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]

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

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Rubi [A]  time = 0.11, antiderivative size = 125, normalized size of antiderivative = 1.00, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.03, size = 78, normalized size = 0.62 \[ -\frac {2 (x (b+c x))^{3/2} \left (A \left (35 b^3-30 b^2 c x+24 b c^2 x^2-16 c^3 x^3\right )+3 b B x \left (15 b^2-12 b c x+8 c^2 x^2\right )\right )}{315 b^4 x^6} \]

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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fricas [A]  time = 0.85, size = 105, normalized size = 0.84 \[ -\frac {2 \, {\left (35 \, A b^{4} + 8 \, {\left (3 \, B b c^{3} - 2 \, A c^{4}\right )} x^{4} - 4 \, {\left (3 \, B b^{2} c^{2} - 2 \, A b c^{3}\right )} x^{3} + 3 \, {\left (3 \, B b^{3} c - 2 \, A b^{2} c^{2}\right )} x^{2} + 5 \, {\left (9 \, B b^{4} + A b^{3} c\right )} x\right )} \sqrt {c x^{2} + b x}}{315 \, b^{4} x^{5}} \]

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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giac [B]  time = 0.20, size = 311, normalized size = 2.49 \[ \frac {2 \, {\left (420 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{6} B c^{2} + 945 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5} B b c^{\frac {3}{2}} + 630 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5} A c^{\frac {5}{2}} + 819 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} B b^{2} c + 1764 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} A b c^{2} + 315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} B b^{3} \sqrt {c} + 1995 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} A b^{2} c^{\frac {3}{2}} + 45 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} B b^{4} + 1125 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} A b^{3} c + 315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} A b^{4} \sqrt {c} + 35 \, A b^{5}\right )}}{315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{9}} \]

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

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maple [A]  time = 0.05, size = 86, normalized size = 0.69 \[ -\frac {2 \left (c x +b \right ) \left (-16 A \,c^{3} x^{3}+24 B b \,c^{2} x^{3}+24 A b \,c^{2} x^{2}-36 B \,b^{2} c \,x^{2}-30 A \,b^{2} c x +45 B \,b^{3} x +35 A \,b^{3}\right ) \sqrt {c \,x^{2}+b x}}{315 b^{4} x^{5}} \]

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)/b^4/x^5

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maxima [A]  time = 0.98, size = 192, normalized size = 1.54 \[ -\frac {16 \, \sqrt {c x^{2} + b x} B c^{3}}{105 \, b^{3} x} + \frac {32 \, \sqrt {c x^{2} + b x} A c^{4}}{315 \, b^{4} x} + \frac {8 \, \sqrt {c x^{2} + b x} B c^{2}}{105 \, b^{2} x^{2}} - \frac {16 \, \sqrt {c x^{2} + b x} A c^{3}}{315 \, b^{3} x^{2}} - \frac {2 \, \sqrt {c x^{2} + b x} B c}{35 \, b x^{3}} + \frac {4 \, \sqrt {c x^{2} + b x} A c^{2}}{105 \, b^{2} x^{3}} - \frac {2 \, \sqrt {c x^{2} + b x} B}{7 \, x^{4}} - \frac {2 \, \sqrt {c x^{2} + b x} A c}{63 \, b x^{4}} - \frac {2 \, \sqrt {c x^{2} + b x} A}{9 \, x^{5}} \]

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]

-16/105*sqrt(c*x^2 + b*x)*B*c^3/(b^3*x) + 32/315*sqrt(c*x^2 + b*x)*A*c^4/(b^4*x) + 8/105*sqrt(c*x^2 + b*x)*B*c
^2/(b^2*x^2) - 16/315*sqrt(c*x^2 + b*x)*A*c^3/(b^3*x^2) - 2/35*sqrt(c*x^2 + b*x)*B*c/(b*x^3) + 4/105*sqrt(c*x^
2 + b*x)*A*c^2/(b^2*x^3) - 2/7*sqrt(c*x^2 + b*x)*B/x^4 - 2/63*sqrt(c*x^2 + b*x)*A*c/(b*x^4) - 2/9*sqrt(c*x^2 +
b*x)*A/x^5

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mupad [B]  time = 1.94, size = 192, normalized size = 1.54 \[ \frac {4\,A\,c^2\,\sqrt {c\,x^2+b\,x}}{105\,b^2\,x^3}-\frac {2\,B\,\sqrt {c\,x^2+b\,x}}{7\,x^4}-\frac {2\,A\,c\,\sqrt {c\,x^2+b\,x}}{63\,b\,x^4}-\frac {2\,B\,c\,\sqrt {c\,x^2+b\,x}}{35\,b\,x^3}-\frac {2\,A\,\sqrt {c\,x^2+b\,x}}{9\,x^5}-\frac {16\,A\,c^3\,\sqrt {c\,x^2+b\,x}}{315\,b^3\,x^2}+\frac {32\,A\,c^4\,\sqrt {c\,x^2+b\,x}}{315\,b^4\,x}+\frac {8\,B\,c^2\,\sqrt {c\,x^2+b\,x}}{105\,b^2\,x^2}-\frac {16\,B\,c^3\,\sqrt {c\,x^2+b\,x}}{105\,b^3\,x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*x + c*x^2)^(1/2)*(A + B*x))/x^6,x)

[Out]

(4*A*c^2*(b*x + c*x^2)^(1/2))/(105*b^2*x^3) - (2*B*(b*x + c*x^2)^(1/2))/(7*x^4) - (2*A*c*(b*x + c*x^2)^(1/2))/
(63*b*x^4) - (2*B*c*(b*x + c*x^2)^(1/2))/(35*b*x^3) - (2*A*(b*x + c*x^2)^(1/2))/(9*x^5) - (16*A*c^3*(b*x + c*x
^2)^(1/2))/(315*b^3*x^2) + (32*A*c^4*(b*x + c*x^2)^(1/2))/(315*b^4*x) + (8*B*c^2*(b*x + c*x^2)^(1/2))/(105*b^2
*x^2) - (16*B*c^3*(b*x + c*x^2)^(1/2))/(105*b^3*x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {x \left (b + c x\right )} \left (A + B x\right )}{x^{6}}\, dx \]

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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