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

Optimal. Leaf size=20 \[ 5-\log \left (e^{2 x}\right ) \log \left (x+e^{5 x} x\right ) \]

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Rubi [A]  time = 1.69, antiderivative size = 13, normalized size of antiderivative = 0.65, number of steps used = 18, number of rules used = 6, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {6742, 2184, 2190, 2279, 2391, 2548} \begin {gather*} -2 x \log \left (\left (e^{5 x}+1\right ) x\right ) \end {gather*}

Antiderivative was successfully verified.

[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 2184

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 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), 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 2279

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 2391

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 2548

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

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 \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 \operatorname {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 \operatorname {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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.25, size = 13, normalized size = 0.65 \begin {gather*} -2 x \log \left (\left (1+e^{5 x}\right ) x\right ) \end {gather*}

Antiderivative was successfully verified.

[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]

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fricas [A]  time = 0.46, size = 12, normalized size = 0.60 \begin {gather*} -2 \, x \log \left (x e^{\left (5 \, x\right )} + x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.19, size = 12, normalized size = 0.60 \begin {gather*} -2 \, x \log \left (x e^{\left (5 \, x\right )} + x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.07, size = 13, normalized size = 0.65

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [B]  time = 0.47, size = 38, normalized size = 1.90 \begin {gather*} -2 \, x \log \relax (x) - 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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 3.16, size = 12, normalized size = 0.60 \begin {gather*} -2\,x\,\ln \left (x\,\left ({\mathrm {e}}^{5\,x}+1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 0.34, size = 14, normalized size = 0.70 \begin {gather*} - 2 x \log {\left (x e^{5 x} + x \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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