\(\int \frac {-2+e^{5 x} (-2-10 x)+(-2-2 e^{5 x}) \log (x+e^{5 x} x)}{1+e^{5 x}} \, dx\) [4246]

Optimal result

Integrand size = 43, antiderivative size = 20 \[ \int \frac {-2+e^{5 x} (-2-10 x)+\left (-2-2 e^{5 x}\right ) \log \left (x+e^{5 x} x\right )}{1+e^{5 x}} \, dx=5-\log \left (e^{2 x}\right ) \log \left (x+e^{5 x} x\right ) \]

[Out]

5-ln(x*exp(5*x)+x)*ln(exp(x)^2)

Rubi [A] (verified)

Time = 1.13 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.65, number of steps used = 18, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {6874, 2215, 2221, 2317, 2438, 2628} \[ \int \frac {-2+e^{5 x} (-2-10 x)+\left (-2-2 e^{5 x}\right ) \log \left (x+e^{5 x} x\right )}{1+e^{5 x}} \, dx=-2 x \log \left (\left (e^{5 x}+1\right ) x\right ) \]

[In]

Int[(-2 + E^(5*x)*(-2 - 10*x) + (-2 - 2*E^(5*x))*Log[x + E^(5*x)*x])/(1 + E^(5*x)),x]

[Out]

-2*x*Log[(1 + E^(5*x))*x]

Rule 2215

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[(c
+ d*x)^(m + 1)/(a*d*(m + 1)), x] - Dist[b/a, Int[(c + d*x)^m*((F^(g*(e + f*x)))^n/(a + b*(F^(g*(e + f*x)))^n))
, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2221

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x]
- Dist[d*(m/(b*f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2628

Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[x*(D[u, x]/u), x], x] /; InverseFunctionFr
eeQ[u, x]

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {2 x}{1+e^x}-\frac {2 \left (-4+3 e^x-2 e^{2 x}+e^{3 x}\right ) x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}}-2 \left (1+5 x+\log \left (\left (1+e^{5 x}\right ) x\right )\right )\right ) \, dx \\ & = 2 \int \frac {x}{1+e^x} \, dx-2 \int \frac {\left (-4+3 e^x-2 e^{2 x}+e^{3 x}\right ) x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}} \, dx-2 \int \left (1+5 x+\log \left (\left (1+e^{5 x}\right ) x\right )\right ) \, dx \\ & = -2 x-4 x^2-2 \int \frac {e^x x}{1+e^x} \, dx-2 \int \left (-\frac {4 x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}}+\frac {3 e^x x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}}-\frac {2 e^{2 x} x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}}+\frac {e^{3 x} x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}}\right ) \, dx-2 \int \log \left (\left (1+e^{5 x}\right ) x\right ) \, dx \\ & = -2 x-4 x^2-2 x \log \left (1+e^x\right )-2 x \log \left (\left (1+e^{5 x}\right ) x\right )-2 \int \frac {e^{3 x} x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}} \, dx+2 \int \frac {1+e^{5 x} (1+5 x)}{1+e^{5 x}} \, dx+2 \int \log \left (1+e^x\right ) \, dx+4 \int \frac {e^{2 x} x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}} \, dx-6 \int \frac {e^x x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}} \, dx+8 \int \frac {x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}} \, dx \\ & = -2 x-4 x^2-2 x \log \left (1+e^x\right )-2 x \log \left (\left (1+e^{5 x}\right ) x\right )-2 \int \frac {e^{3 x} x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}} \, dx+2 \int \left (1+5 x-\frac {x}{1+e^x}+\frac {\left (-4+3 e^x-2 e^{2 x}+e^{3 x}\right ) x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}}\right ) \, dx+2 \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^x\right )+4 \int \frac {e^{2 x} x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}} \, dx-6 \int \frac {e^x x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}} \, dx+8 \int \frac {x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}} \, dx \\ & = x^2-2 x \log \left (1+e^x\right )-2 x \log \left (\left (1+e^{5 x}\right ) x\right )-2 \text {Li}_2\left (-e^x\right )-2 \int \frac {x}{1+e^x} \, dx-2 \int \frac {e^{3 x} x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}} \, dx+2 \int \frac {\left (-4+3 e^x-2 e^{2 x}+e^{3 x}\right ) x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}} \, dx+4 \int \frac {e^{2 x} x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}} \, dx-6 \int \frac {e^x x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}} \, dx+8 \int \frac {x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}} \, dx \\ & = -2 x \log \left (1+e^x\right )-2 x \log \left (\left (1+e^{5 x}\right ) x\right )-2 \text {Li}_2\left (-e^x\right )+2 \int \frac {e^x x}{1+e^x} \, dx-2 \int \frac {e^{3 x} x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}} \, dx+2 \int \left (-\frac {4 x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}}+\frac {3 e^x x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}}-\frac {2 e^{2 x} x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}}+\frac {e^{3 x} x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}}\right ) \, dx+4 \int \frac {e^{2 x} x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}} \, dx-6 \int \frac {e^x x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}} \, dx+8 \int \frac {x}{1-e^x+e^{2 x}-e^{3 x}+e^{4 x}} \, dx \\ & = -2 x \log \left (\left (1+e^{5 x}\right ) x\right )-2 \text {Li}_2\left (-e^x\right )-2 \int \log \left (1+e^x\right ) \, dx \\ & = -2 x \log \left (\left (1+e^{5 x}\right ) x\right )-2 \text {Li}_2\left (-e^x\right )-2 \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^x\right ) \\ & = -2 x \log \left (\left (1+e^{5 x}\right ) x\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.65 \[ \int \frac {-2+e^{5 x} (-2-10 x)+\left (-2-2 e^{5 x}\right ) \log \left (x+e^{5 x} x\right )}{1+e^{5 x}} \, dx=-2 x \log \left (\left (1+e^{5 x}\right ) x\right ) \]

[In]

Integrate[(-2 + E^(5*x)*(-2 - 10*x) + (-2 - 2*E^(5*x))*Log[x + E^(5*x)*x])/(1 + E^(5*x)),x]

[Out]

-2*x*Log[(1 + E^(5*x))*x]

Maple [A] (verified)

Time = 0.99 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.65

method	result	size
norman	\(-2 x \ln \left (x \,{\mathrm e}^{5 x}+x \right )\)	\(13\)
parallelrisch	\(-2 x \ln \left (x \left ({\mathrm e}^{5 x}+1\right )\right )\)	\(13\)
risch	\(-2 x \ln \left ({\mathrm e}^{5 x}+1\right )+i \pi x \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i \left ({\mathrm e}^{5 x}+1\right )\right ) \operatorname {csgn}\left (i x \left ({\mathrm e}^{5 x}+1\right )\right )-i \pi x \,\operatorname {csgn}\left (i x \right ) {\operatorname {csgn}\left (i x \left ({\mathrm e}^{5 x}+1\right )\right )}^{2}-i \pi x \,\operatorname {csgn}\left (i \left ({\mathrm e}^{5 x}+1\right )\right ) {\operatorname {csgn}\left (i x \left ({\mathrm e}^{5 x}+1\right )\right )}^{2}+i \pi x {\operatorname {csgn}\left (i x \left ({\mathrm e}^{5 x}+1\right )\right )}^{3}-2 x \ln \left (x \right )\)	\(117\)

[In]

int(((-2*exp(5*x)-2)*ln(x*exp(5*x)+x)+(-10*x-2)*exp(5*x)-2)/(exp(5*x)+1),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.60 \[ \int \frac {-2+e^{5 x} (-2-10 x)+\left (-2-2 e^{5 x}\right ) \log \left (x+e^{5 x} x\right )}{1+e^{5 x}} \, dx=-2 \, x \log \left (x e^{\left (5 \, x\right )} + x\right ) \]

[In]

integrate(((-2*exp(5*x)-2)*log(x*exp(5*x)+x)+(-10*x-2)*exp(5*x)-2)/(exp(5*x)+1),x, algorithm="fricas")

[Out]

-2*x*log(x*e^(5*x) + x)

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int \frac {-2+e^{5 x} (-2-10 x)+\left (-2-2 e^{5 x}\right ) \log \left (x+e^{5 x} x\right )}{1+e^{5 x}} \, dx=- 2 x \log {\left (x e^{5 x} + x \right )} \]

[In]

integrate(((-2*exp(5*x)-2)*ln(x*exp(5*x)+x)+(-10*x-2)*exp(5*x)-2)/(exp(5*x)+1),x)

[Out]

-2*x*log(x*exp(5*x) + x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 38 vs. \(2 (18) = 36\).

Time = 0.31 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.90 \[ \int \frac {-2+e^{5 x} (-2-10 x)+\left (-2-2 e^{5 x}\right ) \log \left (x+e^{5 x} x\right )}{1+e^{5 x}} \, dx=-2 \, x \log \left (x\right ) - 2 \, x \log \left (e^{\left (4 \, x\right )} - e^{\left (3 \, x\right )} + e^{\left (2 \, x\right )} - e^{x} + 1\right ) - 2 \, x \log \left (e^{x} + 1\right ) \]

[In]

integrate(((-2*exp(5*x)-2)*log(x*exp(5*x)+x)+(-10*x-2)*exp(5*x)-2)/(exp(5*x)+1),x, algorithm="maxima")

[Out]

-2*x*log(x) - 2*x*log(e^(4*x) - e^(3*x) + e^(2*x) - e^x + 1) - 2*x*log(e^x + 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.60 \[ \int \frac {-2+e^{5 x} (-2-10 x)+\left (-2-2 e^{5 x}\right ) \log \left (x+e^{5 x} x\right )}{1+e^{5 x}} \, dx=-2 \, x \log \left (x e^{\left (5 \, x\right )} + x\right ) \]

[In]

integrate(((-2*exp(5*x)-2)*log(x*exp(5*x)+x)+(-10*x-2)*exp(5*x)-2)/(exp(5*x)+1),x, algorithm="giac")

[Out]

-2*x*log(x*e^(5*x) + x)

Mupad [B] (verification not implemented)

Time = 9.26 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.60 \[ \int \frac {-2+e^{5 x} (-2-10 x)+\left (-2-2 e^{5 x}\right ) \log \left (x+e^{5 x} x\right )}{1+e^{5 x}} \, dx=-2\,x\,\ln \left (x\,\left ({\mathrm {e}}^{5\,x}+1\right )\right ) \]

[In]

int(-(log(x + x*exp(5*x))*(2*exp(5*x) + 2) + exp(5*x)*(10*x + 2) + 2)/(exp(5*x) + 1),x)

[Out]

-2*x*log(x*(exp(5*x) + 1))