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

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

[Out]

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

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Rubi [A]  time = 0.0671622, 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} \[ -\frac{a^2 A}{2 x^2}+x \left (2 a B c+2 A b c+b^2 B\right )+\log (x) \left (A \left (2 a c+b^2\right )+2 a b B\right )-\frac{a (a B+2 A b)}{x}+\frac{1}{2} c x^2 (A c+2 b B)+\frac{1}{3} B c^2 x^3 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0509398, size = 86, normalized size = 0.96 \[ -\frac{a^2 (A+2 B x)}{2 x^2}+A \log (x) \left (2 a c+b^2\right )+a \left (2 B c x-\frac{2 A b}{x}\right )+2 a b B \log (x)+b c x (2 A+B x)+\frac{1}{6} c^2 x^2 (3 A+2 B x)+b^2 B x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.007, size = 92, normalized size = 1. \begin{align*}{\frac{B{c}^{2}{x}^{3}}{3}}+{\frac{A{c}^{2}{x}^{2}}{2}}+B{x}^{2}bc+2\,Abcx+2\,aBcx+{b}^{2}Bx+2\,aAc\ln \left ( x \right ) +A{b}^{2}\ln \left ( x \right ) +2\,B\ln \left ( x \right ) ab-{\frac{A{a}^{2}}{2\,{x}^{2}}}-2\,{\frac{Aab}{x}}-{\frac{B{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^3,x)

[Out]

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

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Maxima [A]  time = 1.07231, size = 119, normalized size = 1.32 \begin{align*} \frac{1}{3} \, B c^{2} x^{3} + \frac{1}{2} \,{\left (2 \, B b c + A c^{2}\right )} x^{2} +{\left (B b^{2} + 2 \,{\left (B a + A b\right )} c\right )} x +{\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} \log \left (x\right ) - \frac{A a^{2} + 2 \,{\left (B a^{2} + 2 \, 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)^2/x^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.21603, size = 216, normalized size = 2.4 \begin{align*} \frac{2 \, B c^{2} x^{5} + 3 \,{\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} + 6 \,{\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} x^{2} \log \left (x\right ) - 3 \, A a^{2} - 6 \,{\left (B a^{2} + 2 \, A a b\right )} x}{6 \, 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)^2/x^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.727335, size = 92, normalized size = 1.02 \begin{align*} \frac{B c^{2} x^{3}}{3} + x^{2} \left (\frac{A c^{2}}{2} + B b c\right ) + x \left (2 A b c + 2 B a c + B b^{2}\right ) + \left (2 A a c + A b^{2} + 2 B a b\right ) \log{\left (x \right )} - \frac{A a^{2} + x \left (4 A a b + 2 B a^{2}\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)**2/x**3,x)

[Out]

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

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Giac [A]  time = 1.27496, size = 120, normalized size = 1.33 \begin{align*} \frac{1}{3} \, B c^{2} x^{3} + B b c x^{2} + \frac{1}{2} \, A c^{2} x^{2} + B b^{2} x + 2 \, B a c x + 2 \, A b c x +{\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} \log \left ({\left | x \right |}\right ) - \frac{A a^{2} + 2 \,{\left (B a^{2} + 2 \, 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)^2/x^3,x, algorithm="giac")

[Out]

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