\(\int e^{15/4} (10 x-15 x^2) \, dx\) [9490]

Optimal result

Integrand size = 15, antiderivative size = 20 \[ \int e^{15/4} \left (10 x-15 x^2\right ) \, dx=5 e^{15/4} x \left (x-x^2\right )-\log (4) \]

[Out]

5*(-x^2+x)*x*exp(15/4)-2*ln(2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {12} \[ \int e^{15/4} \left (10 x-15 x^2\right ) \, dx=5 e^{15/4} x^2-5 e^{15/4} x^3 \]

[In]

Int[E^(15/4)*(10*x - 15*x^2),x]

[Out]

5*E^(15/4)*x^2 - 5*E^(15/4)*x^3

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}& = e^{15/4} \int \left (10 x-15 x^2\right ) \, dx \\ & = 5 e^{15/4} x^2-5 e^{15/4} x^3 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int e^{15/4} \left (10 x-15 x^2\right ) \, dx=-5 e^{15/4} \left (-x^2+x^3\right ) \]

[In]

Integrate[E^(15/4)*(10*x - 15*x^2),x]

[Out]

-5*E^(15/4)*(-x^2 + x^3)

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.55

method	result	size
gosper	\(-5 \,{\mathrm e}^{\frac {15}{4}} x^{2} \left (-1+x \right )\)	\(11\)
default	\(5 \,{\mathrm e}^{\frac {15}{4}} \left (-x^{3}+x^{2}\right )\)	\(14\)
parallelrisch	\({\mathrm e}^{\frac {15}{4}} \left (-5 x^{3}+5 x^{2}\right )\)	\(15\)
norman	\(5 \,{\mathrm e}^{\frac {15}{4}} x^{2}-5 \,{\mathrm e}^{\frac {15}{4}} x^{3}\)	\(16\)
risch	\(5 \,{\mathrm e}^{\frac {15}{4}} x^{2}-5 \,{\mathrm e}^{\frac {15}{4}} x^{3}\)	\(16\)

[In]

int((-15*x^2+10*x)*exp(15/4),x,method=_RETURNVERBOSE)

[Out]

-5*exp(15/4)*x^2*(-1+x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.65 \[ \int e^{15/4} \left (10 x-15 x^2\right ) \, dx=-5 \, {\left (x^{3} - x^{2}\right )} e^{\frac {15}{4}} \]

[In]

integrate((-15*x^2+10*x)*exp(15/4),x, algorithm="fricas")

[Out]

-5*(x^3 - x^2)*e^(15/4)

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int e^{15/4} \left (10 x-15 x^2\right ) \, dx=- 5 x^{3} e^{\frac {15}{4}} + 5 x^{2} e^{\frac {15}{4}} \]

[In]

integrate((-15*x**2+10*x)*exp(15/4),x)

[Out]

-5*x**3*exp(15/4) + 5*x**2*exp(15/4)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.65 \[ \int e^{15/4} \left (10 x-15 x^2\right ) \, dx=-5 \, {\left (x^{3} - x^{2}\right )} e^{\frac {15}{4}} \]

[In]

integrate((-15*x^2+10*x)*exp(15/4),x, algorithm="maxima")

[Out]

-5*(x^3 - x^2)*e^(15/4)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.65 \[ \int e^{15/4} \left (10 x-15 x^2\right ) \, dx=-5 \, {\left (x^{3} - x^{2}\right )} e^{\frac {15}{4}} \]

[In]

integrate((-15*x^2+10*x)*exp(15/4),x, algorithm="giac")

[Out]

-5*(x^3 - x^2)*e^(15/4)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.50 \[ \int e^{15/4} \left (10 x-15 x^2\right ) \, dx=-5\,x^2\,{\mathrm {e}}^{15/4}\,\left (x-1\right ) \]

[In]

int(exp(15/4)*(10*x - 15*x^2),x)

[Out]

-5*x^2*exp(15/4)*(x - 1)