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

Optimal. Leaf size=90 \[ -\frac{a^2 A}{x}+\frac{1}{2} x^2 \left (2 a B c+2 A b c+b^2 B\right )+x \left (A \left (2 a c+b^2\right )+2 a b B\right )+a \log (x) (a B+2 A b)+\frac{1}{3} c x^3 (A c+2 b B)+\frac{1}{4} B c^2 x^4 \]

[Out]

-((a^2*A)/x) + (2*a*b*B + A*(b^2 + 2*a*c))*x + ((b^2*B + 2*A*b*c + 2*a*B*c)*x^2)/2 + (c*(2*b*B + A*c)*x^3)/3 +
(B*c^2*x^4)/4 + a*(2*A*b + a*B)*Log[x]

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Rubi [A]  time = 0.0715168, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.048, Rules used = {765} \[ -\frac{a^2 A}{x}+\frac{1}{2} x^2 \left (2 a B c+2 A b c+b^2 B\right )+x \left (A \left (2 a c+b^2\right )+2 a b B\right )+a \log (x) (a B+2 A b)+\frac{1}{3} c x^3 (A c+2 b B)+\frac{1}{4} B c^2 x^4 \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^2*A)/x) + (2*a*b*B + A*(b^2 + 2*a*c))*x + ((b^2*B + 2*A*b*c + 2*a*B*c)*x^2)/2 + (c*(2*b*B + A*c)*x^3)/3 +
(B*c^2*x^4)/4 + a*(2*A*b + a*B)*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 )^2}{x^2} \, dx &=\int \left (A b^2 \left (1+\frac{2 a (b B+A c)}{A b^2}\right )+\frac{a^2 A}{x^2}+\frac{a (2 A b+a B)}{x}+\left (b^2 B+2 A b c+2 a B c\right ) x+c (2 b B+A c) x^2+B c^2 x^3\right ) \, dx\\ &=-\frac{a^2 A}{x}+\left (2 a b B+A \left (b^2+2 a c\right )\right ) x+\frac{1}{2} \left (b^2 B+2 A b c+2 a B c\right ) x^2+\frac{1}{3} c (2 b B+A c) x^3+\frac{1}{4} B c^2 x^4+a (2 A b+a B) \log (x)\\ \end{align*}

Mathematica [A]  time = 0.0595797, size = 87, normalized size = 0.97 \[ -\frac{a^2 A}{x}+a x (2 A c+2 b B+B c x)+a \log (x) (a B+2 A b)+\frac{1}{12} x \left (4 A \left (3 b^2+3 b c x+c^2 x^2\right )+B x \left (6 b^2+8 b c x+3 c^2 x^2\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^2*A)/x) + a*x*(2*b*B + 2*A*c + B*c*x) + (x*(4*A*(3*b^2 + 3*b*c*x + c^2*x^2) + B*x*(6*b^2 + 8*b*c*x + 3*c^
2*x^2)))/12 + a*(2*A*b + a*B)*Log[x]

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Maple [A]  time = 0.005, size = 92, normalized size = 1. \begin{align*}{\frac{B{c}^{2}{x}^{4}}{4}}+{\frac{A{c}^{2}{x}^{3}}{3}}+{\frac{2\,B{x}^{3}bc}{3}}+A{x}^{2}bc+aBc{x}^{2}+{\frac{{b}^{2}B{x}^{2}}{2}}+2\,aAcx+A{b}^{2}x+2\,abBx+2\,A\ln \left ( x \right ) ab+{a}^{2}B\ln \left ( x \right ) -{\frac{A{a}^{2}}{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.16131, size = 119, normalized size = 1.32 \begin{align*} \frac{1}{4} \, B c^{2} x^{4} + \frac{1}{3} \,{\left (2 \, B b c + A c^{2}\right )} x^{3} + \frac{1}{2} \,{\left (B b^{2} + 2 \,{\left (B a + A b\right )} c\right )} x^{2} - \frac{A a^{2}}{x} +{\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} x +{\left (B a^{2} + 2 \, A a b\right )} \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.30279, size = 219, normalized size = 2.43 \begin{align*} \frac{3 \, B c^{2} x^{5} + 4 \,{\left (2 \, B b c + A c^{2}\right )} x^{4} + 6 \,{\left (B b^{2} + 2 \,{\left (B a + A b\right )} c\right )} x^{3} - 12 \, A a^{2} + 12 \,{\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} x^{2} + 12 \,{\left (B a^{2} + 2 \, A a b\right )} x \log \left (x\right )}{12 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*B*c^2*x^5 + 4*(2*B*b*c + A*c^2)*x^4 + 6*(B*b^2 + 2*(B*a + A*b)*c)*x^3 - 12*A*a^2 + 12*(2*B*a*b + A*b^2
+ 2*A*a*c)*x^2 + 12*(B*a^2 + 2*A*a*b)*x*log(x))/x

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Sympy [A]  time = 0.39976, size = 88, normalized size = 0.98 \begin{align*} - \frac{A a^{2}}{x} + \frac{B c^{2} x^{4}}{4} + a \left (2 A b + B a\right ) \log{\left (x \right )} + x^{3} \left (\frac{A c^{2}}{3} + \frac{2 B b c}{3}\right ) + x^{2} \left (A b c + B a c + \frac{B b^{2}}{2}\right ) + x \left (2 A a c + A b^{2} + 2 B a b\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.24888, size = 124, normalized size = 1.38 \begin{align*} \frac{1}{4} \, B c^{2} x^{4} + \frac{2}{3} \, B b c x^{3} + \frac{1}{3} \, A c^{2} x^{3} + \frac{1}{2} \, B b^{2} x^{2} + B a c x^{2} + A b c x^{2} + 2 \, B a b x + A b^{2} x + 2 \, A a c x - \frac{A a^{2}}{x} +{\left (B a^{2} + 2 \, A a b\right )} \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)^2/x^2,x, algorithm="giac")

[Out]

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