\(\int \frac {1}{25} (-9-12 x-3 x^2) \, dx\) [2346]

Optimal result

Integrand size = 14, antiderivative size = 16 \[ \int \frac {1}{25} \left (-9-12 x-3 x^2\right ) \, dx=2-\frac {1}{25} x (3+x)^2+4 \log (2) \]

[Out]

4*ln(2)-1/25*(3+x)^2*x+2

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.25, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {12} \[ \int \frac {1}{25} \left (-9-12 x-3 x^2\right ) \, dx=-\frac {x^3}{25}-\frac {6 x^2}{25}-\frac {9 x}{25} \]

[In]

Int[(-9 - 12*x - 3*x^2)/25,x]

[Out]

(-9*x)/25 - (6*x^2)/25 - x^3/25

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{25} \int \left (-9-12 x-3 x^2\right ) \, dx \\ & = -\frac {9 x}{25}-\frac {6 x^2}{25}-\frac {x^3}{25} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.25 \[ \int \frac {1}{25} \left (-9-12 x-3 x^2\right ) \, dx=-\frac {3}{25} \left (3 x+2 x^2+\frac {x^3}{3}\right ) \]

[In]

Integrate[(-9 - 12*x - 3*x^2)/25,x]

[Out]

(-3*(3*x + 2*x^2 + x^3/3))/25

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75

method	result	size
gosper	\(-\frac {x \left (x^{2}+6 x +9\right )}{25}\)	\(12\)
default	\(-\frac {1}{25} x^{3}-\frac {6}{25} x^{2}-\frac {9}{25} x\)	\(15\)
norman	\(-\frac {1}{25} x^{3}-\frac {6}{25} x^{2}-\frac {9}{25} x\)	\(15\)
risch	\(-\frac {1}{25} x^{3}-\frac {6}{25} x^{2}-\frac {9}{25} x\)	\(15\)
parallelrisch	\(-\frac {1}{25} x^{3}-\frac {6}{25} x^{2}-\frac {9}{25} x\)	\(15\)
parts	\(-\frac {1}{25} x^{3}-\frac {6}{25} x^{2}-\frac {9}{25} x\)	\(15\)

[In]

int(-3/25*x^2-12/25*x-9/25,x,method=_RETURNVERBOSE)

[Out]

-1/25*x*(x^2+6*x+9)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {1}{25} \left (-9-12 x-3 x^2\right ) \, dx=-\frac {1}{25} \, x^{3} - \frac {6}{25} \, x^{2} - \frac {9}{25} \, x \]

[In]

integrate(-3/25*x^2-12/25*x-9/25,x, algorithm="fricas")

[Out]

-1/25*x^3 - 6/25*x^2 - 9/25*x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06 \[ \int \frac {1}{25} \left (-9-12 x-3 x^2\right ) \, dx=- \frac {x^{3}}{25} - \frac {6 x^{2}}{25} - \frac {9 x}{25} \]

[In]

integrate(-3/25*x**2-12/25*x-9/25,x)

[Out]

-x**3/25 - 6*x**2/25 - 9*x/25

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {1}{25} \left (-9-12 x-3 x^2\right ) \, dx=-\frac {1}{25} \, x^{3} - \frac {6}{25} \, x^{2} - \frac {9}{25} \, x \]

[In]

integrate(-3/25*x^2-12/25*x-9/25,x, algorithm="maxima")

[Out]

-1/25*x^3 - 6/25*x^2 - 9/25*x

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {1}{25} \left (-9-12 x-3 x^2\right ) \, dx=-\frac {1}{25} \, x^{3} - \frac {6}{25} \, x^{2} - \frac {9}{25} \, x \]

[In]

integrate(-3/25*x^2-12/25*x-9/25,x, algorithm="giac")

[Out]

-1/25*x^3 - 6/25*x^2 - 9/25*x

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.50 \[ \int \frac {1}{25} \left (-9-12 x-3 x^2\right ) \, dx=-\frac {x\,{\left (x+3\right )}^2}{25} \]

[In]

int(- (12*x)/25 - (3*x^2)/25 - 9/25,x)

[Out]

-(x*(x + 3)^2)/25