3.34.72 \(\int \frac {e^{-x} (124-150 e+330 x+(14+70 e-154 x) \log (\frac {1}{5} (-2-10 e+22 x)))}{1+5 e-11 x} \, dx\)

Optimal. Leaf size=32 \[ 2 e^{-x} \left (1+7 \left (2-\log \left (2 \left (-e+\frac {1}{5} (-1+x)+2 x\right )\right )\right )\right ) \]

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Rubi [A]  time = 0.49, antiderivative size = 31, normalized size of antiderivative = 0.97, number of steps used = 10, number of rules used = 7, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {6741, 6742, 2199, 2194, 2178, 2554, 12} \begin {gather*} 30 e^{-x}-14 e^{-x} \log \left (\frac {22 x}{5}-\frac {2}{5} (1+5 e)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(124 - 150*E + 330*x + (14 + 70*E - 154*x)*Log[(-2 - 10*E + 22*x)/5])/(E^x*(1 + 5*E - 11*x)),x]

[Out]

30/E^x - (14*Log[(-2*(1 + 5*E))/5 + (22*x)/5])/E^x

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$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 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 6741

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-x} \left (124 \left (1-\frac {75 e}{62}\right )+330 x+(14+70 e-154 x) \log \left (\frac {1}{5} (-2-10 e+22 x)\right )\right )}{1+5 e-11 x} \, dx\\ &=\int \left (-\frac {2 e^{-x} (-62+75 e-165 x)}{1+5 e-11 x}+14 e^{-x} \log \left (-\frac {2}{5} (1+5 e)+\frac {22 x}{5}\right )\right ) \, dx\\ &=-\left (2 \int \frac {e^{-x} (-62+75 e-165 x)}{1+5 e-11 x} \, dx\right )+14 \int e^{-x} \log \left (-\frac {2}{5} (1+5 e)+\frac {22 x}{5}\right ) \, dx\\ &=-14 e^{-x} \log \left (-\frac {2}{5} (1+5 e)+\frac {22 x}{5}\right )-2 \int \left (15 e^{-x}-\frac {77 e^{-x}}{1+5 e-11 x}\right ) \, dx-14 \int \frac {11 e^{-x}}{1+5 e-11 x} \, dx\\ &=-14 e^{-x} \log \left (-\frac {2}{5} (1+5 e)+\frac {22 x}{5}\right )-30 \int e^{-x} \, dx\\ &=30 e^{-x}-14 e^{-x} \log \left (-\frac {2}{5} (1+5 e)+\frac {22 x}{5}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 24, normalized size = 0.75 \begin {gather*} 2 e^{-x} \left (15-7 \log \left (-\frac {2}{5} (1+5 e-11 x)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(124 - 150*E + 330*x + (14 + 70*E - 154*x)*Log[(-2 - 10*E + 22*x)/5])/(E^x*(1 + 5*E - 11*x)),x]

[Out]

(2*(15 - 7*Log[(-2*(1 + 5*E - 11*x))/5]))/E^x

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fricas [A]  time = 0.90, size = 23, normalized size = 0.72 \begin {gather*} -14 \, e^{\left (-x\right )} \log \left (\frac {22}{5} \, x - 2 \, e - \frac {2}{5}\right ) + 30 \, e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((70*exp(1)-154*x+14)*log(-2*exp(1)+22/5*x-2/5)-150*exp(1)+330*x+124)/(5*exp(1)-11*x+1)/exp(x),x, al
gorithm="fricas")

[Out]

-14*e^(-x)*log(22/5*x - 2*e - 2/5) + 30*e^(-x)

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giac [A]  time = 0.25, size = 23, normalized size = 0.72 \begin {gather*} -14 \, e^{\left (-x\right )} \log \left (\frac {22}{5} \, x - 2 \, e - \frac {2}{5}\right ) + 30 \, e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((70*exp(1)-154*x+14)*log(-2*exp(1)+22/5*x-2/5)-150*exp(1)+330*x+124)/(5*exp(1)-11*x+1)/exp(x),x, al
gorithm="giac")

[Out]

-14*e^(-x)*log(22/5*x - 2*e - 2/5) + 30*e^(-x)

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maple [A]  time = 0.49, size = 20, normalized size = 0.62

method	result	size
norman	\(\left (30-14 \ln \left (-2 \,{\mathrm e}+\frac {22 x}{5}-\frac {2}{5}\right )\right ) {\mathrm e}^{-x}\)	\(20\)
risch	\(-14 \,{\mathrm e}^{-x} \ln \left (-2 \,{\mathrm e}+\frac {22 x}{5}-\frac {2}{5}\right )+30 \,{\mathrm e}^{-x}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((70*exp(1)-154*x+14)*ln(-2*exp(1)+22/5*x-2/5)-150*exp(1)+330*x+124)/(5*exp(1)-11*x+1)/exp(x),x,method=_RE
TURNVERBOSE)

[Out]

(30-14*ln(-2*exp(1)+22/5*x-2/5))/exp(x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {150}{11} \, e^{\left (-\frac {5}{11} \, e + \frac {10}{11}\right )} E_{1}\left (x - \frac {5}{11} \, e - \frac {1}{11}\right ) + \frac {124}{11} \, e^{\left (-\frac {5}{11} \, e - \frac {1}{11}\right )} E_{1}\left (x - \frac {5}{11} \, e - \frac {1}{11}\right ) + \frac {2 \, {\left (11 \, x {\left (7 \, \log \relax (5) - 7 \, \log \relax (2) + 15\right )} - 7 \, {\left (11 \, x - 5 \, e - 1\right )} \log \left (11 \, x - 5 \, e - 1\right )\right )} e^{\left (-x\right )}}{11 \, x - 5 \, e - 1} - 2 \, \int -\frac {{\left (77 \, {\left (5 \, {\left (\log \relax (5) - \log \relax (2)\right )} e + \log \relax (5) - \log \relax (2) + 11\right )} x - 175 \, {\left (\log \relax (5) - \log \relax (2)\right )} e^{2} + 5 \, {\left (63 \, \log \relax (5) - 63 \, \log \relax (2) + 88\right )} e + 70 \, \log \relax (5) - 70 \, \log \relax (2) + 88\right )} e^{\left (-x\right )}}{121 \, x^{2} - 22 \, x {\left (5 \, e + 1\right )} + 25 \, e^{2} + 10 \, e + 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((70*exp(1)-154*x+14)*log(-2*exp(1)+22/5*x-2/5)-150*exp(1)+330*x+124)/(5*exp(1)-11*x+1)/exp(x),x, al
gorithm="maxima")

[Out]

-150/11*e^(-5/11*e + 10/11)*exp_integral_e(1, x - 5/11*e - 1/11) + 124/11*e^(-5/11*e - 1/11)*exp_integral_e(1,
x - 5/11*e - 1/11) + 2*(11*x*(7*log(5) - 7*log(2) + 15) - 7*(11*x - 5*e - 1)*log(11*x - 5*e - 1))*e^(-x)/(11*
x - 5*e - 1) - 2*integrate(-(77*(5*(log(5) - log(2))*e + log(5) - log(2) + 11)*x - 175*(log(5) - log(2))*e^2 +
5*(63*log(5) - 63*log(2) + 88)*e + 70*log(5) - 70*log(2) + 88)*e^(-x)/(121*x^2 - 22*x*(5*e + 1) + 25*e^2 + 10
*e + 1), x)

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mupad [B]  time = 0.30, size = 20, normalized size = 0.62 \begin {gather*} -2\,{\mathrm {e}}^{-x}\,\left (7\,\ln \left (\frac {22\,x}{5}-2\,\mathrm {e}-\frac {2}{5}\right )-15\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-x)*(330*x - 150*exp(1) + log((22*x)/5 - 2*exp(1) - 2/5)*(70*exp(1) - 154*x + 14) + 124))/(5*exp(1) -
11*x + 1),x)

[Out]

-2*exp(-x)*(7*log((22*x)/5 - 2*exp(1) - 2/5) - 15)

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sympy [A]  time = 0.53, size = 20, normalized size = 0.62 \begin {gather*} \left (30 - 14 \log {\left (\frac {22 x}{5} - 2 e - \frac {2}{5} \right )}\right ) e^{- x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((70*exp(1)-154*x+14)*ln(-2*exp(1)+22/5*x-2/5)-150*exp(1)+330*x+124)/(5*exp(1)-11*x+1)/exp(x),x)

[Out]

(30 - 14*log(22*x/5 - 2*E - 2/5))*exp(-x)

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