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

Optimal. Leaf size=38 \[ x (a B+A b)+a A \log (x)+\frac{1}{2} x^2 (A c+b B)+\frac{1}{3} B c x^3 \]

[Out]

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

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Rubi [A]  time = 0.0198945, antiderivative size = 38, 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} \[ x (a B+A b)+a A \log (x)+\frac{1}{2} x^2 (A c+b B)+\frac{1}{3} B c x^3 \]

Antiderivative was successfully verified.

[In]

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

[Out]

(A*b + a*B)*x + ((b*B + A*c)*x^2)/2 + (B*c*x^3)/3 + a*A*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} \, dx &=\int \left (A b+a B+\frac{a A}{x}+(b B+A c) x+B c x^2\right ) \, dx\\ &=(A b+a B) x+\frac{1}{2} (b B+A c) x^2+\frac{1}{3} B c x^3+a A \log (x)\\ \end{align*}

Mathematica [A]  time = 0.0117722, size = 38, normalized size = 1. \[ x (a B+A b)+a A \log (x)+\frac{1}{2} x^2 (A c+b B)+\frac{1}{3} B c x^3 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.003, size = 36, normalized size = 1. \begin{align*}{\frac{Bc{x}^{3}}{3}}+{\frac{Ac{x}^{2}}{2}}+{\frac{Bb{x}^{2}}{2}}+Abx+aBx+aA\ln \left ( x \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.08282, size = 46, normalized size = 1.21 \begin{align*} \frac{1}{3} \, B c x^{3} + \frac{1}{2} \,{\left (B b + A c\right )} x^{2} + A a \log \left (x\right ) +{\left (B a + A b\right )} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.2809, size = 86, normalized size = 2.26 \begin{align*} \frac{1}{3} \, B c x^{3} + \frac{1}{2} \,{\left (B b + A c\right )} x^{2} + A a \log \left (x\right ) +{\left (B a + A b\right )} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.522803, size = 36, normalized size = 0.95 \begin{align*} A a \log{\left (x \right )} + \frac{B c x^{3}}{3} + x^{2} \left (\frac{A c}{2} + \frac{B b}{2}\right ) + x \left (A b + B a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.16978, size = 49, normalized size = 1.29 \begin{align*} \frac{1}{3} \, B c x^{3} + \frac{1}{2} \, B b x^{2} + \frac{1}{2} \, A c x^{2} + B a x + A b x + A a \log \left ({\left | x \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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