3.521 \(\int \frac{(A+B x) (a^2+2 a b x+b^2 x^2)}{x^7} \, dx\)

Optimal. Leaf size=55 \[ -\frac{a^2 A}{6 x^6}-\frac{a (a B+2 A b)}{5 x^5}-\frac{b (2 a B+A b)}{4 x^4}-\frac{b^2 B}{3 x^3} \]

[Out]

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

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Rubi [A]  time = 0.0238052, 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} \[ -\frac{a^2 A}{6 x^6}-\frac{a (a B+2 A b)}{5 x^5}-\frac{b (2 a B+A b)}{4 x^4}-\frac{b^2 B}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2))/x^7,x]

[Out]

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

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^7} \, dx &=\int \frac{(a+b x)^2 (A+B x)}{x^7} \, dx\\ &=\int \left (\frac{a^2 A}{x^7}+\frac{a (2 A b+a B)}{x^6}+\frac{b (A b+2 a B)}{x^5}+\frac{b^2 B}{x^4}\right ) \, dx\\ &=-\frac{a^2 A}{6 x^6}-\frac{a (2 A b+a B)}{5 x^5}-\frac{b (A b+2 a B)}{4 x^4}-\frac{b^2 B}{3 x^3}\\ \end{align*}

Mathematica [A]  time = 0.0156567, size = 50, normalized size = 0.91 \[ -\frac{2 a^2 (5 A+6 B x)+6 a b x (4 A+5 B x)+5 b^2 x^2 (3 A+4 B x)}{60 x^6} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2))/x^7,x]

[Out]

-(5*b^2*x^2*(3*A + 4*B*x) + 6*a*b*x*(4*A + 5*B*x) + 2*a^2*(5*A + 6*B*x))/(60*x^6)

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Maple [A]  time = 0.004, size = 48, normalized size = 0.9 \begin{align*} -{\frac{A{a}^{2}}{6\,{x}^{6}}}-{\frac{a \left ( 2\,Ab+aB \right ) }{5\,{x}^{5}}}-{\frac{b \left ( Ab+2\,aB \right ) }{4\,{x}^{4}}}-{\frac{{b}^{2}B}{3\,{x}^{3}}} \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^7,x)

[Out]

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

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Maxima [A]  time = 0.982828, size = 69, normalized size = 1.25 \begin{align*} -\frac{20 \, B b^{2} x^{3} + 10 \, A a^{2} + 15 \,{\left (2 \, B a b + A b^{2}\right )} x^{2} + 12 \,{\left (B a^{2} + 2 \, A a b\right )} x}{60 \, x^{6}} \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^7,x, algorithm="maxima")

[Out]

-1/60*(20*B*b^2*x^3 + 10*A*a^2 + 15*(2*B*a*b + A*b^2)*x^2 + 12*(B*a^2 + 2*A*a*b)*x)/x^6

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Fricas [A]  time = 1.31105, size = 120, normalized size = 2.18 \begin{align*} -\frac{20 \, B b^{2} x^{3} + 10 \, A a^{2} + 15 \,{\left (2 \, B a b + A b^{2}\right )} x^{2} + 12 \,{\left (B a^{2} + 2 \, A a b\right )} x}{60 \, x^{6}} \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^7,x, algorithm="fricas")

[Out]

-1/60*(20*B*b^2*x^3 + 10*A*a^2 + 15*(2*B*a*b + A*b^2)*x^2 + 12*(B*a^2 + 2*A*a*b)*x)/x^6

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Sympy [A]  time = 1.27759, size = 54, normalized size = 0.98 \begin{align*} - \frac{10 A a^{2} + 20 B b^{2} x^{3} + x^{2} \left (15 A b^{2} + 30 B a b\right ) + x \left (24 A a b + 12 B a^{2}\right )}{60 x^{6}} \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**7,x)

[Out]

-(10*A*a**2 + 20*B*b**2*x**3 + x**2*(15*A*b**2 + 30*B*a*b) + x*(24*A*a*b + 12*B*a**2))/(60*x**6)

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Giac [A]  time = 1.12546, size = 69, normalized size = 1.25 \begin{align*} -\frac{20 \, B b^{2} x^{3} + 30 \, B a b x^{2} + 15 \, A b^{2} x^{2} + 12 \, B a^{2} x + 24 \, A a b x + 10 \, A a^{2}}{60 \, x^{6}} \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^7,x, algorithm="giac")

[Out]

-1/60*(20*B*b^2*x^3 + 30*B*a*b*x^2 + 15*A*b^2*x^2 + 12*B*a^2*x + 24*A*a*b*x + 10*A*a^2)/x^6