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

Optimal. Leaf size=57 \[ \frac{x^4 (4 a B+A b)}{20 a^2 b (a+b x)^4}+\frac{x^4 (A b-a B)}{5 a b (a+b x)^5} \]

[Out]

((A*b - a*B)*x^4)/(5*a*b*(a + b*x)^5) + ((A*b + 4*a*B)*x^4)/(20*a^2*b*(a + b*x)^4)

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Rubi [A]  time = 0.0144524, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {27, 78, 37} \[ \frac{x^4 (4 a B+A b)}{20 a^2 b (a+b x)^4}+\frac{x^4 (A b-a B)}{5 a b (a+b x)^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A*b - a*B)*x^4)/(5*a*b*(a + b*x)^5) + ((A*b + 4*a*B)*x^4)/(20*a^2*b*(a + b*x)^4)

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 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

\begin{align*} \int \frac{x^3 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx &=\int \frac{x^3 (A+B x)}{(a+b x)^6} \, dx\\ &=\frac{(A b-a B) x^4}{5 a b (a+b x)^5}+\frac{(A b+4 a B) \int \frac{x^3}{(a+b x)^5} \, dx}{5 a b}\\ &=\frac{(A b-a B) x^4}{5 a b (a+b x)^5}+\frac{(A b+4 a B) x^4}{20 a^2 b (a+b x)^4}\\ \end{align*}

Mathematica [A]  time = 0.0312112, size = 76, normalized size = 1.33 \[ -\frac{5 a^2 b^2 x (A+8 B x)+a^3 b (A+20 B x)+4 a^4 B+10 a b^3 x^2 (A+4 B x)+10 b^4 x^3 (A+2 B x)}{20 b^5 (a+b x)^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(4*a^4*B + 10*b^4*x^3*(A + 2*B*x) + 10*a*b^3*x^2*(A + 4*B*x) + 5*a^2*b^2*x*(A + 8*B*x) + a^3*b*(A + 20*B*x))/
(20*b^5*(a + b*x)^5)

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Maple [A]  time = 0.004, size = 102, normalized size = 1.8 \begin{align*} -{\frac{B}{{b}^{5} \left ( bx+a \right ) }}-{\frac{Ab-4\,aB}{2\,{b}^{5} \left ( bx+a \right ) ^{2}}}+{\frac{{a}^{3} \left ( Ab-aB \right ) }{5\,{b}^{5} \left ( bx+a \right ) ^{5}}}-{\frac{{a}^{2} \left ( 3\,Ab-4\,aB \right ) }{4\,{b}^{5} \left ( bx+a \right ) ^{4}}}+{\frac{a \left ( Ab-2\,aB \right ) }{{b}^{5} \left ( bx+a \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.02741, size = 188, normalized size = 3.3 \begin{align*} -\frac{20 \, B b^{4} x^{4} + 4 \, B a^{4} + A a^{3} b + 10 \,{\left (4 \, B a b^{3} + A b^{4}\right )} x^{3} + 10 \,{\left (4 \, B a^{2} b^{2} + A a b^{3}\right )} x^{2} + 5 \,{\left (4 \, B a^{3} b + A a^{2} b^{2}\right )} x}{20 \,{\left (b^{10} x^{5} + 5 \, a b^{9} x^{4} + 10 \, a^{2} b^{8} x^{3} + 10 \, a^{3} b^{7} x^{2} + 5 \, a^{4} b^{6} x + a^{5} b^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/20*(20*B*b^4*x^4 + 4*B*a^4 + A*a^3*b + 10*(4*B*a*b^3 + A*b^4)*x^3 + 10*(4*B*a^2*b^2 + A*a*b^3)*x^2 + 5*(4*B
*a^3*b + A*a^2*b^2)*x)/(b^10*x^5 + 5*a*b^9*x^4 + 10*a^2*b^8*x^3 + 10*a^3*b^7*x^2 + 5*a^4*b^6*x + a^5*b^5)

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Fricas [B]  time = 1.22795, size = 293, normalized size = 5.14 \begin{align*} -\frac{20 \, B b^{4} x^{4} + 4 \, B a^{4} + A a^{3} b + 10 \,{\left (4 \, B a b^{3} + A b^{4}\right )} x^{3} + 10 \,{\left (4 \, B a^{2} b^{2} + A a b^{3}\right )} x^{2} + 5 \,{\left (4 \, B a^{3} b + A a^{2} b^{2}\right )} x}{20 \,{\left (b^{10} x^{5} + 5 \, a b^{9} x^{4} + 10 \, a^{2} b^{8} x^{3} + 10 \, a^{3} b^{7} x^{2} + 5 \, a^{4} b^{6} x + a^{5} b^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/20*(20*B*b^4*x^4 + 4*B*a^4 + A*a^3*b + 10*(4*B*a*b^3 + A*b^4)*x^3 + 10*(4*B*a^2*b^2 + A*a*b^3)*x^2 + 5*(4*B
*a^3*b + A*a^2*b^2)*x)/(b^10*x^5 + 5*a*b^9*x^4 + 10*a^2*b^8*x^3 + 10*a^3*b^7*x^2 + 5*a^4*b^6*x + a^5*b^5)

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Sympy [B]  time = 1.96291, size = 146, normalized size = 2.56 \begin{align*} - \frac{A a^{3} b + 4 B a^{4} + 20 B b^{4} x^{4} + x^{3} \left (10 A b^{4} + 40 B a b^{3}\right ) + x^{2} \left (10 A a b^{3} + 40 B a^{2} b^{2}\right ) + x \left (5 A a^{2} b^{2} + 20 B a^{3} b\right )}{20 a^{5} b^{5} + 100 a^{4} b^{6} x + 200 a^{3} b^{7} x^{2} + 200 a^{2} b^{8} x^{3} + 100 a b^{9} x^{4} + 20 b^{10} x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(A*a**3*b + 4*B*a**4 + 20*B*b**4*x**4 + x**3*(10*A*b**4 + 40*B*a*b**3) + x**2*(10*A*a*b**3 + 40*B*a**2*b**2)
+ x*(5*A*a**2*b**2 + 20*B*a**3*b))/(20*a**5*b**5 + 100*a**4*b**6*x + 200*a**3*b**7*x**2 + 200*a**2*b**8*x**3 +
100*a*b**9*x**4 + 20*b**10*x**5)

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Giac [A]  time = 1.205, size = 126, normalized size = 2.21 \begin{align*} -\frac{20 \, B b^{4} x^{4} + 40 \, B a b^{3} x^{3} + 10 \, A b^{4} x^{3} + 40 \, B a^{2} b^{2} x^{2} + 10 \, A a b^{3} x^{2} + 20 \, B a^{3} b x + 5 \, A a^{2} b^{2} x + 4 \, B a^{4} + A a^{3} b}{20 \,{\left (b x + a\right )}^{5} b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/20*(20*B*b^4*x^4 + 40*B*a*b^3*x^3 + 10*A*b^4*x^3 + 40*B*a^2*b^2*x^2 + 10*A*a*b^3*x^2 + 20*B*a^3*b*x + 5*A*a
^2*b^2*x + 4*B*a^4 + A*a^3*b)/((b*x + a)^5*b^5)