + c*x^2])/x^5,x]

[Out]

(-2*(x*(b + c*x))^(3/2)*(7*b*B*x*(3*b - 2*c*x) + A*(15*b^2 - 12*b*c*x + 8*c^2*x^2)))/(105*b^3*x^5)

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Maple [A]  time = 0.005, size = 62, normalized size = 0.7 \begin{align*} -{\frac{ \left ( 2\,cx+2\,b \right ) \left ( 8\,A{c}^{2}{x}^{2}-14\,B{x}^{2}bc-12\,Abcx+21\,{b}^{2}Bx+15\,A{b}^{2} \right ) }{105\,{x}^{4}{b}^{3}}\sqrt{c{x}^{2}+bx}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)^(1/2)/x^5,x)

[Out]

-2/105*(c*x+b)*(8*A*c^2*x^2-14*B*b*c*x^2-12*A*b*c*x+21*B*b^2*x+15*A*b^2)*(c*x^2+b*x)^(1/2)/x^4/b^3

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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)*(c*x^2+b*x)^(1/2)/x^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.92561, size = 180, normalized size = 2. \begin{align*} -\frac{2 \,{\left (15 \, A b^{3} - 2 \,{\left (7 \, B b c^{2} - 4 \, A c^{3}\right )} x^{3} +{\left (7 \, B b^{2} c - 4 \, A b c^{2}\right )} x^{2} + 3 \,{\left (7 \, B b^{3} + A b^{2} c\right )} x\right )} \sqrt{c x^{2} + b x}}{105 \, b^{3} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(1/2)/x^5,x, algorithm="fricas")

[Out]

-2/105*(15*A*b^3 - 2*(7*B*b*c^2 - 4*A*c^3)*x^3 + (7*B*b^2*c - 4*A*b*c^2)*x^2 + 3*(7*B*b^3 + A*b^2*c)*x)*sqrt(c
*x^2 + b*x)/(b^3*x^4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x \left (b + c x\right )} \left (A + B x\right )}{x^{5}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x)**(1/2)/x**5,x)

[Out]

Integral(sqrt(x*(b + c*x))*(A + B*x)/x**5, x)

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Giac [B]  time = 1.13079, size = 339, normalized size = 3.77 \begin{align*} \frac{2 \,{\left (105 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{5} B c^{\frac{3}{2}} + 175 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{4} B b c + 140 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{4} A c^{2} + 105 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{3} B b^{2} \sqrt{c} + 315 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{3} A b c^{\frac{3}{2}} + 21 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} B b^{3} + 273 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} A b^{2} c + 105 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} A b^{3} \sqrt{c} + 15 \, A b^{4}\right )}}{105 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(1/2)/x^5,x, algorithm="giac")

[Out]

2/105*(105*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*B*c^(3/2) + 175*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*B*b*c + 140*(sq
rt(c)*x - sqrt(c*x^2 + b*x))^4*A*c^2 + 105*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*B*b^2*sqrt(c) + 315*(sqrt(c)*x -
sqrt(c*x^2 + b*x))^3*A*b*c^(3/2) + 21*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*b^3 + 273*(sqrt(c)*x - sqrt(c*x^2 +
b*x))^2*A*b^2*c + 105*(sqrt(c)*x - sqrt(c*x^2 + b*x))*A*b^3*sqrt(c) + 15*A*b^4)/(sqrt(c)*x - sqrt(c*x^2 + b*x)
)^7