\(\int \frac {1-2 x}{(3+5 x)^3} \, dx\) [1229]

Optimal result

Integrand size = 13, antiderivative size = 18 \[ \int \frac {1-2 x}{(3+5 x)^3} \, dx=-\frac {(1-2 x)^2}{22 (3+5 x)^2} \]

[Out]

-1/22*(1-2*x)^2/(3+5*x)^2

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 18, 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} \[ \int \frac {1-2 x}{(3+5 x)^3} \, dx=-\frac {(1-2 x)^2}{22 (5 x+3)^2} \]

[In]

Int[(1 - 2*x)/(3 + 5*x)^3,x]

[Out]

-1/22*(1 - 2*x)^2/(3 + 5*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 {(1-2 x)^2}{22 (3+5 x)^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {1-2 x}{(3+5 x)^3} \, dx=\frac {1+20 x}{50 (3+5 x)^2} \]

[In]

Integrate[(1 - 2*x)/(3 + 5*x)^3,x]

[Out]

(1 + 20*x)/(50*(3 + 5*x)^2)

Maple [A] (verified)

Time = 1.82 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83

method	result	size
gosper	\(\frac {1+20 x}{50 \left (3+5 x \right )^{2}}\)	\(15\)
risch	\(\frac {\frac {2 x}{5}+\frac {1}{50}}{\left (3+5 x \right )^{2}}\)	\(15\)
norman	\(\frac {\frac {1}{3} x -\frac {1}{18} x^{2}}{\left (3+5 x \right )^{2}}\)	\(18\)
parallelrisch	\(\frac {-x^{2}+6 x}{18 \left (3+5 x \right )^{2}}\)	\(19\)
default	\(\frac {2}{25 \left (3+5 x \right )}-\frac {11}{50 \left (3+5 x \right )^{2}}\)	\(20\)
meijerg	\(\frac {x \left (\frac {5 x}{3}+2\right )}{54 \left (1+\frac {5 x}{3}\right )^{2}}-\frac {x^{2}}{27 \left (1+\frac {5 x}{3}\right )^{2}}\)	\(29\)

[In]

int((1-2*x)/(3+5*x)^3,x,method=_RETURNVERBOSE)

[Out]

1/50*(1+20*x)/(3+5*x)^2

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \frac {1-2 x}{(3+5 x)^3} \, dx=\frac {20 \, x + 1}{50 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]

[In]

integrate((1-2*x)/(3+5*x)^3,x, algorithm="fricas")

[Out]

1/50*(20*x + 1)/(25*x^2 + 30*x + 9)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {1-2 x}{(3+5 x)^3} \, dx=- \frac {- 20 x - 1}{1250 x^{2} + 1500 x + 450} \]

[In]

integrate((1-2*x)/(3+5*x)**3,x)

[Out]

-(-20*x - 1)/(1250*x**2 + 1500*x + 450)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \frac {1-2 x}{(3+5 x)^3} \, dx=\frac {20 \, x + 1}{50 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]

[In]

integrate((1-2*x)/(3+5*x)^3,x, algorithm="maxima")

[Out]

1/50*(20*x + 1)/(25*x^2 + 30*x + 9)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {1-2 x}{(3+5 x)^3} \, dx=\frac {20 \, x + 1}{50 \, {\left (5 \, x + 3\right )}^{2}} \]

[In]

integrate((1-2*x)/(3+5*x)^3,x, algorithm="giac")

[Out]

1/50*(20*x + 1)/(5*x + 3)^2

Mupad [B] (verification not implemented)

Time = 1.31 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {1-2 x}{(3+5 x)^3} \, dx=\frac {20\,x+1}{50\,{\left (5\,x+3\right )}^2} \]

[In]

int(-(2*x - 1)/(5*x + 3)^3,x)

[Out]

(20*x + 1)/(50*(5*x + 3)^2)