\(\int \frac {1}{4} (-267+510 x+e^3 (-4+170 x)) \, dx\) [2393]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 18, antiderivative size = 18 \[ \int \frac {1}{4} \left (-267+510 x+e^3 (-4+170 x)\right ) \, dx=\left (-1+\frac {85 x}{4}\right ) \left (-3+3 x+e^3 x\right ) \]

[Out]

(85/4*x-1)*(x*exp(3)-3+3*x)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.50, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {12} \[ \int \frac {1}{4} \left (-267+510 x+e^3 (-4+170 x)\right ) \, dx=\frac {255 x^2}{4}+\frac {1}{340} e^3 (2-85 x)^2-\frac {267 x}{4} \]

[In]

Int[(-267 + 510*x + E^3*(-4 + 170*x))/4,x]

[Out]

(E^3*(2 - 85*x)^2)/340 - (267*x)/4 + (255*x^2)/4

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}{4} \int \left (-267+510 x+e^3 (-4+170 x)\right ) \, dx \\ & = \frac {1}{340} e^3 (2-85 x)^2-\frac {267 x}{4}+\frac {255 x^2}{4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.50 \[ \int \frac {1}{4} \left (-267+510 x+e^3 (-4+170 x)\right ) \, dx=\frac {1}{4} \left (-267 x-4 e^3 x+255 x^2+85 e^3 x^2\right ) \]

[In]

Integrate[(-267 + 510*x + E^3*(-4 + 170*x))/4,x]

[Out]

(-267*x - 4*E^3*x + 255*x^2 + 85*E^3*x^2)/4

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00

method result size
gosper \(\frac {x \left (85 x \,{\mathrm e}^{3}-4 \,{\mathrm e}^{3}+255 x -267\right )}{4}\) \(18\)
norman \(\left (-{\mathrm e}^{3}-\frac {267}{4}\right ) x +\left (\frac {85 \,{\mathrm e}^{3}}{4}+\frac {255}{4}\right ) x^{2}\) \(20\)
default \(\frac {85 x^{2} {\mathrm e}^{3}}{4}-x \,{\mathrm e}^{3}+\frac {255 x^{2}}{4}-\frac {267 x}{4}\) \(22\)
risch \(\frac {85 x^{2} {\mathrm e}^{3}}{4}-x \,{\mathrm e}^{3}+\frac {255 x^{2}}{4}-\frac {267 x}{4}\) \(22\)
parallelrisch \(\frac {85 x^{2} {\mathrm e}^{3}}{4}-x \,{\mathrm e}^{3}+\frac {255 x^{2}}{4}-\frac {267 x}{4}\) \(22\)
parts \(\frac {85 x^{2} {\mathrm e}^{3}}{4}-x \,{\mathrm e}^{3}+\frac {255 x^{2}}{4}-\frac {267 x}{4}\) \(22\)

[In]

int(1/4*(170*x-4)*exp(3)+255/2*x-267/4,x,method=_RETURNVERBOSE)

[Out]

1/4*x*(85*x*exp(3)-4*exp(3)+255*x-267)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.22 \[ \int \frac {1}{4} \left (-267+510 x+e^3 (-4+170 x)\right ) \, dx=\frac {255}{4} \, x^{2} + \frac {1}{4} \, {\left (85 \, x^{2} - 4 \, x\right )} e^{3} - \frac {267}{4} \, x \]

[In]

integrate(1/4*(170*x-4)*exp(3)+255/2*x-267/4,x, algorithm="fricas")

[Out]

255/4*x^2 + 1/4*(85*x^2 - 4*x)*e^3 - 267/4*x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.22 \[ \int \frac {1}{4} \left (-267+510 x+e^3 (-4+170 x)\right ) \, dx=x^{2} \cdot \left (\frac {255}{4} + \frac {85 e^{3}}{4}\right ) + x \left (- \frac {267}{4} - e^{3}\right ) \]

[In]

integrate(1/4*(170*x-4)*exp(3)+255/2*x-267/4,x)

[Out]

x**2*(255/4 + 85*exp(3)/4) + x*(-267/4 - exp(3))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.22 \[ \int \frac {1}{4} \left (-267+510 x+e^3 (-4+170 x)\right ) \, dx=\frac {255}{4} \, x^{2} + \frac {1}{4} \, {\left (85 \, x^{2} - 4 \, x\right )} e^{3} - \frac {267}{4} \, x \]

[In]

integrate(1/4*(170*x-4)*exp(3)+255/2*x-267/4,x, algorithm="maxima")

[Out]

255/4*x^2 + 1/4*(85*x^2 - 4*x)*e^3 - 267/4*x

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.22 \[ \int \frac {1}{4} \left (-267+510 x+e^3 (-4+170 x)\right ) \, dx=\frac {255}{4} \, x^{2} + \frac {1}{4} \, {\left (85 \, x^{2} - 4 \, x\right )} e^{3} - \frac {267}{4} \, x \]

[In]

integrate(1/4*(170*x-4)*exp(3)+255/2*x-267/4,x, algorithm="giac")

[Out]

255/4*x^2 + 1/4*(85*x^2 - 4*x)*e^3 - 267/4*x

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {1}{4} \left (-267+510 x+e^3 (-4+170 x)\right ) \, dx=x^2\,\left (\frac {85\,{\mathrm {e}}^3}{4}+\frac {255}{4}\right )-x\,\left ({\mathrm {e}}^3+\frac {267}{4}\right ) \]

[In]

int((255*x)/2 + (exp(3)*(170*x - 4))/4 - 267/4,x)

[Out]

x^2*((85*exp(3))/4 + 255/4) - x*(exp(3) + 267/4)