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

Optimal. Leaf size=94 \[ \frac{8 (2 a+b x) (A b-2 a B)}{3 \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2}}-\frac{2 x^2 (-2 a B-x (b B-2 A c)+A b)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0375209, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {804, 636} \[ \frac{8 (2 a+b x) (A b-2 a B)}{3 \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2}}-\frac{2 x^2 (-2 a B-x (b B-2 A c)+A b)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 804

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*(b*f - 2*a*g + (2*c*f - b*g)*x))/((p + 1)*(b^2 - 4*a*c)), x] - Dist[(m
*(b*(e*f + d*g) - 2*(c*d*f + a*e*g)))/((p + 1)*(b^2 - 4*a*c)), 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] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[Sim
plify[m + 2*p + 3], 0] && LtQ[p, -1]

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

Mathematica [A]  time = 0.289217, size = 110, normalized size = 1.17 \[ \frac{2 \left (8 a^2 (A b-3 B x (b+c x))-16 a^3 B+2 a x \left (A \left (6 b^2+6 b c x+4 c^2 x^2\right )-3 b B x (b+2 c x)\right )+b^2 x^2 (3 A b+2 A c x+b B x)\right )}{3 \left (b^2-4 a c\right )^2 (a+x (b+c x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-16*a^3*B + b^2*x^2*(3*A*b + b*B*x + 2*A*c*x) + 8*a^2*(A*b - 3*B*x*(b + c*x)) + 2*a*x*(-3*b*B*x*(b + 2*c*x
) + A*(6*b^2 + 6*b*c*x + 4*c^2*x^2))))/(3*(b^2 - 4*a*c)^2*(a + x*(b + c*x))^(3/2))

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Maple [A]  time = 0.009, size = 141, normalized size = 1.5 \begin{align*}{\frac{16\,aA{c}^{2}{x}^{3}+4\,A{x}^{3}{b}^{2}c-24\,B{x}^{3}abc+2\,{b}^{3}B{x}^{3}+24\,A{x}^{2}abc+6\,A{b}^{3}{x}^{2}-48\,{a}^{2}Bc{x}^{2}-12\,B{x}^{2}a{b}^{2}+24\,Aa{b}^{2}x-48\,B{a}^{2}bx+16\,Ab{a}^{2}-32\,B{a}^{3}}{48\,{a}^{2}{c}^{2}-24\,a{b}^{2}c+3\,{b}^{4}} \left ( c{x}^{2}+bx+a \right ) ^{-{\frac{3}{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+a)^(5/2),x)

[Out]

2/3/(c*x^2+b*x+a)^(3/2)*(8*A*a*c^2*x^3+2*A*b^2*c*x^3-12*B*a*b*c*x^3+B*b^3*x^3+12*A*a*b*c*x^2+3*A*b^3*x^2-24*B*
a^2*c*x^2-6*B*a*b^2*x^2+12*A*a*b^2*x-24*B*a^2*b*x+8*A*a^2*b-16*B*a^3)/(16*a^2*c^2-8*a*b^2*c+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(x^2*(B*x+A)/(c*x^2+b*x+a)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 7.08546, size = 522, normalized size = 5.55 \begin{align*} -\frac{2 \,{\left (16 \, B a^{3} - 8 \, A a^{2} b -{\left (B b^{3} + 8 \, A a c^{2} - 2 \,{\left (6 \, B a b - A b^{2}\right )} c\right )} x^{3} + 3 \,{\left (2 \, B a b^{2} - A b^{3} + 4 \,{\left (2 \, B a^{2} - A a b\right )} c\right )} x^{2} + 12 \,{\left (2 \, B a^{2} b - A a b^{2}\right )} x\right )} \sqrt{c x^{2} + b x + a}}{3 \,{\left (a^{2} b^{4} - 8 \, a^{3} b^{2} c + 16 \, a^{4} c^{2} +{\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} x^{4} + 2 \,{\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} x^{3} +{\left (b^{6} - 6 \, a b^{4} c + 32 \, a^{3} c^{3}\right )} x^{2} + 2 \,{\left (a b^{5} - 8 \, a^{2} b^{3} c + 16 \, a^{3} b c^{2}\right )} x\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+a)^(5/2),x, algorithm="fricas")

[Out]

-2/3*(16*B*a^3 - 8*A*a^2*b - (B*b^3 + 8*A*a*c^2 - 2*(6*B*a*b - A*b^2)*c)*x^3 + 3*(2*B*a*b^2 - A*b^3 + 4*(2*B*a
^2 - A*a*b)*c)*x^2 + 12*(2*B*a^2*b - A*a*b^2)*x)*sqrt(c*x^2 + b*x + a)/(a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2 + (
b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*x^4 + 2*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x^3 + (b^6 - 6*a*b^4*c + 32*a
^3*c^3)*x^2 + 2*(a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.26811, size = 296, normalized size = 3.15 \begin{align*} \frac{{\left ({\left (\frac{{\left (B b^{3} - 12 \, B a b c + 2 \, A b^{2} c + 8 \, A a c^{2}\right )} x}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}} - \frac{3 \,{\left (2 \, B a b^{2} - A b^{3} + 8 \, B a^{2} c - 4 \, A a b c\right )}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )} x - \frac{12 \,{\left (2 \, B a^{2} b - A a b^{2}\right )}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )} x - \frac{8 \,{\left (2 \, B a^{3} - A a^{2} b\right )}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}}{3 \,{\left (c x^{2} + b x + a\right )}^{\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+a)^(5/2),x, algorithm="giac")

[Out]

1/3*((((B*b^3 - 12*B*a*b*c + 2*A*b^2*c + 8*A*a*c^2)*x/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4) - 3*(2*B*a*b^2 - A*
b^3 + 8*B*a^2*c - 4*A*a*b*c)/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4))*x - 12*(2*B*a^2*b - A*a*b^2)/(b^4*c^2 - 8*a
*b^2*c^3 + 16*a^2*c^4))*x - 8*(2*B*a^3 - A*a^2*b)/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4))/(c*x^2 + b*x + a)^(3/2
)