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

Optimal. Leaf size=17 \[ \frac {e^x (4+x) \log (3)}{2 \log (\log (2))} \]

________________________________________________________________________________________

Rubi [A]  time = 0.01, antiderivative size = 32, normalized size of antiderivative = 1.88, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {12, 2176, 2194} \begin {gather*} \frac {e^x (x+5) \log (3)}{2 \log (\log (2))}-\frac {e^x \log (3)}{2 \log (\log (2))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\log (3) \int e^x (5+x) \, dx}{2 \log (\log (2))}\\ &=\frac {e^x (5+x) \log (3)}{2 \log (\log (2))}-\frac {\log (3) \int e^x \, dx}{2 \log (\log (2))}\\ &=-\frac {e^x \log (3)}{2 \log (\log (2))}+\frac {e^x (5+x) \log (3)}{2 \log (\log (2))}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 17, normalized size = 1.00 \begin {gather*} \frac {e^x (4+x) \log (3)}{2 \log (\log (2))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(E^x*(4 + x)*Log[3])/(2*Log[Log[2]])

________________________________________________________________________________________

fricas [A]  time = 1.36, size = 14, normalized size = 0.82 \begin {gather*} \frac {{\left (x + 4\right )} e^{x} \log \relax (3)}{2 \, \log \left (\log \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(x + 4)*e^x*log(3)/log(log(2))

________________________________________________________________________________________

giac [A]  time = 0.20, size = 14, normalized size = 0.82 \begin {gather*} \frac {{\left (x + 4\right )} e^{x} \log \relax (3)}{2 \, \log \left (\log \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(x + 4)*e^x*log(3)/log(log(2))

________________________________________________________________________________________

maple [A]  time = 0.04, size = 15, normalized size = 0.88

method	result	size
gosper	\(\frac {\left (4+x \right ) \ln \relax (3) {\mathrm e}^{x}}{2 \ln \left (\ln \relax (2)\right )}\)	\(15\)
risch	\(\frac {\left (4+x \right ) \ln \relax (3) {\mathrm e}^{x}}{2 \ln \left (\ln \relax (2)\right )}\)	\(15\)
default	\(\frac {\ln \relax (3) \left ({\mathrm e}^{x} x +4 \,{\mathrm e}^{x}\right )}{2 \ln \left (\ln \relax (2)\right )}\)	\(19\)
norman	\(\frac {2 \ln \relax (3) {\mathrm e}^{x}}{\ln \left (\ln \relax (2)\right )}+\frac {\ln \relax (3) x \,{\mathrm e}^{x}}{2 \ln \left (\ln \relax (2)\right )}\)	\(25\)
meijerg	\(-\frac {5 \ln \relax (3) \left (1-{\mathrm e}^{x}\right )}{2 \ln \left (\ln \relax (2)\right )}+\frac {\ln \relax (3) \left (1-\frac {\left (-2 x +2\right ) {\mathrm e}^{x}}{2}\right )}{2 \ln \left (\ln \relax (2)\right )}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(5+x)*ln(3)*exp(x)/ln(ln(2)),x,method=_RETURNVERBOSE)

[Out]

1/2*(4+x)*ln(3)*exp(x)/ln(ln(2))

________________________________________________________________________________________

maxima [A]  time = 0.35, size = 20, normalized size = 1.18 \begin {gather*} \frac {{\left ({\left (x - 1\right )} e^{x} + 5 \, e^{x}\right )} \log \relax (3)}{2 \, \log \left (\log \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.07, size = 19, normalized size = 1.12 \begin {gather*} \frac {{\mathrm {e}}^x\,\ln \left (81\right )+x\,{\mathrm {e}}^x\,\ln \relax (3)}{2\,\ln \left (\ln \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*log(3)*(x + 5))/(2*log(log(2))),x)

[Out]

(exp(x)*log(81) + x*exp(x)*log(3))/(2*log(log(2)))

________________________________________________________________________________________

sympy [A]  time = 0.11, size = 19, normalized size = 1.12 \begin {gather*} \frac {\left (x \log {\relax (3 )} + 4 \log {\relax (3 )}\right ) e^{x}}{2 \log {\left (\log {\relax (2 )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(5+x)*ln(3)*exp(x)/ln(ln(2)),x)

[Out]

(x*log(3) + 4*log(3))*exp(x)/(2*log(log(2)))

________________________________________________________________________________________