\(\int \frac {e^x (2 x+x^2)-4 x \log (3)+2 x \log (3) \log (\log (3))}{\log (3)} \, dx\) [6834]

Optimal result

Integrand size = 30, antiderivative size = 24 \[ \int \frac {e^x \left (2 x+x^2\right )-4 x \log (3)+2 x \log (3) \log (\log (3))}{\log (3)} \, dx=-6-x \left (2 x-x \left (\frac {e^x}{\log (3)}+\log (\log (3))\right )\right ) \]

[Out]

-6-(2*x-x*(exp(x)/ln(3)+ln(ln(3))))*x

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6, 12, 1607, 2227, 2207, 2225} \[ \int \frac {e^x \left (2 x+x^2\right )-4 x \log (3)+2 x \log (3) \log (\log (3))}{\log (3)} \, dx=\frac {e^x x^2}{\log (3)}-x^2 (2-\log (\log (3))) \]

[In]

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

[Out]

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

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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 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 2227

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] &&  !TrueQ[$UseGamma]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^x \left (2 x+x^2\right )+x \log (3) (-4+2 \log (\log (3)))}{\log (3)} \, dx \\ & = \frac {\int \left (e^x \left (2 x+x^2\right )+x \log (3) (-4+2 \log (\log (3)))\right ) \, dx}{\log (3)} \\ & = -x^2 (2-\log (\log (3)))+\frac {\int e^x \left (2 x+x^2\right ) \, dx}{\log (3)} \\ & = -x^2 (2-\log (\log (3)))+\frac {\int e^x x (2+x) \, dx}{\log (3)} \\ & = -x^2 (2-\log (\log (3)))+\frac {\int \left (2 e^x x+e^x x^2\right ) \, dx}{\log (3)} \\ & = -x^2 (2-\log (\log (3)))+\frac {\int e^x x^2 \, dx}{\log (3)}+\frac {2 \int e^x x \, dx}{\log (3)} \\ & = \frac {2 e^x x}{\log (3)}+\frac {e^x x^2}{\log (3)}-x^2 (2-\log (\log (3)))-\frac {2 \int e^x \, dx}{\log (3)}-\frac {2 \int e^x x \, dx}{\log (3)} \\ & = -\frac {2 e^x}{\log (3)}+\frac {e^x x^2}{\log (3)}-x^2 (2-\log (\log (3)))+\frac {2 \int e^x \, dx}{\log (3)} \\ & = \frac {e^x x^2}{\log (3)}-x^2 (2-\log (\log (3))) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {e^x \left (2 x+x^2\right )-4 x \log (3)+2 x \log (3) \log (\log (3))}{\log (3)} \, dx=\frac {x^2 \left (e^x+\log (3) (-2+\log (\log (3)))\right )}{\log (3)} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88

method	result	size
norman	\(\left (\ln \left (\ln \left (3\right )\right )-2\right ) x^{2}+\frac {x^{2} {\mathrm e}^{x}}{\ln \left (3\right )}\)	\(21\)
risch	\(x^{2} \ln \left (\ln \left (3\right )\right )-2 x^{2}+\frac {x^{2} {\mathrm e}^{x}}{\ln \left (3\right )}\)	\(24\)
parts	\(x^{2} \ln \left (\ln \left (3\right )\right )-2 x^{2}+\frac {x^{2} {\mathrm e}^{x}}{\ln \left (3\right )}\)	\(24\)
default	\(\frac {{\mathrm e}^{x} x^{2}-2 x^{2} \ln \left (3\right )+\ln \left (3\right ) \ln \left (\ln \left (3\right )\right ) x^{2}}{\ln \left (3\right )}\)	\(29\)
parallelrisch	\(\frac {{\mathrm e}^{x} x^{2}-2 x^{2} \ln \left (3\right )+\ln \left (3\right ) \ln \left (\ln \left (3\right )\right ) x^{2}}{\ln \left (3\right )}\)	\(29\)

[In]

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

[Out]

(ln(ln(3))-2)*x^2+x^2/ln(3)*exp(x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {e^x \left (2 x+x^2\right )-4 x \log (3)+2 x \log (3) \log (\log (3))}{\log (3)} \, dx=\frac {x^{2} \log \left (3\right ) \log \left (\log \left (3\right )\right ) + x^{2} e^{x} - 2 \, x^{2} \log \left (3\right )}{\log \left (3\right )} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {e^x \left (2 x+x^2\right )-4 x \log (3)+2 x \log (3) \log (\log (3))}{\log (3)} \, dx=\frac {x^{2} e^{x}}{\log {\left (3 \right )}} + x^{2} \left (-2 + \log {\left (\log {\left (3 \right )} \right )}\right ) \]

[In]

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

[Out]

x**2*exp(x)/log(3) + x**2*(-2 + log(log(3)))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {e^x \left (2 x+x^2\right )-4 x \log (3)+2 x \log (3) \log (\log (3))}{\log (3)} \, dx=\frac {x^{2} \log \left (3\right ) \log \left (\log \left (3\right )\right ) + x^{2} e^{x} - 2 \, x^{2} \log \left (3\right )}{\log \left (3\right )} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {e^x \left (2 x+x^2\right )-4 x \log (3)+2 x \log (3) \log (\log (3))}{\log (3)} \, dx=\frac {x^{2} \log \left (3\right ) \log \left (\log \left (3\right )\right ) + x^{2} e^{x} - 2 \, x^{2} \log \left (3\right )}{\log \left (3\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {e^x \left (2 x+x^2\right )-4 x \log (3)+2 x \log (3) \log (\log (3))}{\log (3)} \, dx=\frac {x^2\,\left ({\mathrm {e}}^x-\ln \left (9\right )+\ln \left (3\right )\,\ln \left (\ln \left (3\right )\right )\right )}{\ln \left (3\right )} \]

[In]

int((exp(x)*(2*x + x^2) - 4*x*log(3) + 2*x*log(3)*log(log(3)))/log(3),x)

[Out]

(x^2*(exp(x) - log(9) + log(3)*log(log(3))))/log(3)