3.863 \(\int \frac{(A+B x) (a+b x+c x^2)^2}{x^6} \, dx\)

Optimal. Leaf size=95 \[ -\frac{a^2 A}{5 x^5}-\frac{2 a B c+2 A b c+b^2 B}{2 x^2}-\frac{A \left (2 a c+b^2\right )+2 a b B}{3 x^3}-\frac{a (a B+2 A b)}{4 x^4}-\frac{c (A c+2 b B)}{x}+B c^2 \log (x) \]

[Out]

-(a^2*A)/(5*x^5) - (a*(2*A*b + a*B))/(4*x^4) - (2*a*b*B + A*(b^2 + 2*a*c))/(3*x^3) - (b^2*B + 2*A*b*c + 2*a*B*
c)/(2*x^2) - (c*(2*b*B + A*c))/x + B*c^2*Log[x]

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Rubi [A]  time = 0.0544118, antiderivative size = 95, 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} \[ -\frac{a^2 A}{5 x^5}-\frac{2 a B c+2 A b c+b^2 B}{2 x^2}-\frac{A \left (2 a c+b^2\right )+2 a b B}{3 x^3}-\frac{a (a B+2 A b)}{4 x^4}-\frac{c (A c+2 b B)}{x}+B c^2 \log (x) \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + b*x + c*x^2)^2)/x^6,x]

[Out]

-(a^2*A)/(5*x^5) - (a*(2*A*b + a*B))/(4*x^4) - (2*a*b*B + A*(b^2 + 2*a*c))/(3*x^3) - (b^2*B + 2*A*b*c + 2*a*B*
c)/(2*x^2) - (c*(2*b*B + A*c))/x + B*c^2*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^6} \, dx &=\int \left (\frac{a^2 A}{x^6}+\frac{a (2 A b+a B)}{x^5}+\frac{2 a b B+A \left (b^2+2 a c\right )}{x^4}+\frac{b^2 B+2 A b c+2 a B c}{x^3}+\frac{c (2 b B+A c)}{x^2}+\frac{B c^2}{x}\right ) \, dx\\ &=-\frac{a^2 A}{5 x^5}-\frac{a (2 A b+a B)}{4 x^4}-\frac{2 a b B+A \left (b^2+2 a c\right )}{3 x^3}-\frac{b^2 B+2 A b c+2 a B c}{2 x^2}-\frac{c (2 b B+A c)}{x}+B c^2 \log (x)\\ \end{align*}

Mathematica [A]  time = 0.0622706, size = 92, normalized size = 0.97 \[ B c^2 \log (x)-\frac{3 a^2 (4 A+5 B x)+10 a x \left (3 A b+4 A c x+4 b B x+6 B c x^2\right )+10 x^2 \left (2 A \left (b^2+3 b c x+3 c^2 x^2\right )+3 b B x (b+4 c x)\right )}{60 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a + b*x + c*x^2)^2)/x^6,x]

[Out]

-(3*a^2*(4*A + 5*B*x) + 10*a*x*(3*A*b + 4*b*B*x + 4*A*c*x + 6*B*c*x^2) + 10*x^2*(3*b*B*x*(b + 4*c*x) + 2*A*(b^
2 + 3*b*c*x + 3*c^2*x^2)))/(60*x^5) + B*c^2*Log[x]

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Maple [A]  time = 0.007, size = 102, normalized size = 1.1 \begin{align*} B{c}^{2}\ln \left ( x \right ) -{\frac{2\,aAc}{3\,{x}^{3}}}-{\frac{A{b}^{2}}{3\,{x}^{3}}}-{\frac{2\,abB}{3\,{x}^{3}}}-{\frac{Abc}{{x}^{2}}}-{\frac{aBc}{{x}^{2}}}-{\frac{{b}^{2}B}{2\,{x}^{2}}}-{\frac{A{c}^{2}}{x}}-2\,{\frac{Bcb}{x}}-{\frac{A{a}^{2}}{5\,{x}^{5}}}-{\frac{Aab}{2\,{x}^{4}}}-{\frac{B{a}^{2}}{4\,{x}^{4}}} \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^6,x)

[Out]

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

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Maxima [A]  time = 1.09443, size = 124, normalized size = 1.31 \begin{align*} B c^{2} \log \left (x\right ) - \frac{60 \,{\left (2 \, B b c + A c^{2}\right )} x^{4} + 30 \,{\left (B b^{2} + 2 \,{\left (B a + A b\right )} c\right )} x^{3} + 12 \, A a^{2} + 20 \,{\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} x^{2} + 15 \,{\left (B a^{2} + 2 \, A a b\right )} x}{60 \, x^{5}} \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^6,x, algorithm="maxima")

[Out]

B*c^2*log(x) - 1/60*(60*(2*B*b*c + A*c^2)*x^4 + 30*(B*b^2 + 2*(B*a + A*b)*c)*x^3 + 12*A*a^2 + 20*(2*B*a*b + A*
b^2 + 2*A*a*c)*x^2 + 15*(B*a^2 + 2*A*a*b)*x)/x^5

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Fricas [A]  time = 1.20106, size = 225, normalized size = 2.37 \begin{align*} \frac{60 \, B c^{2} x^{5} \log \left (x\right ) - 60 \,{\left (2 \, B b c + A c^{2}\right )} x^{4} - 30 \,{\left (B b^{2} + 2 \,{\left (B a + A b\right )} c\right )} x^{3} - 12 \, A a^{2} - 20 \,{\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} x^{2} - 15 \,{\left (B a^{2} + 2 \, A a b\right )} x}{60 \, x^{5}} \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^6,x, algorithm="fricas")

[Out]

1/60*(60*B*c^2*x^5*log(x) - 60*(2*B*b*c + A*c^2)*x^4 - 30*(B*b^2 + 2*(B*a + A*b)*c)*x^3 - 12*A*a^2 - 20*(2*B*a
*b + A*b^2 + 2*A*a*c)*x^2 - 15*(B*a^2 + 2*A*a*b)*x)/x^5

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Sympy [A]  time = 10.8477, size = 99, normalized size = 1.04 \begin{align*} B c^{2} \log{\left (x \right )} - \frac{12 A a^{2} + x^{4} \left (60 A c^{2} + 120 B b c\right ) + x^{3} \left (60 A b c + 60 B a c + 30 B b^{2}\right ) + x^{2} \left (40 A a c + 20 A b^{2} + 40 B a b\right ) + x \left (30 A a b + 15 B a^{2}\right )}{60 x^{5}} \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**6,x)

[Out]

B*c**2*log(x) - (12*A*a**2 + x**4*(60*A*c**2 + 120*B*b*c) + x**3*(60*A*b*c + 60*B*a*c + 30*B*b**2) + x**2*(40*
A*a*c + 20*A*b**2 + 40*B*a*b) + x*(30*A*a*b + 15*B*a**2))/(60*x**5)

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Giac [A]  time = 1.25355, size = 126, normalized size = 1.33 \begin{align*} B c^{2} \log \left ({\left | x \right |}\right ) - \frac{60 \,{\left (2 \, B b c + A c^{2}\right )} x^{4} + 30 \,{\left (B b^{2} + 2 \, B a c + 2 \, A b c\right )} x^{3} + 12 \, A a^{2} + 20 \,{\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} x^{2} + 15 \,{\left (B a^{2} + 2 \, A a b\right )} x}{60 \, x^{5}} \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^6,x, algorithm="giac")

[Out]

B*c^2*log(abs(x)) - 1/60*(60*(2*B*b*c + A*c^2)*x^4 + 30*(B*b^2 + 2*B*a*c + 2*A*b*c)*x^3 + 12*A*a^2 + 20*(2*B*a
*b + A*b^2 + 2*A*a*c)*x^2 + 15*(B*a^2 + 2*A*a*b)*x)/x^5