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

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

[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])

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Rubi [A]  time = 0.0837657, antiderivative size = 93, 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, 613} \[ \frac{8 c (b+2 c x) (5 b B-6 A c)}{15 b^4 \sqrt{b x+c x^2}}-\frac{2 (5 b B-6 A c)}{15 b^2 x \sqrt{b x+c x^2}}-\frac{2 A}{5 b x^2 \sqrt{b x+c x^2}} \]

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 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 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]

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*}

Mathematica [A]  time = 0.0259445, size = 75, normalized size = 0.81 \[ -\frac{2 \left (3 A \left (-2 b^2 c x+b^3+8 b c^2 x^2+16 c^3 x^3\right )+5 b B x \left (b^2-4 b c x-8 c^2 x^2\right )\right )}{15 b^4 x^2 \sqrt{x (b+c x)}} \]

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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Maple [A]  time = 0.005, size = 86, normalized size = 0.9 \begin{align*} -{\frac{ \left ( 2\,cx+2\,b \right ) \left ( 48\,A{x}^{3}{c}^{3}-40\,B{x}^{3}b{c}^{2}+24\,A{x}^{2}b{c}^{2}-20\,B{x}^{2}{b}^{2}c-6\,A{b}^{2}cx+5\,{b}^{3}Bx+3\,A{b}^{3} \right ) }{15\,x{b}^{4}} \left ( c{x}^{2}+bx \right ) ^{-{\frac{3}{2}}}} \end{align*}

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.80933, size = 196, normalized size = 2.11 \begin{align*} -\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{align*}

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

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

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)