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

Optimal. Leaf size=45 \[ \frac{a (A b-a B)}{b^3 (a+b x)}+\frac{(A b-2 a B) \log (a+b x)}{b^3}+\frac{B x}{b^2} \]

[Out]

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

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Rubi [A]  time = 0.0362801, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.08, Rules used = {27, 77} \[ \frac{a (A b-a B)}{b^3 (a+b x)}+\frac{(A b-2 a B) \log (a+b x)}{b^3}+\frac{B x}{b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*x)/b^2 + (a*(A*b - a*B))/(b^3*(a + b*x)) + ((A*b - 2*a*B)*Log[a + b*x])/b^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)}{a^2+2 a b x+b^2 x^2} \, dx &=\int \frac{x (A+B x)}{(a+b x)^2} \, dx\\ &=\int \left (\frac{B}{b^2}+\frac{a (-A b+a B)}{b^2 (a+b x)^2}+\frac{A b-2 a B}{b^2 (a+b x)}\right ) \, dx\\ &=\frac{B x}{b^2}+\frac{a (A b-a B)}{b^3 (a+b x)}+\frac{(A b-2 a B) \log (a+b x)}{b^3}\\ \end{align*}

Mathematica [A]  time = 0.0241257, size = 41, normalized size = 0.91 \[ \frac{\frac{a (A b-a B)}{a+b x}+(A b-2 a B) \log (a+b x)+b B x}{b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.005, size = 61, normalized size = 1.4 \begin{align*}{\frac{Bx}{{b}^{2}}}+{\frac{aA}{{b}^{2} \left ( bx+a \right ) }}-{\frac{B{a}^{2}}{{b}^{3} \left ( bx+a \right ) }}+{\frac{\ln \left ( bx+a \right ) A}{{b}^{2}}}-2\,{\frac{\ln \left ( bx+a \right ) aB}{{b}^{3}}} \end{align*}

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),x)

[Out]

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

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Maxima [A]  time = 1.00007, size = 72, normalized size = 1.6 \begin{align*} -\frac{B a^{2} - A a b}{b^{4} x + a b^{3}} + \frac{B x}{b^{2}} - \frac{{\left (2 \, B a - A b\right )} \log \left (b x + a\right )}{b^{3}} \end{align*}

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),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.2461, size = 149, normalized size = 3.31 \begin{align*} \frac{B b^{2} x^{2} + B a b x - B a^{2} + A a b -{\left (2 \, B a^{2} - A a b +{\left (2 \, B a b - A b^{2}\right )} x\right )} \log \left (b x + a\right )}{b^{4} x + a b^{3}} \end{align*}

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),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.464968, size = 44, normalized size = 0.98 \begin{align*} \frac{B x}{b^{2}} - \frac{- A a b + B a^{2}}{a b^{3} + b^{4} x} - \frac{\left (- A b + 2 B a\right ) \log{\left (a + b x \right )}}{b^{3}} \end{align*}

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),x)

[Out]

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

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Giac [A]  time = 1.17415, size = 69, normalized size = 1.53 \begin{align*} \frac{B x}{b^{2}} - \frac{{\left (2 \, B a - A b\right )} \log \left ({\left | b x + a \right |}\right )}{b^{3}} - \frac{B a^{2} - A a b}{{\left (b x + a\right )} b^{3}} \end{align*}

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),x, algorithm="giac")

[Out]

B*x/b^2 - (2*B*a - A*b)*log(abs(b*x + a))/b^3 - (B*a^2 - A*a*b)/((b*x + a)*b^3)