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

Optimal. Leaf size=55 \[ -\frac{a^2 A}{5 x^5}-\frac{a (a B+2 A b)}{4 x^4}-\frac{b (2 a B+A b)}{3 x^3}-\frac{b^2 B}{2 x^2} \]

[Out]

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

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Rubi [A]  time = 0.0235665, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.08, Rules used = {27, 76} \[ -\frac{a^2 A}{5 x^5}-\frac{a (a B+2 A b)}{4 x^4}-\frac{b (2 a B+A b)}{3 x^3}-\frac{b^2 B}{2 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

Rubi steps

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

Mathematica [A]  time = 0.0154738, size = 50, normalized size = 0.91 \[ -\frac{3 a^2 (4 A+5 B x)+10 a b x (3 A+4 B x)+10 b^2 x^2 (2 A+3 B x)}{60 x^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(10*b^2*x^2*(2*A + 3*B*x) + 10*a*b*x*(3*A + 4*B*x) + 3*a^2*(4*A + 5*B*x))/(60*x^5)

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Maple [A]  time = 0.006, size = 48, normalized size = 0.9 \begin{align*} -{\frac{A{a}^{2}}{5\,{x}^{5}}}-{\frac{a \left ( 2\,Ab+aB \right ) }{4\,{x}^{4}}}-{\frac{b \left ( Ab+2\,aB \right ) }{3\,{x}^{3}}}-{\frac{{b}^{2}B}{2\,{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)/x^6,x)

[Out]

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

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Maxima [A]  time = 0.995954, size = 69, normalized size = 1.25 \begin{align*} -\frac{30 \, B b^{2} x^{3} + 12 \, A a^{2} + 20 \,{\left (2 \, B a b + A b^{2}\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)*(b^2*x^2+2*a*b*x+a^2)/x^6,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.2413, size = 120, normalized size = 2.18 \begin{align*} -\frac{30 \, B b^{2} x^{3} + 12 \, A a^{2} + 20 \,{\left (2 \, B a b + A b^{2}\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)*(b^2*x^2+2*a*b*x+a^2)/x^6,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.10722, size = 54, normalized size = 0.98 \begin{align*} - \frac{12 A a^{2} + 30 B b^{2} x^{3} + x^{2} \left (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)*(b**2*x**2+2*a*b*x+a**2)/x**6,x)

[Out]

-(12*A*a**2 + 30*B*b**2*x**3 + x**2*(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 = 1.10557, size = 69, normalized size = 1.25 \begin{align*} -\frac{30 \, B b^{2} x^{3} + 40 \, B a b x^{2} + 20 \, A b^{2} x^{2} + 15 \, B a^{2} x + 30 \, A a b x + 12 \, A a^{2}}{60 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)/x^6,x, algorithm="giac")

[Out]

-1/60*(30*B*b^2*x^3 + 40*B*a*b*x^2 + 20*A*b^2*x^2 + 15*B*a^2*x + 30*A*a*b*x + 12*A*a^2)/x^5