\(\int \frac {1}{10} (-9+54 x-108 x^2+e (-18+48 x)) \, dx\) [7815]

Optimal result

Integrand size = 21, antiderivative size = 27 \[ \int \frac {1}{10} \left (-9+54 x-108 x^2+e (-18+48 x)\right ) \, dx=\frac {3}{10} \left ((1-2 x)^2+x\right ) \left (-x+\log \left (e^{2 e-2 x}\right )\right ) \]

[Out]

3*(ln(exp(exp(1)-x)^2)-x)*(1/10*x+1/10*(1-2*x)^2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {12} \[ \int \frac {1}{10} \left (-9+54 x-108 x^2+e (-18+48 x)\right ) \, dx=-\frac {18 x^3}{5}+\frac {27 x^2}{10}-\frac {9 x}{10}+\frac {3}{80} e (3-8 x)^2 \]

[In]

Int[(-9 + 54*x - 108*x^2 + E*(-18 + 48*x))/10,x]

[Out]

(3*E*(3 - 8*x)^2)/80 - (9*x)/10 + (27*x^2)/10 - (18*x^3)/5

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}{10} \int \left (-9+54 x-108 x^2+e (-18+48 x)\right ) \, dx \\ & = \frac {3}{80} e (3-8 x)^2-\frac {9 x}{10}+\frac {27 x^2}{10}-\frac {18 x^3}{5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {1}{10} \left (-9+54 x-108 x^2+e (-18+48 x)\right ) \, dx=\frac {3}{10} \left (-3 x-6 e x+9 x^2+8 e x^2-12 x^3\right ) \]

[In]

Integrate[(-9 + 54*x - 108*x^2 + E*(-18 + 48*x))/10,x]

[Out]

(3*(-3*x - 6*E*x + 9*x^2 + 8*E*x^2 - 12*x^3))/10

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85

method	result	size
gosper	\(\frac {3 x \left (8 x \,{\mathrm e}-12 x^{2}-6 \,{\mathrm e}+9 x -3\right )}{10}\)	\(23\)
norman	\(\left (-\frac {9 \,{\mathrm e}}{5}-\frac {9}{10}\right ) x +\left (\frac {12 \,{\mathrm e}}{5}+\frac {27}{10}\right ) x^{2}-\frac {18 x^{3}}{5}\)	\(25\)
default	\(\frac {12 x^{2} {\mathrm e}}{5}-\frac {18 x^{3}}{5}-\frac {9 x \,{\mathrm e}}{5}+\frac {27 x^{2}}{10}-\frac {9 x}{10}\)	\(27\)
risch	\(\frac {12 x^{2} {\mathrm e}}{5}-\frac {18 x^{3}}{5}-\frac {9 x \,{\mathrm e}}{5}+\frac {27 x^{2}}{10}-\frac {9 x}{10}\)	\(27\)
parallelrisch	\(\frac {12 x^{2} {\mathrm e}}{5}-\frac {18 x^{3}}{5}-\frac {9 x \,{\mathrm e}}{5}+\frac {27 x^{2}}{10}-\frac {9 x}{10}\)	\(27\)
parts	\(\frac {12 x^{2} {\mathrm e}}{5}-\frac {18 x^{3}}{5}-\frac {9 x \,{\mathrm e}}{5}+\frac {27 x^{2}}{10}-\frac {9 x}{10}\)	\(27\)

[In]

int(1/10*(48*x-18)*exp(1)-54/5*x^2+27/5*x-9/10,x,method=_RETURNVERBOSE)

[Out]

3/10*x*(8*x*exp(1)-12*x^2-6*exp(1)+9*x-3)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {1}{10} \left (-9+54 x-108 x^2+e (-18+48 x)\right ) \, dx=-\frac {18}{5} \, x^{3} + \frac {27}{10} \, x^{2} + \frac {3}{5} \, {\left (4 \, x^{2} - 3 \, x\right )} e - \frac {9}{10} \, x \]

[In]

integrate(1/10*(48*x-18)*exp(1)-54/5*x^2+27/5*x-9/10,x, algorithm="fricas")

[Out]

-18/5*x^3 + 27/10*x^2 + 3/5*(4*x^2 - 3*x)*e - 9/10*x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {1}{10} \left (-9+54 x-108 x^2+e (-18+48 x)\right ) \, dx=- \frac {18 x^{3}}{5} + x^{2} \cdot \left (\frac {27}{10} + \frac {12 e}{5}\right ) + x \left (- \frac {9 e}{5} - \frac {9}{10}\right ) \]

[In]

integrate(1/10*(48*x-18)*exp(1)-54/5*x**2+27/5*x-9/10,x)

[Out]

-18*x**3/5 + x**2*(27/10 + 12*E/5) + x*(-9*E/5 - 9/10)

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {1}{10} \left (-9+54 x-108 x^2+e (-18+48 x)\right ) \, dx=-\frac {18}{5} \, x^{3} + \frac {27}{10} \, x^{2} + \frac {3}{5} \, {\left (4 \, x^{2} - 3 \, x\right )} e - \frac {9}{10} \, x \]

[In]

integrate(1/10*(48*x-18)*exp(1)-54/5*x^2+27/5*x-9/10,x, algorithm="maxima")

[Out]

-18/5*x^3 + 27/10*x^2 + 3/5*(4*x^2 - 3*x)*e - 9/10*x

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {1}{10} \left (-9+54 x-108 x^2+e (-18+48 x)\right ) \, dx=-\frac {18}{5} \, x^{3} + \frac {27}{10} \, x^{2} + \frac {3}{5} \, {\left (4 \, x^{2} - 3 \, x\right )} e - \frac {9}{10} \, x \]

[In]

integrate(1/10*(48*x-18)*exp(1)-54/5*x^2+27/5*x-9/10,x, algorithm="giac")

[Out]

-18/5*x^3 + 27/10*x^2 + 3/5*(4*x^2 - 3*x)*e - 9/10*x

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {1}{10} \left (-9+54 x-108 x^2+e (-18+48 x)\right ) \, dx=-\frac {18\,x^3}{5}+\left (\frac {12\,\mathrm {e}}{5}+\frac {27}{10}\right )\,x^2+\left (-\frac {9\,\mathrm {e}}{5}-\frac {9}{10}\right )\,x \]

[In]

int((27*x)/5 - (54*x^2)/5 + (exp(1)*(48*x - 18))/10 - 9/10,x)

[Out]

x^2*((12*exp(1))/5 + 27/10) - (18*x^3)/5 - x*((9*exp(1))/5 + 9/10)