\(\int \frac {x^2 (A+B x)}{(a+b x)^2} \, dx\) [186]

Optimal result

Integrand size = 16, antiderivative size = 69 \[ \int \frac {x^2 (A+B x)}{(a+b x)^2} \, dx=\frac {(A b-2 a B) x}{b^3}+\frac {B x^2}{2 b^2}-\frac {a^2 (A b-a B)}{b^4 (a+b x)}-\frac {a (2 A b-3 a B) \log (a+b x)}{b^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 69, 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 = {78} \[ \int \frac {x^2 (A+B x)}{(a+b x)^2} \, dx=-\frac {a^2 (A b-a B)}{b^4 (a+b x)}-\frac {a (2 A b-3 a B) \log (a+b x)}{b^4}+\frac {x (A b-2 a B)}{b^3}+\frac {B x^2}{2 b^2} \]

[In]

Int[(x^2*(A + B*x))/(a + b*x)^2,x]

[Out]

((A*b - 2*a*B)*x)/b^3 + (B*x^2)/(2*b^2) - (a^2*(A*b - a*B))/(b^4*(a + b*x)) - (a*(2*A*b - 3*a*B)*Log[a + b*x])
/b^4

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {A b-2 a B}{b^3}+\frac {B x}{b^2}-\frac {a^2 (-A b+a B)}{b^3 (a+b x)^2}+\frac {a (-2 A b+3 a B)}{b^3 (a+b x)}\right ) \, dx \\ & = \frac {(A b-2 a B) x}{b^3}+\frac {B x^2}{2 b^2}-\frac {a^2 (A b-a B)}{b^4 (a+b x)}-\frac {a (2 A b-3 a B) \log (a+b x)}{b^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.96 \[ \int \frac {x^2 (A+B x)}{(a+b x)^2} \, dx=\frac {2 b (A b-2 a B) x+b^2 B x^2+\frac {2 a^2 (-A b+a B)}{a+b x}+2 a (-2 A b+3 a B) \log (a+b x)}{2 b^4} \]

[In]

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

[Out]

(2*b*(A*b - 2*a*B)*x + b^2*B*x^2 + (2*a^2*(-(A*b) + a*B))/(a + b*x) + 2*a*(-2*A*b + 3*a*B)*Log[a + b*x])/(2*b^
4)

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.97

method	result	size
default	\(\frac {\frac {1}{2} b B \,x^{2}+A b x -2 B a x}{b^{3}}-\frac {a \left (2 A b -3 B a \right ) \ln \left (b x +a \right )}{b^{4}}-\frac {a^{2} \left (A b -B a \right )}{b^{4} \left (b x +a \right )}\)	\(67\)
norman	\(\frac {\frac {B \,x^{3}}{2 b}-\frac {a \left (2 a b A -3 a^{2} B \right )}{b^{4}}+\frac {\left (2 A b -3 B a \right ) x^{2}}{2 b^{2}}}{b x +a}-\frac {a \left (2 A b -3 B a \right ) \ln \left (b x +a \right )}{b^{4}}\)	\(76\)
risch	\(\frac {B \,x^{2}}{2 b^{2}}+\frac {A x}{b^{2}}-\frac {2 B a x}{b^{3}}-\frac {2 a \ln \left (b x +a \right ) A}{b^{3}}+\frac {3 a^{2} \ln \left (b x +a \right ) B}{b^{4}}-\frac {a^{2} A}{b^{3} \left (b x +a \right )}+\frac {a^{3} B}{b^{4} \left (b x +a \right )}\)	\(84\)
parallelrisch	\(-\frac {-b^{3} B \,x^{3}+4 A \ln \left (b x +a \right ) x a \,b^{2}-2 A \,b^{3} x^{2}-6 B \ln \left (b x +a \right ) x \,a^{2} b +3 B a \,b^{2} x^{2}+4 A \ln \left (b x +a \right ) a^{2} b -6 B \ln \left (b x +a \right ) a^{3}+4 a^{2} b A -6 a^{3} B}{2 b^{4} \left (b x +a \right )}\)	\(108\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.64 \[ \int \frac {x^2 (A+B x)}{(a+b x)^2} \, dx=\frac {B b^{3} x^{3} + 2 \, B a^{3} - 2 \, A a^{2} b - {\left (3 \, B a b^{2} - 2 \, A b^{3}\right )} x^{2} - 2 \, {\left (2 \, B a^{2} b - A a b^{2}\right )} x + 2 \, {\left (3 \, B a^{3} - 2 \, A a^{2} b + {\left (3 \, B a^{2} b - 2 \, A a b^{2}\right )} x\right )} \log \left (b x + a\right )}{2 \, {\left (b^{5} x + a b^{4}\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.99 \[ \int \frac {x^2 (A+B x)}{(a+b x)^2} \, dx=\frac {B x^{2}}{2 b^{2}} + \frac {a \left (- 2 A b + 3 B a\right ) \log {\left (a + b x \right )}}{b^{4}} + x \left (\frac {A}{b^{2}} - \frac {2 B a}{b^{3}}\right ) + \frac {- A a^{2} b + B a^{3}}{a b^{4} + b^{5} x} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.07 \[ \int \frac {x^2 (A+B x)}{(a+b x)^2} \, dx=\frac {B a^{3} - A a^{2} b}{b^{5} x + a b^{4}} + \frac {B b x^{2} - 2 \, {\left (2 \, B a - A b\right )} x}{2 \, b^{3}} + \frac {{\left (3 \, B a^{2} - 2 \, A a b\right )} \log \left (b x + a\right )}{b^{4}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.61 \[ \int \frac {x^2 (A+B x)}{(a+b x)^2} \, dx=\frac {{\left (b x + a\right )}^{2} {\left (B - \frac {2 \, {\left (3 \, B a b - A b^{2}\right )}}{{\left (b x + a\right )} b}\right )}}{2 \, b^{4}} - \frac {{\left (3 \, B a^{2} - 2 \, A a b\right )} \log \left (\frac {{\left | b x + a \right |}}{{\left (b x + a\right )}^{2} {\left | b \right |}}\right )}{b^{4}} + \frac {\frac {B a^{3} b^{2}}{b x + a} - \frac {A a^{2} b^{3}}{b x + a}}{b^{6}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.12 \[ \int \frac {x^2 (A+B x)}{(a+b x)^2} \, dx=x\,\left (\frac {A}{b^2}-\frac {2\,B\,a}{b^3}\right )+\frac {B\,x^2}{2\,b^2}+\frac {B\,a^3-A\,a^2\,b}{b\,\left (x\,b^4+a\,b^3\right )}+\frac {\ln \left (a+b\,x\right )\,\left (3\,B\,a^2-2\,A\,a\,b\right )}{b^4} \]

[In]

int((x^2*(A + B*x))/(a + b*x)^2,x)

[Out]

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