3.63.93 \(\int \frac {3 e^{x+x^2}+e^{x+x^2} (-3+15 x-6 x^2) \log (x)+e^x (-1+5 x) \log ^2(x)+(-3 e^{x+x^2} x \log (x)-e^x x \log ^2(x)) \log (\frac {e^{-1/e} (3 e^{x^2} x+x \log (x))}{\log (x)})}{3 e^{x^2} x \log (x)+x \log ^2(x)} \, dx\)

Optimal. Leaf size=31 \[ e^x \left (5-\log \left (e^{-1/e} \left (x+\frac {3 e^{x^2} x}{\log (x)}\right )\right )\right ) \]

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Rubi [A]  time = 5.57, antiderivative size = 34, normalized size of antiderivative = 1.10, number of steps used = 27, number of rules used = 7, integrand size = 113, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {6741, 6742, 2199, 2194, 2178, 2176, 2555} \begin {gather*} (1+5 e) e^{x-1}-e^x \log \left (\frac {x \left (3 e^{x^2}+\log (x)\right )}{\log (x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3*E^(x + x^2) + E^(x + x^2)*(-3 + 15*x - 6*x^2)*Log[x] + E^x*(-1 + 5*x)*Log[x]^2 + (-3*E^(x + x^2)*x*Log[
x] - E^x*x*Log[x]^2)*Log[(3*E^x^2*x + x*Log[x])/(E^E^(-1)*Log[x])])/(3*E^x^2*x*Log[x] + x*Log[x]^2),x]

[Out]

E^(-1 + x)*(1 + 5*E) - E^x*Log[(x*(3*E^x^2 + Log[x]))/Log[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 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 2199

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

Rule 2555

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

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

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 {3 e^{x+x^2}+e^{x+x^2} \left (-3+15 x-6 x^2\right ) \log (x)+e^x (-1+5 x) \log ^2(x)+\left (-3 e^{x+x^2} x \log (x)-e^x x \log ^2(x)\right ) \log \left (\frac {e^{-1/e} \left (3 e^{x^2} x+x \log (x)\right )}{\log (x)}\right )}{x \log (x) \left (3 e^{x^2}+\log (x)\right )} \, dx\\ &=\int \left (\frac {e^x \left (-1+2 x^2 \log (x)\right )}{x \left (3 e^{x^2}+\log (x)\right )}+\frac {e^{-1+x} \left (e-e \log (x)+(1+5 e) x \log (x)-2 e x^2 \log (x)-e x \log (x) \log \left (x+\frac {3 e^{x^2} x}{\log (x)}\right )\right )}{x \log (x)}\right ) \, dx\\ &=\int \frac {e^x \left (-1+2 x^2 \log (x)\right )}{x \left (3 e^{x^2}+\log (x)\right )} \, dx+\int \frac {e^{-1+x} \left (e-e \log (x)+(1+5 e) x \log (x)-2 e x^2 \log (x)-e x \log (x) \log \left (x+\frac {3 e^{x^2} x}{\log (x)}\right )\right )}{x \log (x)} \, dx\\ &=\int \left (-\frac {e^x}{x \left (3 e^{x^2}+\log (x)\right )}+\frac {2 e^x x \log (x)}{3 e^{x^2}+\log (x)}\right ) \, dx+\int \left (\frac {e^{-1+x} \left (e-e \log (x)+(1+5 e) x \log (x)-2 e x^2 \log (x)\right )}{x \log (x)}-e^x \log \left (\frac {x \left (3 e^{x^2}+\log (x)\right )}{\log (x)}\right )\right ) \, dx\\ &=2 \int \frac {e^x x \log (x)}{3 e^{x^2}+\log (x)} \, dx-\int \frac {e^x}{x \left (3 e^{x^2}+\log (x)\right )} \, dx+\int \frac {e^{-1+x} \left (e-e \log (x)+(1+5 e) x \log (x)-2 e x^2 \log (x)\right )}{x \log (x)} \, dx-\int e^x \log \left (\frac {x \left (3 e^{x^2}+\log (x)\right )}{\log (x)}\right ) \, dx\\ &=-e^x \log \left (\frac {x \left (3 e^{x^2}+\log (x)\right )}{\log (x)}\right )+2 \int \frac {e^x x \log (x)}{3 e^{x^2}+\log (x)} \, dx+\int \left (\frac {e^{-1+x} \left (-e+(1+5 e) x-2 e x^2\right )}{x}+\frac {e^x}{x \log (x)}\right ) \, dx-\int \frac {e^x}{x \left (3 e^{x^2}+\log (x)\right )} \, dx+\int \frac {e^x \left (-3 e^{x^2}+3 e^{x^2} \left (1+2 x^2\right ) \log (x)+\log ^2(x)\right )}{x \log (x) \left (3 e^{x^2}+\log (x)\right )} \, dx\\ &=-e^x \log \left (\frac {x \left (3 e^{x^2}+\log (x)\right )}{\log (x)}\right )+2 \int \frac {e^x x \log (x)}{3 e^{x^2}+\log (x)} \, dx+\int \frac {e^{-1+x} \left (-e+(1+5 e) x-2 e x^2\right )}{x} \, dx+\int \frac {e^x}{x \log (x)} \, dx-\int \frac {e^x}{x \left (3 e^{x^2}+\log (x)\right )} \, dx+\int \left (-\frac {e^x \left (-1+2 x^2 \log (x)\right )}{x \left (3 e^{x^2}+\log (x)\right )}+\frac {e^x \left (-1+\log (x)+2 x^2 \log (x)\right )}{x \log (x)}\right ) \, dx\\ &=-e^x \log \left (\frac {x \left (3 e^{x^2}+\log (x)\right )}{\log (x)}\right )+2 \int \frac {e^x x \log (x)}{3 e^{x^2}+\log (x)} \, dx+\int \left (e^{-1+x} (1+5 e)-\frac {e^x}{x}-2 e^x x\right ) \, dx+\int \frac {e^x}{x \log (x)} \, dx-\int \frac {e^x}{x \left (3 e^{x^2}+\log (x)\right )} \, dx-\int \frac {e^x \left (-1+2 x^2 \log (x)\right )}{x \left (3 e^{x^2}+\log (x)\right )} \, dx+\int \frac {e^x \left (-1+\log (x)+2 x^2 \log (x)\right )}{x \log (x)} \, dx\\ &=-e^x \log \left (\frac {x \left (3 e^{x^2}+\log (x)\right )}{\log (x)}\right )-2 \int e^x x \, dx+2 \int \frac {e^x x \log (x)}{3 e^{x^2}+\log (x)} \, dx+(1+5 e) \int e^{-1+x} \, dx-\int \frac {e^x}{x} \, dx+\int \left (\frac {e^x \left (1+2 x^2\right )}{x}-\frac {e^x}{x \log (x)}\right ) \, dx+\int \frac {e^x}{x \log (x)} \, dx-\int \frac {e^x}{x \left (3 e^{x^2}+\log (x)\right )} \, dx-\int \left (-\frac {e^x}{x \left (3 e^{x^2}+\log (x)\right )}+\frac {2 e^x x \log (x)}{3 e^{x^2}+\log (x)}\right ) \, dx\\ &=e^{-1+x} (1+5 e)-2 e^x x-\text {Ei}(x)-e^x \log \left (\frac {x \left (3 e^{x^2}+\log (x)\right )}{\log (x)}\right )+2 \int e^x \, dx+\int \frac {e^x \left (1+2 x^2\right )}{x} \, dx\\ &=2 e^x+e^{-1+x} (1+5 e)-2 e^x x-\text {Ei}(x)-e^x \log \left (\frac {x \left (3 e^{x^2}+\log (x)\right )}{\log (x)}\right )+\int \left (\frac {e^x}{x}+2 e^x x\right ) \, dx\\ &=2 e^x+e^{-1+x} (1+5 e)-2 e^x x-\text {Ei}(x)-e^x \log \left (\frac {x \left (3 e^{x^2}+\log (x)\right )}{\log (x)}\right )+2 \int e^x x \, dx+\int \frac {e^x}{x} \, dx\\ &=2 e^x+e^{-1+x} (1+5 e)-e^x \log \left (\frac {x \left (3 e^{x^2}+\log (x)\right )}{\log (x)}\right )-2 \int e^x \, dx\\ &=e^{-1+x} (1+5 e)-e^x \log \left (\frac {x \left (3 e^{x^2}+\log (x)\right )}{\log (x)}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 29, normalized size = 0.94 \begin {gather*} e^{-1+x} \left (1+5 e-e \log \left (x+\frac {3 e^{x^2} x}{\log (x)}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3*E^(x + x^2) + E^(x + x^2)*(-3 + 15*x - 6*x^2)*Log[x] + E^x*(-1 + 5*x)*Log[x]^2 + (-3*E^(x + x^2)*
x*Log[x] - E^x*x*Log[x]^2)*Log[(3*E^x^2*x + x*Log[x])/(E^E^(-1)*Log[x])])/(3*E^x^2*x*Log[x] + x*Log[x]^2),x]

[Out]

E^(-1 + x)*(1 + 5*E - E*Log[x + (3*E^x^2*x)/Log[x]])

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fricas [A]  time = 0.93, size = 47, normalized size = 1.52 \begin {gather*} -{\left (e^{\left (x^{2} + x\right )} \log \left (\frac {{\left (3 \, x e^{\left (x^{2}\right )} + x \log \relax (x)\right )} e^{\left (-e^{\left (-1\right )}\right )}}{\log \relax (x)}\right ) - 5 \, e^{\left (x^{2} + x\right )}\right )} e^{\left (-x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*exp(x)*log(x)^2-3*x*exp(x)*exp(x^2)*log(x))*log((x*log(x)+3*exp(x^2)*x)/exp(1/exp(1))/log(x))+(
5*x-1)*exp(x)*log(x)^2+(-6*x^2+15*x-3)*exp(x)*exp(x^2)*log(x)+3*exp(x)*exp(x^2))/(x*log(x)^2+3*x*exp(x^2)*log(
x)),x, algorithm="fricas")

[Out]

-(e^(x^2 + x)*log((3*x*e^(x^2) + x*log(x))*e^(-e^(-1))/log(x)) - 5*e^(x^2 + x))*e^(-x^2)

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giac [A]  time = 0.27, size = 46, normalized size = 1.48 \begin {gather*} -{\left (e^{\left (x + 1\right )} \log \relax (x) + e^{\left (x + 1\right )} \log \left (3 \, e^{\left (x^{2}\right )} + \log \relax (x)\right ) - e^{\left (x + 1\right )} \log \left (\log \relax (x)\right ) - 5 \, e^{\left (x + 1\right )} - e^{x}\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*exp(x)*log(x)^2-3*x*exp(x)*exp(x^2)*log(x))*log((x*log(x)+3*exp(x^2)*x)/exp(1/exp(1))/log(x))+(
5*x-1)*exp(x)*log(x)^2+(-6*x^2+15*x-3)*exp(x)*exp(x^2)*log(x)+3*exp(x)*exp(x^2))/(x*log(x)^2+3*x*exp(x^2)*log(
x)),x, algorithm="giac")

[Out]

-(e^(x + 1)*log(x) + e^(x + 1)*log(3*e^(x^2) + log(x)) - e^(x + 1)*log(log(x)) - 5*e^(x + 1) - e^x)*e^(-1)

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maple [C]  time = 0.32, size = 333, normalized size = 10.74

method	result	size
risch	\(-{\mathrm e}^{x} \ln \left ({\mathrm e}^{x^{2}}+\frac {\ln \relax (x )}{3}\right )+\frac {\left (i \pi \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{x^{2}}+\frac {\ln \relax (x )}{3}\right )}{\ln \relax (x )}\right )^{3} {\mathrm e}+i \pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{x^{2}}+\frac {\ln \relax (x )}{3}\right )\right ) \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{x^{2}}+\frac {\ln \relax (x )}{3}\right )}{\ln \relax (x )}\right ) \mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right ) {\mathrm e}-i \pi \,\mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{x^{2}}+\frac {\ln \relax (x )}{3}\right )}{\ln \relax (x )}\right ) \mathrm {csgn}\left (\frac {i x \left ({\mathrm e}^{x^{2}}+\frac {\ln \relax (x )}{3}\right )}{\ln \relax (x )}\right )^{2} {\mathrm e}+i \pi \mathrm {csgn}\left (\frac {i x \left ({\mathrm e}^{x^{2}}+\frac {\ln \relax (x )}{3}\right )}{\ln \relax (x )}\right )^{3} {\mathrm e}-i \pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{x^{2}}+\frac {\ln \relax (x )}{3}\right )\right ) \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{x^{2}}+\frac {\ln \relax (x )}{3}\right )}{\ln \relax (x )}\right )^{2} {\mathrm e}+i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{x^{2}}+\frac {\ln \relax (x )}{3}\right )}{\ln \relax (x )}\right ) \mathrm {csgn}\left (\frac {i x \left ({\mathrm e}^{x^{2}}+\frac {\ln \relax (x )}{3}\right )}{\ln \relax (x )}\right ) {\mathrm e}-i \pi \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{x^{2}}+\frac {\ln \relax (x )}{3}\right )}{\ln \relax (x )}\right )^{2} \mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right ) {\mathrm e}-i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i x \left ({\mathrm e}^{x^{2}}+\frac {\ln \relax (x )}{3}\right )}{\ln \relax (x )}\right )^{2} {\mathrm e}+2-2 \ln \relax (3) {\mathrm e}-2 \,{\mathrm e} \ln \relax (x )+2 \ln \left (\ln \relax (x )\right ) {\mathrm e}+10 \,{\mathrm e}\right ) {\mathrm e}^{x -1}}{2}\)	\(333\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x*exp(x)*ln(x)^2-3*x*exp(x)*exp(x^2)*ln(x))*ln((x*ln(x)+3*exp(x^2)*x)/exp(1/exp(1))/ln(x))+(5*x-1)*exp(
x)*ln(x)^2+(-6*x^2+15*x-3)*exp(x)*exp(x^2)*ln(x)+3*exp(x)*exp(x^2))/(x*ln(x)^2+3*x*exp(x^2)*ln(x)),x,method=_R
ETURNVERBOSE)

[Out]

-exp(x)*ln(exp(x^2)+1/3*ln(x))+1/2*(I*Pi*csgn(I/ln(x)*(exp(x^2)+1/3*ln(x)))^3*exp(1)+I*Pi*csgn(I*(exp(x^2)+1/3
*ln(x)))*csgn(I/ln(x)*(exp(x^2)+1/3*ln(x)))*csgn(I/ln(x))*exp(1)-I*Pi*csgn(I/ln(x)*(exp(x^2)+1/3*ln(x)))*csgn(
I*x/ln(x)*(exp(x^2)+1/3*ln(x)))^2*exp(1)+I*Pi*csgn(I*x/ln(x)*(exp(x^2)+1/3*ln(x)))^3*exp(1)-I*Pi*csgn(I*(exp(x
^2)+1/3*ln(x)))*csgn(I/ln(x)*(exp(x^2)+1/3*ln(x)))^2*exp(1)+I*Pi*csgn(I*x)*csgn(I/ln(x)*(exp(x^2)+1/3*ln(x)))*
csgn(I*x/ln(x)*(exp(x^2)+1/3*ln(x)))*exp(1)-I*Pi*csgn(I/ln(x)*(exp(x^2)+1/3*ln(x)))^2*csgn(I/ln(x))*exp(1)-I*P
i*csgn(I*x)*csgn(I*x/ln(x)*(exp(x^2)+1/3*ln(x)))^2*exp(1)+2-2*ln(3)*exp(1)-2*exp(1)*ln(x)+2*ln(ln(x))*exp(1)+1
0*exp(1))*exp(x-1)

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maxima [A]  time = 0.44, size = 43, normalized size = 1.39 \begin {gather*} -{\left ({\left (e \log \relax (x) - 5 \, e - 1\right )} e^{x} + e^{\left (x + 1\right )} \log \left (3 \, e^{\left (x^{2}\right )} + \log \relax (x)\right ) - e^{\left (x + 1\right )} \log \left (\log \relax (x)\right )\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*exp(x)*log(x)^2-3*x*exp(x)*exp(x^2)*log(x))*log((x*log(x)+3*exp(x^2)*x)/exp(1/exp(1))/log(x))+(
5*x-1)*exp(x)*log(x)^2+(-6*x^2+15*x-3)*exp(x)*exp(x^2)*log(x)+3*exp(x)*exp(x^2))/(x*log(x)^2+3*x*exp(x^2)*log(
x)),x, algorithm="maxima")

[Out]

-((e*log(x) - 5*e - 1)*e^x + e^(x + 1)*log(3*e^(x^2) + log(x)) - e^(x + 1)*log(log(x)))*e^(-1)

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mupad [B]  time = 4.56, size = 29, normalized size = 0.94 \begin {gather*} -{\mathrm {e}}^x\,\left (\ln \left (\frac {{\mathrm {e}}^{-{\mathrm {e}}^{-1}}\,\left (3\,x\,{\mathrm {e}}^{x^2}+x\,\ln \relax (x)\right )}{\ln \relax (x)}\right )-5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*exp(x^2)*exp(x) - log((exp(-exp(-1))*(3*x*exp(x^2) + x*log(x)))/log(x))*(x*exp(x)*log(x)^2 + 3*x*exp(x^
2)*exp(x)*log(x)) + exp(x)*log(x)^2*(5*x - 1) - exp(x^2)*exp(x)*log(x)*(6*x^2 - 15*x + 3))/(x*log(x)^2 + 3*x*e
xp(x^2)*log(x)),x)

[Out]

-exp(x)*(log((exp(-exp(-1))*(3*x*exp(x^2) + x*log(x)))/log(x)) - 5)

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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(((-x*exp(x)*ln(x)**2-3*x*exp(x)*exp(x**2)*ln(x))*ln((x*ln(x)+3*exp(x**2)*x)/exp(1/exp(1))/ln(x))+(5*
x-1)*exp(x)*ln(x)**2+(-6*x**2+15*x-3)*exp(x)*exp(x**2)*ln(x)+3*exp(x)*exp(x**2))/(x*ln(x)**2+3*x*exp(x**2)*ln(
x)),x)

[Out]

Timed out

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