3.74.85 \(\int \frac {18 e+e^x (e (-12-12 x)+18 x+9 x^2)+6 e \log (5)}{2 e} \, dx\)

Optimal. Leaf size=24 \[ 1+3 x \left (3-e^x \left (2-\frac {3 x}{2 e}\right )+\log (5)\right ) \]

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Rubi [A]  time = 0.07, antiderivative size = 33, normalized size of antiderivative = 1.38, number of steps used = 11, number of rules used = 4, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {12, 2196, 2176, 2194} \begin {gather*} \frac {9}{2} e^{x-1} x^2+6 e^x-6 e^x (x+1)+3 x (3+\log (5)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(18*E + E^x*(E*(-12 - 12*x) + 18*x + 9*x^2) + 6*E*Log[5])/(2*E),x]

[Out]

6*E^x + (9*E^(-1 + x)*x^2)/2 - 6*E^x*(1 + x) + 3*x*(3 + Log[5])

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 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} &=\frac {\int \left (18 e+e^x \left (e (-12-12 x)+18 x+9 x^2\right )+6 e \log (5)\right ) \, dx}{2 e}\\ &=3 x (3+\log (5))+\frac {\int e^x \left (e (-12-12 x)+18 x+9 x^2\right ) \, dx}{2 e}\\ &=3 x (3+\log (5))+\frac {\int \left (18 e^x x+9 e^x x^2-12 e^{1+x} (1+x)\right ) \, dx}{2 e}\\ &=3 x (3+\log (5))+\frac {9 \int e^x x^2 \, dx}{2 e}-\frac {6 \int e^{1+x} (1+x) \, dx}{e}+\frac {9 \int e^x x \, dx}{e}\\ &=9 e^{-1+x} x+\frac {9}{2} e^{-1+x} x^2-6 e^x (1+x)+3 x (3+\log (5))+\frac {6 \int e^{1+x} \, dx}{e}-\frac {9 \int e^x \, dx}{e}-\frac {9 \int e^x x \, dx}{e}\\ &=-9 e^{-1+x}+6 e^x+\frac {9}{2} e^{-1+x} x^2-6 e^x (1+x)+3 x (3+\log (5))+\frac {9 \int e^x \, dx}{e}\\ &=6 e^x+\frac {9}{2} e^{-1+x} x^2-6 e^x (1+x)+3 x (3+\log (5))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 26, normalized size = 1.08 \begin {gather*} 9 x-6 e^x x+\frac {9}{2} e^{-1+x} x^2+x \log (125) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(18*E + E^x*(E*(-12 - 12*x) + 18*x + 9*x^2) + 6*E*Log[5])/(2*E),x]

[Out]

9*x - 6*E^x*x + (9*E^(-1 + x)*x^2)/2 + x*Log[125]

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fricas [A]  time = 0.99, size = 31, normalized size = 1.29 \begin {gather*} \frac {3}{2} \, {\left (2 \, x e \log \relax (5) + 6 \, x e + {\left (3 \, x^{2} - 4 \, x e\right )} e^{x}\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(((-12*x-12)*exp(1)+9*x^2+18*x)*exp(x)+6*exp(1)*log(5)+18*exp(1))/exp(1),x, algorithm="fricas")

[Out]

3/2*(2*x*e*log(5) + 6*x*e + (3*x^2 - 4*x*e)*e^x)*e^(-1)

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giac [A]  time = 0.12, size = 31, normalized size = 1.29 \begin {gather*} \frac {3}{2} \, {\left (3 \, x^{2} e^{x} + 2 \, x e \log \relax (5) + 6 \, x e - 4 \, x e^{\left (x + 1\right )}\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(((-12*x-12)*exp(1)+9*x^2+18*x)*exp(x)+6*exp(1)*log(5)+18*exp(1))/exp(1),x, algorithm="giac")

[Out]

3/2*(3*x^2*e^x + 2*x*e*log(5) + 6*x*e - 4*x*e^(x + 1))*e^(-1)

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maple [A]  time = 0.03, size = 26, normalized size = 1.08

method	result	size
norman	\(\left (3 \ln \relax (5)+9\right ) x -6 \,{\mathrm e}^{x} x +\frac {9 x^{2} {\mathrm e}^{-1} {\mathrm e}^{x}}{2}\)	\(26\)
risch	\(3 x \ln \relax (5)+9 x +\frac {\left (-12 x \,{\mathrm e}+9 x^{2}\right ) {\mathrm e}^{x -1}}{2}\)	\(27\)
default	\(\frac {{\mathrm e}^{-1} \left (9 \,{\mathrm e}^{x} x^{2}-12 \,{\mathrm e} \,{\mathrm e}^{x}-12 \,{\mathrm e} \left ({\mathrm e}^{x} x -{\mathrm e}^{x}\right )+6 x \,{\mathrm e} \ln \relax (5)+18 x \,{\mathrm e}\right )}{2}\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(((-12*x-12)*exp(1)+9*x^2+18*x)*exp(x)+6*exp(1)*ln(5)+18*exp(1))/exp(1),x,method=_RETURNVERBOSE)

[Out]

(3*ln(5)+9)*x-6*exp(x)*x+9/2*x^2/exp(1)*exp(x)

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maxima [A]  time = 0.49, size = 31, normalized size = 1.29 \begin {gather*} \frac {3}{2} \, {\left (2 \, x e \log \relax (5) + 6 \, x e + {\left (3 \, x^{2} - 4 \, x e\right )} e^{x}\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(((-12*x-12)*exp(1)+9*x^2+18*x)*exp(x)+6*exp(1)*log(5)+18*exp(1))/exp(1),x, algorithm="maxima")

[Out]

3/2*(2*x*e*log(5) + 6*x*e + (3*x^2 - 4*x*e)*e^x)*e^(-1)

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mupad [B]  time = 0.10, size = 23, normalized size = 0.96 \begin {gather*} 9\,x+3\,x\,\ln \relax (5)-6\,x\,{\mathrm {e}}^x+\frac {9\,x^2\,{\mathrm {e}}^{-1}\,{\mathrm {e}}^x}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-1)*(9*exp(1) + 3*exp(1)*log(5) + (exp(x)*(18*x + 9*x^2 - exp(1)*(12*x + 12)))/2),x)

[Out]

9*x + 3*x*log(5) - 6*x*exp(x) + (9*x^2*exp(-1)*exp(x))/2

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sympy [A]  time = 0.13, size = 27, normalized size = 1.12 \begin {gather*} x \left (3 \log {\relax (5 )} + 9\right ) + \frac {\left (9 x^{2} - 12 e x\right ) e^{x}}{2 e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(((-12*x-12)*exp(1)+9*x**2+18*x)*exp(x)+6*exp(1)*ln(5)+18*exp(1))/exp(1),x)

[Out]

x*(3*log(5) + 9) + (9*x**2 - 12*E*x)*exp(-1)*exp(x)/2

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