\(\int \frac {A+B x}{(a+b x)^3} \, dx\) [197]

Optimal result

Integrand size = 13, antiderivative size = 28 \[ \int \frac {A+B x}{(a+b x)^3} \, dx=-\frac {(A+B x)^2}{2 (A b-a B) (a+b x)^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {37} \[ \int \frac {A+B x}{(a+b x)^3} \, dx=-\frac {(A+B x)^2}{2 (a+b x)^2 (A b-a B)} \]

[In]

Int[(A + B*x)/(a + b*x)^3,x]

[Out]

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

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {(A+B x)^2}{2 (A b-a B) (a+b x)^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {A+B x}{(a+b x)^3} \, dx=-\frac {A b+B (a+2 b x)}{2 b^2 (a+b x)^2} \]

[In]

Integrate[(A + B*x)/(a + b*x)^3,x]

[Out]

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

Maple [A] (verified)

Time = 0.96 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89

method	result	size
gosper	\(-\frac {2 b B x +A b +B a}{2 b^{2} \left (b x +a \right )^{2}}\)	\(25\)
parallelrisch	\(-\frac {2 b B x +A b +B a}{2 b^{2} \left (b x +a \right )^{2}}\)	\(25\)
norman	\(\frac {-\frac {B x}{b}-\frac {A b +B a}{2 b^{2}}}{\left (b x +a \right )^{2}}\)	\(29\)
risch	\(\frac {-\frac {B x}{b}-\frac {A b +B a}{2 b^{2}}}{\left (b x +a \right )^{2}}\)	\(29\)
default	\(-\frac {A b -B a}{2 b^{2} \left (b x +a \right )^{2}}-\frac {B}{\left (b x +a \right ) b^{2}}\)	\(35\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \frac {A+B x}{(a+b x)^3} \, dx=-\frac {2 \, B b x + B a + A b}{2 \, {\left (b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int \frac {A+B x}{(a+b x)^3} \, dx=\frac {- A b - B a - 2 B b x}{2 a^{2} b^{2} + 4 a b^{3} x + 2 b^{4} x^{2}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \frac {A+B x}{(a+b x)^3} \, dx=-\frac {2 \, B b x + B a + A b}{2 \, {\left (b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2}\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {A+B x}{(a+b x)^3} \, dx=-\frac {2 \, B b x + B a + A b}{2 \, {\left (b x + a\right )}^{2} b^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int \frac {A+B x}{(a+b x)^3} \, dx=-\frac {\frac {A\,b+B\,a}{2\,b^2}+\frac {B\,x}{b}}{a^2+2\,a\,b\,x+b^2\,x^2} \]

[In]

int((A + B*x)/(a + b*x)^3,x)

[Out]

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