\(\int \frac {2 e^x x^3+10 x^4+e^x x (e^x+5 x) (e^x+5 x+(5 x+2 e^x x+5 x^2) \log (x))}{e^x x^2+5 x^3} \, dx\) [4220]

Optimal result

Integrand size = 67, antiderivative size = 17 \[ \int \frac {2 e^x x^3+10 x^4+e^x x \left (e^x+5 x\right ) \left (e^x+5 x+\left (5 x+2 e^x x+5 x^2\right ) \log (x)\right )}{e^x x^2+5 x^3} \, dx=x^2+e^x \left (e^x+5 x\right ) \log (x) \]

[Out]

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

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(58\) vs. \(2(17)=34\).

Time = 0.22 (sec) , antiderivative size = 58, normalized size of antiderivative = 3.41, number of steps used = 14, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.134, Rules used = {6820, 2225, 2209, 6874, 2207, 2634, 2294, 2208, 14} \[ \int \frac {2 e^x x^3+10 x^4+e^x x \left (e^x+5 x\right ) \left (e^x+5 x+\left (5 x+2 e^x x+5 x^2\right ) \log (x)\right )}{e^x x^2+5 x^3} \, dx=x^2+\frac {25 x}{2}+5 e^x+\frac {e^{2 x}}{2 x}-\frac {\left (5 x+e^x\right )^2}{2 x}+5 e^x x \log (x)+e^{2 x} \log (x) \]

[In]

Int[(2*E^x*x^3 + 10*x^4 + E^x*x*(E^x + 5*x)*(E^x + 5*x + (5*x + 2*E^x*x + 5*x^2)*Log[x]))/(E^x*x^2 + 5*x^3),x]

[Out]

5*E^x + E^(2*x)/(2*x) + (25*x)/2 + x^2 - (E^x + 5*x)^2/(2*x) + E^(2*x)*Log[x] + 5*E^x*x*Log[x]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2207

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] &&  !TrueQ[$UseGamma]

Rule 2208

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(c + d*x)^(m
+ 1)*((b*F^(g*(e + f*x)))^n/(d*(m + 1))), x] - Dist[f*g*n*(Log[F]/(d*(m + 1))), 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] && LtQ[m, -1] && IntegerQ[2*m] &&  !TrueQ[$UseGamm
a]

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

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 2294

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

Rule 2634

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 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, 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 (5 e^x+\frac {e^{2 x}}{x}+2 x+e^x \left (5+2 e^x+5 x\right ) \log (x)\right ) \, dx \\ & = x^2+5 \int e^x \, dx+\int \frac {e^{2 x}}{x} \, dx+\int e^x \left (5+2 e^x+5 x\right ) \log (x) \, dx \\ & = 5 e^x+x^2+\text {Ei}(2 x)+e^{2 x} \log (x)+5 e^x x \log (x)-\int \frac {e^x \left (e^x+5 x\right )}{x} \, dx \\ & = 5 e^x+x^2-\frac {\left (e^x+5 x\right )^2}{2 x}+\text {Ei}(2 x)+e^{2 x} \log (x)+5 e^x x \log (x)-\frac {1}{2} \int \frac {\left (e^x+5 x\right )^2}{x^2} \, dx+5 \int \frac {e^x+5 x}{x} \, dx \\ & = 5 e^x+x^2-\frac {\left (e^x+5 x\right )^2}{2 x}+\text {Ei}(2 x)+e^{2 x} \log (x)+5 e^x x \log (x)-\frac {1}{2} \int \left (25+\frac {e^{2 x}}{x^2}+\frac {10 e^x}{x}\right ) \, dx+5 \int \left (5+\frac {e^x}{x}\right ) \, dx \\ & = 5 e^x+\frac {25 x}{2}+x^2-\frac {\left (e^x+5 x\right )^2}{2 x}+\text {Ei}(2 x)+e^{2 x} \log (x)+5 e^x x \log (x)-\frac {1}{2} \int \frac {e^{2 x}}{x^2} \, dx \\ & = 5 e^x+\frac {e^{2 x}}{2 x}+\frac {25 x}{2}+x^2-\frac {\left (e^x+5 x\right )^2}{2 x}+\text {Ei}(2 x)+e^{2 x} \log (x)+5 e^x x \log (x)-\int \frac {e^{2 x}}{x} \, dx \\ & = 5 e^x+\frac {e^{2 x}}{2 x}+\frac {25 x}{2}+x^2-\frac {\left (e^x+5 x\right )^2}{2 x}+e^{2 x} \log (x)+5 e^x x \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {2 e^x x^3+10 x^4+e^x x \left (e^x+5 x\right ) \left (e^x+5 x+\left (5 x+2 e^x x+5 x^2\right ) \log (x)\right )}{e^x x^2+5 x^3} \, dx=x^2+e^x \left (e^x+5 x\right ) \log (x) \]

[In]

Integrate[(2*E^x*x^3 + 10*x^4 + E^x*x*(E^x + 5*x)*(E^x + 5*x + (5*x + 2*E^x*x + 5*x^2)*Log[x]))/(E^x*x^2 + 5*x
^3),x]

[Out]

x^2 + E^x*(E^x + 5*x)*Log[x]

Maple [A] (verified)

Time = 2.20 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06

method	result	size
risch	\(\left (5 \,{\mathrm e}^{x} x +{\mathrm e}^{2 x}\right ) \ln \left (x \right )+x^{2}\)	\(18\)
parallelrisch	\(-\frac {-10 x^{3}-10 \ln \left (x \right ) {\mathrm e}^{\ln \left (5 x +{\mathrm e}^{x}\right )+x +\ln \left (x \right )}}{10 x}\)	\(28\)

[In]

int((((2*exp(x)*x+5*x^2+5*x)*ln(x)+5*x+exp(x))*exp(ln(5*x+exp(x))+x+ln(x))+2*exp(x)*x^3+10*x^4)/(exp(x)*x^2+5*
x^3),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {2 e^x x^3+10 x^4+e^x x \left (e^x+5 x\right ) \left (e^x+5 x+\left (5 x+2 e^x x+5 x^2\right ) \log (x)\right )}{e^x x^2+5 x^3} \, dx=x^{2} + {\left (5 \, x e^{x} + e^{\left (2 \, x\right )}\right )} \log \left (x\right ) \]

[In]

integrate((((2*exp(x)*x+5*x^2+5*x)*log(x)+5*x+exp(x))*exp(log(5*x+exp(x))+x+log(x))+2*exp(x)*x^3+10*x^4)/(exp(
x)*x^2+5*x^3),x, algorithm="fricas")

[Out]

x^2 + (5*x*e^x + e^(2*x))*log(x)

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.18 \[ \int \frac {2 e^x x^3+10 x^4+e^x x \left (e^x+5 x\right ) \left (e^x+5 x+\left (5 x+2 e^x x+5 x^2\right ) \log (x)\right )}{e^x x^2+5 x^3} \, dx=x^{2} + 5 x e^{x} \log {\left (x \right )} + e^{2 x} \log {\left (x \right )} \]

[In]

integrate((((2*exp(x)*x+5*x**2+5*x)*ln(x)+5*x+exp(x))*exp(ln(5*x+exp(x))+x+ln(x))+2*exp(x)*x**3+10*x**4)/(exp(
x)*x**2+5*x**3),x)

[Out]

x**2 + 5*x*exp(x)*log(x) + exp(2*x)*log(x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int \frac {2 e^x x^3+10 x^4+e^x x \left (e^x+5 x\right ) \left (e^x+5 x+\left (5 x+2 e^x x+5 x^2\right ) \log (x)\right )}{e^x x^2+5 x^3} \, dx=5 \, x e^{x} \log \left (x\right ) + x^{2} + e^{\left (2 \, x\right )} \log \left (x\right ) \]

[In]

integrate((((2*exp(x)*x+5*x^2+5*x)*log(x)+5*x+exp(x))*exp(log(5*x+exp(x))+x+log(x))+2*exp(x)*x^3+10*x^4)/(exp(
x)*x^2+5*x^3),x, algorithm="maxima")

[Out]

5*x*e^x*log(x) + x^2 + e^(2*x)*log(x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int \frac {2 e^x x^3+10 x^4+e^x x \left (e^x+5 x\right ) \left (e^x+5 x+\left (5 x+2 e^x x+5 x^2\right ) \log (x)\right )}{e^x x^2+5 x^3} \, dx=5 \, x e^{x} \log \left (x\right ) + x^{2} + e^{\left (2 \, x\right )} \log \left (x\right ) \]

[In]

integrate((((2*exp(x)*x+5*x^2+5*x)*log(x)+5*x+exp(x))*exp(log(5*x+exp(x))+x+log(x))+2*exp(x)*x^3+10*x^4)/(exp(
x)*x^2+5*x^3),x, algorithm="giac")

[Out]

5*x*e^x*log(x) + x^2 + e^(2*x)*log(x)

Mupad [B] (verification not implemented)

Time = 10.83 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int \frac {2 e^x x^3+10 x^4+e^x x \left (e^x+5 x\right ) \left (e^x+5 x+\left (5 x+2 e^x x+5 x^2\right ) \log (x)\right )}{e^x x^2+5 x^3} \, dx={\mathrm {e}}^{2\,x}\,\ln \left (x\right )+x^2+5\,x\,{\mathrm {e}}^x\,\ln \left (x\right ) \]

[In]

int((2*x^3*exp(x) + exp(x + log(5*x + exp(x)) + log(x))*(5*x + exp(x) + log(x)*(5*x + 2*x*exp(x) + 5*x^2)) + 1
0*x^4)/(x^2*exp(x) + 5*x^3),x)

[Out]

exp(2*x)*log(x) + x^2 + 5*x*exp(x)*log(x)