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

Optimal. Leaf size=20 \[ 5 e^{15/4} x \left (x-x^2\right )-\log (4) \]

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Rubi [A]  time = 0.00, antiderivative size = 21, normalized size of antiderivative = 1.05, number of steps used = 2, number of rules used = 1, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {12} \begin {gather*} 5 e^{15/4} x^2-5 e^{15/4} x^3 \end {gather*}

Antiderivative was successfully verified.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 16, normalized size = 0.80 \begin {gather*} -5 e^{15/4} \left (-x^2+x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(15/4)*(10*x - 15*x^2),x]

[Out]

-5*E^(15/4)*(-x^2 + x^3)

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fricas [A]  time = 0.85, size = 13, normalized size = 0.65 \begin {gather*} -5 \, {\left (x^{3} - x^{2}\right )} e^{\frac {15}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5*(x^3 - x^2)*e^(15/4)

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giac [A]  time = 1.64, size = 13, normalized size = 0.65 \begin {gather*} -5 \, {\left (x^{3} - x^{2}\right )} e^{\frac {15}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5*(x^3 - x^2)*e^(15/4)

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maple [A]  time = 0.02, size = 11, normalized size = 0.55

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.35, size = 13, normalized size = 0.65 \begin {gather*} -5 \, {\left (x^{3} - x^{2}\right )} e^{\frac {15}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5*(x^3 - x^2)*e^(15/4)

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mupad [B]  time = 0.04, size = 10, normalized size = 0.50 \begin {gather*} -5\,x^2\,{\mathrm {e}}^{15/4}\,\left (x-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(15/4)*(10*x - 15*x^2),x)

[Out]

-5*x^2*exp(15/4)*(x - 1)

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sympy [A]  time = 0.05, size = 19, normalized size = 0.95 \begin {gather*} - 5 x^{3} e^{\frac {15}{4}} + 5 x^{2} e^{\frac {15}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5*x**3*exp(15/4) + 5*x**2*exp(15/4)

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