3.104.29 \(\int (-1+5 e^{1+\frac {1}{4} (3+20 e^{1+x})+x}) \, dx\)

Optimal. Leaf size=17 \[ e^{\frac {3}{4}+5 e^{1+x}}-x \]

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Rubi [A]  time = 0.02, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2282, 2194} \begin {gather*} e^{5 e^{x+1}+\frac {3}{4}}-x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-1 + 5*E^(1 + (3 + 20*E^(1 + x))/4 + x),x]

[Out]

E^(3/4 + 5*E^(1 + x)) - 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]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-x+5 \int e^{1+\frac {1}{4} \left (3+20 e^{1+x}\right )+x} \, dx\\ &=-x+5 \operatorname {Subst}\left (\int e^{\frac {7}{4}+5 e x} \, dx,x,e^x\right )\\ &=e^{\frac {3}{4}+5 e^{1+x}}-x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 17, normalized size = 1.00 \begin {gather*} e^{\frac {3}{4}+5 e^{1+x}}-x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-1 + 5*E^(1 + (3 + 20*E^(1 + x))/4 + x),x]

[Out]

E^(3/4 + 5*E^(1 + x)) - x

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fricas [B]  time = 0.61, size = 27, normalized size = 1.59 \begin {gather*} -{\left (x e^{\left (x + 1\right )} - e^{\left (x + 5 \, e^{\left (x + 1\right )} + \frac {7}{4}\right )}\right )} e^{\left (-x - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x*e^(x + 1) - e^(x + 5*e^(x + 1) + 7/4))*e^(-x - 1)

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giac [A]  time = 0.14, size = 13, normalized size = 0.76 \begin {gather*} -x + e^{\left (5 \, e^{\left (x + 1\right )} + \frac {3}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x + e^(5*e^(x + 1) + 3/4)

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maple [A]  time = 0.06, size = 14, normalized size = 0.82

method	result	size
default	\({\mathrm e}^{5 \,{\mathrm e}^{x +1}+\frac {3}{4}}-x\)	\(14\)
norman	\({\mathrm e}^{5 \,{\mathrm e}^{x +1}+\frac {3}{4}}-x\)	\(14\)
risch	\({\mathrm e}^{5 \,{\mathrm e}^{x +1}+\frac {3}{4}}-x\)	\(14\)
derivativedivides	\({\mathrm e}^{5 \,{\mathrm e}^{x +1}+\frac {3}{4}}-\ln \left (20 \,{\mathrm e}^{x +1}\right )\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 13, normalized size = 0.76 \begin {gather*} -x + e^{\left (5 \, e^{\left (x + 1\right )} + \frac {3}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x + e^(5*e^(x + 1) + 3/4)

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mupad [B]  time = 0.14, size = 14, normalized size = 0.82 \begin {gather*} {\mathrm {e}}^{5\,{\mathrm {e}}^{x+1}}\,{\mathrm {e}}^{3/4}-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(5*exp(x + 1)*exp(5*exp(x + 1) + 3/4) - 1,x)

[Out]

exp(5*exp(x + 1))*exp(3/4) - x

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sympy [A]  time = 0.11, size = 12, normalized size = 0.71 \begin {gather*} - x + e^{5 e^{x + 1} + \frac {3}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x + exp(5*exp(x + 1) + 3/4)

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