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

Optimal. Leaf size=114 \[ \frac{2 (b+2 c x) \left (4 a B c-4 A b c+b^2 B\right )}{3 c \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2}}-\frac{2 \left (x \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)\right )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.04445, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {777, 613} \[ \frac{2 (b+2 c x) \left (4 a B c-4 A b c+b^2 B\right )}{3 c \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2}}-\frac{2 \left (x \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)\right )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 777

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((2
*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (b^2*e*g - b*c*(e*f + d*g) + 2*c*(c*d*f - a*e*g))*x)*(a + b*x + c*x^2)^
(p + 1))/(c*(p + 1)*(b^2 - 4*a*c)), x] - Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p
+ 3))/(c*(p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && N
eQ[b^2 - 4*a*c, 0] && LtQ[p, -1]

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{x (A+B x)}{\left (a+b x+c x^2\right )^{5/2}} \, dx &=-\frac{2 \left (a (b B-2 A c)+\left (b^2 B-A b c-2 a B c\right ) x\right )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac{\left (b^2 B-4 A b c+4 a B c\right ) \int \frac{1}{\left (a+b x+c x^2\right )^{3/2}} \, dx}{3 c \left (b^2-4 a c\right )}\\ &=-\frac{2 \left (a (b B-2 A c)+\left (b^2 B-A b c-2 a B c\right ) x\right )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}+\frac{2 \left (b^2 B-4 A b c+4 a B c\right ) (b+2 c x)}{3 c \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.181436, size = 114, normalized size = 1. \[ \frac{2 \left (8 a^2 (b B-A c)-2 a A b (b+6 c x)+4 a B x \left (3 b^2+3 b c x+2 c^2 x^2\right )+b x \left (b B x (3 b+2 c x)-A \left (3 b^2+12 b c x+8 c^2 x^2\right )\right )\right )}{3 \left (b^2-4 a c\right )^2 (a+x (b+c x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(8*a^2*(b*B - A*c) - 2*a*A*b*(b + 6*c*x) + 4*a*B*x*(3*b^2 + 3*b*c*x + 2*c^2*x^2) + b*x*(b*B*x*(3*b + 2*c*x)
- A*(3*b^2 + 12*b*c*x + 8*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 = 0.005, size = 138, normalized size = 1.2 \begin{align*} -{\frac{16\,A{x}^{3}b{c}^{2}-16\,aB{c}^{2}{x}^{3}-4\,B{x}^{3}{b}^{2}c+24\,A{x}^{2}{b}^{2}c-24\,B{x}^{2}abc-6\,{b}^{3}B{x}^{2}+24\,Aabcx+6\,A{b}^{3}x-24\,Ba{b}^{2}x+16\,A{a}^{2}c+4\,Aa{b}^{2}-16\,B{a}^{2}b}{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*(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*b*c^2*x^3-8*B*a*c^2*x^3-2*B*b^2*c*x^3+12*A*b^2*c*x^2-12*B*a*b*c*x^2-3*B*b^3*x^2+
12*A*a*b*c*x+3*A*b^3*x-12*B*a*b^2*x+8*A*a^2*c+2*A*a*b^2-8*B*a^2*b)/(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*(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 = 7.85058, size = 516, normalized size = 4.53 \begin{align*} \frac{2 \,{\left (8 \, B a^{2} b - 2 \, A a b^{2} - 8 \, A a^{2} c + 2 \,{\left (B b^{2} c + 4 \,{\left (B a - A b\right )} c^{2}\right )} x^{3} + 3 \,{\left (B b^{3} + 4 \,{\left (B a b - A b^{2}\right )} c\right )} x^{2} + 3 \,{\left (4 \, B a b^{2} - A b^{3} - 4 \, A a b c\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*(B*x+A)/(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")

[Out]

2/3*(8*B*a^2*b - 2*A*a*b^2 - 8*A*a^2*c + 2*(B*b^2*c + 4*(B*a - A*b)*c^2)*x^3 + 3*(B*b^3 + 4*(B*a*b - A*b^2)*c)
*x^2 + 3*(4*B*a*b^2 - A*b^3 - 4*A*a*b*c)*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*(B*x+A)/(c*x**2+b*x+a)**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 1.21078, size = 297, normalized size = 2.61 \begin{align*} \frac{{\left ({\left (\frac{2 \,{\left (B b^{2} c + 4 \, B a c^{2} - 4 \, A b c^{2}\right )} x}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}} + \frac{3 \,{\left (B b^{3} + 4 \, B a b c - 4 \, A b^{2} c\right )}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )} x + \frac{3 \,{\left (4 \, B a b^{2} - A b^{3} - 4 \, A a b c\right )}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )} x + \frac{2 \,{\left (4 \, B a^{2} b - A a b^{2} - 4 \, A a^{2} c\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*(B*x+A)/(c*x^2+b*x+a)^(5/2),x, algorithm="giac")

[Out]

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