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

Optimal. Leaf size=44 \[ -\frac{a^2 A}{x}+b x (2 a B+A b)+a \log (x) (a B+2 A b)+\frac{1}{2} b^2 B x^2 \]

[Out]

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

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Rubi [A]  time = 0.0247861, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.062, Rules used = {76} \[ -\frac{a^2 A}{x}+b x (2 a B+A b)+a \log (x) (a B+2 A b)+\frac{1}{2} b^2 B x^2 \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^2*(A + B*x))/x^2,x]

[Out]

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

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)^2 (A+B x)}{x^2} \, dx &=\int \left (b (A b+2 a B)+\frac{a^2 A}{x^2}+\frac{a (2 A b+a B)}{x}+b^2 B x\right ) \, dx\\ &=-\frac{a^2 A}{x}+b (A b+2 a B) x+\frac{1}{2} b^2 B x^2+a (2 A b+a B) \log (x)\\ \end{align*}

Mathematica [A]  time = 0.0209799, size = 43, normalized size = 0.98 \[ -\frac{a^2 A}{x}+a \log (x) (a B+2 A b)+2 a b B x+\frac{1}{2} b^2 x (2 A+B x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.005, size = 46, normalized size = 1.1 \begin{align*}{\frac{{b}^{2}B{x}^{2}}{2}}+{b}^{2}Ax+2\,abBx+2\,A\ln \left ( x \right ) ab+B\ln \left ( x \right ){a}^{2}-{\frac{A{a}^{2}}{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.06216, size = 62, normalized size = 1.41 \begin{align*} \frac{1}{2} \, B b^{2} x^{2} - \frac{A a^{2}}{x} +{\left (2 \, B a b + A b^{2}\right )} x +{\left (B a^{2} + 2 \, A a b\right )} \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.90792, size = 116, normalized size = 2.64 \begin{align*} \frac{B b^{2} x^{3} - 2 \, A a^{2} + 2 \,{\left (2 \, B a b + A b^{2}\right )} x^{2} + 2 \,{\left (B a^{2} + 2 \, A a b\right )} x \log \left (x\right )}{2 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.407185, size = 42, normalized size = 0.95 \begin{align*} - \frac{A a^{2}}{x} + \frac{B b^{2} x^{2}}{2} + a \left (2 A b + B a\right ) \log{\left (x \right )} + x \left (A b^{2} + 2 B a b\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**2*(B*x+A)/x**2,x)

[Out]

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

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Giac [A]  time = 1.22677, size = 62, normalized size = 1.41 \begin{align*} \frac{1}{2} \, B b^{2} x^{2} + 2 \, B a b x + A b^{2} x - \frac{A a^{2}}{x} +{\left (B a^{2} + 2 \, A a b\right )} \log \left ({\left | x \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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