3.70.77 \(\int \frac {e^x (-8+x)+x+(-8+x) \log (2)+(-8+x) \log (8-x)}{-8+x} \, dx\)

Optimal. Leaf size=16 \[ 14+e^x+x (\log (2)+\log (8-x)) \]

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Rubi [A]  time = 0.09, antiderivative size = 29, normalized size of antiderivative = 1.81, number of steps used = 7, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {6688, 2194, 43, 2389, 2295} \begin {gather*} e^x+x \log (2)-(8-x) \log (8-x)+8 \log (8-x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(-8 + x) + x + (-8 + x)*Log[2] + (-8 + x)*Log[8 - x])/(-8 + x),x]

[Out]

E^x + x*Log[2] + 8*Log[8 - x] - (8 - x)*Log[8 - x]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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]

Rule 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (e^x+\frac {x}{-8+x}+\log (2)+\log (8-x)\right ) \, dx\\ &=x \log (2)+\int e^x \, dx+\int \frac {x}{-8+x} \, dx+\int \log (8-x) \, dx\\ &=e^x+x \log (2)+\int \left (1+\frac {8}{-8+x}\right ) \, dx-\operatorname {Subst}(\int \log (x) \, dx,x,8-x)\\ &=e^x+x \log (2)+8 \log (8-x)-(8-x) \log (8-x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 27, normalized size = 1.69 \begin {gather*} e^x+x \log (2)-(8-x) \log (8-x)+8 \log (-8+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(-8 + x) + x + (-8 + x)*Log[2] + (-8 + x)*Log[8 - x])/(-8 + x),x]

[Out]

E^x + x*Log[2] - (8 - x)*Log[8 - x] + 8*Log[-8 + x]

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fricas [A]  time = 0.97, size = 15, normalized size = 0.94 \begin {gather*} x \log \relax (2) + x \log \left (-x + 8\right ) + e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8+x)*log(8-x)+(-8+x)*exp(x)+(-8+x)*log(2)+x)/(-8+x),x, algorithm="fricas")

[Out]

x*log(2) + x*log(-x + 8) + e^x

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giac [A]  time = 0.13, size = 15, normalized size = 0.94 \begin {gather*} x \log \relax (2) + x \log \left (-x + 8\right ) + e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8+x)*log(8-x)+(-8+x)*exp(x)+(-8+x)*log(2)+x)/(-8+x),x, algorithm="giac")

[Out]

x*log(2) + x*log(-x + 8) + e^x

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maple [A]  time = 0.09, size = 16, normalized size = 1.00

method	result	size
norman	\(x \ln \relax (2)+\ln \left (8-x \right ) x +{\mathrm e}^{x}\)	\(16\)
risch	\(x \ln \relax (2)+\ln \left (8-x \right ) x +{\mathrm e}^{x}\)	\(16\)
default	\(x \ln \relax (2)+8 \ln \left (-8+x \right )+{\mathrm e}^{x}-\ln \left (8-x \right ) \left (8-x \right )+8\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-8+x)*ln(8-x)+(-8+x)*exp(x)+(-8+x)*ln(2)+x)/(-8+x),x,method=_RETURNVERBOSE)

[Out]

x*ln(2)+ln(8-x)*x+exp(x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 8 \, e^{8} E_{1}\left (-x + 8\right ) + {\left (x + 8 \, \log \left (x - 8\right )\right )} \log \relax (2) - 8 \, \log \relax (2) \log \left (x - 8\right ) - 4 \, \log \left (x - 8\right )^{2} + {\left (x + 8 \, \log \left (x - 8\right )\right )} \log \left (-x + 8\right ) - 4 \, \log \left (-x + 8\right )^{2} + \frac {x e^{x}}{x - 8} + 8 \, \int \frac {e^{x}}{x^{2} - 16 \, x + 64}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8+x)*log(8-x)+(-8+x)*exp(x)+(-8+x)*log(2)+x)/(-8+x),x, algorithm="maxima")

[Out]

8*e^8*exp_integral_e(1, -x + 8) + (x + 8*log(x - 8))*log(2) - 8*log(2)*log(x - 8) - 4*log(x - 8)^2 + (x + 8*lo
g(x - 8))*log(-x + 8) - 4*log(-x + 8)^2 + x*e^x/(x - 8) + 8*integrate(e^x/(x^2 - 16*x + 64), x)

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mupad [B]  time = 4.13, size = 11, normalized size = 0.69 \begin {gather*} {\mathrm {e}}^x+x\,\ln \left (16-2\,x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + exp(x)*(x - 8) + log(2)*(x - 8) + log(8 - x)*(x - 8))/(x - 8),x)

[Out]

exp(x) + x*log(16 - 2*x)

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sympy [A]  time = 0.29, size = 14, normalized size = 0.88 \begin {gather*} x \log {\left (8 - x \right )} + x \log {\relax (2 )} + e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8+x)*ln(8-x)+(-8+x)*exp(x)+(-8+x)*ln(2)+x)/(-8+x),x)

[Out]

x*log(8 - x) + x*log(2) + exp(x)

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