\(\int \frac {(a+b x)^2 (A+B x)}{x^3} \, dx\) [98]

Optimal result

Integrand size = 16, antiderivative size = 44 \[ \int \frac {(a+b x)^2 (A+B x)}{x^3} \, dx=-\frac {a^2 A}{2 x^2}-\frac {a (2 A b+a B)}{x}+b^2 B x+b (A b+2 a B) \log (x) \]

[Out]

-1/2*a^2*A/x^2-a*(2*A*b+B*a)/x+b^2*B*x+b*(A*b+2*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} \[ \int \frac {(a+b x)^2 (A+B x)}{x^3} \, dx=-\frac {a^2 A}{2 x^2}-\frac {a (a B+2 A b)}{x}+b \log (x) (2 a B+A b)+b^2 B x \]

[In]

Int[((a + b*x)^2*(A + B*x))/x^3,x]

[Out]

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

Time = 0.02 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.98 \[ \int \frac {(a+b x)^2 (A+B x)}{x^3} \, dx=-\frac {2 a A b}{x}+b^2 B x-\frac {a^2 (A+2 B x)}{2 x^2}+b (A b+2 a B) \log (x) \]

[In]

Integrate[((a + b*x)^2*(A + B*x))/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]

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.98

method	result	size
default	\(-\frac {a^{2} A}{2 x^{2}}-\frac {a \left (2 A b +B a \right )}{x}+b^{2} B x +b \left (A b +2 B a \right ) \ln \left (x \right )\)	\(43\)
risch	\(b^{2} B x +\frac {\left (-2 a b A -a^{2} B \right ) x -\frac {a^{2} A}{2}}{x^{2}}+A \ln \left (x \right ) b^{2}+2 B \ln \left (x \right ) a b\)	\(47\)
norman	\(\frac {\left (-2 a b A -a^{2} B \right ) x +b^{2} B \,x^{3}-\frac {a^{2} A}{2}}{x^{2}}+\left (b^{2} A +2 a b B \right ) \ln \left (x \right )\)	\(49\)
parallelrisch	\(\frac {2 A \ln \left (x \right ) x^{2} b^{2}+4 B \ln \left (x \right ) x^{2} a b +2 b^{2} B \,x^{3}-4 a A b x -2 a^{2} B x -a^{2} A}{2 x^{2}}\)	\(56\)

[In]

int((b*x+a)^2*(B*x+A)/x^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.20 \[ \int \frac {(a+b x)^2 (A+B x)}{x^3} \, dx=\frac {2 \, B b^{2} x^{3} + 2 \, {\left (2 \, B a b + A b^{2}\right )} x^{2} \log \left (x\right ) - A a^{2} - 2 \, {\left (B a^{2} + 2 \, A a b\right )} x}{2 \, x^{2}} \]

[In]

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

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.05 \[ \int \frac {(a+b x)^2 (A+B x)}{x^3} \, dx=B b^{2} x + b \left (A b + 2 B a\right ) \log {\left (x \right )} + \frac {- A a^{2} + x \left (- 4 A a b - 2 B a^{2}\right )}{2 x^{2}} \]

[In]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.05 \[ \int \frac {(a+b x)^2 (A+B x)}{x^3} \, dx=B b^{2} x + {\left (2 \, B a b + A b^{2}\right )} \log \left (x\right ) - \frac {A a^{2} + 2 \, {\left (B a^{2} + 2 \, A a b\right )} x}{2 \, x^{2}} \]

[In]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.07 \[ \int \frac {(a+b x)^2 (A+B x)}{x^3} \, dx=B b^{2} x + {\left (2 \, B a b + A b^{2}\right )} \log \left ({\left | x \right |}\right ) - \frac {A a^{2} + 2 \, {\left (B a^{2} + 2 \, A a b\right )} x}{2 \, x^{2}} \]

[In]

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

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.05 \[ \int \frac {(a+b x)^2 (A+B x)}{x^3} \, dx=\ln \left (x\right )\,\left (A\,b^2+2\,B\,a\,b\right )-\frac {\frac {A\,a^2}{2}+x\,\left (B\,a^2+2\,A\,b\,a\right )}{x^2}+B\,b^2\,x \]

[In]

int(((A + B*x)*(a + b*x)^2)/x^3,x)

[Out]

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