3.16.65 \(\int \frac {-25+e^{e^x+x} (-4+x) \log ^2(20-5 x)+e^x (-8+2 x) \log ^2(20-5 x)}{(-4+x) \log ^2(20-5 x)} \, dx\)

Optimal. Leaf size=24 \[ 3+e^{e^x}+2 e^x+\frac {25}{\log (5 (4-x))} \]

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Rubi [A]  time = 0.36, antiderivative size = 23, normalized size of antiderivative = 0.96, number of steps used = 9, number of rules used = 7, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {6688, 2282, 2194, 2390, 12, 2302, 30} \begin {gather*} e^{e^x}+2 e^x+\frac {25}{\log (5 (4-x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-25 + E^(E^x + x)*(-4 + x)*Log[20 - 5*x]^2 + E^x*(-8 + 2*x)*Log[20 - 5*x]^2)/((-4 + x)*Log[20 - 5*x]^2),x
]

[Out]

E^E^x + 2*E^x + 25/Log[5*(4 - 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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2302

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

Rule 2390

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[((f*x)/d)^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

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 \left (2+e^{e^x}\right )-\frac {25}{(-4+x) \log ^2(20-5 x)}\right ) \, dx\\ &=-\left (25 \int \frac {1}{(-4+x) \log ^2(20-5 x)} \, dx\right )+\int e^x \left (2+e^{e^x}\right ) \, dx\\ &=5 \operatorname {Subst}\left (\int -\frac {5}{x \log ^2(x)} \, dx,x,20-5 x\right )+\operatorname {Subst}\left (\int \left (2+e^x\right ) \, dx,x,e^x\right )\\ &=2 e^x-25 \operatorname {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,20-5 x\right )+\operatorname {Subst}\left (\int e^x \, dx,x,e^x\right )\\ &=e^{e^x}+2 e^x-25 \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (-5 (-4+x))\right )\\ &=e^{e^x}+2 e^x+\frac {25}{\log (5 (4-x))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 21, normalized size = 0.88 \begin {gather*} e^{e^x}+2 e^x+\frac {25}{\log (-5 (-4+x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-25 + E^(E^x + x)*(-4 + x)*Log[20 - 5*x]^2 + E^x*(-8 + 2*x)*Log[20 - 5*x]^2)/((-4 + x)*Log[20 - 5*x
]^2),x]

[Out]

E^E^x + 2*E^x + 25/Log[-5*(-4 + x)]

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fricas [B]  time = 0.87, size = 42, normalized size = 1.75 \begin {gather*} \frac {{\left (2 \, e^{\left (2 \, x\right )} \log \left (-5 \, x + 20\right ) + e^{\left (x + e^{x}\right )} \log \left (-5 \, x + 20\right ) + 25 \, e^{x}\right )} e^{\left (-x\right )}}{\log \left (-5 \, x + 20\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-4)*exp(x)*log(-5*x+20)^2*exp(exp(x))+(2*x-8)*exp(x)*log(-5*x+20)^2-25)/(x-4)/log(-5*x+20)^2,x, a
lgorithm="fricas")

[Out]

(2*e^(2*x)*log(-5*x + 20) + e^(x + e^x)*log(-5*x + 20) + 25*e^x)*e^(-x)/log(-5*x + 20)

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giac [B]  time = 0.31, size = 42, normalized size = 1.75 \begin {gather*} \frac {{\left (2 \, e^{\left (2 \, x\right )} \log \left (-5 \, x + 20\right ) + e^{\left (x + e^{x}\right )} \log \left (-5 \, x + 20\right ) + 25 \, e^{x}\right )} e^{\left (-x\right )}}{\log \left (-5 \, x + 20\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-4)*exp(x)*log(-5*x+20)^2*exp(exp(x))+(2*x-8)*exp(x)*log(-5*x+20)^2-25)/(x-4)/log(-5*x+20)^2,x, a
lgorithm="giac")

[Out]

(2*e^(2*x)*log(-5*x + 20) + e^(x + e^x)*log(-5*x + 20) + 25*e^x)*e^(-x)/log(-5*x + 20)

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maple [A]  time = 0.28, size = 19, normalized size = 0.79

method	result	size
default	\(2 \,{\mathrm e}^{x}+{\mathrm e}^{{\mathrm e}^{x}}+\frac {25}{\ln \left (-5 x +20\right )}\)	\(19\)
risch	\(2 \,{\mathrm e}^{x}+{\mathrm e}^{{\mathrm e}^{x}}+\frac {25}{\ln \left (-5 x +20\right )}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x-4)*exp(x)*ln(-5*x+20)^2*exp(exp(x))+(2*x-8)*exp(x)*ln(-5*x+20)^2-25)/(x-4)/ln(-5*x+20)^2,x,method=_RET
URNVERBOSE)

[Out]

2*exp(x)+exp(exp(x))+25/ln(-5*x+20)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 8 \, e^{4} E_{1}\left (-x + 4\right ) + \frac {2 \, x e^{x} + {\left (x - 4\right )} e^{\left (e^{x}\right )}}{x - 4} + \frac {25}{i \, \pi + \log \relax (5) + \log \left (x - 4\right )} + 8 \, \int \frac {e^{x}}{x^{2} - 8 \, x + 16}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-4)*exp(x)*log(-5*x+20)^2*exp(exp(x))+(2*x-8)*exp(x)*log(-5*x+20)^2-25)/(x-4)/log(-5*x+20)^2,x, a
lgorithm="maxima")

[Out]

8*e^4*exp_integral_e(1, -x + 4) + (2*x*e^x + (x - 4)*e^(e^x))/(x - 4) + 25/(I*pi + log(5) + log(x - 4)) + 8*in
tegrate(e^x/(x^2 - 8*x + 16), x)

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mupad [B]  time = 0.14, size = 18, normalized size = 0.75 \begin {gather*} {\mathrm {e}}^{{\mathrm {e}}^x}+2\,{\mathrm {e}}^x+\frac {25}{\ln \left (20-5\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*log(20 - 5*x)^2*(2*x - 8) + exp(exp(x))*exp(x)*log(20 - 5*x)^2*(x - 4) - 25)/(log(20 - 5*x)^2*(x -
4)),x)

[Out]

exp(exp(x)) + 2*exp(x) + 25/log(20 - 5*x)

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sympy [A]  time = 0.44, size = 17, normalized size = 0.71 \begin {gather*} 2 e^{x} + e^{e^{x}} + \frac {25}{\log {\left (20 - 5 x \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-4)*exp(x)*ln(-5*x+20)**2*exp(exp(x))+(2*x-8)*exp(x)*ln(-5*x+20)**2-25)/(x-4)/ln(-5*x+20)**2,x)

[Out]

2*exp(x) + exp(exp(x)) + 25/log(20 - 5*x)

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