\(\int \frac {-5+(-10+e^x (-8-4 x)-5 x) \log (4+2 x)}{(10+5 x) \log (4+2 x)} \, dx\) [8234]

Optimal result

Integrand size = 39, antiderivative size = 23 \[ \int \frac {-5+\left (-10+e^x (-8-4 x)-5 x\right ) \log (4+2 x)}{(10+5 x) \log (4+2 x)} \, dx=4-\frac {4}{5} \left (1+e^x\right )-x-\log (\log (4+2 x)) \]

[Out]

16/5-ln(ln(4+2*x))-x-4/5*exp(x)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {6820, 2225, 2437, 12, 2339, 29} \[ \int \frac {-5+\left (-10+e^x (-8-4 x)-5 x\right ) \log (4+2 x)}{(10+5 x) \log (4+2 x)} \, dx=-x-\frac {4 e^x}{5}-\log (\log (2 (x+2))) \]

[In]

Int[(-5 + (-10 + E^x*(-8 - 4*x) - 5*x)*Log[4 + 2*x])/((10 + 5*x)*Log[4 + 2*x]),x]

[Out]

(-4*E^x)/5 - x - Log[Log[2*(2 + 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 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2225

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 2339

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 2437

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 6820

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

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {-5+\left (-10+e^x (-8-4 x)-5 x\right ) \log (4+2 x)}{(10+5 x) \log (4+2 x)} \, dx=-\frac {4 e^x}{5}-x-\log (\log (2 (2+x))) \]

[In]

Integrate[(-5 + (-10 + E^x*(-8 - 4*x) - 5*x)*Log[4 + 2*x])/((10 + 5*x)*Log[4 + 2*x]),x]

[Out]

(-4*E^x)/5 - x - Log[Log[2*(2 + x)]]

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78

method	result	size
default	\(-x -\ln \left (\ln \left (4+2 x \right )\right )-\frac {4 \,{\mathrm e}^{x}}{5}\)	\(18\)
norman	\(-x -\ln \left (\ln \left (4+2 x \right )\right )-\frac {4 \,{\mathrm e}^{x}}{5}\)	\(18\)
risch	\(-x -\ln \left (\ln \left (4+2 x \right )\right )-\frac {4 \,{\mathrm e}^{x}}{5}\)	\(18\)
parts	\(-x -\ln \left (\ln \left (4+2 x \right )\right )-\frac {4 \,{\mathrm e}^{x}}{5}\)	\(18\)
parallelrisch	\(4-\ln \left (\ln \left (4+2 x \right )\right )-x -\frac {4 \,{\mathrm e}^{x}}{5}\)	\(19\)

[In]

int((((-4*x-8)*exp(x)-5*x-10)*ln(4+2*x)-5)/(5*x+10)/ln(4+2*x),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {-5+\left (-10+e^x (-8-4 x)-5 x\right ) \log (4+2 x)}{(10+5 x) \log (4+2 x)} \, dx=-x - \frac {4}{5} \, e^{x} - \log \left (\log \left (2 \, x + 4\right )\right ) \]

[In]

integrate((((-4*x-8)*exp(x)-5*x-10)*log(4+2*x)-5)/(5*x+10)/log(4+2*x),x, algorithm="fricas")

[Out]

-x - 4/5*e^x - log(log(2*x + 4))

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {-5+\left (-10+e^x (-8-4 x)-5 x\right ) \log (4+2 x)}{(10+5 x) \log (4+2 x)} \, dx=- x - \frac {4 e^{x}}{5} - \log {\left (\log {\left (2 x + 4 \right )} \right )} \]

[In]

integrate((((-4*x-8)*exp(x)-5*x-10)*ln(4+2*x)-5)/(5*x+10)/ln(4+2*x),x)

[Out]

-x - 4*exp(x)/5 - log(log(2*x + 4))

Maxima [F]

\[ \int \frac {-5+\left (-10+e^x (-8-4 x)-5 x\right ) \log (4+2 x)}{(10+5 x) \log (4+2 x)} \, dx=\int { -\frac {{\left (4 \, {\left (x + 2\right )} e^{x} + 5 \, x + 10\right )} \log \left (2 \, x + 4\right ) + 5}{5 \, {\left (x + 2\right )} \log \left (2 \, x + 4\right )} \,d x } \]

[In]

integrate((((-4*x-8)*exp(x)-5*x-10)*log(4+2*x)-5)/(5*x+10)/log(4+2*x),x, algorithm="maxima")

[Out]

8/5*e^(-2)*exp_integral_e(1, -x - 2) + 2*(log(2) + log(x + 2))*log(log(2) + log(x + 2)) - 2*log(2*x + 4)*log(l
og(2) + log(x + 2)) - x - 4/5*x*e^x/(x + 2) + 8/5*integrate(e^x/(x^2 + 4*x + 4), x) - log(log(2) + log(x + 2))

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {-5+\left (-10+e^x (-8-4 x)-5 x\right ) \log (4+2 x)}{(10+5 x) \log (4+2 x)} \, dx=-x - \frac {4}{5} \, e^{x} - \log \left (\log \left (2 \, x + 4\right )\right ) \]

[In]

integrate((((-4*x-8)*exp(x)-5*x-10)*log(4+2*x)-5)/(5*x+10)/log(4+2*x),x, algorithm="giac")

[Out]

-x - 4/5*e^x - log(log(2*x + 4))

Mupad [B] (verification not implemented)

Time = 13.73 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {-5+\left (-10+e^x (-8-4 x)-5 x\right ) \log (4+2 x)}{(10+5 x) \log (4+2 x)} \, dx=-x-\ln \left (\ln \left (2\,x+4\right )\right )-\frac {4\,{\mathrm {e}}^x}{5} \]

[In]

int(-(log(2*x + 4)*(5*x + exp(x)*(4*x + 8) + 10) + 5)/(log(2*x + 4)*(5*x + 10)),x)

[Out]

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