3.647 \(\int \frac{x^2 (A+B x)}{(a^2+2 a b x+b^2 x^2)^3} \, dx\)

Optimal. Leaf size=87 \[ -\frac{a^2 (A b-a B)}{5 b^4 (a+b x)^5}+\frac{a (2 A b-3 a B)}{4 b^4 (a+b x)^4}-\frac{A b-3 a B}{3 b^4 (a+b x)^3}-\frac{B}{2 b^4 (a+b x)^2} \]

[Out]

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

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Rubi [A]  time = 0.0656881, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.074, Rules used = {27, 77} \[ -\frac{a^2 (A b-a B)}{5 b^4 (a+b x)^5}+\frac{a (2 A b-3 a B)}{4 b^4 (a+b x)^4}-\frac{A b-3 a B}{3 b^4 (a+b x)^3}-\frac{B}{2 b^4 (a+b x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int \frac{x^2 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx &=\int \frac{x^2 (A+B x)}{(a+b x)^6} \, dx\\ &=\int \left (-\frac{a^2 (-A b+a B)}{b^3 (a+b x)^6}+\frac{a (-2 A b+3 a B)}{b^3 (a+b x)^5}+\frac{A b-3 a B}{b^3 (a+b x)^4}+\frac{B}{b^3 (a+b x)^3}\right ) \, dx\\ &=-\frac{a^2 (A b-a B)}{5 b^4 (a+b x)^5}+\frac{a (2 A b-3 a B)}{4 b^4 (a+b x)^4}-\frac{A b-3 a B}{3 b^4 (a+b x)^3}-\frac{B}{2 b^4 (a+b x)^2}\\ \end{align*}

Mathematica [A]  time = 0.0236723, size = 63, normalized size = 0.72 \[ -\frac{a^2 b (2 A+15 B x)+3 a^3 B+10 a b^2 x (A+3 B x)+10 b^3 x^2 (2 A+3 B x)}{60 b^4 (a+b x)^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(3*a^3*B + 10*a*b^2*x*(A + 3*B*x) + 10*b^3*x^2*(2*A + 3*B*x) + a^2*b*(2*A + 15*B*x))/(60*b^4*(a + b*x)^5)

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Maple [A]  time = 0.006, size = 80, normalized size = 0.9 \begin{align*}{\frac{a \left ( 2\,Ab-3\,aB \right ) }{4\,{b}^{4} \left ( bx+a \right ) ^{4}}}-{\frac{{a}^{2} \left ( Ab-aB \right ) }{5\,{b}^{4} \left ( bx+a \right ) ^{5}}}-{\frac{B}{2\,{b}^{4} \left ( bx+a \right ) ^{2}}}-{\frac{Ab-3\,aB}{3\,{b}^{4} \left ( bx+a \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a*(2*A*b-3*B*a)/b^4/(b*x+a)^4-1/5*a^2*(A*b-B*a)/b^4/(b*x+a)^5-1/2*B/b^4/(b*x+a)^2-1/3*(A*b-3*B*a)/b^4/(b*x
+a)^3

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Maxima [A]  time = 1.06455, size = 161, normalized size = 1.85 \begin{align*} -\frac{30 \, B b^{3} x^{3} + 3 \, B a^{3} + 2 \, A a^{2} b + 10 \,{\left (3 \, B a b^{2} + 2 \, A b^{3}\right )} x^{2} + 5 \,{\left (3 \, B a^{2} b + 2 \, A a b^{2}\right )} x}{60 \,{\left (b^{9} x^{5} + 5 \, a b^{8} x^{4} + 10 \, a^{2} b^{7} x^{3} + 10 \, a^{3} b^{6} x^{2} + 5 \, a^{4} b^{5} x + a^{5} b^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(30*B*b^3*x^3 + 3*B*a^3 + 2*A*a^2*b + 10*(3*B*a*b^2 + 2*A*b^3)*x^2 + 5*(3*B*a^2*b + 2*A*a*b^2)*x)/(b^9*x
^5 + 5*a*b^8*x^4 + 10*a^2*b^7*x^3 + 10*a^3*b^6*x^2 + 5*a^4*b^5*x + a^5*b^4)

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Fricas [A]  time = 1.25498, size = 252, normalized size = 2.9 \begin{align*} -\frac{30 \, B b^{3} x^{3} + 3 \, B a^{3} + 2 \, A a^{2} b + 10 \,{\left (3 \, B a b^{2} + 2 \, A b^{3}\right )} x^{2} + 5 \,{\left (3 \, B a^{2} b + 2 \, A a b^{2}\right )} x}{60 \,{\left (b^{9} x^{5} + 5 \, a b^{8} x^{4} + 10 \, a^{2} b^{7} x^{3} + 10 \, a^{3} b^{6} x^{2} + 5 \, a^{4} b^{5} x + a^{5} b^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(30*B*b^3*x^3 + 3*B*a^3 + 2*A*a^2*b + 10*(3*B*a*b^2 + 2*A*b^3)*x^2 + 5*(3*B*a^2*b + 2*A*a*b^2)*x)/(b^9*x
^5 + 5*a*b^8*x^4 + 10*a^2*b^7*x^3 + 10*a^3*b^6*x^2 + 5*a^4*b^5*x + a^5*b^4)

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Sympy [A]  time = 1.34, size = 124, normalized size = 1.43 \begin{align*} - \frac{2 A a^{2} b + 3 B a^{3} + 30 B b^{3} x^{3} + x^{2} \left (20 A b^{3} + 30 B a b^{2}\right ) + x \left (10 A a b^{2} + 15 B a^{2} b\right )}{60 a^{5} b^{4} + 300 a^{4} b^{5} x + 600 a^{3} b^{6} x^{2} + 600 a^{2} b^{7} x^{3} + 300 a b^{8} x^{4} + 60 b^{9} x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*A*a**2*b + 3*B*a**3 + 30*B*b**3*x**3 + x**2*(20*A*b**3 + 30*B*a*b**2) + x*(10*A*a*b**2 + 15*B*a**2*b))/(60
*a**5*b**4 + 300*a**4*b**5*x + 600*a**3*b**6*x**2 + 600*a**2*b**7*x**3 + 300*a*b**8*x**4 + 60*b**9*x**5)

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Giac [A]  time = 1.16206, size = 95, normalized size = 1.09 \begin{align*} -\frac{30 \, B b^{3} x^{3} + 30 \, B a b^{2} x^{2} + 20 \, A b^{3} x^{2} + 15 \, B a^{2} b x + 10 \, A a b^{2} x + 3 \, B a^{3} + 2 \, A a^{2} b}{60 \,{\left (b x + a\right )}^{5} b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(30*B*b^3*x^3 + 30*B*a*b^2*x^2 + 20*A*b^3*x^2 + 15*B*a^2*b*x + 10*A*a*b^2*x + 3*B*a^3 + 2*A*a^2*b)/((b*x
+ a)^5*b^4)