3.24.14 \(\int \frac {1}{4} (1+e^{-1+x} (1-9 x-2 x^2+x^3+(-5-3 x+x^2) \log (2))) \, dx\)

Optimal. Leaf size=28 \[ \frac {1}{4} \left (x-e^{-1+x} (-x+(5-x) x (x+\log (2)))\right ) \]

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Rubi [B]  time = 0.17, antiderivative size = 66, normalized size of antiderivative = 2.36, number of steps used = 22, number of rules used = 4, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {12, 2196, 2194, 2176} \begin {gather*} \frac {1}{4} e^{x-1} x^3-\frac {5}{4} e^{x-1} x^2+\frac {1}{4} e^{x-1} x^2 \log (2)+\frac {1}{4} e^{x-1} x+\frac {x}{4}-\frac {5}{4} e^{x-1} x \log (2) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + E^(-1 + x)*(1 - 9*x - 2*x^2 + x^3 + (-5 - 3*x + x^2)*Log[2]))/4,x]

[Out]

x/4 + (E^(-1 + x)*x)/4 - (5*E^(-1 + x)*x^2)/4 + (E^(-1 + x)*x^3)/4 - (5*E^(-1 + x)*x*Log[2])/4 + (E^(-1 + x)*x
^2*Log[2])/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 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 {1}{4} \int \left (1+e^{-1+x} \left (1-9 x-2 x^2+x^3+\left (-5-3 x+x^2\right ) \log (2)\right )\right ) \, dx\\ &=\frac {x}{4}+\frac {1}{4} \int e^{-1+x} \left (1-9 x-2 x^2+x^3+\left (-5-3 x+x^2\right ) \log (2)\right ) \, dx\\ &=\frac {x}{4}+\frac {1}{4} \int \left (e^{-1+x}-9 e^{-1+x} x-2 e^{-1+x} x^2+e^{-1+x} x^3+e^{-1+x} \left (-5-3 x+x^2\right ) \log (2)\right ) \, dx\\ &=\frac {x}{4}+\frac {1}{4} \int e^{-1+x} \, dx+\frac {1}{4} \int e^{-1+x} x^3 \, dx-\frac {1}{2} \int e^{-1+x} x^2 \, dx-\frac {9}{4} \int e^{-1+x} x \, dx+\frac {1}{4} \log (2) \int e^{-1+x} \left (-5-3 x+x^2\right ) \, dx\\ &=\frac {e^{-1+x}}{4}+\frac {x}{4}-\frac {9}{4} e^{-1+x} x-\frac {1}{2} e^{-1+x} x^2+\frac {1}{4} e^{-1+x} x^3-\frac {3}{4} \int e^{-1+x} x^2 \, dx+\frac {9}{4} \int e^{-1+x} \, dx+\frac {1}{4} \log (2) \int \left (-5 e^{-1+x}-3 e^{-1+x} x+e^{-1+x} x^2\right ) \, dx+\int e^{-1+x} x \, dx\\ &=\frac {5 e^{-1+x}}{2}+\frac {x}{4}-\frac {5}{4} e^{-1+x} x-\frac {5}{4} e^{-1+x} x^2+\frac {1}{4} e^{-1+x} x^3+\frac {3}{2} \int e^{-1+x} x \, dx+\frac {1}{4} \log (2) \int e^{-1+x} x^2 \, dx-\frac {1}{4} (3 \log (2)) \int e^{-1+x} x \, dx-\frac {1}{4} (5 \log (2)) \int e^{-1+x} \, dx-\int e^{-1+x} \, dx\\ &=\frac {3 e^{-1+x}}{2}+\frac {x}{4}+\frac {1}{4} e^{-1+x} x-\frac {5}{4} e^{-1+x} x^2+\frac {1}{4} e^{-1+x} x^3-\frac {5}{4} e^{-1+x} \log (2)-\frac {3}{4} e^{-1+x} x \log (2)+\frac {1}{4} e^{-1+x} x^2 \log (2)-\frac {3}{2} \int e^{-1+x} \, dx-\frac {1}{2} \log (2) \int e^{-1+x} x \, dx+\frac {1}{4} (3 \log (2)) \int e^{-1+x} \, dx\\ &=\frac {x}{4}+\frac {1}{4} e^{-1+x} x-\frac {5}{4} e^{-1+x} x^2+\frac {1}{4} e^{-1+x} x^3-\frac {1}{2} e^{-1+x} \log (2)-\frac {5}{4} e^{-1+x} x \log (2)+\frac {1}{4} e^{-1+x} x^2 \log (2)+\frac {1}{2} \log (2) \int e^{-1+x} \, dx\\ &=\frac {x}{4}+\frac {1}{4} e^{-1+x} x-\frac {5}{4} e^{-1+x} x^2+\frac {1}{4} e^{-1+x} x^3-\frac {5}{4} e^{-1+x} x \log (2)+\frac {1}{4} e^{-1+x} x^2 \log (2)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 40, normalized size = 1.43 \begin {gather*} \frac {1}{4} \left (x+e^x \left (\frac {x^3}{e}+\frac {x (1-5 \log (2))}{e}+\frac {x^2 (-5+\log (2))}{e}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + E^(-1 + x)*(1 - 9*x - 2*x^2 + x^3 + (-5 - 3*x + x^2)*Log[2]))/4,x]

[Out]

(x + E^x*(x^3/E + (x*(1 - 5*Log[2]))/E + (x^2*(-5 + Log[2]))/E))/4

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fricas [A]  time = 0.60, size = 30, normalized size = 1.07 \begin {gather*} \frac {1}{4} \, {\left (x^{3} - 5 \, x^{2} + {\left (x^{2} - 5 \, x\right )} \log \relax (2) + x\right )} e^{\left (x - 1\right )} + \frac {1}{4} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((x^2-3*x-5)*log(2)+x^3-2*x^2-9*x+1)*exp(x-1)+1/4,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.18, size = 31, normalized size = 1.11 \begin {gather*} \frac {1}{4} \, {\left (x^{3} + x^{2} \log \relax (2) - 5 \, x^{2} - 5 \, x \log \relax (2) + x\right )} e^{\left (x - 1\right )} + \frac {1}{4} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((x^2-3*x-5)*log(2)+x^3-2*x^2-9*x+1)*exp(x-1)+1/4,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.08, size = 32, normalized size = 1.14

method	result	size
risch	\(\frac {\left (x^{2} \ln \relax (2)+x^{3}-5 x \ln \relax (2)-5 x^{2}+x \right ) {\mathrm e}^{x -1}}{4}+\frac {x}{4}\)	\(32\)
norman	\(\left (-\frac {5}{4}+\frac {\ln \relax (2)}{4}\right ) x^{2} {\mathrm e}^{x -1}+\left (-\frac {5 \ln \relax (2)}{4}+\frac {1}{4}\right ) x \,{\mathrm e}^{x -1}+\frac {x}{4}+\frac {x^{3} {\mathrm e}^{x -1}}{4}\)	\(40\)
default	\(\frac {x}{4}-\frac {{\mathrm e}^{x -1} \left (x -1\right )^{2}}{2}-\frac {3 \,{\mathrm e}^{x -1} \left (x -1\right )}{2}-\frac {3 \,{\mathrm e}^{x -1}}{4}+\frac {{\mathrm e}^{x -1} \left (x -1\right )^{3}}{4}+\frac {{\mathrm e}^{x -1} \ln \relax (2) \left (x -1\right )^{2}}{4}-\frac {3 \,{\mathrm e}^{x -1} \ln \relax (2) \left (x -1\right )}{4}-\ln \relax (2) {\mathrm e}^{x -1}\)	\(74\)
derivativedivides	\(\frac {x}{4}-\frac {1}{4}-\frac {{\mathrm e}^{x -1} \left (x -1\right )^{2}}{2}-\frac {3 \,{\mathrm e}^{x -1} \left (x -1\right )}{2}-\frac {3 \,{\mathrm e}^{x -1}}{4}+\frac {{\mathrm e}^{x -1} \left (x -1\right )^{3}}{4}+\frac {{\mathrm e}^{x -1} \ln \relax (2) \left (x -1\right )^{2}}{4}-\frac {3 \,{\mathrm e}^{x -1} \ln \relax (2) \left (x -1\right )}{4}-\ln \relax (2) {\mathrm e}^{x -1}\)	\(75\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*((x^2-3*x-5)*ln(2)+x^3-2*x^2-9*x+1)*exp(x-1)+1/4,x,method=_RETURNVERBOSE)

[Out]

1/4*(x^2*ln(2)+x^3-5*x*ln(2)-5*x^2+x)*exp(x-1)+1/4*x

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maxima [B]  time = 0.49, size = 87, normalized size = 3.11 \begin {gather*} \frac {1}{4} \, {\left (x^{2} - 2 \, x + 2\right )} e^{\left (x - 1\right )} \log \relax (2) - \frac {3}{4} \, {\left (x - 1\right )} e^{\left (x - 1\right )} \log \relax (2) + \frac {1}{4} \, {\left (x^{3} - 3 \, x^{2} + 6 \, x - 6\right )} e^{\left (x - 1\right )} - \frac {1}{2} \, {\left (x^{2} - 2 \, x + 2\right )} e^{\left (x - 1\right )} - \frac {9}{4} \, {\left (x - 1\right )} e^{\left (x - 1\right )} - \frac {5}{4} \, e^{\left (x - 1\right )} \log \relax (2) + \frac {1}{4} \, x + \frac {1}{4} \, e^{\left (x - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((x^2-3*x-5)*log(2)+x^3-2*x^2-9*x+1)*exp(x-1)+1/4,x, algorithm="maxima")

[Out]

1/4*(x^2 - 2*x + 2)*e^(x - 1)*log(2) - 3/4*(x - 1)*e^(x - 1)*log(2) + 1/4*(x^3 - 3*x^2 + 6*x - 6)*e^(x - 1) -
1/2*(x^2 - 2*x + 2)*e^(x - 1) - 9/4*(x - 1)*e^(x - 1) - 5/4*e^(x - 1)*log(2) + 1/4*x + 1/4*e^(x - 1)

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mupad [B]  time = 0.08, size = 37, normalized size = 1.32 \begin {gather*} \frac {x}{4}+\frac {x^3\,{\mathrm {e}}^{x-1}}{4}+\frac {x^2\,{\mathrm {e}}^{x-1}\,\left (\ln \relax (2)-5\right )}{4}-\frac {x\,{\mathrm {e}}^{x-1}\,\left (\ln \left (32\right )-1\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4 - (exp(x - 1)*(9*x + log(2)*(3*x - x^2 + 5) + 2*x^2 - x^3 - 1))/4,x)

[Out]

x/4 + (x^3*exp(x - 1))/4 + (x^2*exp(x - 1)*(log(2) - 5))/4 - (x*exp(x - 1)*(log(32) - 1))/4

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sympy [A]  time = 0.16, size = 32, normalized size = 1.14 \begin {gather*} \frac {x}{4} + \frac {\left (x^{3} - 5 x^{2} + x^{2} \log {\relax (2 )} - 5 x \log {\relax (2 )} + x\right ) e^{x - 1}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((x**2-3*x-5)*ln(2)+x**3-2*x**2-9*x+1)*exp(x-1)+1/4,x)

[Out]

x/4 + (x**3 - 5*x**2 + x**2*log(2) - 5*x*log(2) + x)*exp(x - 1)/4

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