3.862 \(\int \frac{(A+B x) (a+b x+c x^2)^2}{x^5} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0592905, 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} \[ -\frac{a^2 A}{4 x^4}-\frac{A \left (2 a c+b^2\right )+2 a b B}{2 x^2}-\frac{2 a B c+2 A b c+b^2 B}{x}-\frac{a (a B+2 A b)}{3 x^3}+c \log (x) (A c+2 b B)+B c^2 x \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + b*x + c*x^2)^2)/x^5,x]

[Out]

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

Mathematica [A]  time = 0.0510006, size = 92, normalized size = 1.02 \[ -\frac{a^2 (3 A+4 B x)+4 a x (A (2 b+3 c x)+3 B x (b+2 c x))+6 x^2 \left (A b (b+4 c x)+2 B x \left (b^2-c^2 x^2\right )\right )-12 c x^4 \log (x) (A c+2 b B)}{12 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a + b*x + c*x^2)^2)/x^5,x]

[Out]

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

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Maple [A]  time = 0.005, size = 98, normalized size = 1.1 \begin{align*} B{c}^{2}x+A\ln \left ( x \right ){c}^{2}+2\,B\ln \left ( x \right ) bc-{\frac{2\,Aab}{3\,{x}^{3}}}-{\frac{B{a}^{2}}{3\,{x}^{3}}}-{\frac{aAc}{{x}^{2}}}-{\frac{A{b}^{2}}{2\,{x}^{2}}}-{\frac{abB}{{x}^{2}}}-2\,{\frac{Abc}{x}}-2\,{\frac{aBc}{x}}-{\frac{{b}^{2}B}{x}}-{\frac{A{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^5,x)

[Out]

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

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Maxima [A]  time = 1.09526, size = 120, normalized size = 1.33 \begin{align*} B c^{2} x +{\left (2 \, B b c + A c^{2}\right )} \log \left (x\right ) - \frac{12 \,{\left (B b^{2} + 2 \,{\left (B a + A b\right )} c\right )} x^{3} + 3 \, A a^{2} + 6 \,{\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} x^{2} + 4 \,{\left (B a^{2} + 2 \, A a b\right )} x}{12 \, x^{4}} \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^5,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.27931, size = 221, normalized size = 2.46 \begin{align*} \frac{12 \, B c^{2} x^{5} + 12 \,{\left (2 \, B b c + A c^{2}\right )} x^{4} \log \left (x\right ) - 12 \,{\left (B b^{2} + 2 \,{\left (B a + A b\right )} c\right )} x^{3} - 3 \, A a^{2} - 6 \,{\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} x^{2} - 4 \,{\left (B a^{2} + 2 \, A a b\right )} x}{12 \, x^{4}} \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^5,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 5.17766, size = 94, normalized size = 1.04 \begin{align*} B c^{2} x + c \left (A c + 2 B b\right ) \log{\left (x \right )} - \frac{3 A a^{2} + x^{3} \left (24 A b c + 24 B a c + 12 B b^{2}\right ) + x^{2} \left (12 A a c + 6 A b^{2} + 12 B a b\right ) + x \left (8 A a b + 4 B a^{2}\right )}{12 x^{4}} \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**5,x)

[Out]

B*c**2*x + c*(A*c + 2*B*b)*log(x) - (3*A*a**2 + x**3*(24*A*b*c + 24*B*a*c + 12*B*b**2) + x**2*(12*A*a*c + 6*A*
b**2 + 12*B*a*b) + x*(8*A*a*b + 4*B*a**2))/(12*x**4)

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Giac [A]  time = 1.27123, size = 122, normalized size = 1.36 \begin{align*} B c^{2} x +{\left (2 \, B b c + A c^{2}\right )} \log \left ({\left | x \right |}\right ) - \frac{12 \,{\left (B b^{2} + 2 \, B a c + 2 \, A b c\right )} x^{3} + 3 \, A a^{2} + 6 \,{\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} x^{2} + 4 \,{\left (B a^{2} + 2 \, A a b\right )} x}{12 \, x^{4}} \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^5,x, algorithm="giac")

[Out]

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