3.30.57 \(\int \frac {1}{5} (9 e^x (1+24 e^x+12 x)+45 e^x \log (\log (4))) \, dx\)

Optimal. Leaf size=20 \[ 9 e^x \left (5+\frac {12}{5} \left (-3+e^x+x\right )+\log (\log (4))\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 33, normalized size of antiderivative = 1.65, number of steps used = 9, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 6742, 2194, 2176} \begin {gather*} \frac {108 e^x x}{5}-\frac {99 e^x}{5}+\frac {108 e^{2 x}}{5}+9 e^x \log (\log (4)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(9*E^x*(1 + 24*E^x + 12*x) + 45*E^x*Log[Log[4]])/5,x]

[Out]

(-99*E^x)/5 + (108*E^(2*x))/5 + (108*E^x*x)/5 + 9*E^x*Log[Log[4]]

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]

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \left (9 e^x \left (1+24 e^x+12 x\right )+45 e^x \log (\log (4))\right ) \, dx\\ &=\frac {9}{5} \int e^x \left (1+24 e^x+12 x\right ) \, dx+(9 \log (\log (4))) \int e^x \, dx\\ &=9 e^x \log (\log (4))+\frac {9}{5} \int \left (e^x+24 e^{2 x}+12 e^x x\right ) \, dx\\ &=9 e^x \log (\log (4))+\frac {9 \int e^x \, dx}{5}+\frac {108}{5} \int e^x x \, dx+\frac {216}{5} \int e^{2 x} \, dx\\ &=\frac {9 e^x}{5}+\frac {108 e^{2 x}}{5}+\frac {108 e^x x}{5}+9 e^x \log (\log (4))-\frac {108 \int e^x \, dx}{5}\\ &=-\frac {99 e^x}{5}+\frac {108 e^{2 x}}{5}+\frac {108 e^x x}{5}+9 e^x \log (\log (4))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 22, normalized size = 1.10 \begin {gather*} \frac {9}{5} e^x \left (-11+12 e^x+12 x+5 \log (\log (4))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(9*E^x*(1 + 24*E^x + 12*x) + 45*E^x*Log[Log[4]])/5,x]

[Out]

(9*E^x*(-11 + 12*E^x + 12*x + 5*Log[Log[4]]))/5

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fricas [A]  time = 0.48, size = 39, normalized size = 1.95 \begin {gather*} \frac {1}{5} \, {\left (12 \, x - 11\right )} e^{\left (x + 2 \, \log \relax (3)\right )} + e^{\left (x + 2 \, \log \relax (3)\right )} \log \left (2 \, \log \relax (2)\right ) + \frac {4}{15} \, e^{\left (2 \, x + 4 \, \log \relax (3)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*log(3)+x)*log(2*log(2))+1/5*(24*exp(x)+12*x+1)*exp(2*log(3)+x),x, algorithm="fricas")

[Out]

1/5*(12*x - 11)*e^(x + 2*log(3)) + e^(x + 2*log(3))*log(2*log(2)) + 4/15*e^(2*x + 4*log(3))

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giac [A]  time = 0.26, size = 29, normalized size = 1.45 \begin {gather*} \frac {9}{5} \, {\left (12 \, x - 11\right )} e^{x} + e^{\left (x + 2 \, \log \relax (3)\right )} \log \left (2 \, \log \relax (2)\right ) + \frac {108}{5} \, e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*log(3)+x)*log(2*log(2))+1/5*(24*exp(x)+12*x+1)*exp(2*log(3)+x),x, algorithm="giac")

[Out]

9/5*(12*x - 11)*e^x + e^(x + 2*log(3))*log(2*log(2)) + 108/5*e^(2*x)

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maple [A]  time = 0.04, size = 27, normalized size = 1.35

method	result	size
norman	\(\left (-\frac {99}{5}+9 \ln \relax (2)+9 \ln \left (\ln \relax (2)\right )\right ) {\mathrm e}^{x}+\frac {108 \,{\mathrm e}^{2 x}}{5}+\frac {108 \,{\mathrm e}^{x} x}{5}\)	\(27\)
default	\({\mathrm e}^{2 \ln \relax (3)+x} \ln \left (2 \ln \relax (2)\right )+\frac {108 \,{\mathrm e}^{2 x}}{5}+\frac {108 \,{\mathrm e}^{x} x}{5}-\frac {99 \,{\mathrm e}^{x}}{5}\)	\(30\)
risch	\(9 \,{\mathrm e}^{x} \ln \relax (2)+9 \,{\mathrm e}^{x} \ln \left (\ln \relax (2)\right )+\frac {108 \,{\mathrm e}^{2 x}}{5}+\frac {9 \left (-11+12 x \right ) {\mathrm e}^{x}}{5}\)	\(30\)
meijerg	\(-\frac {54 \left (-2 x +2\right ) {\mathrm e}^{x}}{5}-\left (9 \ln \left (2 \ln \relax (2)\right )+\frac {9}{5}\right ) \left (1-{\mathrm e}^{x}\right )+\frac {108 \,{\mathrm e}^{2 x}}{5}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*ln(3)+x)*ln(2*ln(2))+1/5*(24*exp(x)+12*x+1)*exp(2*ln(3)+x),x,method=_RETURNVERBOSE)

[Out]

(-99/5+9*ln(2)+9*ln(ln(2)))*exp(x)+108/5*exp(x)^2+108/5*exp(x)*x

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maxima [A]  time = 0.44, size = 27, normalized size = 1.35 \begin {gather*} \frac {108}{5} \, {\left (x - 1\right )} e^{x} + 9 \, e^{x} \log \left (2 \, \log \relax (2)\right ) + \frac {108}{5} \, e^{\left (2 \, x\right )} + \frac {9}{5} \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*log(3)+x)*log(2*log(2))+1/5*(24*exp(x)+12*x+1)*exp(2*log(3)+x),x, algorithm="maxima")

[Out]

108/5*(x - 1)*e^x + 9*e^x*log(2*log(2)) + 108/5*e^(2*x) + 9/5*e^x

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mupad [B]  time = 0.06, size = 18, normalized size = 0.90 \begin {gather*} \frac {{\mathrm {e}}^x\,\left (108\,x+\ln \left ({\ln \relax (4)}^{45}\right )+108\,{\mathrm {e}}^x-99\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(2*log(2))*exp(x + 2*log(3)) + (exp(x + 2*log(3))*(12*x + 24*exp(x) + 1))/5,x)

[Out]

(exp(x)*(108*x + log(log(4)^45) + 108*exp(x) - 99))/5

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sympy [A]  time = 0.13, size = 29, normalized size = 1.45 \begin {gather*} \frac {\left (540 x - 495 + 225 \log {\left (\log {\relax (2 )} \right )} + 225 \log {\relax (2 )}\right ) e^{x}}{25} + \frac {108 e^{2 x}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*ln(3)+x)*ln(2*ln(2))+1/5*(24*exp(x)+12*x+1)*exp(2*ln(3)+x),x)

[Out]

(540*x - 495 + 225*log(log(2)) + 225*log(2))*exp(x)/25 + 108*exp(2*x)/5

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