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

Optimal. Leaf size=93 \[ \frac {8 c (b+2 c x) (5 b B-6 A c)}{15 b^4 \sqrt {b x+c x^2}}-\frac {2 (5 b B-6 A c)}{15 b^2 x \sqrt {b x+c x^2}}-\frac {2 A}{5 b x^2 \sqrt {b x+c x^2}} \]

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Rubi [A]  time = 0.08, antiderivative size = 93, normalized size of antiderivative = 1.00, 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, 613} \begin {gather*} \frac {8 c (b+2 c x) (5 b B-6 A c)}{15 b^4 \sqrt {b x+c x^2}}-\frac {2 (5 b B-6 A c)}{15 b^2 x \sqrt {b x+c x^2}}-\frac {2 A}{5 b x^2 \sqrt {b x+c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A)/(5*b*x^2*Sqrt[b*x + c*x^2]) - (2*(5*b*B - 6*A*c))/(15*b^2*x*Sqrt[b*x + c*x^2]) + (8*c*(5*b*B - 6*A*c)*(
b + 2*c*x))/(15*b^4*Sqrt[b*x + c*x^2])

Rule 613

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[(-2*(b + 2*c*x))/((b^2 - 4*a*c)*Sqrt[a + b*x
+ c*x^2]), x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 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}{x^2 \left (b x+c x^2\right )^{3/2}} \, dx &=-\frac {2 A}{5 b x^2 \sqrt {b x+c x^2}}+\frac {\left (2 \left (\frac {1}{2} (b B-2 A c)-2 (-b B+A c)\right )\right ) \int \frac {1}{x \left (b x+c x^2\right )^{3/2}} \, dx}{5 b}\\ &=-\frac {2 A}{5 b x^2 \sqrt {b x+c x^2}}-\frac {2 (5 b B-6 A c)}{15 b^2 x \sqrt {b x+c x^2}}-\frac {(4 c (5 b B-6 A c)) \int \frac {1}{\left (b x+c x^2\right )^{3/2}} \, dx}{15 b^2}\\ &=-\frac {2 A}{5 b x^2 \sqrt {b x+c x^2}}-\frac {2 (5 b B-6 A c)}{15 b^2 x \sqrt {b x+c x^2}}+\frac {8 c (5 b B-6 A c) (b+2 c x)}{15 b^4 \sqrt {b x+c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 75, normalized size = 0.81 \begin {gather*} -\frac {2 \left (3 A \left (b^3-2 b^2 c x+8 b c^2 x^2+16 c^3 x^3\right )+5 b B x \left (b^2-4 b c x-8 c^2 x^2\right )\right )}{15 b^4 x^2 \sqrt {x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(5*b*B*x*(b^2 - 4*b*c*x - 8*c^2*x^2) + 3*A*(b^3 - 2*b^2*c*x + 8*b*c^2*x^2 + 16*c^3*x^3)))/(15*b^4*x^2*Sqrt
[x*(b + c*x)])

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IntegrateAlgebraic [A]  time = 0.36, size = 91, normalized size = 0.98 \begin {gather*} \frac {2 \sqrt {b x+c x^2} \left (-3 A b^3+6 A b^2 c x-24 A b c^2 x^2-48 A c^3 x^3-5 b^3 B x+20 b^2 B c x^2+40 b B c^2 x^3\right )}{15 b^4 x^3 (b+c x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[b*x + c*x^2]*(-3*A*b^3 - 5*b^3*B*x + 6*A*b^2*c*x + 20*b^2*B*c*x^2 - 24*A*b*c^2*x^2 + 40*b*B*c^2*x^3 -
48*A*c^3*x^3))/(15*b^4*x^3*(b + c*x))

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fricas [A]  time = 0.41, size = 93, normalized size = 1.00 \begin {gather*} -\frac {2 \, {\left (3 \, A b^{3} - 8 \, {\left (5 \, B b c^{2} - 6 \, A c^{3}\right )} x^{3} - 4 \, {\left (5 \, B b^{2} c - 6 \, A b c^{2}\right )} x^{2} + {\left (5 \, B b^{3} - 6 \, A b^{2} c\right )} x\right )} \sqrt {c x^{2} + b x}}{15 \, {\left (b^{4} c x^{4} + b^{5} x^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(3*A*b^3 - 8*(5*B*b*c^2 - 6*A*c^3)*x^3 - 4*(5*B*b^2*c - 6*A*b*c^2)*x^2 + (5*B*b^3 - 6*A*b^2*c)*x)*sqrt(c
*x^2 + b*x)/(b^4*c*x^4 + b^5*x^3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {B x + A}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*x + A)/((c*x^2 + b*x)^(3/2)*x^2), x)

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maple [A]  time = 0.05, size = 86, normalized size = 0.92 \begin {gather*} -\frac {2 \left (c x +b \right ) \left (48 A \,c^{3} x^{3}-40 B b \,c^{2} x^{3}+24 A b \,c^{2} x^{2}-20 B \,b^{2} c \,x^{2}-6 A \,b^{2} c x +5 B \,b^{3} x +3 A \,b^{3}\right )}{15 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} b^{4} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^2/(c*x^2+b*x)^(3/2),x)

[Out]

-2/15*(c*x+b)*(48*A*c^3*x^3-40*B*b*c^2*x^3+24*A*b*c^2*x^2-20*B*b^2*c*x^2-6*A*b^2*c*x+5*B*b^3*x+3*A*b^3)/x/b^4/
(c*x^2+b*x)^(3/2)

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maxima [A]  time = 0.92, size = 142, normalized size = 1.53 \begin {gather*} \frac {16 \, B c^{2} x}{3 \, \sqrt {c x^{2} + b x} b^{3}} - \frac {32 \, A c^{3} x}{5 \, \sqrt {c x^{2} + b x} b^{4}} + \frac {8 \, B c}{3 \, \sqrt {c x^{2} + b x} b^{2}} - \frac {16 \, A c^{2}}{5 \, \sqrt {c x^{2} + b x} b^{3}} - \frac {2 \, B}{3 \, \sqrt {c x^{2} + b x} b x} + \frac {4 \, A c}{5 \, \sqrt {c x^{2} + b x} b^{2} x} - \frac {2 \, A}{5 \, \sqrt {c x^{2} + b x} b x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

16/3*B*c^2*x/(sqrt(c*x^2 + b*x)*b^3) - 32/5*A*c^3*x/(sqrt(c*x^2 + b*x)*b^4) + 8/3*B*c/(sqrt(c*x^2 + b*x)*b^2)
- 16/5*A*c^2/(sqrt(c*x^2 + b*x)*b^3) - 2/3*B/(sqrt(c*x^2 + b*x)*b*x) + 4/5*A*c/(sqrt(c*x^2 + b*x)*b^2*x) - 2/5
*A/(sqrt(c*x^2 + b*x)*b*x^2)

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mupad [B]  time = 1.23, size = 87, normalized size = 0.94 \begin {gather*} -\frac {2\,\sqrt {c\,x^2+b\,x}\,\left (5\,B\,b^3\,x+3\,A\,b^3-20\,B\,b^2\,c\,x^2-6\,A\,b^2\,c\,x-40\,B\,b\,c^2\,x^3+24\,A\,b\,c^2\,x^2+48\,A\,c^3\,x^3\right )}{15\,b^4\,x^3\,\left (b+c\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^2*(b*x + c*x^2)^(3/2)),x)

[Out]

-(2*(b*x + c*x^2)^(1/2)*(3*A*b^3 + 48*A*c^3*x^3 + 5*B*b^3*x - 6*A*b^2*c*x + 24*A*b*c^2*x^2 - 20*B*b^2*c*x^2 -
40*B*b*c^2*x^3))/(15*b^4*x^3*(b + c*x))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A + B x}{x^{2} \left (x \left (b + c x\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x**2/(c*x**2+b*x)**(3/2),x)

[Out]

Integral((A + B*x)/(x**2*(x*(b + c*x))**(3/2)), x)

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