3.77.78 \(\int \frac {1}{3} (1+2 x+e^{5+x} (1+x)-\log (\frac {x}{4})) \, dx\)

Optimal. Leaf size=26 \[ 2+\frac {1}{3} \left (-1+x \left (2+e^{5+x}+x-\log \left (\frac {x}{4}\right )\right )\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 45, normalized size of antiderivative = 1.73, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {12, 2176, 2194, 2295} \begin {gather*} \frac {x^2}{3}+\frac {2 x}{3}-\frac {e^{x+5}}{3}+\frac {1}{3} e^{x+5} (x+1)-\frac {1}{3} x \log \left (\frac {x}{4}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + 2*x + E^(5 + x)*(1 + x) - Log[x/4])/3,x]

[Out]

-1/3*E^(5 + x) + (2*x)/3 + x^2/3 + (E^(5 + x)*(1 + x))/3 - (x*Log[x/4])/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]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{3} \int \left (1+2 x+e^{5+x} (1+x)-\log \left (\frac {x}{4}\right )\right ) \, dx\\ &=\frac {x}{3}+\frac {x^2}{3}+\frac {1}{3} \int e^{5+x} (1+x) \, dx-\frac {1}{3} \int \log \left (\frac {x}{4}\right ) \, dx\\ &=\frac {2 x}{3}+\frac {x^2}{3}+\frac {1}{3} e^{5+x} (1+x)-\frac {1}{3} x \log \left (\frac {x}{4}\right )-\frac {1}{3} \int e^{5+x} \, dx\\ &=-\frac {e^{5+x}}{3}+\frac {2 x}{3}+\frac {x^2}{3}+\frac {1}{3} e^{5+x} (1+x)-\frac {1}{3} x \log \left (\frac {x}{4}\right )\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 27, normalized size = 1.04 \begin {gather*} \frac {1}{3} \left (2 x+e^{5+x} x+x^2+x \log (4)-x \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + 2*x + E^(5 + x)*(1 + x) - Log[x/4])/3,x]

[Out]

(2*x + E^(5 + x)*x + x^2 + x*Log[4] - x*Log[x])/3

________________________________________________________________________________________

fricas [A]  time = 0.65, size = 23, normalized size = 0.88 \begin {gather*} \frac {1}{3} \, x^{2} + \frac {1}{3} \, x e^{\left (x + 5\right )} - \frac {1}{3} \, x \log \left (\frac {1}{4} \, x\right ) + \frac {2}{3} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-1/3*log(1/4*x)+1/3*(x+1)*exp(5+x)+2/3*x+1/3,x, algorithm="fricas")

[Out]

1/3*x^2 + 1/3*x*e^(x + 5) - 1/3*x*log(1/4*x) + 2/3*x

________________________________________________________________________________________

giac [A]  time = 0.14, size = 23, normalized size = 0.88 \begin {gather*} \frac {1}{3} \, x^{2} + \frac {1}{3} \, x e^{\left (x + 5\right )} - \frac {1}{3} \, x \log \left (\frac {1}{4} \, x\right ) + \frac {2}{3} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-1/3*log(1/4*x)+1/3*(x+1)*exp(5+x)+2/3*x+1/3,x, algorithm="giac")

[Out]

1/3*x^2 + 1/3*x*e^(x + 5) - 1/3*x*log(1/4*x) + 2/3*x

________________________________________________________________________________________

maple [A]  time = 0.02, size = 24, normalized size = 0.92

method	result	size
norman	\(\frac {2 x}{3}+\frac {x^{2}}{3}+\frac {x \,{\mathrm e}^{5+x}}{3}-\frac {x \ln \left (\frac {x}{4}\right )}{3}\)	\(24\)
risch	\(\frac {2 x}{3}+\frac {x^{2}}{3}+\frac {x \,{\mathrm e}^{5+x}}{3}-\frac {x \ln \left (\frac {x}{4}\right )}{3}\)	\(24\)
default	\(\frac {x^{2}}{3}+\frac {2 x}{3}+\frac {{\mathrm e}^{5+x} \left (5+x \right )}{3}-\frac {5 \,{\mathrm e}^{5+x}}{3}-\frac {x \ln \left (\frac {x}{4}\right )}{3}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/3*ln(1/4*x)+1/3*(x+1)*exp(5+x)+2/3*x+1/3,x,method=_RETURNVERBOSE)

[Out]

2/3*x+1/3*x^2+1/3*x*exp(5+x)-1/3*x*ln(1/4*x)

________________________________________________________________________________________

maxima [A]  time = 0.36, size = 23, normalized size = 0.88 \begin {gather*} \frac {1}{3} \, x^{2} + \frac {1}{3} \, x e^{\left (x + 5\right )} - \frac {1}{3} \, x \log \left (\frac {1}{4} \, x\right ) + \frac {2}{3} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-1/3*log(1/4*x)+1/3*(x+1)*exp(5+x)+2/3*x+1/3,x, algorithm="maxima")

[Out]

1/3*x^2 + 1/3*x*e^(x + 5) - 1/3*x*log(1/4*x) + 2/3*x

________________________________________________________________________________________

mupad [B]  time = 5.79, size = 16, normalized size = 0.62 \begin {gather*} \frac {x\,\left (x+{\mathrm {e}}^{x+5}-\ln \left (\frac {x}{4}\right )+2\right )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x)/3 - log(x/4)/3 + (exp(x + 5)*(x + 1))/3 + 1/3,x)

[Out]

(x*(x + exp(x + 5) - log(x/4) + 2))/3

________________________________________________________________________________________

sympy [A]  time = 0.26, size = 26, normalized size = 1.00 \begin {gather*} \frac {x^{2}}{3} + \frac {x e^{x + 5}}{3} - \frac {x \log {\left (\frac {x}{4} \right )}}{3} + \frac {2 x}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-1/3*ln(1/4*x)+1/3*(x+1)*exp(5+x)+2/3*x+1/3,x)

[Out]

x**2/3 + x*exp(x + 5)/3 - x*log(x/4)/3 + 2*x/3

________________________________________________________________________________________