3.64.47 \(\int \frac {e^{2 x} (4+x)+(e^{2 x} (5+x)+e^x (25+5 x)) \log (5+e^x)}{5+e^x} \, dx\)

Optimal. Leaf size=13 \[ e^x (4+x) \log \left (5+e^x\right ) \]

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Rubi [A]  time = 0.40, antiderivative size = 25, normalized size of antiderivative = 1.92, number of steps used = 20, number of rules used = 10, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {6688, 6742, 2254, 2176, 2194, 2184, 2190, 2279, 2391, 2554} \begin {gather*} e^x (x+5) \log \left (e^x+5\right )-e^x \log \left (e^x+5\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(E^x*Log[5 + E^x]) + E^x*(5 + x)*Log[5 + E^x]

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

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 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 2254

Int[((a_.) + (b_.)*(F_)^(u_))^(p_.)*((c_.) + (d_.)*(F_)^(v_))^(q_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> W
ith[{w = ExpandIntegrand[(e + f*x)^m, (a + b*F^u)^p*(c + d*F^v)^q, x]}, Int[w, x] /; SumQ[w]] /; FreeQ[{F, a,
b, c, d, e, f, m}, x] && IntegersQ[p, q] && LinearQ[{u, v}, x] && RationalQ[Simplify[u/v]]

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 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, 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 e^x \left (\frac {e^x (4+x)}{5+e^x}+(5+x) \log \left (5+e^x\right )\right ) \, dx\\ &=\int \left (\frac {e^{2 x} (4+x)}{5+e^x}+e^x (5+x) \log \left (5+e^x\right )\right ) \, dx\\ &=\int \frac {e^{2 x} (4+x)}{5+e^x} \, dx+\int e^x (5+x) \log \left (5+e^x\right ) \, dx\\ &=-e^x \log \left (5+e^x\right )+e^x (5+x) \log \left (5+e^x\right )-\int \frac {e^{2 x} (4+x)}{5+e^x} \, dx+\int \left (-5 (4+x)+e^x (4+x)+\frac {25 (4+x)}{5+e^x}\right ) \, dx\\ &=-\frac {5}{2} (4+x)^2-e^x \log \left (5+e^x\right )+e^x (5+x) \log \left (5+e^x\right )+25 \int \frac {4+x}{5+e^x} \, dx+\int e^x (4+x) \, dx-\int \left (-5 (4+x)+e^x (4+x)+\frac {25 (4+x)}{5+e^x}\right ) \, dx\\ &=e^x (4+x)+\frac {5}{2} (4+x)^2-e^x \log \left (5+e^x\right )+e^x (5+x) \log \left (5+e^x\right )-5 \int \frac {e^x (4+x)}{5+e^x} \, dx-25 \int \frac {4+x}{5+e^x} \, dx-\int e^x \, dx-\int e^x (4+x) \, dx\\ &=-e^x-5 (4+x) \log \left (1+\frac {e^x}{5}\right )-e^x \log \left (5+e^x\right )+e^x (5+x) \log \left (5+e^x\right )+5 \int \frac {e^x (4+x)}{5+e^x} \, dx+5 \int \log \left (1+\frac {e^x}{5}\right ) \, dx+\int e^x \, dx\\ &=-e^x \log \left (5+e^x\right )+e^x (5+x) \log \left (5+e^x\right )-5 \int \log \left (1+\frac {e^x}{5}\right ) \, dx+5 \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {x}{5}\right )}{x} \, dx,x,e^x\right )\\ &=-e^x \log \left (5+e^x\right )+e^x (5+x) \log \left (5+e^x\right )-5 \text {Li}_2\left (-\frac {e^x}{5}\right )-5 \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {x}{5}\right )}{x} \, dx,x,e^x\right )\\ &=-e^x \log \left (5+e^x\right )+e^x (5+x) \log \left (5+e^x\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 13, normalized size = 1.00 \begin {gather*} e^x (4+x) \log \left (5+e^x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^x*(4 + x)*Log[5 + E^x]

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fricas [A]  time = 0.64, size = 11, normalized size = 0.85 \begin {gather*} {\left (x + 4\right )} e^{x} \log \left (e^{x} + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5+x)*exp(x)^2+(25+5*x)*exp(x))*log(exp(x)+5)+(4+x)*exp(x)^2)/(exp(x)+5),x, algorithm="fricas")

[Out]

(x + 4)*e^x*log(e^x + 5)

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giac [A]  time = 0.24, size = 19, normalized size = 1.46 \begin {gather*} x e^{x} \log \left (e^{x} + 5\right ) + 4 \, e^{x} \log \left (e^{x} + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5+x)*exp(x)^2+(25+5*x)*exp(x))*log(exp(x)+5)+(4+x)*exp(x)^2)/(exp(x)+5),x, algorithm="giac")

[Out]

x*e^x*log(e^x + 5) + 4*e^x*log(e^x + 5)

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maple [A]  time = 0.03, size = 12, normalized size = 0.92

method	result	size
risch	\(\ln \left ({\mathrm e}^{x}+5\right ) \left (4+x \right ) {\mathrm e}^{x}\)	\(12\)
norman	\({\mathrm e}^{x} \ln \left ({\mathrm e}^{x}+5\right ) x +4 \,{\mathrm e}^{x} \ln \left ({\mathrm e}^{x}+5\right )\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((5+x)*exp(x)^2+(25+5*x)*exp(x))*ln(exp(x)+5)+(4+x)*exp(x)^2)/(exp(x)+5),x,method=_RETURNVERBOSE)

[Out]

ln(exp(x)+5)*(4+x)*exp(x)

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maxima [A]  time = 0.40, size = 22, normalized size = 1.69 \begin {gather*} {\left ({\left (x + 4\right )} e^{x} + 20\right )} \log \left (e^{x} + 5\right ) - 20 \, \log \left (e^{x} + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5+x)*exp(x)^2+(25+5*x)*exp(x))*log(exp(x)+5)+(4+x)*exp(x)^2)/(exp(x)+5),x, algorithm="maxima")

[Out]

((x + 4)*e^x + 20)*log(e^x + 5) - 20*log(e^x + 5)

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mupad [B]  time = 4.25, size = 78, normalized size = 6.00 \begin {gather*} \frac {40\,{\mathrm {e}}^{2\,x}\,\ln \left ({\mathrm {e}}^x+5\right )+4\,{\mathrm {e}}^{3\,x}\,\ln \left ({\mathrm {e}}^x+5\right )+100\,{\mathrm {e}}^x\,\ln \left ({\mathrm {e}}^x+5\right )+25\,x\,{\mathrm {e}}^x\,\ln \left ({\mathrm {e}}^x+5\right )+10\,x\,{\mathrm {e}}^{2\,x}\,\ln \left ({\mathrm {e}}^x+5\right )+x\,{\mathrm {e}}^{3\,x}\,\ln \left ({\mathrm {e}}^x+5\right )}{{\mathrm {e}}^{2\,x}+10\,{\mathrm {e}}^x+25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(exp(x) + 5)*(exp(2*x)*(x + 5) + exp(x)*(5*x + 25)) + exp(2*x)*(x + 4))/(exp(x) + 5),x)

[Out]

(40*exp(2*x)*log(exp(x) + 5) + 4*exp(3*x)*log(exp(x) + 5) + 100*exp(x)*log(exp(x) + 5) + 25*x*exp(x)*log(exp(x
) + 5) + 10*x*exp(2*x)*log(exp(x) + 5) + x*exp(3*x)*log(exp(x) + 5))/(exp(2*x) + 10*exp(x) + 25)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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