3.30.64 \(\int \frac {3+e^{-1+e^3+x} (1+x)}{\log (75)} \, dx\)

Optimal. Leaf size=18 \[ -3+\frac {\left (3+e^{-1+e^3+x}\right ) x}{\log (75)} \]

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Rubi [B]  time = 0.02, antiderivative size = 38, normalized size of antiderivative = 2.11, number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {12, 2176, 2194} \begin {gather*} \frac {3 x}{\log (75)}-\frac {e^{x+e^3-1}}{\log (75)}+\frac {e^{x+e^3-1} (x+1)}{\log (75)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3 + E^(-1 + E^3 + x)*(1 + x))/Log[75],x]

[Out]

-(E^(-1 + E^3 + x)/Log[75]) + (3*x)/Log[75] + (E^(-1 + E^3 + x)*(1 + x))/Log[75]

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 (3+e^{-1+e^3+x} (1+x)\right ) \, dx}{\log (75)}\\ &=\frac {3 x}{\log (75)}+\frac {\int e^{-1+e^3+x} (1+x) \, dx}{\log (75)}\\ &=\frac {3 x}{\log (75)}+\frac {e^{-1+e^3+x} (1+x)}{\log (75)}-\frac {\int e^{-1+e^3+x} \, dx}{\log (75)}\\ &=-\frac {e^{-1+e^3+x}}{\log (75)}+\frac {3 x}{\log (75)}+\frac {e^{-1+e^3+x} (1+x)}{\log (75)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 19, normalized size = 1.06 \begin {gather*} \frac {3 x+e^{-1+e^3+x} x}{\log (75)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3 + E^(-1 + E^3 + x)*(1 + x))/Log[75],x]

[Out]

(3*x + E^(-1 + E^3 + x)*x)/Log[75]

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fricas [A]  time = 1.35, size = 17, normalized size = 0.94 \begin {gather*} \frac {x e^{\left (x + e^{3} - 1\right )} + 3 \, x}{\log \left (75\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+1)*exp(exp(3)+x-1)+3)/log(75),x, algorithm="fricas")

[Out]

(x*e^(x + e^3 - 1) + 3*x)/log(75)

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giac [A]  time = 0.21, size = 17, normalized size = 0.94 \begin {gather*} \frac {x e^{\left (x + e^{3} - 1\right )} + 3 \, x}{\log \left (75\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+1)*exp(exp(3)+x-1)+3)/log(75),x, algorithm="giac")

[Out]

(x*e^(x + e^3 - 1) + 3*x)/log(75)

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maple [A]  time = 0.07, size = 21, normalized size = 1.17




method result size



norman \(\frac {x \,{\mathrm e}^{{\mathrm e}^{3}+x -1}}{\ln \left (75\right )}+\frac {3 x}{\ln \left (75\right )}\) \(21\)
risch \(\frac {{\mathrm e}^{{\mathrm e}^{3}+x -1} x}{\ln \relax (3)+2 \ln \relax (5)}+\frac {3 x}{\ln \relax (3)+2 \ln \relax (5)}\) \(31\)
default \(\frac {3 x +{\mathrm e}^{{\mathrm e}^{3}+x -1} \left ({\mathrm e}^{3}+x -1\right )+{\mathrm e}^{{\mathrm e}^{3}+x -1}-{\mathrm e}^{{\mathrm e}^{3}+x -1} {\mathrm e}^{3}}{\ln \left (75\right )}\) \(38\)
derivativedivides \(\frac {3 \,{\mathrm e}^{3}+3 x -3+{\mathrm e}^{{\mathrm e}^{3}+x -1} \left ({\mathrm e}^{3}+x -1\right )+{\mathrm e}^{{\mathrm e}^{3}+x -1}-{\mathrm e}^{{\mathrm e}^{3}+x -1} {\mathrm e}^{3}}{\ln \left (75\right )}\) \(43\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x+1)*exp(exp(3)+x-1)+3)/ln(75),x,method=_RETURNVERBOSE)

[Out]

1/ln(75)*x*exp(exp(3)+x-1)+3/ln(75)*x

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maxima [A]  time = 0.56, size = 31, normalized size = 1.72 \begin {gather*} \frac {{\left (x e^{\left (e^{3}\right )} - e^{\left (e^{3}\right )}\right )} e^{\left (x - 1\right )} + 3 \, x + e^{\left (x + e^{3} - 1\right )}}{\log \left (75\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+1)*exp(exp(3)+x-1)+3)/log(75),x, algorithm="maxima")

[Out]

((x*e^(e^3) - e^(e^3))*e^(x - 1) + 3*x + e^(x + e^3 - 1))/log(75)

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mupad [B]  time = 1.74, size = 19, normalized size = 1.06 \begin {gather*} \frac {x\,{\mathrm {e}}^{-1}\,\left (3\,\mathrm {e}+{\mathrm {e}}^{{\mathrm {e}}^3}\,{\mathrm {e}}^x\right )}{\ln \left (75\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x + exp(3) - 1)*(x + 1) + 3)/log(75),x)

[Out]

(x*exp(-1)*(3*exp(1) + exp(exp(3))*exp(x)))/log(75)

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sympy [A]  time = 0.10, size = 19, normalized size = 1.06 \begin {gather*} \frac {x e^{x - 1 + e^{3}}}{\log {\left (75 \right )}} + \frac {3 x}{\log {\left (75 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+1)*exp(exp(3)+x-1)+3)/ln(75),x)

[Out]

x*exp(x - 1 + exp(3))/log(75) + 3*x/log(75)

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