3.57.69 \(\int \frac {10 e^x x-25 x^2+e^{2 x} (2 x+x^2)+(50 x+30 e^x x-25 x^2+e^{2 x} (6 x+3 x^2)) \log (x)+(20 e^x x+e^{2 x} (6 x+3 x^2)) \log ^2(x)+e^{2 x} (2 x+x^2) \log ^3(x)}{e^x+3 e^x \log (x)+3 e^x \log ^2(x)+e^x \log ^3(x)} \, dx\)

Optimal. Leaf size=23 \[ e^{-x} \left (e^x x+\frac {5 x}{1+\log (x)}\right )^2 \]

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Rubi [C]  time = 1.80, antiderivative size = 96, normalized size of antiderivative = 4.17, number of steps used = 18, number of rules used = 11, integrand size = 133, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {6688, 6742, 2196, 2176, 2194, 2306, 2309, 2178, 2366, 6482, 2288} \begin {gather*} \frac {20 \text {Ei}(2 (\log (x)+1))}{e^2}-\frac {40 (\log (x)+1) \text {Ei}(2 (\log (x)+1))}{e^2}+\frac {20 (2 \log (x)+1) \text {Ei}(2 (\log (x)+1))}{e^2}+e^x x^2+20 x^2-\frac {10 x^2 (2 \log (x)+1)}{\log (x)+1}+\frac {25 e^{-x} x (x+x \log (x))}{(\log (x)+1)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(10*E^x*x - 25*x^2 + E^(2*x)*(2*x + x^2) + (50*x + 30*E^x*x - 25*x^2 + E^(2*x)*(6*x + 3*x^2))*Log[x] + (20
*E^x*x + E^(2*x)*(6*x + 3*x^2))*Log[x]^2 + E^(2*x)*(2*x + x^2)*Log[x]^3)/(E^x + 3*E^x*Log[x] + 3*E^x*Log[x]^2
+ E^x*Log[x]^3),x]

[Out]

20*x^2 + E^x*x^2 + (20*ExpIntegralEi[2*(1 + Log[x])])/E^2 - (40*ExpIntegralEi[2*(1 + Log[x])]*(1 + Log[x]))/E^
2 + (20*ExpIntegralEi[2*(1 + Log[x])]*(1 + 2*Log[x]))/E^2 - (10*x^2*(1 + 2*Log[x]))/(1 + Log[x]) + (25*x*(x +
x*Log[x]))/(E^x*(1 + Log[x])^3)

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 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 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 2306

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

Rule 2309

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[E^((m + 1)*x)*(a
+ b*x)^p, x], x, Log[c*x]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[m]

Rule 2366

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(f_.)*(x_)^(r_.)]*(e_.))*((g_.)*(x_))^(m_.), x_Sy
mbol] :> With[{u = IntHide[(g*x)^m*(a + b*Log[c*x^n])^p, x]}, Dist[d + e*Log[f*x^r], u, x] - Dist[e*r, Int[Sim
plifyIntegrand[u/x, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, r}, x] &&  !(EqQ[p, 1] && EqQ[a, 0] &&
NeQ[d, 0])

Rule 6482

Int[ExpIntegralEi[(a_.) + (b_.)*(x_)], x_Symbol] :> Simp[((a + b*x)*ExpIntegralEi[a + b*x])/b, x] - Simp[E^(a
+ b*x)/b, x] /; FreeQ[{a, b}, 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 \frac {e^{-x} x \left (5+e^x+e^x \log (x)\right ) \left (-5 x+e^x (2+x)+\left (-5 (-2+x)+2 e^x (2+x)\right ) \log (x)+e^x (2+x) \log ^2(x)\right )}{(1+\log (x))^3} \, dx\\ &=\int \left (e^x x (2+x)+\frac {10 x (1+2 \log (x))}{(1+\log (x))^2}-\frac {25 e^{-x} x (x-2 \log (x)+x \log (x))}{(1+\log (x))^3}\right ) \, dx\\ &=10 \int \frac {x (1+2 \log (x))}{(1+\log (x))^2} \, dx-25 \int \frac {e^{-x} x (x-2 \log (x)+x \log (x))}{(1+\log (x))^3} \, dx+\int e^x x (2+x) \, dx\\ &=\frac {20 \text {Ei}(2 (1+\log (x))) (1+2 \log (x))}{e^2}-\frac {10 x^2 (1+2 \log (x))}{1+\log (x)}+\frac {25 e^{-x} x (x+x \log (x))}{(1+\log (x))^3}-20 \int \left (\frac {2 \text {Ei}(2 (1+\log (x)))}{e^2 x}-\frac {x}{1+\log (x)}\right ) \, dx+\int \left (2 e^x x+e^x x^2\right ) \, dx\\ &=\frac {20 \text {Ei}(2 (1+\log (x))) (1+2 \log (x))}{e^2}-\frac {10 x^2 (1+2 \log (x))}{1+\log (x)}+\frac {25 e^{-x} x (x+x \log (x))}{(1+\log (x))^3}+2 \int e^x x \, dx+20 \int \frac {x}{1+\log (x)} \, dx-\frac {40 \int \frac {\text {Ei}(2 (1+\log (x)))}{x} \, dx}{e^2}+\int e^x x^2 \, dx\\ &=2 e^x x+e^x x^2+\frac {20 \text {Ei}(2 (1+\log (x))) (1+2 \log (x))}{e^2}-\frac {10 x^2 (1+2 \log (x))}{1+\log (x)}+\frac {25 e^{-x} x (x+x \log (x))}{(1+\log (x))^3}-2 \int e^x \, dx-2 \int e^x x \, dx+20 \operatorname {Subst}\left (\int \frac {e^{2 x}}{1+x} \, dx,x,\log (x)\right )-\frac {40 \operatorname {Subst}(\int \text {Ei}(2 (1+x)) \, dx,x,\log (x))}{e^2}\\ &=-2 e^x+e^x x^2+\frac {20 \text {Ei}(2 (1+\log (x)))}{e^2}+\frac {20 \text {Ei}(2 (1+\log (x))) (1+2 \log (x))}{e^2}-\frac {10 x^2 (1+2 \log (x))}{1+\log (x)}+\frac {25 e^{-x} x (x+x \log (x))}{(1+\log (x))^3}+2 \int e^x \, dx-\frac {20 \operatorname {Subst}(\int \text {Ei}(x) \, dx,x,2+2 \log (x))}{e^2}\\ &=20 x^2+e^x x^2+\frac {20 \text {Ei}(2 (1+\log (x)))}{e^2}-\frac {40 \text {Ei}(2+2 \log (x)) (1+\log (x))}{e^2}+\frac {20 \text {Ei}(2 (1+\log (x))) (1+2 \log (x))}{e^2}-\frac {10 x^2 (1+2 \log (x))}{1+\log (x)}+\frac {25 e^{-x} x (x+x \log (x))}{(1+\log (x))^3}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.80, size = 28, normalized size = 1.22 \begin {gather*} \frac {e^{-x} x^2 \left (5+e^x+e^x \log (x)\right )^2}{(1+\log (x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10*E^x*x - 25*x^2 + E^(2*x)*(2*x + x^2) + (50*x + 30*E^x*x - 25*x^2 + E^(2*x)*(6*x + 3*x^2))*Log[x]
+ (20*E^x*x + E^(2*x)*(6*x + 3*x^2))*Log[x]^2 + E^(2*x)*(2*x + x^2)*Log[x]^3)/(E^x + 3*E^x*Log[x] + 3*E^x*Log
[x]^2 + E^x*Log[x]^3),x]

[Out]

(x^2*(5 + E^x + E^x*Log[x])^2)/(E^x*(1 + Log[x])^2)

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fricas [B]  time = 0.54, size = 72, normalized size = 3.13 \begin {gather*} \frac {x^{2} e^{\left (2 \, x\right )} \log \relax (x)^{2} + x^{2} e^{\left (2 \, x\right )} + 10 \, x^{2} e^{x} + 25 \, x^{2} + 2 \, {\left (x^{2} e^{\left (2 \, x\right )} + 5 \, x^{2} e^{x}\right )} \log \relax (x)}{e^{x} \log \relax (x)^{2} + 2 \, e^{x} \log \relax (x) + e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+2*x)*exp(x)^2*log(x)^3+((3*x^2+6*x)*exp(x)^2+20*exp(x)*x)*log(x)^2+((3*x^2+6*x)*exp(x)^2+30*ex
p(x)*x-25*x^2+50*x)*log(x)+(x^2+2*x)*exp(x)^2+10*exp(x)*x-25*x^2)/(exp(x)*log(x)^3+3*exp(x)*log(x)^2+3*exp(x)*
log(x)+exp(x)),x, algorithm="fricas")

[Out]

(x^2*e^(2*x)*log(x)^2 + x^2*e^(2*x) + 10*x^2*e^x + 25*x^2 + 2*(x^2*e^(2*x) + 5*x^2*e^x)*log(x))/(e^x*log(x)^2
+ 2*e^x*log(x) + e^x)

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giac [B]  time = 0.18, size = 60, normalized size = 2.61 \begin {gather*} \frac {x^{2} e^{x} \log \relax (x)^{2} + 2 \, x^{2} e^{x} \log \relax (x) + 25 \, x^{2} e^{\left (-x\right )} + x^{2} e^{x} + 10 \, x^{2} \log \relax (x) + 10 \, x^{2}}{\log \relax (x)^{2} + 2 \, \log \relax (x) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+2*x)*exp(x)^2*log(x)^3+((3*x^2+6*x)*exp(x)^2+20*exp(x)*x)*log(x)^2+((3*x^2+6*x)*exp(x)^2+30*ex
p(x)*x-25*x^2+50*x)*log(x)+(x^2+2*x)*exp(x)^2+10*exp(x)*x-25*x^2)/(exp(x)*log(x)^3+3*exp(x)*log(x)^2+3*exp(x)*
log(x)+exp(x)),x, algorithm="giac")

[Out]

(x^2*e^x*log(x)^2 + 2*x^2*e^x*log(x) + 25*x^2*e^(-x) + x^2*e^x + 10*x^2*log(x) + 10*x^2)/(log(x)^2 + 2*log(x)
+ 1)

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maple [A]  time = 0.06, size = 35, normalized size = 1.52

method	result	size
risch	\({\mathrm e}^{x} x^{2}+\frac {5 x^{2} \left (2 \,{\mathrm e}^{x} \ln \relax (x )+2 \,{\mathrm e}^{x}+5\right ) {\mathrm e}^{-x}}{\left (\ln \relax (x )+1\right )^{2}}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^2+2*x)*exp(x)^2*ln(x)^3+((3*x^2+6*x)*exp(x)^2+20*exp(x)*x)*ln(x)^2+((3*x^2+6*x)*exp(x)^2+30*exp(x)*x-2
5*x^2+50*x)*ln(x)+(x^2+2*x)*exp(x)^2+10*exp(x)*x-25*x^2)/(exp(x)*ln(x)^3+3*exp(x)*ln(x)^2+3*exp(x)*ln(x)+exp(x
)),x,method=_RETURNVERBOSE)

[Out]

exp(x)*x^2+5*x^2*(2*exp(x)*ln(x)+2*exp(x)+5)/(ln(x)+1)^2*exp(-x)

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maxima [B]  time = 0.44, size = 57, normalized size = 2.48 \begin {gather*} \frac {25 \, x^{2} e^{\left (-x\right )} + 10 \, x^{2} \log \relax (x) + 10 \, x^{2} + {\left (x^{2} \log \relax (x)^{2} + 2 \, x^{2} \log \relax (x) + x^{2}\right )} e^{x}}{\log \relax (x)^{2} + 2 \, \log \relax (x) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+2*x)*exp(x)^2*log(x)^3+((3*x^2+6*x)*exp(x)^2+20*exp(x)*x)*log(x)^2+((3*x^2+6*x)*exp(x)^2+30*ex
p(x)*x-25*x^2+50*x)*log(x)+(x^2+2*x)*exp(x)^2+10*exp(x)*x-25*x^2)/(exp(x)*log(x)^3+3*exp(x)*log(x)^2+3*exp(x)*
log(x)+exp(x)),x, algorithm="maxima")

[Out]

(25*x^2*e^(-x) + 10*x^2*log(x) + 10*x^2 + (x^2*log(x)^2 + 2*x^2*log(x) + x^2)*e^x)/(log(x)^2 + 2*log(x) + 1)

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mupad [B]  time = 3.66, size = 169, normalized size = 7.35 \begin {gather*} x^2\,{\mathrm {e}}^x+20\,x^2\,\ln \relax (x)-\frac {10\,x^2\,{\ln \relax (x)}^2-\frac {5\,x^2\,{\mathrm {e}}^{-x}\,\left (5\,x-2\,{\mathrm {e}}^x\right )}{2}+\frac {5\,x^2\,{\mathrm {e}}^{-x}\,\ln \relax (x)\,\left (6\,{\mathrm {e}}^x-5\,x+10\right )}{2}}{{\ln \relax (x)}^2+2\,\ln \relax (x)+1}+{\mathrm {e}}^{-x}\,\left (\frac {25\,x^4}{2}-\frac {125\,x^3}{2}+50\,x^2\right )+40\,x^2-\frac {20\,x^2\,{\ln \relax (x)}^2+\frac {25\,x\,{\mathrm {e}}^{-x}\,\left (2\,x+2\,x\,{\mathrm {e}}^x-4\,x^2+x^3\right )}{2}+\frac {25\,x\,{\mathrm {e}}^{-x}\,\ln \relax (x)\,\left (4\,x+4\,x\,{\mathrm {e}}^x-5\,x^2+x^3\right )}{2}}{\ln \relax (x)+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x)*(50*x + exp(2*x)*(6*x + 3*x^2) + 30*x*exp(x) - 25*x^2) + exp(2*x)*(2*x + x^2) + 10*x*exp(x) + log(
x)^2*(exp(2*x)*(6*x + 3*x^2) + 20*x*exp(x)) - 25*x^2 + exp(2*x)*log(x)^3*(2*x + x^2))/(exp(x) + 3*exp(x)*log(x
) + 3*exp(x)*log(x)^2 + exp(x)*log(x)^3),x)

[Out]

x^2*exp(x) + 20*x^2*log(x) - (10*x^2*log(x)^2 - (5*x^2*exp(-x)*(5*x - 2*exp(x)))/2 + (5*x^2*exp(-x)*log(x)*(6*
exp(x) - 5*x + 10))/2)/(2*log(x) + log(x)^2 + 1) + exp(-x)*(50*x^2 - (125*x^3)/2 + (25*x^4)/2) + 40*x^2 - (20*
x^2*log(x)^2 + (25*x*exp(-x)*(2*x + 2*x*exp(x) - 4*x^2 + x^3))/2 + (25*x*exp(-x)*log(x)*(4*x + 4*x*exp(x) - 5*
x^2 + x^3))/2)/(log(x) + 1)

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sympy [B]  time = 0.38, size = 53, normalized size = 2.30 \begin {gather*} \frac {10 x^{2}}{\log {\relax (x )} + 1} + \frac {25 x^{2} e^{- x} + \left (x^{2} \log {\relax (x )}^{2} + 2 x^{2} \log {\relax (x )} + x^{2}\right ) e^{x}}{\log {\relax (x )}^{2} + 2 \log {\relax (x )} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**2+2*x)*exp(x)**2*ln(x)**3+((3*x**2+6*x)*exp(x)**2+20*exp(x)*x)*ln(x)**2+((3*x**2+6*x)*exp(x)**2
+30*exp(x)*x-25*x**2+50*x)*ln(x)+(x**2+2*x)*exp(x)**2+10*exp(x)*x-25*x**2)/(exp(x)*ln(x)**3+3*exp(x)*ln(x)**2+
3*exp(x)*ln(x)+exp(x)),x)

[Out]

10*x**2/(log(x) + 1) + (25*x**2*exp(-x) + (x**2*log(x)**2 + 2*x**2*log(x) + x**2)*exp(x))/(log(x)**2 + 2*log(x
) + 1)

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