3.861 \(\int \frac{(A+B x) (a+b x+c x^2)^2}{x^4} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0613583, 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} \[ -\frac{a^2 A}{3 x^3}-\frac{A \left (2 a c+b^2\right )+2 a b B}{x}+\log (x) \left (2 a B c+2 A b c+b^2 B\right )-\frac{a (a B+2 A b)}{2 x^2}+c x (A c+2 b B)+\frac{1}{2} B c^2 x^2 \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + b*x + c*x^2)^2)/x^4,x]

[Out]

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

Mathematica [A]  time = 0.0532036, size = 90, normalized size = 1. \[ -\frac{a^2 (2 A+3 B x)}{6 x^3}+\log (x) \left (2 a B c+2 A b c+b^2 B\right )-\frac{a (A b+2 A c x+2 b B x)}{x^2}-\frac{A b^2}{x}+A c^2 x+2 b B c x+\frac{1}{2} B c^2 x^2 \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a + b*x + c*x^2)^2)/x^4,x]

[Out]

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

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Maple [A]  time = 0.007, size = 95, normalized size = 1.1 \begin{align*}{\frac{B{c}^{2}{x}^{2}}{2}}+A{c}^{2}x+2\,Bcbx+2\,A\ln \left ( x \right ) bc+2\,B\ln \left ( x \right ) ac+{b}^{2}B\ln \left ( x \right ) -{\frac{A{a}^{2}}{3\,{x}^{3}}}-{\frac{Aab}{{x}^{2}}}-{\frac{B{a}^{2}}{2\,{x}^{2}}}-2\,{\frac{aAc}{x}}-{\frac{A{b}^{2}}{x}}-2\,{\frac{abB}{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^4,x)

[Out]

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

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Maxima [A]  time = 1.16202, size = 120, normalized size = 1.33 \begin{align*} \frac{1}{2} \, B c^{2} x^{2} +{\left (2 \, B b c + A c^{2}\right )} x +{\left (B b^{2} + 2 \,{\left (B a + A b\right )} c\right )} \log \left (x\right ) - \frac{2 \, A a^{2} + 6 \,{\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} x^{2} + 3 \,{\left (B a^{2} + 2 \, A a b\right )} x}{6 \, x^{3}} \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^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.36503, size = 216, normalized size = 2.4 \begin{align*} \frac{3 \, B c^{2} x^{5} + 6 \,{\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} \log \left (x\right ) - 2 \, A a^{2} - 6 \,{\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} x^{2} - 3 \,{\left (B a^{2} + 2 \, A a b\right )} x}{6 \, x^{3}} \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^4,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.66425, size = 95, normalized size = 1.06 \begin{align*} \frac{B c^{2} x^{2}}{2} + x \left (A c^{2} + 2 B b c\right ) + \left (2 A b c + 2 B a c + B b^{2}\right ) \log{\left (x \right )} - \frac{2 A a^{2} + x^{2} \left (12 A a c + 6 A b^{2} + 12 B a b\right ) + x \left (6 A a b + 3 B a^{2}\right )}{6 x^{3}} \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**4,x)

[Out]

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

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Giac [A]  time = 1.27497, size = 120, normalized size = 1.33 \begin{align*} \frac{1}{2} \, B c^{2} x^{2} + 2 \, B b c x + A c^{2} x +{\left (B b^{2} + 2 \, B a c + 2 \, A b c\right )} \log \left ({\left | x \right |}\right ) - \frac{2 \, A a^{2} + 6 \,{\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} x^{2} + 3 \,{\left (B a^{2} + 2 \, A a b\right )} x}{6 \, x^{3}} \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^4,x, algorithm="giac")

[Out]

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