3.177 \(\int \frac{x (A+B x)}{a+b x} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0294363, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071, Rules used = {77} \[ \frac{x (A b-a B)}{b^2}-\frac{a (A b-a B) \log (a+b x)}{b^3}+\frac{B x^2}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[(x*(A + B*x))/(a + b*x),x]

[Out]

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

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

Mathematica [A]  time = 0.0142685, size = 41, normalized size = 0.91 \[ \frac{b x (-2 a B+2 A b+b B x)+2 a (a B-A b) \log (a+b x)}{2 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*(A + B*x))/(a + b*x),x]

[Out]

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

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Maple [A]  time = 0.001, size = 52, normalized size = 1.2 \begin{align*}{\frac{B{x}^{2}}{2\,b}}+{\frac{Ax}{b}}-{\frac{Bax}{{b}^{2}}}-{\frac{a\ln \left ( bx+a \right ) A}{{b}^{2}}}+{\frac{{a}^{2}\ln \left ( bx+a \right ) B}{{b}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.00934, size = 61, normalized size = 1.36 \begin{align*} \frac{B b x^{2} - 2 \,{\left (B a - A b\right )} x}{2 \, b^{2}} + \frac{{\left (B a^{2} - 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*x+a),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.63176, size = 103, normalized size = 2.29 \begin{align*} \frac{B b^{2} x^{2} - 2 \,{\left (B a b - A b^{2}\right )} x + 2 \,{\left (B a^{2} - A a b\right )} \log \left (b x + a\right )}{2 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.531168, size = 37, normalized size = 0.82 \begin{align*} \frac{B x^{2}}{2 b} + \frac{a \left (- A b + B a\right ) \log{\left (a + b x \right )}}{b^{3}} - \frac{x \left (- A b + B a\right )}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.25031, size = 61, normalized size = 1.36 \begin{align*} \frac{B b x^{2} - 2 \, B a x + 2 \, A b x}{2 \, b^{2}} + \frac{{\left (B a^{2} - A a b\right )} \log \left ({\left | b x + a \right |}\right )}{b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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