3.1.29 \(\int (10 e)^x \, dx\) [29]

Optimal. Leaf size=12 \[ \frac {(10 e)^x}{1+\log (10)} \]

[Out]

(10*exp(1))^x/(1+ln(10))

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Rubi [A]
time = 0.00, antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2225} \begin {gather*} \frac {(10 e)^x}{1+\log (10)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(10*E)^x,x]

[Out]

(10*E)^x/(1 + Log[10])

Rule 2225

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 {align*} \int (10 e)^x \, dx &=\frac {(10 e)^x}{1+\log (10)}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 12, normalized size = 1.00 \begin {gather*} \frac {(10 e)^x}{\log (10 e)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10*E)^x,x]

[Out]

(10*E)^x/Log[10*E]

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Mathics [A]
time = 1.75, size = 12, normalized size = 1.00 \begin {gather*} \frac {\left (10 E\right )^x}{1+\text {Log}\left [10\right ]} \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[10^x*E^x,x]')

[Out]

(10 E) ^ x / (1 + Log[10])

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Maple [A]
time = 0.02, size = 15, normalized size = 1.25

method	result	size
gosper	\(\frac {\left (10 \,{\mathrm e}\right )^{x}}{\ln \left (10 \,{\mathrm e}\right )}\)	\(15\)
derivativedivides	\(\frac {\left (10 \,{\mathrm e}\right )^{x}}{\ln \left (10 \,{\mathrm e}\right )}\)	\(15\)
default	\(\frac {\left (10 \,{\mathrm e}\right )^{x}}{\ln \left (10 \,{\mathrm e}\right )}\)	\(15\)
norman	\(\frac {{\mathrm e}^{x \ln \left (10 \,{\mathrm e}\right )}}{1+\ln \left (10\right )}\)	\(16\)
risch	\(\frac {5^{x} 2^{x} {\mathrm e}^{x}}{1+\ln \left (2\right )+\ln \left (5\right )}\)	\(18\)
meijerg	\(-\frac {1-{\mathrm e}^{x \left (1+\ln \left (10\right )\right )}}{1+\ln \left (10\right )}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*exp(1))^x,x,method=_RETURNVERBOSE)

[Out]

1/ln(10*exp(1))*(10*exp(1))^x

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Maxima [A]
time = 0.25, size = 14, normalized size = 1.17 \begin {gather*} \frac {\left (10 \, e\right )^{x}}{\log \left (10 \, e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*exp(1))^x,x, algorithm="maxima")

[Out]

(10*e)^x/log(10*e)

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Fricas [A]
time = 0.32, size = 14, normalized size = 1.17 \begin {gather*} \frac {e^{\left (x \log \left (10\right ) + x\right )}}{\log \left (10\right ) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*exp(1))^x,x, algorithm="fricas")

[Out]

e^(x*log(10) + x)/(log(10) + 1)

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Sympy [A]
time = 0.05, size = 10, normalized size = 0.83 \begin {gather*} \frac {\left (10 e\right )^{x}}{1 + \log {\left (10 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*E)**x,x)

[Out]

(10*E)**x/(1 + log(10))

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Giac [A]
time = 0.00, size = 12, normalized size = 1.00 \begin {gather*} \frac {\left (10 \mathrm {e}\right )^{x}}{\ln \left (10\right )+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*exp(1))^x,x)

[Out]

(10*e)^x/(log(10) + 1)

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Mupad [B]
time = 0.07, size = 12, normalized size = 1.00 \begin {gather*} \frac {{10}^x\,{\mathrm {e}}^x}{\ln \left (10\right )+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*exp(1))^x,x)

[Out]

(10^x*exp(x))/(log(10) + 1)

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