3.91.32 \(\int \frac {1-e^x+2 x \log (3)+2 x \log (4 e^6)}{\log (3)+\log (4 e^6)} \, dx\)

Optimal. Leaf size=23 \[ x^2+\frac {-e^x+x}{\log (3)+\log \left (4 e^6\right )} \]

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Rubi [A]  time = 0.01, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {6, 12, 2194} \begin {gather*} x^2+\frac {x}{6+\log (12)}-\frac {e^x}{6+\log (12)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - E^x + 2*x*Log[3] + 2*x*Log[4*E^6])/(Log[3] + Log[4*E^6]),x]

[Out]

x^2 - E^x/(6 + Log[12]) + x/(6 + Log[12])

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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} &=\int \frac {1-e^x+2 x \left (\log (3)+\log \left (4 e^6\right )\right )}{\log (3)+\log \left (4 e^6\right )} \, dx\\ &=\frac {\int \left (1-e^x+2 x \left (\log (3)+\log \left (4 e^6\right )\right )\right ) \, dx}{6+\log (12)}\\ &=x^2+\frac {x}{6+\log (12)}-\frac {\int e^x \, dx}{6+\log (12)}\\ &=x^2-\frac {e^x}{6+\log (12)}+\frac {x}{6+\log (12)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 28, normalized size = 1.22 \begin {gather*} \frac {-e^x+x+6 x^2+\frac {1}{2} x^2 \log (144)}{6+\log (12)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 - E^x + 2*x*Log[3] + 2*x*Log[4*E^6])/(Log[3] + Log[4*E^6]),x]

[Out]

(-E^x + x + 6*x^2 + (x^2*Log[144])/2)/(6 + Log[12])

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fricas [A]  time = 0.43, size = 35, normalized size = 1.52 \begin {gather*} \frac {x^{2} \log \relax (3) + 2 \, x^{2} \log \relax (2) + 6 \, x^{2} + x - e^{x}}{\log \relax (3) + 2 \, \log \relax (2) + 6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*log(4*exp(6))-exp(x)+2*x*log(3)+1)/(log(4*exp(6))+log(3)),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.13, size = 32, normalized size = 1.39 \begin {gather*} \frac {x^{2} \log \relax (3) + x^{2} \log \left (4 \, e^{6}\right ) + x - e^{x}}{\log \relax (3) + \log \left (4 \, e^{6}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*log(4*exp(6))-exp(x)+2*x*log(3)+1)/(log(4*exp(6))+log(3)),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 31, normalized size = 1.35

method	result	size
norman	\(x^{2}+\frac {x}{6+2 \ln \relax (2)+\ln \relax (3)}-\frac {{\mathrm e}^{x}}{6+2 \ln \relax (2)+\ln \relax (3)}\)	\(31\)
default	\(\frac {x +x^{2} \ln \relax (3)+x^{2} \ln \left (4 \,{\mathrm e}^{6}\right )-{\mathrm e}^{x}}{\ln \left (4 \,{\mathrm e}^{6}\right )+\ln \relax (3)}\)	\(33\)
risch	\(\frac {2 x^{2} \ln \relax (2)}{6+2 \ln \relax (2)+\ln \relax (3)}+\frac {x^{2} \ln \relax (3)}{6+2 \ln \relax (2)+\ln \relax (3)}+\frac {6 x^{2}}{6+2 \ln \relax (2)+\ln \relax (3)}+\frac {x}{6+2 \ln \relax (2)+\ln \relax (3)}-\frac {{\mathrm e}^{x}}{6+2 \ln \relax (2)+\ln \relax (3)}\)	\(76\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x*ln(4*exp(6))-exp(x)+2*x*ln(3)+1)/(ln(4*exp(6))+ln(3)),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.34, size = 32, normalized size = 1.39 \begin {gather*} \frac {x^{2} \log \relax (3) + x^{2} \log \left (4 \, e^{6}\right ) + x - e^{x}}{\log \relax (3) + \log \left (4 \, e^{6}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*log(4*exp(6))-exp(x)+2*x*log(3)+1)/(log(4*exp(6))+log(3)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.10, size = 22, normalized size = 0.96 \begin {gather*} \frac {x-{\mathrm {e}}^x+x^2\,\left (\ln \left (12\right )+6\right )}{\ln \left (12\,{\mathrm {e}}^6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x*log(4*exp(6)) - exp(x) + 2*x*log(3) + 1)/(log(4*exp(6)) + log(3)),x)

[Out]

(x - exp(x) + x^2*(log(12) + 6))/log(12*exp(6))

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sympy [A]  time = 0.13, size = 27, normalized size = 1.17 \begin {gather*} x^{2} + \frac {x}{\log {\relax (3 )} + 2 \log {\relax (2 )} + 6} - \frac {e^{x}}{\log {\relax (3 )} + 2 \log {\relax (2 )} + 6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*ln(4*exp(6))-exp(x)+2*x*ln(3)+1)/(ln(4*exp(6))+ln(3)),x)

[Out]

x**2 + x/(log(3) + 2*log(2) + 6) - exp(x)/(log(3) + 2*log(2) + 6)

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