\(\int \frac {-45-15 e x+10 e \log (3+e x)}{3+e x} \, dx\) [681]

Optimal result

Integrand size = 23, antiderivative size = 15 \[ \int \frac {-45-15 e x+10 e \log (3+e x)}{3+e x} \, dx=5 \left (-20-3 x+\log ^2(3+e x)\right ) \]

[Out]

-15*x+5*ln(x*exp(1)+3)^2-100

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {6873, 12, 6874, 2437, 2338} \[ \int \frac {-45-15 e x+10 e \log (3+e x)}{3+e x} \, dx=5 \log ^2(e x+3)-15 x \]

[In]

Int[(-45 - 15*E*x + 10*E*Log[3 + E*x])/(3 + E*x),x]

[Out]

-15*x + 5*Log[3 + E*x]^2

Rule 12

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

Rule 2338

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, 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 6873

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

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {5 (-9-3 e x+2 e \log (3+e x))}{3+e x} \, dx \\ & = 5 \int \frac {-9-3 e x+2 e \log (3+e x)}{3+e x} \, dx \\ & = 5 \int \left (-3+\frac {2 e \log (3+e x)}{3+e x}\right ) \, dx \\ & = -15 x+(10 e) \int \frac {\log (3+e x)}{3+e x} \, dx \\ & = -15 x+10 \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,3+e x\right ) \\ & = -15 x+5 \log ^2(3+e x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.07 \[ \int \frac {-45-15 e x+10 e \log (3+e x)}{3+e x} \, dx=-5 \left (3 x-\log ^2(3+e x)\right ) \]

[In]

Integrate[(-45 - 15*E*x + 10*E*Log[3 + E*x])/(3 + E*x),x]

[Out]

-5*(3*x - Log[3 + E*x]^2)

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.07

method	result	size
norman	\(-15 x +5 \ln \left (x \,{\mathrm e}+3\right )^{2}\)	\(16\)
risch	\(-15 x +5 \ln \left (x \,{\mathrm e}+3\right )^{2}\)	\(16\)
parts	\(-15 x +5 \ln \left (x \,{\mathrm e}+3\right )^{2}\)	\(16\)
derivativedivides	\(5 \,{\mathrm e}^{-1} \left ({\mathrm e} \ln \left (x \,{\mathrm e}+3\right )^{2}-3 x \,{\mathrm e}-9\right )\)	\(26\)
default	\(5 \,{\mathrm e}^{-1} \left ({\mathrm e} \ln \left (x \,{\mathrm e}+3\right )^{2}-3 x \,{\mathrm e}-9\right )\)	\(26\)

[In]

int((10*exp(1)*ln(x*exp(1)+3)-15*x*exp(1)-45)/(x*exp(1)+3),x,method=_RETURNVERBOSE)

[Out]

-15*x+5*ln(x*exp(1)+3)^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {-45-15 e x+10 e \log (3+e x)}{3+e x} \, dx=5 \, \log \left (x e + 3\right )^{2} - 15 \, x \]

[In]

integrate((10*exp(1)*log(x*exp(1)+3)-15*x*exp(1)-45)/(x*exp(1)+3),x, algorithm="fricas")

[Out]

5*log(x*e + 3)^2 - 15*x

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int \frac {-45-15 e x+10 e \log (3+e x)}{3+e x} \, dx=- 15 x + 5 \log {\left (e x + 3 \right )}^{2} \]

[In]

integrate((10*exp(1)*ln(x*exp(1)+3)-15*x*exp(1)-45)/(x*exp(1)+3),x)

[Out]

-15*x + 5*log(E*x + 3)**2

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 43 vs. \(2 (16) = 32\).

Time = 0.19 (sec) , antiderivative size = 43, normalized size of antiderivative = 2.87 \[ \int \frac {-45-15 e x+10 e \log (3+e x)}{3+e x} \, dx=-15 \, {\left (x e^{\left (-1\right )} - 3 \, e^{\left (-2\right )} \log \left (x e + 3\right )\right )} e - 45 \, e^{\left (-1\right )} \log \left (x e + 3\right ) + 5 \, \log \left (x e + 3\right )^{2} \]

[In]

integrate((10*exp(1)*log(x*exp(1)+3)-15*x*exp(1)-45)/(x*exp(1)+3),x, algorithm="maxima")

[Out]

-15*(x*e^(-1) - 3*e^(-2)*log(x*e + 3))*e - 45*e^(-1)*log(x*e + 3) + 5*log(x*e + 3)^2

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {-45-15 e x+10 e \log (3+e x)}{3+e x} \, dx=5 \, \log \left (x e + 3\right )^{2} - 15 \, x \]

[In]

integrate((10*exp(1)*log(x*exp(1)+3)-15*x*exp(1)-45)/(x*exp(1)+3),x, algorithm="giac")

[Out]

5*log(x*e + 3)^2 - 15*x

Mupad [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {-45-15 e x+10 e \log (3+e x)}{3+e x} \, dx=5\,{\ln \left (x\,\mathrm {e}+3\right )}^2-15\,x \]

[In]

int(-(15*x*exp(1) - 10*log(x*exp(1) + 3)*exp(1) + 45)/(x*exp(1) + 3),x)

[Out]

5*log(x*exp(1) + 3)^2 - 15*x