3.40.74 \(\int \frac {1}{3} (e^x (1+x)-5^x \log (5)) \, dx\)

Optimal. Leaf size=27 \[ \frac {1}{3} \left (-3-5^x-e^{12-\frac {2}{e^2}}+e^x x\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 25, normalized size of antiderivative = 0.93, number of steps used = 5, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {12, 2176, 2194} \begin {gather*} \frac {1}{3} e^x (x+1)-\frac {5^x}{3}-\frac {e^x}{3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(1 + x) - 5^x*Log[5])/3,x]

[Out]

-1/3*5^x - E^x/3 + (E^x*(1 + 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]]

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

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} &=\frac {1}{3} \int \left (e^x (1+x)-5^x \log (5)\right ) \, dx\\ &=\frac {1}{3} \int e^x (1+x) \, dx-\frac {1}{3} \log (5) \int 5^x \, dx\\ &=-\frac {5^x}{3}+\frac {1}{3} e^x (1+x)-\frac {\int e^x \, dx}{3}\\ &=-\frac {5^x}{3}-\frac {e^x}{3}+\frac {1}{3} e^x (1+x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 15, normalized size = 0.56 \begin {gather*} \frac {1}{3} \left (-5^x+e^x x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(1 + x) - 5^x*Log[5])/3,x]

[Out]

(-5^x + E^x*x)/3

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fricas [A]  time = 0.50, size = 11, normalized size = 0.41 \begin {gather*} \frac {1}{3} \, x e^{x} - \frac {1}{3} \cdot 5^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-1/3*log(5)*exp(x*log(5))+1/3*(x+1)*exp(x),x, algorithm="fricas")

[Out]

1/3*x*e^x - 1/3*5^x

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giac [A]  time = 0.15, size = 11, normalized size = 0.41 \begin {gather*} \frac {1}{3} \, x e^{x} - \frac {1}{3} \cdot 5^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-1/3*log(5)*exp(x*log(5))+1/3*(x+1)*exp(x),x, algorithm="giac")

[Out]

1/3*x*e^x - 1/3*5^x

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maple [A]  time = 0.04, size = 12, normalized size = 0.44

method	result	size
risch	\(\frac {{\mathrm e}^{x} x}{3}-\frac {5^{x}}{3}\)	\(12\)
default	\(\frac {{\mathrm e}^{x} x}{3}-\frac {{\mathrm e}^{x \ln \relax (5)}}{3}\)	\(14\)
norman	\(\frac {{\mathrm e}^{x} x}{3}-\frac {{\mathrm e}^{x \ln \relax (5)}}{3}\)	\(14\)
meijerg	\(-\frac {{\mathrm e}^{x \ln \relax (5)}}{3}+\frac {{\mathrm e}^{x}}{3}+\frac {1}{3}-\frac {\left (-2 x +2\right ) {\mathrm e}^{x}}{6}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/3*ln(5)*exp(x*ln(5))+1/3*(x+1)*exp(x),x,method=_RETURNVERBOSE)

[Out]

1/3*exp(x)*x-1/3*5^x

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maxima [A]  time = 0.42, size = 17, normalized size = 0.63 \begin {gather*} \frac {1}{3} \, {\left (x - 1\right )} e^{x} - \frac {1}{3} \cdot 5^{x} + \frac {1}{3} \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-1/3*log(5)*exp(x*log(5))+1/3*(x+1)*exp(x),x, algorithm="maxima")

[Out]

1/3*(x - 1)*e^x - 1/3*5^x + 1/3*e^x

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mupad [B]  time = 2.50, size = 11, normalized size = 0.41 \begin {gather*} \frac {x\,{\mathrm {e}}^x}{3}-\frac {5^x}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(x + 1))/3 - (exp(x*log(5))*log(5))/3,x)

[Out]

(x*exp(x))/3 - 5^x/3

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sympy [A]  time = 0.11, size = 14, normalized size = 0.52 \begin {gather*} \frac {x e^{x}}{3} - \frac {e^{x \log {\relax (5 )}}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-1/3*ln(5)*exp(x*ln(5))+1/3*(x+1)*exp(x),x)

[Out]

x*exp(x)/3 - exp(x*log(5))/3

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