3.68.69 \(\int (50-10 e^{1+2 x}) \, dx\)

Optimal. Leaf size=20 \[ 5 \left (1-e^{1+2 x}+5 (-24+2 x)\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 13, normalized size of antiderivative = 0.65, number of steps used = 2, number of rules used = 1, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2194} \begin {gather*} 50 x-5 e^{2 x+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[50 - 10*E^(1 + 2*x),x]

[Out]

-5*E^(1 + 2*x) + 50*x

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=50 x-10 \int e^{1+2 x} \, dx\\ &=-5 e^{1+2 x}+50 x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 17, normalized size = 0.85 \begin {gather*} -10 \left (\frac {1}{2} e^{1+2 x}-5 x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[50 - 10*E^(1 + 2*x),x]

[Out]

-10*(E^(1 + 2*x)/2 - 5*x)

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fricas [A]  time = 0.60, size = 12, normalized size = 0.60 \begin {gather*} 50 \, x - 5 \, e^{\left (2 \, x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-10*exp(2*x+1)+50,x, algorithm="fricas")

[Out]

50*x - 5*e^(2*x + 1)

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giac [A]  time = 0.23, size = 12, normalized size = 0.60 \begin {gather*} 50 \, x - 5 \, e^{\left (2 \, x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-10*exp(2*x+1)+50,x, algorithm="giac")

[Out]

50*x - 5*e^(2*x + 1)

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maple [A]  time = 0.02, size = 13, normalized size = 0.65

method	result	size
default	\(50 x -5 \,{\mathrm e}^{2 x +1}\)	\(13\)
norman	\(50 x -5 \,{\mathrm e}^{2 x +1}\)	\(13\)
risch	\(50 x -5 \,{\mathrm e}^{2 x +1}\)	\(13\)
derivativedivides	\(-5 \,{\mathrm e}^{2 x +1}+25 \ln \left ({\mathrm e}^{2 x +1}\right )\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-10*exp(2*x+1)+50,x,method=_RETURNVERBOSE)

[Out]

50*x-5*exp(2*x+1)

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maxima [A]  time = 0.40, size = 12, normalized size = 0.60 \begin {gather*} 50 \, x - 5 \, e^{\left (2 \, x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-10*exp(2*x+1)+50,x, algorithm="maxima")

[Out]

50*x - 5*e^(2*x + 1)

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mupad [B]  time = 0.06, size = 12, normalized size = 0.60 \begin {gather*} 50\,x-5\,{\mathrm {e}}^{2\,x}\,\mathrm {e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(50 - 10*exp(2*x + 1),x)

[Out]

50*x - 5*exp(2*x)*exp(1)

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sympy [A]  time = 0.09, size = 10, normalized size = 0.50 \begin {gather*} 50 x - 5 e^{2 x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-10*exp(2*x+1)+50,x)

[Out]

50*x - 5*exp(2*x + 1)

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