\(\int \frac {(a+b x)^2 (A+B x)}{x^2} \, dx\) [97]

Optimal result

Integrand size = 16, antiderivative size = 44 \[ \int \frac {(a+b x)^2 (A+B x)}{x^2} \, dx=-\frac {a^2 A}{x}+b (A b+2 a B) x+\frac {1}{2} b^2 B x^2+a (2 A b+a B) \log (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {77} \[ \int \frac {(a+b x)^2 (A+B x)}{x^2} \, dx=-\frac {a^2 A}{x}+b x (2 a B+A b)+a \log (x) (a B+2 A b)+\frac {1}{2} b^2 B x^2 \]

[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 77

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*} \text {integral}& = \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] (verified)

Time = 0.02 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.98 \[ \int \frac {(a+b x)^2 (A+B x)}{x^2} \, dx=-\frac {a^2 A}{x}+2 a b B x+\frac {1}{2} b^2 x (2 A+B x)+a (2 A b+a B) \log (x) \]

[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]

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00

method	result	size
default	\(\frac {b^{2} B \,x^{2}}{2}+A \,b^{2} x +2 B a b x +a \left (2 A b +B a \right ) \ln \left (x \right )-\frac {a^{2} A}{x}\)	\(44\)
risch	\(\frac {b^{2} B \,x^{2}}{2}+A \,b^{2} x +2 B a b x -\frac {a^{2} A}{x}+2 A \ln \left (x \right ) a b +B \ln \left (x \right ) a^{2}\)	\(46\)
norman	\(\frac {\left (b^{2} A +2 a b B \right ) x^{2}-a^{2} A +\frac {b^{2} B \,x^{3}}{2}}{x}+\left (2 a b A +a^{2} B \right ) \ln \left (x \right )\)	\(51\)
parallelrisch	\(\frac {b^{2} B \,x^{3}+4 A \ln \left (x \right ) x a b +2 A \,b^{2} x^{2}+2 B \ln \left (x \right ) x \,a^{2}+4 B a b \,x^{2}-2 a^{2} A}{2 x}\)	\(55\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.18 \[ \int \frac {(a+b x)^2 (A+B x)}{x^2} \, dx=\frac {B b^{2} x^{3} - 2 \, A a^{2} + 2 \, {\left (2 \, B a b + A b^{2}\right )} x^{2} + 2 \, {\left (B a^{2} + 2 \, A a b\right )} x \log \left (x\right )}{2 \, x} \]

[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

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.95 \[ \int \frac {(a+b x)^2 (A+B x)}{x^2} \, dx=- \frac {A a^{2}}{x} + \frac {B b^{2} x^{2}}{2} + a \left (2 A b + B a\right ) \log {\left (x \right )} + x \left (A b^{2} + 2 B a b\right ) \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.05 \[ \int \frac {(a+b x)^2 (A+B x)}{x^2} \, dx=\frac {1}{2} \, B b^{2} x^{2} - \frac {A a^{2}}{x} + {\left (2 \, B a b + A b^{2}\right )} x + {\left (B a^{2} + 2 \, A a b\right )} \log \left (x\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.05 \[ \int \frac {(a+b x)^2 (A+B x)}{x^2} \, dx=\frac {1}{2} \, B b^{2} x^{2} + 2 \, B a b x + A b^{2} x - \frac {A a^{2}}{x} + {\left (B a^{2} + 2 \, A a b\right )} \log \left ({\left | x \right |}\right ) \]

[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))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.05 \[ \int \frac {(a+b x)^2 (A+B x)}{x^2} \, dx=\ln \left (x\right )\,\left (B\,a^2+2\,A\,b\,a\right )+x\,\left (A\,b^2+2\,B\,a\,b\right )-\frac {A\,a^2}{x}+\frac {B\,b^2\,x^2}{2} \]

[In]

int(((A + B*x)*(a + b*x)^2)/x^2,x)

[Out]

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