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

Optimal. Leaf size=24 \[ -6-x \left (2 x-x \left (\frac {e^x}{\log (3)}+\log (\log (3))\right )\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6, 12, 1593, 2196, 2176, 2194} \begin {gather*} \frac {e^x x^2}{\log (3)}-x^2 (2-\log (\log (3))) \end {gather*}

Antiderivative was successfully verified.

[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 1593

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 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 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

Rubi steps

\begin {gather*} \begin {aligned} \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 {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 20, normalized size = 0.83 \begin {gather*} \frac {x^2 \left (e^x+\log (3) (-2+\log (\log (3)))\right )}{\log (3)} \end {gather*}

Antiderivative was successfully verified.

[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]

________________________________________________________________________________________

fricas [A]  time = 0.98, size = 28, normalized size = 1.17 \begin {gather*} \frac {x^{2} \log \relax (3) \log \left (\log \relax (3)\right ) + x^{2} e^{x} - 2 \, x^{2} \log \relax (3)}{\log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

giac [A]  time = 0.14, size = 28, normalized size = 1.17 \begin {gather*} \frac {x^{2} \log \relax (3) \log \left (\log \relax (3)\right ) + x^{2} e^{x} - 2 \, x^{2} \log \relax (3)}{\log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maple [A]  time = 0.03, size = 21, normalized size = 0.88

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maxima [A]  time = 0.35, size = 28, normalized size = 1.17 \begin {gather*} \frac {x^{2} \log \relax (3) \log \left (\log \relax (3)\right ) + x^{2} e^{x} - 2 \, x^{2} \log \relax (3)}{\log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

mupad [B]  time = 0.05, size = 21, normalized size = 0.88 \begin {gather*} \frac {x^2\,\left ({\mathrm {e}}^x-\ln \relax (9)+\ln \relax (3)\,\ln \left (\ln \relax (3)\right )\right )}{\ln \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

sympy [A]  time = 0.16, size = 19, normalized size = 0.79 \begin {gather*} \frac {x^{2} e^{x}}{\log {\relax (3 )}} + x^{2} \left (-2 + \log {\left (\log {\relax (3 )} \right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

________________________________________________________________________________________