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

Optimal. Leaf size=67 \[ \frac{2 x (2 A c+b B)}{3 b^2 c \sqrt{b x+c x^2}}-\frac{2 x^2 (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}} \]

[Out]

(-2*(b*B - A*c)*x^2)/(3*b*c*(b*x + c*x^2)^(3/2)) + (2*(b*B + 2*A*c)*x)/(3*b^2*c*Sqrt[b*x + c*x^2])

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Rubi [A]  time = 0.0542218, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {788, 636} \[ \frac{2 x (2 A c+b B)}{3 b^2 c \sqrt{b x+c x^2}}-\frac{2 x^2 (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*B - A*c)*x^2)/(3*b*c*(b*x + c*x^2)^(3/2)) + (2*(b*B + 2*A*c)*x)/(3*b^2*c*Sqrt[b*x + c*x^2])

Rule 788

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((g*(c*d - b*e) + c*e*f)*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(c*(p + 1)*(2*c*d - b*e)), x] - Dist[(e*(m*(g
*(c*d - b*e) + c*e*f) + e*(p + 1)*(2*c*f - b*g)))/(c*(p + 1)*(2*c*d - b*e)), Int[(d + e*x)^(m - 1)*(a + b*x +
c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2,
0] && LtQ[p, -1] && GtQ[m, 0]

Rule 636

Int[((d_.) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(-2*(b*d - 2*a*e + (2*c*
d - b*e)*x))/((b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2]), x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] &&
NeQ[b^2 - 4*a*c, 0]

Rubi steps

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

Mathematica [A]  time = 0.0155197, size = 35, normalized size = 0.52 \[ \frac{2 x^2 (3 A b+2 A c x+b B x)}{3 b^2 (x (b+c x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.006, size = 39, normalized size = 0.6 \begin{align*}{\frac{2\,{x}^{3} \left ( cx+b \right ) \left ( 2\,Acx+bBx+3\,Ab \right ) }{3\,{b}^{2}} \left ( c{x}^{2}+bx \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*x^3*(c*x+b)*(2*A*c*x+B*b*x+3*A*b)/b^2/(c*x^2+b*x)^(5/2)

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Maxima [B]  time = 1.02029, size = 181, normalized size = 2.7 \begin{align*} -\frac{B x^{2}}{{\left (c x^{2} + b x\right )}^{\frac{3}{2}} c} + \frac{4 \, A x}{3 \, \sqrt{c x^{2} + b x} b^{2}} - \frac{B b x}{3 \,{\left (c x^{2} + b x\right )}^{\frac{3}{2}} c^{2}} - \frac{2 \, A x}{3 \,{\left (c x^{2} + b x\right )}^{\frac{3}{2}} c} + \frac{2 \, B x}{3 \, \sqrt{c x^{2} + b x} b c} + \frac{B}{3 \, \sqrt{c x^{2} + b x} c^{2}} + \frac{2 \, A}{3 \, \sqrt{c x^{2} + b x} b c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-B*x^2/((c*x^2 + b*x)^(3/2)*c) + 4/3*A*x/(sqrt(c*x^2 + b*x)*b^2) - 1/3*B*b*x/((c*x^2 + b*x)^(3/2)*c^2) - 2/3*A
*x/((c*x^2 + b*x)^(3/2)*c) + 2/3*B*x/(sqrt(c*x^2 + b*x)*b*c) + 1/3*B/(sqrt(c*x^2 + b*x)*c^2) + 2/3*A/(sqrt(c*x
^2 + b*x)*b*c)

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Fricas [A]  time = 1.94466, size = 109, normalized size = 1.63 \begin{align*} \frac{2 \, \sqrt{c x^{2} + b x}{\left (3 \, A b +{\left (B b + 2 \, A c\right )} x\right )}}{3 \,{\left (b^{2} c^{2} x^{2} + 2 \, b^{3} c x + b^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*sqrt(c*x^2 + b*x)*(3*A*b + (B*b + 2*A*c)*x)/(b^2*c^2*x^2 + 2*b^3*c*x + b^4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2} \left (A + B x\right )}{\left (x \left (b + c x\right )\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.1674, size = 161, normalized size = 2.4 \begin{align*} \frac{2 \,{\left (3 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} B c + 3 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} B b \sqrt{c} + 3 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} A c^{\frac{3}{2}} + B b^{2} + 2 \, A b c\right )}}{3 \,{\left ({\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} + b\right )}^{3} c^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(3*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*c + 3*(sqrt(c)*x - sqrt(c*x^2 + b*x))*B*b*sqrt(c) + 3*(sqrt(c)*x -
sqrt(c*x^2 + b*x))*A*c^(3/2) + B*b^2 + 2*A*b*c)/(((sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) + b)^3*c^(3/2))