3.39.31 \(\int \frac {1}{2} (-5+60 e^3) \, dx\)

Optimal. Leaf size=17 \[ 1-\frac {5 x}{2}+6 e^3 (5+5 x) \]

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Rubi [A]  time = 0.00, antiderivative size = 12, normalized size of antiderivative = 0.71, number of steps used = 1, number of rules used = 1, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {8} \begin {gather*} -\frac {5}{2} \left (1-12 e^3\right ) x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-5 + 60*E^3)/2,x]

[Out]

(-5*(1 - 12*E^3)*x)/2

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\frac {5}{2} \left (1-12 e^3\right ) x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 12, normalized size = 0.71 \begin {gather*} -\frac {5 x}{2}+30 e^3 x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5 + 60*E^3)/2,x]

[Out]

(-5*x)/2 + 30*E^3*x

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fricas [A]  time = 0.88, size = 12, normalized size = 0.71 \begin {gather*} 10 \, x e^{\left (\log \relax (3) + 3\right )} - \frac {5}{2} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(10*exp(3+log(3))-5/2,x, algorithm="fricas")

[Out]

10*x*e^(log(3) + 3) - 5/2*x

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giac [A]  time = 0.22, size = 12, normalized size = 0.71 \begin {gather*} \frac {5}{2} \, x {\left (4 \, e^{\left (\log \relax (3) + 3\right )} - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(10*exp(3+log(3))-5/2,x, algorithm="giac")

[Out]

5/2*x*(4*e^(log(3) + 3) - 1)

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maple [A]  time = 0.01, size = 9, normalized size = 0.53

method	result	size
norman	\(\left (30 \,{\mathrm e}^{3}-\frac {5}{2}\right ) x\)	\(9\)
risch	\(30 x \,{\mathrm e}^{3}-\frac {5 x}{2}\)	\(10\)
default	\(\left (10 \,{\mathrm e}^{3+\ln \relax (3)}-\frac {5}{2}\right ) x\)	\(12\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(10*exp(3+ln(3))-5/2,x,method=_RETURNVERBOSE)

[Out]

(30*exp(3)-5/2)*x

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maxima [A]  time = 0.46, size = 9, normalized size = 0.53 \begin {gather*} \frac {5}{2} \, x {\left (12 \, e^{3} - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(10*exp(3+log(3))-5/2,x, algorithm="maxima")

[Out]

5/2*x*(12*e^3 - 1)

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mupad [B]  time = 0.00, size = 11, normalized size = 0.65 \begin {gather*} x\,\left (10\,{\mathrm {e}}^{\ln \relax (3)+3}-\frac {5}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(10*exp(log(3) + 3) - 5/2,x)

[Out]

x*(10*exp(log(3) + 3) - 5/2)

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sympy [A]  time = 0.05, size = 8, normalized size = 0.47 \begin {gather*} x \left (- \frac {5}{2} + 30 e^{3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(10*exp(3+ln(3))-5/2,x)

[Out]

x*(-5/2 + 30*exp(3))

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