3.517 \(\int \frac{(A+B x) (a^2+2 a b x+b^2 x^2)}{x^3} \, dx\)

Optimal. Leaf size=44 \[ -\frac{a^2 A}{2 x^2}-\frac{a (a B+2 A b)}{x}+b \log (x) (2 a B+A b)+b^2 B x \]

[Out]

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

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Rubi [A]  time = 0.0244103, antiderivative size = 44, 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} \[ -\frac{a^2 A}{2 x^2}-\frac{a (a B+2 A b)}{x}+b \log (x) (2 a B+A b)+b^2 B x \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2))/x^3,x]

[Out]

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

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

Mathematica [A]  time = 0.0246844, size = 43, normalized size = 0.98 \[ -\frac{a^2 (A+2 B x)}{2 x^2}+b \log (x) (2 a B+A b)-\frac{2 a A b}{x}+b^2 B x \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2))/x^3,x]

[Out]

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

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Maple [A]  time = 0.006, size = 48, normalized size = 1.1 \begin{align*}{b}^{2}Bx+A{b}^{2}\ln \left ( x \right ) +2\,B\ln \left ( x \right ) ab-{\frac{A{a}^{2}}{2\,{x}^{2}}}-2\,{\frac{Aab}{x}}-{\frac{B{a}^{2}}{x}} \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^3,x)

[Out]

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

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Maxima [A]  time = 1.02099, size = 62, normalized size = 1.41 \begin{align*} B b^{2} x +{\left (2 \, B a b + A b^{2}\right )} \log \left (x\right ) - \frac{A a^{2} + 2 \,{\left (B a^{2} + 2 \, A a b\right )} x}{2 \, x^{2}} \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^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.30302, size = 119, normalized size = 2.7 \begin{align*} \frac{2 \, B b^{2} x^{3} + 2 \,{\left (2 \, B a b + A b^{2}\right )} x^{2} \log \left (x\right ) - A a^{2} - 2 \,{\left (B a^{2} + 2 \, A a b\right )} x}{2 \, x^{2}} \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^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.515756, size = 44, normalized size = 1. \begin{align*} B b^{2} x + b \left (A b + 2 B a\right ) \log{\left (x \right )} - \frac{A a^{2} + x \left (4 A a b + 2 B a^{2}\right )}{2 x^{2}} \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**3,x)

[Out]

B*b**2*x + b*(A*b + 2*B*a)*log(x) - (A*a**2 + x*(4*A*a*b + 2*B*a**2))/(2*x**2)

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Giac [A]  time = 1.14319, size = 63, normalized size = 1.43 \begin{align*} B b^{2} x +{\left (2 \, B a b + A b^{2}\right )} \log \left ({\left | x \right |}\right ) - \frac{A a^{2} + 2 \,{\left (B a^{2} + 2 \, A a b\right )} x}{2 \, x^{2}} \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^3,x, algorithm="giac")

[Out]

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