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

Optimal. Leaf size=36 \[ -\frac{a B+A b}{x}-\frac{a A}{2 x^2}+\log (x) (A c+b B)+B c x \]

[Out]

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

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Rubi [A]  time = 0.0237494, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.053, Rules used = {765} \[ -\frac{a B+A b}{x}-\frac{a A}{2 x^2}+\log (x) (A c+b B)+B c x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 765

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand
Integrand[(e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, e, f, g, m}, x] && IntegerQ[p] && (
GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

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

Mathematica [A]  time = 0.0197833, size = 37, normalized size = 1.03 \[ \frac{-a B-A b}{x}-\frac{a A}{2 x^2}+\log (x) (A c+b B)+B c x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.005, size = 37, normalized size = 1. \begin{align*} Bcx+Ac\ln \left ( x \right ) +B\ln \left ( x \right ) b-{\frac{aA}{2\,{x}^{2}}}-{\frac{Ab}{x}}-{\frac{aB}{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.0563, size = 46, normalized size = 1.28 \begin{align*} B c x +{\left (B b + A c\right )} \log \left (x\right ) - \frac{A a + 2 \,{\left (B a + A b\right )} x}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.27495, size = 97, normalized size = 2.69 \begin{align*} \frac{2 \, B c x^{3} + 2 \,{\left (B b + A c\right )} x^{2} \log \left (x\right ) - A a - 2 \,{\left (B a + A b\right )} x}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.784262, size = 34, normalized size = 0.94 \begin{align*} B c x + \left (A c + B b\right ) \log{\left (x \right )} - \frac{A a + x \left (2 A b + 2 B a\right )}{2 x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.17604, size = 47, normalized size = 1.31 \begin{align*} B c x +{\left (B b + A c\right )} \log \left ({\left | x \right |}\right ) - \frac{A a + 2 \,{\left (B a + A b\right )} x}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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