3.76 \(\int \frac{(A+B x) \sqrt{b x+c x^2}}{x^6} \, dx\)

Optimal. Leaf size=125 \[ -\frac{16 c^2 \left (b x+c x^2\right )^{3/2} (3 b B-2 A c)}{315 b^4 x^3}+\frac{8 c \left (b x+c x^2\right )^{3/2} (3 b B-2 A c)}{105 b^3 x^4}-\frac{2 \left (b x+c x^2\right )^{3/2} (3 b B-2 A c)}{21 b^2 x^5}-\frac{2 A \left (b x+c x^2\right )^{3/2}}{9 b x^6} \]

[Out]

(-2*A*(b*x + c*x^2)^(3/2))/(9*b*x^6) - (2*(3*b*B - 2*A*c)*(b*x + c*x^2)^(3/2))/(21*b^2*x^5) + (8*c*(3*b*B - 2*
A*c)*(b*x + c*x^2)^(3/2))/(105*b^3*x^4) - (16*c^2*(3*b*B - 2*A*c)*(b*x + c*x^2)^(3/2))/(315*b^4*x^3)

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Rubi [A]  time = 0.10906, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {792, 658, 650} \[ -\frac{16 c^2 \left (b x+c x^2\right )^{3/2} (3 b B-2 A c)}{315 b^4 x^3}+\frac{8 c \left (b x+c x^2\right )^{3/2} (3 b B-2 A c)}{105 b^3 x^4}-\frac{2 \left (b x+c x^2\right )^{3/2} (3 b B-2 A c)}{21 b^2 x^5}-\frac{2 A \left (b x+c x^2\right )^{3/2}}{9 b x^6} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*Sqrt[b*x + c*x^2])/x^6,x]

[Out]

(-2*A*(b*x + c*x^2)^(3/2))/(9*b*x^6) - (2*(3*b*B - 2*A*c)*(b*x + c*x^2)^(3/2))/(21*b^2*x^5) + (8*c*(3*b*B - 2*
A*c)*(b*x + c*x^2)^(3/2))/(105*b^3*x^4) - (16*c^2*(3*b*B - 2*A*c)*(b*x + c*x^2)^(3/2))/(315*b^4*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^6} \, dx &=-\frac{2 A \left (b x+c x^2\right )^{3/2}}{9 b x^6}+\frac{\left (2 \left (-6 (-b B+A c)+\frac{3}{2} (-b B+2 A c)\right )\right ) \int \frac{\sqrt{b x+c x^2}}{x^5} \, dx}{9 b}\\ &=-\frac{2 A \left (b x+c x^2\right )^{3/2}}{9 b x^6}-\frac{2 (3 b B-2 A c) \left (b x+c x^2\right )^{3/2}}{21 b^2 x^5}-\frac{(4 c (3 b B-2 A c)) \int \frac{\sqrt{b x+c x^2}}{x^4} \, dx}{21 b^2}\\ &=-\frac{2 A \left (b x+c x^2\right )^{3/2}}{9 b x^6}-\frac{2 (3 b B-2 A c) \left (b x+c x^2\right )^{3/2}}{21 b^2 x^5}+\frac{8 c (3 b B-2 A c) \left (b x+c x^2\right )^{3/2}}{105 b^3 x^4}+\frac{\left (8 c^2 (3 b B-2 A c)\right ) \int \frac{\sqrt{b x+c x^2}}{x^3} \, dx}{105 b^3}\\ &=-\frac{2 A \left (b x+c x^2\right )^{3/2}}{9 b x^6}-\frac{2 (3 b B-2 A c) \left (b x+c x^2\right )^{3/2}}{21 b^2 x^5}+\frac{8 c (3 b B-2 A c) \left (b x+c x^2\right )^{3/2}}{105 b^3 x^4}-\frac{16 c^2 (3 b B-2 A c) \left (b x+c x^2\right )^{3/2}}{315 b^4 x^3}\\ \end{align*}

Mathematica [A]  time = 0.0327115, size = 78, normalized size = 0.62 \[ -\frac{2 (x (b+c x))^{3/2} \left (A \left (-30 b^2 c x+35 b^3+24 b c^2 x^2-16 c^3 x^3\right )+3 b B x \left (15 b^2-12 b c x+8 c^2 x^2\right )\right )}{315 b^4 x^6} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*Sqrt[b*x + c*x^2])/x^6,x]

[Out]

(-2*(x*(b + c*x))^(3/2)*(3*b*B*x*(15*b^2 - 12*b*c*x + 8*c^2*x^2) + A*(35*b^3 - 30*b^2*c*x + 24*b*c^2*x^2 - 16*
c^3*x^3)))/(315*b^4*x^6)

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Maple [A]  time = 0.007, size = 86, normalized size = 0.7 \begin{align*} -{\frac{ \left ( 2\,cx+2\,b \right ) \left ( -16\,A{x}^{3}{c}^{3}+24\,B{x}^{3}b{c}^{2}+24\,A{x}^{2}b{c}^{2}-36\,B{x}^{2}{b}^{2}c-30\,A{b}^{2}cx+45\,{b}^{3}Bx+35\,A{b}^{3} \right ) }{315\,{x}^{5}{b}^{4}}\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^6,x)

[Out]

-2/315*(c*x+b)*(-16*A*c^3*x^3+24*B*b*c^2*x^3+24*A*b*c^2*x^2-36*B*b^2*c*x^2-30*A*b^2*c*x+45*B*b^3*x+35*A*b^3)*(
c*x^2+b*x)^(1/2)/x^5/b^4

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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^6,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.8555, size = 231, normalized size = 1.85 \begin{align*} -\frac{2 \,{\left (35 \, A b^{4} + 8 \,{\left (3 \, B b c^{3} - 2 \, A c^{4}\right )} x^{4} - 4 \,{\left (3 \, B b^{2} c^{2} - 2 \, A b c^{3}\right )} x^{3} + 3 \,{\left (3 \, B b^{3} c - 2 \, A b^{2} c^{2}\right )} x^{2} + 5 \,{\left (9 \, B b^{4} + A b^{3} c\right )} x\right )} \sqrt{c x^{2} + b x}}{315 \, b^{4} x^{5}} \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^6,x, algorithm="fricas")

[Out]

-2/315*(35*A*b^4 + 8*(3*B*b*c^3 - 2*A*c^4)*x^4 - 4*(3*B*b^2*c^2 - 2*A*b*c^3)*x^3 + 3*(3*B*b^3*c - 2*A*b^2*c^2)
*x^2 + 5*(9*B*b^4 + A*b^3*c)*x)*sqrt(c*x^2 + b*x)/(b^4*x^5)

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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^{6}}\, 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**6,x)

[Out]

Integral(sqrt(x*(b + c*x))*(A + B*x)/x**6, x)

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Giac [B]  time = 1.181, size = 420, normalized size = 3.36 \begin{align*} \frac{2 \,{\left (420 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{6} B c^{2} + 945 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{5} B b c^{\frac{3}{2}} + 630 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{5} A c^{\frac{5}{2}} + 819 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{4} B b^{2} c + 1764 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{4} A b c^{2} + 315 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{3} B b^{3} \sqrt{c} + 1995 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{3} A b^{2} c^{\frac{3}{2}} + 45 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} B b^{4} + 1125 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} A b^{3} c + 315 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} A b^{4} \sqrt{c} + 35 \, A b^{5}\right )}}{315 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{9}} \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^6,x, algorithm="giac")

[Out]

2/315*(420*(sqrt(c)*x - sqrt(c*x^2 + b*x))^6*B*c^2 + 945*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*B*b*c^(3/2) + 630*(
sqrt(c)*x - sqrt(c*x^2 + b*x))^5*A*c^(5/2) + 819*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*B*b^2*c + 1764*(sqrt(c)*x -
sqrt(c*x^2 + b*x))^4*A*b*c^2 + 315*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*B*b^3*sqrt(c) + 1995*(sqrt(c)*x - sqrt(c
*x^2 + b*x))^3*A*b^2*c^(3/2) + 45*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*b^4 + 1125*(sqrt(c)*x - sqrt(c*x^2 + b*x
))^2*A*b^3*c + 315*(sqrt(c)*x - sqrt(c*x^2 + b*x))*A*b^4*sqrt(c) + 35*A*b^5)/(sqrt(c)*x - sqrt(c*x^2 + b*x))^9