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

Optimal. Leaf size=17 \[ x^2+e^x \left (e^x+5 x\right ) \log (x) \]

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Rubi [B]  time = 0.34, antiderivative size = 58, normalized size of antiderivative = 3.41, number of steps used = 14, number of rules used = 9, integrand size = 67, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.134, Rules used = {6688, 2194, 2178, 6742, 2176, 2554, 2262, 2177, 14} \begin {gather*} 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) \end {gather*}

Antiderivative was successfully verified.

[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 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 2177

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] &&  !$UseGamma ===
True

Rule 2178

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

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 2262

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 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 \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 17, normalized size = 1.00 \begin {gather*} x^2+e^x \left (e^x+5 x\right ) \log (x) \end {gather*}

Antiderivative was successfully verified.

[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]

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fricas [A]  time = 1.66, size = 17, normalized size = 1.00 \begin {gather*} x^{2} + {\left (5 \, x e^{x} + e^{\left (2 \, x\right )}\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.14, size = 18, normalized size = 1.06 \begin {gather*} 5 \, x e^{x} \log \relax (x) + x^{2} + e^{\left (2 \, x\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.03, size = 18, normalized size = 1.06

method	result	size
risch	\(\left (5 \,{\mathrm e}^{x} x +{\mathrm e}^{2 x}\right ) \ln \relax (x )+x^{2}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[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

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maxima [A]  time = 0.38, size = 18, normalized size = 1.06 \begin {gather*} 5 \, x e^{x} \log \relax (x) + x^{2} + e^{\left (2 \, x\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 3.36, size = 18, normalized size = 1.06 \begin {gather*} {\mathrm {e}}^{2\,x}\,\ln \relax (x)+x^2+5\,x\,{\mathrm {e}}^x\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.36, size = 20, normalized size = 1.18 \begin {gather*} x^{2} + 5 x e^{x} \log {\relax (x )} + e^{2 x} \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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