\(\int \frac {x (A+B x)}{(a+b x)^3} \, dx\) [196]

Optimal result

Integrand size = 14, antiderivative size = 55 \[ \int \frac {x (A+B x)}{(a+b x)^3} \, dx=\frac {a (A b-a B)}{2 b^3 (a+b x)^2}-\frac {A b-2 a B}{b^3 (a+b x)}+\frac {B \log (a+b x)}{b^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {78} \[ \int \frac {x (A+B x)}{(a+b x)^3} \, dx=-\frac {A b-2 a B}{b^3 (a+b x)}+\frac {a (A b-a B)}{2 b^3 (a+b x)^2}+\frac {B \log (a+b x)}{b^3} \]

[In]

Int[(x*(A + B*x))/(a + b*x)^3,x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.98 \[ \int \frac {x (A+B x)}{(a+b x)^3} \, dx=\frac {3 a^2 B-2 A b^2 x-a b (A-4 B x)+2 B (a+b x)^2 \log (a+b x)}{2 b^3 (a+b x)^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.91

method	result	size
norman	\(\frac {-\frac {a \left (A b -3 B a \right )}{2 b^{3}}-\frac {\left (A b -2 B a \right ) x}{b^{2}}}{\left (b x +a \right )^{2}}+\frac {B \ln \left (b x +a \right )}{b^{3}}\)	\(50\)
risch	\(\frac {-\frac {a \left (A b -3 B a \right )}{2 b^{3}}-\frac {\left (A b -2 B a \right ) x}{b^{2}}}{\left (b x +a \right )^{2}}+\frac {B \ln \left (b x +a \right )}{b^{3}}\)	\(50\)
default	\(\frac {B \ln \left (b x +a \right )}{b^{3}}+\frac {a \left (A b -B a \right )}{2 b^{3} \left (b x +a \right )^{2}}-\frac {A b -2 B a}{b^{3} \left (b x +a \right )}\)	\(54\)
parallelrisch	\(-\frac {-2 B \ln \left (b x +a \right ) x^{2} b^{2}-4 B \ln \left (b x +a \right ) x a b +2 A \,b^{2} x -2 B \ln \left (b x +a \right ) a^{2}-4 B a b x +a b A -3 a^{2} B}{2 b^{3} \left (b x +a \right )^{2}}\)	\(76\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.47 \[ \int \frac {x (A+B x)}{(a+b x)^3} \, dx=\frac {3 \, B a^{2} - A a b + 2 \, {\left (2 \, B a b - A b^{2}\right )} x + 2 \, {\left (B b^{2} x^{2} + 2 \, B a b x + B a^{2}\right )} \log \left (b x + a\right )}{2 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} \]

[In]

integrate(x*(B*x+A)/(b*x+a)^3,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.15 \[ \int \frac {x (A+B x)}{(a+b x)^3} \, dx=\frac {B \log {\left (a + b x \right )}}{b^{3}} + \frac {- A a b + 3 B a^{2} + x \left (- 2 A b^{2} + 4 B a b\right )}{2 a^{2} b^{3} + 4 a b^{4} x + 2 b^{5} x^{2}} \]

[In]

integrate(x*(B*x+A)/(b*x+a)**3,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.18 \[ \int \frac {x (A+B x)}{(a+b x)^3} \, dx=\frac {3 \, B a^{2} - A a b + 2 \, {\left (2 \, B a b - A b^{2}\right )} x}{2 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} + \frac {B \log \left (b x + a\right )}{b^{3}} \]

[In]

integrate(x*(B*x+A)/(b*x+a)^3,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.98 \[ \int \frac {x (A+B x)}{(a+b x)^3} \, dx=\frac {B \log \left ({\left | b x + a \right |}\right )}{b^{3}} + \frac {2 \, {\left (2 \, B a - A b\right )} x + \frac {3 \, B a^{2} - A a b}{b}}{2 \, {\left (b x + a\right )}^{2} b^{2}} \]

[In]

integrate(x*(B*x+A)/(b*x+a)^3,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.15 \[ \int \frac {x (A+B x)}{(a+b x)^3} \, dx=\frac {\frac {3\,B\,a^2-A\,a\,b}{2\,b^3}-\frac {x\,\left (A\,b-2\,B\,a\right )}{b^2}}{a^2+2\,a\,b\,x+b^2\,x^2}+\frac {B\,\ln \left (a+b\,x\right )}{b^3} \]

[In]

int((x*(A + B*x))/(a + b*x)^3,x)

[Out]

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