3.60.67 \(\int \frac {1}{4} (1+2 x+e^x (1+x)+(-4-8 x) \log (9)) \, dx\)

Optimal. Leaf size=21 \[ \frac {1}{4} x \left (1+e^x+x\right )-\left (x+x^2\right ) \log (9) \]

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Rubi [B]  time = 0.02, antiderivative size = 43, normalized size of antiderivative = 2.05, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 2176, 2194} \begin {gather*} \frac {x^2}{4}+\frac {x}{4}-\frac {e^x}{4}+\frac {1}{4} e^x (x+1)-\frac {1}{4} (2 x+1)^2 \log (9) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + 2*x + E^x*(1 + x) + (-4 - 8*x)*Log[9])/4,x]

[Out]

-1/4*E^x + x/4 + x^2/4 + (E^x*(1 + x))/4 - ((1 + 2*x)^2*Log[9])/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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int \left (1+2 x+e^x (1+x)+(-4-8 x) \log (9)\right ) \, dx\\ &=\frac {x}{4}+\frac {x^2}{4}-\frac {1}{4} (1+2 x)^2 \log (9)+\frac {1}{4} \int e^x (1+x) \, dx\\ &=\frac {x}{4}+\frac {x^2}{4}+\frac {1}{4} e^x (1+x)-\frac {1}{4} (1+2 x)^2 \log (9)-\frac {\int e^x \, dx}{4}\\ &=-\frac {e^x}{4}+\frac {x}{4}+\frac {x^2}{4}+\frac {1}{4} e^x (1+x)-\frac {1}{4} (1+2 x)^2 \log (9)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 26, normalized size = 1.24 \begin {gather*} \frac {1}{4} \left (x+e^x x+x^2-4 x \log (9)-4 x^2 \log (9)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + 2*x + E^x*(1 + x) + (-4 - 8*x)*Log[9])/4,x]

[Out]

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

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fricas [A]  time = 0.67, size = 23, normalized size = 1.10 \begin {gather*} \frac {1}{4} \, x^{2} + \frac {1}{4} \, x e^{x} - 2 \, {\left (x^{2} + x\right )} \log \relax (3) + \frac {1}{4} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(x+1)*exp(x)+1/2*(-8*x-4)*log(3)+1/2*x+1/4,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.17, size = 23, normalized size = 1.10 \begin {gather*} \frac {1}{4} \, x^{2} + \frac {1}{4} \, x e^{x} - 2 \, {\left (x^{2} + x\right )} \log \relax (3) + \frac {1}{4} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(x+1)*exp(x)+1/2*(-8*x-4)*log(3)+1/2*x+1/4,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.03, size = 25, normalized size = 1.19

method	result	size
norman	\(\left (-2 \ln \relax (3)+\frac {1}{4}\right ) x +\left (-2 \ln \relax (3)+\frac {1}{4}\right ) x^{2}+\frac {{\mathrm e}^{x} x}{4}\)	\(25\)
default	\(\frac {x}{4}+\frac {x^{2}}{4}-2 x^{2} \ln \relax (3)-2 x \ln \relax (3)+\frac {{\mathrm e}^{x} x}{4}\)	\(27\)
risch	\(\frac {x}{4}+\frac {x^{2}}{4}-2 x^{2} \ln \relax (3)-2 x \ln \relax (3)+\frac {{\mathrm e}^{x} x}{4}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(x+1)*exp(x)+1/2*(-8*x-4)*ln(3)+1/2*x+1/4,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.36, size = 23, normalized size = 1.10 \begin {gather*} \frac {1}{4} \, x^{2} + \frac {1}{4} \, x e^{x} - 2 \, {\left (x^{2} + x\right )} \log \relax (3) + \frac {1}{4} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(x+1)*exp(x)+1/2*(-8*x-4)*log(3)+1/2*x+1/4,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.23, size = 26, normalized size = 1.24 \begin {gather*} \frac {x\,{\mathrm {e}}^x}{4}-x^2\,\left (2\,\ln \relax (3)-\frac {1}{4}\right )-x\,\left (2\,\ln \relax (3)-\frac {1}{4}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/2 - (log(3)*(8*x + 4))/2 + (exp(x)*(x + 1))/4 + 1/4,x)

[Out]

(x*exp(x))/4 - x^2*(2*log(3) - 1/4) - x*(2*log(3) - 1/4)

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sympy [A]  time = 0.10, size = 27, normalized size = 1.29 \begin {gather*} x^{2} \left (\frac {1}{4} - 2 \log {\relax (3 )}\right ) + \frac {x e^{x}}{4} + x \left (\frac {1}{4} - 2 \log {\relax (3 )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(x+1)*exp(x)+1/2*(-8*x-4)*ln(3)+1/2*x+1/4,x)

[Out]

x**2*(1/4 - 2*log(3)) + x*exp(x)/4 + x*(1/4 - 2*log(3))

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