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

Optimal. Leaf size=61 \[ -\frac {A b-2 a B}{4 b^3 (a+b x)^4}+\frac {a (A b-a B)}{5 b^3 (a+b x)^5}-\frac {B}{3 b^3 (a+b x)^3} \]

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Rubi [A]  time = 0.04, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {27, 77} \begin {gather*} -\frac {A b-2 a B}{4 b^3 (a+b x)^4}+\frac {a (A b-a B)}{5 b^3 (a+b x)^5}-\frac {B}{3 b^3 (a+b x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.02, size = 46, normalized size = 0.75 \begin {gather*} -\frac {2 a^2 B+a b (3 A+10 B x)+5 b^2 x (3 A+4 B x)}{60 b^3 (a+b x)^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.40, size = 95, normalized size = 1.56 \begin {gather*} -\frac {20 \, B b^{2} x^{2} + 2 \, B a^{2} + 3 \, A a b + 5 \, {\left (2 \, B a b + 3 \, A b^{2}\right )} x}{60 \, {\left (b^{8} x^{5} + 5 \, a b^{7} x^{4} + 10 \, a^{2} b^{6} x^{3} + 10 \, a^{3} b^{5} x^{2} + 5 \, a^{4} b^{4} x + a^{5} b^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(20*B*b^2*x^2 + 2*B*a^2 + 3*A*a*b + 5*(2*B*a*b + 3*A*b^2)*x)/(b^8*x^5 + 5*a*b^7*x^4 + 10*a^2*b^6*x^3 + 1
0*a^3*b^5*x^2 + 5*a^4*b^4*x + a^5*b^3)

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giac [A]  time = 0.18, size = 46, normalized size = 0.75 \begin {gather*} -\frac {20 \, B b^{2} x^{2} + 10 \, B a b x + 15 \, A b^{2} x + 2 \, B a^{2} + 3 \, A a b}{60 \, {\left (b x + a\right )}^{5} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 56, normalized size = 0.92 \begin {gather*} -\frac {B}{3 \left (b x +a \right )^{3} b^{3}}+\frac {\left (A b -B a \right ) a}{5 \left (b x +a \right )^{5} b^{3}}-\frac {A b -2 B a}{4 \left (b x +a \right )^{4} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.63, size = 95, normalized size = 1.56 \begin {gather*} -\frac {20 \, B b^{2} x^{2} + 2 \, B a^{2} + 3 \, A a b + 5 \, {\left (2 \, B a b + 3 \, A b^{2}\right )} x}{60 \, {\left (b^{8} x^{5} + 5 \, a b^{7} x^{4} + 10 \, a^{2} b^{6} x^{3} + 10 \, a^{3} b^{5} x^{2} + 5 \, a^{4} b^{4} x + a^{5} b^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(20*B*b^2*x^2 + 2*B*a^2 + 3*A*a*b + 5*(2*B*a*b + 3*A*b^2)*x)/(b^8*x^5 + 5*a*b^7*x^4 + 10*a^2*b^6*x^3 + 1
0*a^3*b^5*x^2 + 5*a^4*b^4*x + a^5*b^3)

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mupad [B]  time = 1.10, size = 93, normalized size = 1.52 \begin {gather*} -\frac {\frac {B\,x^2}{3\,b}+\frac {a\,\left (3\,A\,b+2\,B\,a\right )}{60\,b^3}+\frac {x\,\left (3\,A\,b+2\,B\,a\right )}{12\,b^2}}{a^5+5\,a^4\,b\,x+10\,a^3\,b^2\,x^2+10\,a^2\,b^3\,x^3+5\,a\,b^4\,x^4+b^5\,x^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((B*x^2)/(3*b) + (a*(3*A*b + 2*B*a))/(60*b^3) + (x*(3*A*b + 2*B*a))/(12*b^2))/(a^5 + b^5*x^5 + 5*a*b^4*x^4 +
10*a^3*b^2*x^2 + 10*a^2*b^3*x^3 + 5*a^4*b*x)

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sympy [A]  time = 0.68, size = 100, normalized size = 1.64 \begin {gather*} \frac {- 3 A a b - 2 B a^{2} - 20 B b^{2} x^{2} + x \left (- 15 A b^{2} - 10 B a b\right )}{60 a^{5} b^{3} + 300 a^{4} b^{4} x + 600 a^{3} b^{5} x^{2} + 600 a^{2} b^{6} x^{3} + 300 a b^{7} x^{4} + 60 b^{8} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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