\(\int (-240-8 e^{\frac {6}{5+e^3}}+32 x) \, dx\) [7193]

Optimal result

Integrand size = 18, antiderivative size = 20 \[ \int \left (-240-8 e^{\frac {6}{5+e^3}}+32 x\right ) \, dx=\left (-30-e^{\frac {6}{5+e^3}}+4 x\right )^2 \]

[Out]

(4*x-30-exp(3/(exp(3)+5))^2)^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10, number of steps used = 1, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \left (-240-8 e^{\frac {6}{5+e^3}}+32 x\right ) \, dx=16 x^2-8 \left (30+e^{\frac {6}{5+e^3}}\right ) x \]

[In]

Int[-240 - 8*E^(6/(5 + E^3)) + 32*x,x]

[Out]

-8*(30 + E^(6/(5 + E^3)))*x + 16*x^2

Rubi steps \begin{align*} \text {integral}& = -8 \left (30+e^{\frac {6}{5+e^3}}\right ) x+16 x^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15 \[ \int \left (-240-8 e^{\frac {6}{5+e^3}}+32 x\right ) \, dx=-240 x-8 e^{\frac {6}{5+e^3}} x+16 x^2 \]

[In]

Integrate[-240 - 8*E^(6/(5 + E^3)) + 32*x,x]

[Out]

-240*x - 8*E^(6/(5 + E^3))*x + 16*x^2

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00

method	result	size
gosper	\(-8 x \left ({\mathrm e}^{\frac {6}{{\mathrm e}^{3}+5}}-2 x +30\right )\)	\(20\)
norman	\(\left (-8 \,{\mathrm e}^{\frac {6}{{\mathrm e}^{3}+5}}-240\right ) x +16 x^{2}\)	\(22\)
risch	\(-8 \,{\mathrm e}^{\frac {6}{{\mathrm e}^{3}+5}} x +16 x^{2}-240 x\)	\(22\)
default	\(-8 \,{\mathrm e}^{\frac {6}{{\mathrm e}^{3}+5}} x +16 x^{2}-240 x\)	\(24\)
parallelrisch	\(\left (-8 \,{\mathrm e}^{\frac {6}{{\mathrm e}^{3}+5}}-240\right ) x +16 x^{2}\)	\(24\)
parts	\(-8 \,{\mathrm e}^{\frac {6}{{\mathrm e}^{3}+5}} x +16 x^{2}-240 x\)	\(24\)

[In]

int(-8*exp(3/(exp(3)+5))^2+32*x-240,x,method=_RETURNVERBOSE)

[Out]

-8*x*(exp(3/(exp(3)+5))^2-2*x+30)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \left (-240-8 e^{\frac {6}{5+e^3}}+32 x\right ) \, dx=16 \, x^{2} - 8 \, x e^{\left (\frac {6}{e^{3} + 5}\right )} - 240 \, x \]

[In]

integrate(-8*exp(3/(exp(3)+5))^2+32*x-240,x, algorithm="fricas")

[Out]

16*x^2 - 8*x*e^(6/(e^3 + 5)) - 240*x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \left (-240-8 e^{\frac {6}{5+e^3}}+32 x\right ) \, dx=16 x^{2} + x \left (-240 - 8 e^{\frac {6}{5 + e^{3}}}\right ) \]

[In]

integrate(-8*exp(3/(exp(3)+5))**2+32*x-240,x)

[Out]

16*x**2 + x*(-240 - 8*exp(6/(5 + exp(3))))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \left (-240-8 e^{\frac {6}{5+e^3}}+32 x\right ) \, dx=16 \, x^{2} - 8 \, x e^{\left (\frac {6}{e^{3} + 5}\right )} - 240 \, x \]

[In]

integrate(-8*exp(3/(exp(3)+5))^2+32*x-240,x, algorithm="maxima")

[Out]

16*x^2 - 8*x*e^(6/(e^3 + 5)) - 240*x

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \left (-240-8 e^{\frac {6}{5+e^3}}+32 x\right ) \, dx=16 \, x^{2} - 8 \, x e^{\left (\frac {6}{e^{3} + 5}\right )} - 240 \, x \]

[In]

integrate(-8*exp(3/(exp(3)+5))^2+32*x-240,x, algorithm="giac")

[Out]

16*x^2 - 8*x*e^(6/(e^3 + 5)) - 240*x

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \left (-240-8 e^{\frac {6}{5+e^3}}+32 x\right ) \, dx=16\,x^2-x\,\left (8\,{\mathrm {e}}^{\frac {6}{{\mathrm {e}}^3+5}}+240\right ) \]

[In]

int(32*x - 8*exp(6/(exp(3) + 5)) - 240,x)

[Out]

16*x^2 - x*(8*exp(6/(exp(3) + 5)) + 240)