3.85.47 \(\int \frac {16+32 e^{2 x}+e^x (-72-8 x)+2 x}{81 \log ^2(3)} \, dx\)

Optimal. Leaf size=18 \[ \frac {\left (8-4 e^x+x\right )^2}{81 \log ^2(3)} \]

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Rubi [B]  time = 0.02, antiderivative size = 59, normalized size of antiderivative = 3.28, number of steps used = 5, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {12, 2194, 2176} \begin {gather*} \frac {x^2}{81 \log ^2(3)}+\frac {16 x}{81 \log ^2(3)}+\frac {8 e^x}{81 \log ^2(3)}+\frac {16 e^{2 x}}{81 \log ^2(3)}-\frac {8 e^x (x+9)}{81 \log ^2(3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(16 + 32*E^(2*x) + E^x*(-72 - 8*x) + 2*x)/(81*Log[3]^2),x]

[Out]

(8*E^x)/(81*Log[3]^2) + (16*E^(2*x))/(81*Log[3]^2) + (16*x)/(81*Log[3]^2) + x^2/(81*Log[3]^2) - (8*E^x*(9 + x)
)/(81*Log[3]^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 {\int \left (16+32 e^{2 x}+e^x (-72-8 x)+2 x\right ) \, dx}{81 \log ^2(3)}\\ &=\frac {16 x}{81 \log ^2(3)}+\frac {x^2}{81 \log ^2(3)}+\frac {\int e^x (-72-8 x) \, dx}{81 \log ^2(3)}+\frac {32 \int e^{2 x} \, dx}{81 \log ^2(3)}\\ &=\frac {16 e^{2 x}}{81 \log ^2(3)}+\frac {16 x}{81 \log ^2(3)}+\frac {x^2}{81 \log ^2(3)}-\frac {8 e^x (9+x)}{81 \log ^2(3)}+\frac {8 \int e^x \, dx}{81 \log ^2(3)}\\ &=\frac {8 e^x}{81 \log ^2(3)}+\frac {16 e^{2 x}}{81 \log ^2(3)}+\frac {16 x}{81 \log ^2(3)}+\frac {x^2}{81 \log ^2(3)}-\frac {8 e^x (9+x)}{81 \log ^2(3)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 20, normalized size = 1.11 \begin {gather*} \frac {\left (-8+4 e^x-x\right )^2}{81 \log ^2(3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(16 + 32*E^(2*x) + E^x*(-72 - 8*x) + 2*x)/(81*Log[3]^2),x]

[Out]

(-8 + 4*E^x - x)^2/(81*Log[3]^2)

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fricas [A]  time = 0.70, size = 26, normalized size = 1.44 \begin {gather*} \frac {x^{2} - 8 \, {\left (x + 8\right )} e^{x} + 16 \, x + 16 \, e^{\left (2 \, x\right )}}{81 \, \log \relax (3)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/81*(32*exp(x)^2+(-8*x-72)*exp(x)+2*x+16)/log(3)^2,x, algorithm="fricas")

[Out]

1/81*(x^2 - 8*(x + 8)*e^x + 16*x + 16*e^(2*x))/log(3)^2

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giac [A]  time = 0.17, size = 26, normalized size = 1.44 \begin {gather*} \frac {x^{2} - 8 \, {\left (x + 8\right )} e^{x} + 16 \, x + 16 \, e^{\left (2 \, x\right )}}{81 \, \log \relax (3)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/81*(32*exp(x)^2+(-8*x-72)*exp(x)+2*x+16)/log(3)^2,x, algorithm="giac")

[Out]

1/81*(x^2 - 8*(x + 8)*e^x + 16*x + 16*e^(2*x))/log(3)^2

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maple [A]  time = 0.04, size = 29, normalized size = 1.61

method	result	size
default	\(\frac {16 x -8 \,{\mathrm e}^{x} x -64 \,{\mathrm e}^{x}+x^{2}+16 \,{\mathrm e}^{2 x}}{81 \ln \relax (3)^{2}}\)	\(29\)
risch	\(\frac {x^{2}}{81 \ln \relax (3)^{2}}+\frac {16 \,{\mathrm e}^{2 x}}{81 \ln \relax (3)^{2}}+\frac {16 x}{81 \ln \relax (3)^{2}}+\frac {\left (-8 x -64\right ) {\mathrm e}^{x}}{81 \ln \relax (3)^{2}}\)	\(41\)
norman	\(\frac {\frac {16 x}{81 \ln \relax (3)}+\frac {x^{2}}{81 \ln \relax (3)}-\frac {64 \,{\mathrm e}^{x}}{81 \ln \relax (3)}+\frac {16 \,{\mathrm e}^{2 x}}{81 \ln \relax (3)}-\frac {8 x \,{\mathrm e}^{x}}{81 \ln \relax (3)}}{\ln \relax (3)}\)	\(50\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/81*(32*exp(x)^2+(-8*x-72)*exp(x)+2*x+16)/ln(3)^2,x,method=_RETURNVERBOSE)

[Out]

1/81/ln(3)^2*(16*x-8*exp(x)*x-64*exp(x)+x^2+16*exp(x)^2)

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maxima [A]  time = 0.38, size = 26, normalized size = 1.44 \begin {gather*} \frac {x^{2} - 8 \, {\left (x + 8\right )} e^{x} + 16 \, x + 16 \, e^{\left (2 \, x\right )}}{81 \, \log \relax (3)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/81*(32*exp(x)^2+(-8*x-72)*exp(x)+2*x+16)/log(3)^2,x, algorithm="maxima")

[Out]

1/81*(x^2 - 8*(x + 8)*e^x + 16*x + 16*e^(2*x))/log(3)^2

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mupad [B]  time = 0.06, size = 28, normalized size = 1.56 \begin {gather*} \frac {16\,x+16\,{\mathrm {e}}^{2\,x}-64\,{\mathrm {e}}^x-8\,x\,{\mathrm {e}}^x+x^2}{81\,{\ln \relax (3)}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x)/81 + (32*exp(2*x))/81 - (exp(x)*(8*x + 72))/81 + 16/81)/log(3)^2,x)

[Out]

(16*x + 16*exp(2*x) - 64*exp(x) - 8*x*exp(x) + x^2)/(81*log(3)^2)

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sympy [B]  time = 0.17, size = 58, normalized size = 3.22 \begin {gather*} \frac {x^{2}}{81 \log {\relax (3 )}^{2}} + \frac {16 x}{81 \log {\relax (3 )}^{2}} + \frac {\left (- 648 x \log {\relax (3 )}^{2} - 5184 \log {\relax (3 )}^{2}\right ) e^{x} + 1296 e^{2 x} \log {\relax (3 )}^{2}}{6561 \log {\relax (3 )}^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/81*(32*exp(x)**2+(-8*x-72)*exp(x)+2*x+16)/ln(3)**2,x)

[Out]

x**2/(81*log(3)**2) + 16*x/(81*log(3)**2) + ((-648*x*log(3)**2 - 5184*log(3)**2)*exp(x) + 1296*exp(2*x)*log(3)
**2)/(6561*log(3)**4)

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