3.9.79 \(\int \frac {e+20 x+e^x (100+100 x)}{e} \, dx\)

Optimal. Leaf size=17 \[ x+\frac {10 \left (10 e^x x+x^2\right )}{e} \]

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 27, normalized size of antiderivative = 1.59, number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {12, 2176, 2194} \begin {gather*} \frac {10 x^2}{e}+x-100 e^{x-1}+100 e^{x-1} (x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E + 20*x + E^x*(100 + 100*x))/E,x]

[Out]

-100*E^(-1 + x) + x + (10*x^2)/E + 100*E^(-1 + x)*(1 + 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 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 (e+20 x+e^x (100+100 x)\right ) \, dx}{e}\\ &=x+\frac {10 x^2}{e}+\frac {\int e^x (100+100 x) \, dx}{e}\\ &=x+\frac {10 x^2}{e}+100 e^{-1+x} (1+x)-\frac {100 \int e^x \, dx}{e}\\ &=-100 e^{-1+x}+x+\frac {10 x^2}{e}+100 e^{-1+x} (1+x)\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 19, normalized size = 1.12 \begin {gather*} \frac {e x+100 e^x x+10 x^2}{e} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E + 20*x + E^x*(100 + 100*x))/E,x]

[Out]

(E*x + 100*E^x*x + 10*x^2)/E

________________________________________________________________________________________

fricas [A]  time = 0.62, size = 18, normalized size = 1.06 \begin {gather*} {\left (10 \, x^{2} + x e + 100 \, x e^{x}\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((100*x+100)*exp(x)+exp(1)+20*x)/exp(1),x, algorithm="fricas")

[Out]

(10*x^2 + x*e + 100*x*e^x)*e^(-1)

________________________________________________________________________________________

giac [A]  time = 0.39, size = 18, normalized size = 1.06 \begin {gather*} {\left (10 \, x^{2} + x e + 100 \, x e^{x}\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((100*x+100)*exp(x)+exp(1)+20*x)/exp(1),x, algorithm="giac")

[Out]

(10*x^2 + x*e + 100*x*e^x)*e^(-1)

________________________________________________________________________________________

maple [A]  time = 0.03, size = 17, normalized size = 1.00

method	result	size
risch	\(x +10 x^{2} {\mathrm e}^{-1}+100 x \,{\mathrm e}^{x -1}\)	\(17\)
default	\({\mathrm e}^{-1} \left (100 \,{\mathrm e}^{x} x +10 x^{2}+x \,{\mathrm e}\right )\)	\(21\)
norman	\(x +10 x^{2} {\mathrm e}^{-1}+100 x \,{\mathrm e}^{-1} {\mathrm e}^{x}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((100*x+100)*exp(x)+exp(1)+20*x)/exp(1),x,method=_RETURNVERBOSE)

[Out]

x+10*x^2*exp(-1)+100*x*exp(x-1)

________________________________________________________________________________________

maxima [A]  time = 0.39, size = 18, normalized size = 1.06 \begin {gather*} {\left (10 \, x^{2} + x e + 100 \, x e^{x}\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((100*x+100)*exp(x)+exp(1)+20*x)/exp(1),x, algorithm="maxima")

[Out]

(10*x^2 + x*e + 100*x*e^x)*e^(-1)

________________________________________________________________________________________

mupad [B]  time = 0.06, size = 14, normalized size = 0.82 \begin {gather*} x\,{\mathrm {e}}^{-1}\,\left (10\,x+\mathrm {e}+100\,{\mathrm {e}}^x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-1)*(20*x + exp(1) + exp(x)*(100*x + 100)),x)

[Out]

x*exp(-1)*(10*x + exp(1) + 100*exp(x))

________________________________________________________________________________________

sympy [A]  time = 0.10, size = 19, normalized size = 1.12 \begin {gather*} \frac {10 x^{2}}{e} + \frac {100 x e^{x}}{e} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((100*x+100)*exp(x)+exp(1)+20*x)/exp(1),x)

[Out]

10*x**2*exp(-1) + 100*x*exp(-1)*exp(x) + x

________________________________________________________________________________________